TOPIC 1 Investigating data distributions • EXERCISE 1.3
1
Topic 1 — Investigating data distributions
13 This type of data is both categorical and ordinal.
14 This numerical data is discrete data.
15 This categorical data is ordinal data.
1.2 Types of data
1.2 Exercise
1 The option that is not numerical data is option D. The
finishing positions in the Melbourne Cup are not numerical
data.
2 The option that is not numerical data is option D. Salaries are
not categorical data.
3 Numerical:
a. Heights in centimetres of a group of children
b. Diameters in millimetres of a tray of ball bearings
c. Numbers of visitors to a display each day
Categorical:
d. The modes of transport that students in Year 12 take
to school
e. The 10 most-watched television programs in a week
f. The occupations of a group of 30-year-olds
4 Categorical:
c. Species of fish
d. Blood groups
e. Years of birth
f. Countries of birth
g. Tax brackets
1.3 Stem plots
1
to fixed values, so it is
continuous.
This is not restricted
to fixed values, so it is
continuous.
c The numbers of visitors at This can only have whole
a display each day
numbers, so it is discrete.
d The modes of transport
that students in Year 11
take to school
e The 10 most-watched
television programs in a
week
f The occupations of a
group of 30-year-olds
1 The variables age and preferred travel destination are both
categorical variables. Categories have been given for both.
The correct answer is A.
2 A discrete variable means whole countable numbers. This is
only true for number of wings, so only 1 of these variables is a
discrete numerical variable.
The correct answer is A.
3 Number of moths is a categorical variable that can be ordered,
so it is classified as ordinal. Trap type is a categorical variable
that is not ordered, so it is classified as nominal.
The correct answer is E.
1.3 Exercise
5 The option that is not discrete data is option C. The average
temperature in March is not discrete data.
6 The option that is not continuous data is option B. The number
of shots missed in a basketball game is not continuous data.
7 a The heights, in
This is not restricted
centimetres, of a group of
children
b The diameters, in
millimetres, of a tray of
ball bearings
1.2 Exam questions
This data is not numerical.
This data is not numerical.
This data is not numerical.
8 The total attendance at Carlton football matches is an example
of a numerical variable. The answer is C.
9 The weight measured is numerical data. Since the weight of
the truck could take any value in a range, it is continuous.
Therefore, the answer is C.
10 This type of data in a menu is categorical and nominal.
11 The measurement of distance to other stars in the universe is
continuous. Therefore, the correct option is B.
12 The data collected, colours of flowers, is nominal. Therefore,
the correct option is A.
Stem
1
2
3
4
Leaf
6 8
1 5 8 8 9
0 1 3 5 8 9
2 8 9
Key: 1 6 = 16
Leaf
Stem
2
0
5
1 8 9
1
2
3 7 9
3
1 2 5 6 7 9
1 2 3 5
4
5
2
Key: 0 5 = $5
The busker’s earnings are inconsistent.
Stem
Leaf
3
86
8
87
7
88
0
89
8
90
2 4 8 9
91
92
0 1 2 6
93
94
3 9
95
96
97
0
98
3 5 9
Key: 86 8 = 86.8%
P df_Fol i o: 1
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
2
TOPIC 1 Investigating data distributions • EXERCISE 1.3
Stem
Leaf
18
5 7 9
19
1 5 6 6 7 9
20
1 3 3 5 9
21
7
1
22
Key: 18 5 = 1.85 cm
Stem
Leaf
5
3
7 9
2 9 9
4
5
1 1 2 3 7 8 9
1 3 3 8
6
Key: 3 7 = 37 years
It seems to be an activity for older people.
Stem
Leaf
6
0
4
1
2
2 7 8 9 9
0 1 4 6
3
1 1
4
Key: 2 5 = 25
Data from stem plot is:
4, 22, 27, 28, 29, 29, 30, 31, 34, 36, 41, 41
Therefore, the answer is C.
7 Lowest = 19
Highest = 39
Use stems from 1 to 3; split stems into halves because of
clustered data.
4
∗
Leaf
Stem
Leaf
Stem
1
9
2
2∗
5 8 8 9 9 9
0 0 2 2 2 3 3 4
3
3∗
5 5 7 8 9
Key: 2 5 = 25 years
Ages are spread, considerably, therefore not all parents
are young.
8 Lowest = 18
Highest = 77
Use stems from 1 to 7 and split stems into halves because of
the clustered data.
∗
1
2
2∗
3
3∗
4
5
5∗
6
6∗
7
7∗
8
9
7
3
6
0
0
5
7
6
1
2
Key: 2 1 = 21 hit outs
The three teams with the highest performing ruckmen were
the GWS Giants (77), Gold Coast Suns (55) and Port
Adelaide Power (54).
9 Lowest = 330
Highest = 450
Leaf
Stem
33
0
34
35
0 0 1
36
5
37
3
38
0 0
39
0 0 5
0 6
40
41
0 5
42
1 3
0 0
43
44
45
0
Key: 33 0 = $330
The stem plot shows a fairly even spread of rental prices with
no obvious outliers.
Stem
Leaf
10 a
2
2∗
3
3∗
b
11 a
8
9
6
4
9
b
0 2 2
5 6 8 8
3 3 3
7 7 8 9 9 9
Key: 2 0 = 20 points
Stem
Leaf
2
0
2 2
2
2
5
2
6
2
8 8
3
3
3 3 3
3
3
7 7
3
8 9 9 9
Key: 2 0 = 20 cm
Leaf
Stem
4
3 7 7 8 8 9 9 9
5
0 0 0 0 1 2 2 3
Key: 4 3 = 43 cm
Stem
Leaf
4
4∗
5
5∗
3
7 7 7 8 8 9 9 9
0 0 0 0 1 2 2 3
Key: 4 3 = 43 cm
7
P df_Fol i o: 2
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.3
c
Stem
4
4
4
4
4
5
5
5
5
5
Leaf
14
3
7 7
8 8 9 9 9
0 0 0 0 1
2 2 3
Key: 4 3 = 43 cm
12 a 1, 2, 5, 8, 12, 13, 13, 16, 16, 17, 21, 23, 24, 25, 25, 26,
27, 30, 32
b 10, 11, 23, 23, 30, 35, 39, 41, 42, 47, 55, 62
c 101, 102, 115, 118, 122, 123, 123, 136, 136, 137, 141,
143, 144, 155, 155, 156, 157
d 50, 51, 53, 53, 54, 55, 55, 56, 56, 57, 59
e 1, 4, 5, 8, 10, 12, 16, 19, 19, 21, 21, 25, 29
13 Data 23 15 18 17 17 19 22
19 20 16 20 21 19 23
17 19 21 23 20 21
a Stems 1 and 2
Stem
1
2
Leaf
5 6 7 7 7 8 9 9 9 9
0 0 0 1 1 1 2 3 3 3
Stem
Leaf
Stem
Leaf
7
2 8
8
3 3 5 7 8 8
9
0 1 2 2 3 4 8 9
10
0 2 4
11
2
Key: 7 2 = 72 shots
Stem
15
Key: 6 0 = 60%
b
5 6 7 7 7 8 9 9 9 9
0 0 0 1 1 1 2 3 3 3
Key: 1 5 = 15 mm
c Stems 1 and 2 split into fifths
Stem
1
1
1
1
1
2
2
2
2
2
Leaf
5
6 7 7 7
8 9 9 9 9
0 0 0 1 1 1
2 3 3 3
Key: 1 5 = 15 mm
The screw lengths are clustered and evenly spread across
the range of 15 mm to 23 mm.
P df_Fol i o: 3
Leaf
6
6∗
7 8 9
7
1 1 2 2 3 3 3 3 4 4
5 5 5 6 6 7
7∗
8
8∗
6
Key: 7 1 = 71 net score
The handicapper has done a good job as most of the net scores
are around the same scores, that is, in the 70s.
Stem
Leaf
16 a
6
0 3 9
7
0 1 3 5 6 7 8
8
0 1 3 4 7 8 9 9
1 3 7 8
9
Key: 1 5 = 15 mm
b Stems 1 and 2 split into halves
1
1∗
2
2∗
3
17 a
Stem
6
6∗
7
7∗
8
8∗
9
9∗
Leaf
0 3
9
0 1 3
5 6 7 8
0 1 3 4
7 8 9 9
1 3
7 8
Key: 6 0 = 60%
Computer 1
Stem
Computer 2
5
8 2
6 3
6
1 0
2 1
5
0
34
35
36
37
38
39
40
41
0 2 6 8
2 3 5 5 7 8
1 2
Key: 34 0 = 340 minutes
b Computer 1 lasts longer but is not as consistent.
Computer 2 is more consistent but doesn’t last as long.
Year 8
Stem
Year 10
18 a
9 8
14
7 5 5 5 3 1 0
15
2 4 6 8 9
8 6 5 4 3 2 1 0
16
0 4 5 7 7 9
5 2 1
17
2 3 4 6 7 8 8
18
2 5
Key: 14 8 = 148 cm
b As you would expect, Year 10 students are taller than Year
8 students; however, there is a large overlap in the heights.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
4
TOPIC 1 Investigating data distributions • EXERCISE 1.4
1.3 Exam questions
0
3
Number of
students
1 Mode means the most frequent score. Therefore, 2.8°C is the
modal temperature.
The correct answer is A.
2 There are 19 countries with more than 22% of homes
connected to broadband Internet in 2007.
The correct answer is C.
3 Given there are 20 pieces of data, the median is halfway
between the 10th and 11th data pieces.
Q1 is then halfway between the 5th and 6th pieces of data.
That is, Q1 = 28.
Q3 is then halfway between the 15th and 16th pieces of data.
That is, Q3 = 41.
IQR = Q3 − Q1 = 41 − 28 = 13
Lower bound = 28 − 1.5 × 13 = 8.5
Upper bound = 41 + 1.5 × 13 = 60.5
Any number less than 8.5 or greater than 60.5 is
considered an outlier.
Therefore, there are two outliers: 7 and 8.
The correct answer is C.
150 − 159
Class interval
170 − 179
180 − 189
Frequency
Number of
students
Hours
∣∣∣∣
Frequency
∣∣∣∣ ∣∣∣∣
4
Tally
0
1
2
∣
9
∣∣∣∣ ∣
3
∣∣∣
3
4
5
6
7
0
2
∣∣
8
∣∣
9
10
5 a
∣∣∣∣
Frequency
∣∣∣∣ ∣∣∣∣ ∣∣
4
∣∣∣∣ ∣∣∣∣ ∣∣
12
∣∣
10
Total
40
12
2
Class
Frequency
1–
2–
3–
4–
5–
6–
1
2
2
6
5
1
6
∣∣
Tally
0 150 160 170 180 190 200
Height (cm)
1
∣∣∣∣ ∣∣∣
8
Number of students
∣∣∣∣
4
9 10
12
10
8
6
4
2
x
Frequency
2
8
∣∣∣∣ ∣∣∣∣
190 − 199
y
5
6
7
8
9
10
Number of questions completed
4 5 6 7
Hours/week
15 16 17 18 19 20 21 22 23 24
Number of hours spent on
homework
1.4 Exercise
0
3
160 − 169
1.4 Dot plots, frequency tables, histograms,
bar charts and logarithmic scales
1
2
5
4
3
2
1
0
4
1
2
1
40
6
5
4
3
2
1
0
1
2
3
4
Class
5
6
7
10
8
6
4
2
0
1
2
3
4
5
6
7
Number of hours
8
9
10
P df_Fol i o: 4
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.4
Class interval
Frequency
10–
15–
20–
25–
30–
35–
3
9
10
10
10
1
b
Frequency
b
8
7
6
5
4
3
2
1
7
0
10
15
20 25 30
Class interval
Score
Frequency
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1
2
1
1
1
2
2
2
1
1
1
35
40
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7
Fish
Frequency
8
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Fatalities
9
Team
Goals
Adelaide
48
Brisbane
55
Carlton
58
Collingwood
34
Essendon
41
Fremantle
37
Geelong
62
Gold Coast
47
GWS
45
3
Hawthorn
33
0
2
Melbourne
59
North
Melbourne
42
2
1
0
0.
3
0.
4
0.
5
0.
6
0.
7
0.
8
0.
9
1.
0
1.
1
1.
2
1.
3
Frequency
2 3 4 5 6 7 8 9 10 11 12
Number of dogs
c Check your histogram against that shown in the solution to
part b.
10
8
6
4
2
Frequency
Frequency
0
c
Score
6 a
5
Number of dogs
Tally
2
∣∣
Frequency
∣∣∣
2
∣∣
3
∣∣
2
∣
2
∣∣∣∣
1
∣∣
5
∣∣∣∣ ∣∣∣
2
∣∣∣
8
3
4
5
6
7
8
9
10
11
12
∣∣
30
Percentage
48
× 100 = 5.72%
839
55
× 100 = 6.56%
839
58
× 100 = 6.91%
839
34
× 100 = 4.05%
839
41
× 100 = 4.89%
839
37
× 100 = 4.41%
839
62
× 100 = 7.39%
839
47
× 100 = 5.60%
839
45
× 100 = 5.36%
839
33
× 100 = 3.93%
839
59
× 100 = 7.03%
839
42
× 100 = 5.00%
839
(continued)
P df_Fol i o: 5
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.4
Team
Goals
Port Adelaide
48
Richmond
51
St Kilda
38
Sydney
51
West Coast
42
Western
Bulldogs
48
All clubs
839
Percentage
Log (weight (kg))
48
× 100 = 5.72%
839
51
× 100 = 6.08%
839
38
× 100 = 4.53%
839
51
× 100 = 6.08%
839
42
× 100 = 5.01%
839
48
× 100 = 5.72%
839
Frequency
6
∼ 100% (99.9%)
Ca Co Es Fr Ge GC
Ha Me NM PA Ri St Syd WCWB
10 The statement seems untrue as there are similar participation
rates for all groups.
18–24 years 14.2%
25–34 years 21.0%
35–44 years 20.3%
45–54 years 18.1%
55–64 years 14%
65 and over 12.5%
Number of families
11
24
21
18
15
12
9
6
3
0
1
2
3
4
5
6
7
Number of children
8
12 Calculate the logarithmic values of all of the weights.
Weight (kg)
Log (weight (kg))
18
55
63
70 000
175
9
0.4
200
4
725
1.26
1.74
1.80
4.85
2.24
0.95
−0.40
2.30
0.60
2.86
Group into class intervals.
P df_Fol i o: 6
0−1
1−2
2−3
3−4
4−5
2
3
3
0
1
9
1
4
3
2
1
0
–1
GWS
Ad Br
−1−0
Frequency
0
1
2
3
Log (weight (kg))
4
5
13 log10 207 = 2.316
This places the gorilla between the jaguar and the horse.
The correct answer is D.
14 log10 64 000 = 4.81
The correct answer is D.
15 a 106 = 1 000 000
The correct answer is B.
b 105 = 100 000
The correct answer is A.
c 104 = 10 000
105 = 100 000
The correct answer is D.
16 log10 2407 = 3.382
The correct answer is A.
17 log10 1088 = 3.037
The correct answer is B.
108.1
= 101.2
18
106.9
= 15.85
The earthquake of magnitude 8.1 is approximately 16 times
greater than the earthquake of magnitude 6.9.
19 As the pH value decreases, the acidity of the liquid increases.
100.7 = 5.01
The acidity has increased by approximately 5 times.
20 a
Resident
Country
departures in
2019 (×1000)
New Zealand
921.1
USA
492.3
UK
420.3
Indonesia
380.7
China
277.3
Thailand
404.1
Fiji
236.2
Percentage
921.1
× 100 = 24.5%
3753.9
492.3
× 100 = 13.1%
3753.9
420.3
× 100 = 11.2%
3753.9
380.7
× 100 = 10.1%
3753.9
277.3
× 100 = 7.4%
3753.9
404.1
× 100 = 10.8%
3753.9
236.2
× 100 = 6.3%
3753.9
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.5
217.8
Hong Kong
213.1
Malaysia
191.0
All countries
3753.9
217.8
× 100 = 5.8%
3753.9
213.1
× 100 = 5.7%
3753.9
191.0
× 100 = 5.1%
3753.9
100%
b
NZ 24.5%
Indonesia 10.1%
Fiji 6.3%
Malaysia 5.1%
US 13.1%
China 7.4%
Singapore 5.8%
UK 11.2%
Thailand 10.8%
HK 5.7%
1.4 Exam questions
1 The contention: there is an association between preferred
travel destination and age.
The answer will need to reference the different age groups
and preferred travel destination.
Option D fits the contention.
The correct answer is D.
2 Segmented bar charts show bars stacked on top of one another
to give a single bar with several parts. The lengths are
determined by frequencies.
The only option that is broken into parts is D: below average,
average, above average rainfall.
The correct answer is D.
3 log10 10 000 = 4
Therefore, the number of species with a weight of less than 4
is 17.
17
× 100 = 85%
20
The correct answer is E.
1.5 Describing stem plots and histograms
1.5 Exercise
1 Most of the data occurs at stem 1 and 2 and it tails off as the
stems increase. Therefore, the data is positively skewed.
2 Most of the data occurs in between the ages of 25 and 30,
which is the middle bar. However, the frequencies are greater
at older ages compared to younger; therefore, the data is
negatively skewed.
3
Stem
Leaf
0
1
0
2
0
4 4 5
0
6 6 6 7
0
8 8 8 8 9 9
1
0 0 0 1 1 1 1
1
2 2 2 3 3 3
1
4 4 5 5
1
6 7 7
1
8 9
P df_Fol i o: 7
Key: 1 8 = 18
The distribution of the data shown in the stem plot could be
described as symmetric.
The correct answer is E.
4
Frequency
Singapore
x
The distribution of the data shown in the histogram could be
described as positively skewed.
The correct answer is C.
5 a
Stem
0
1
2
3
4
5
6
7
Leaf
1 3
2 4 7
3 4 4 7 8
2 5 7 9 9 9 9
1 3 6 7
0 4
4 7
1
Key: 1 2 = 12
The data is approximately symmetric.
Stem
Leaf
b
1
3
6
2
3
3 8
4
2 6 8 8 9
5
4 7 7 7 8 9 9
0 2 2 4 5
6
Key: 10 4 = 104
The data is negatively skewed.
Stem
Leaf
c
2
3 5 5 6 7 8 9 9
3
0 2 2 3 4 6 6 7 8 8
2 2 4 5 6 6 6 7 9
4
5
0 3 3 5 6
6
2 4
7
5 9
8
2
9
7
10
Key: 10 4 = 104
The data is positively skewed.
Stem
Leaf
d
1
1∗
5
2
1 4
5 7 8 8 9
2∗
3
1 2 2 3 3 4 4
3∗
5 5 5 6
4
3 4
4∗
Key: 2 4 = 24
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
7
TOPIC 1 Investigating data distributions • EXERCISE 1.5
e
d
The data is approximately symmetric.
Leaf
Stem
3
3
8 9
4
0 0 1 1 1
2 3 3 3 3
4
4
4 5 5 5
4
6 7
8
4
Frequency
8
Key: 4 3 = 0.43
The data is approximately symmetric.
Leaf
Stem
f
60
2 5 8
61
1 3 3 6 7 8 9
0 1 2 4 6 7 8 8 9
62
63
2 2 4 5 7 8
3 6 7
64
65
4 5 8
66
3 5
67
4
x
Negatively skewed, no outliers
Frequency
e
x
Negatively skewed, no outliers
Key: 62 3 = 623
The data is positively skewed.
f
Frequency
Frequency
6 a
x
Positively skewed, no outliers
x
7
Approximately symmetric, no outliers
Frequency
Frequency
b
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Number of enquiries
The data is negatively skewed.
x
Approximately symmetric, no outliers
c
Frequency
8
Approximately symmetric, no outliers
P df_Fol i o: 8
x
8
7
6
5
4
3
2
1
Stem
Leaf
0
0 0 1 1
0
2 2 3 3 3 3 3 3 3 3
0
4 4 5 5 5 5 5
0
6 6 6 6 7
0
8 8 8 9
1
0 0 1
1
4 4
1
5 5
1
7
Key: 1 4 = 14 nights
The stem plot is positively skewed. The distribution shows
that the majority of flight attendants in this group stay away
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.6
b The data is positively skewed.
c The car sales are at their greatest in the months of June,
July and November.
d A reason for the most cars being sold at this time could be
that this is when the end of financial year sales occur.
between 2 and 5 nights. Fewer flight attendants stay away for
the larger number of nights.
9
Stem
Leaf
0
4
0∗
5 7 9
1 2 4 4
1
1∗
5 6 6 7 8 9
1 2 2 3
2
∗
2
6 7
Key: 0 4 = 4 kg
9
1.5 Exam questions
a The distribution is approximately symmetrical.
b Most dogs weigh in the range of 10 to 19 kg, with only a
few dogs heavier or lighter.
Frequency
1
Frequency
10 a
8
7
6
5
4
3
2
1
0
This histogram represents negatively distributed data.
The correct answer is B.
2 The data trails off on the positive end, so the distribution is
positively skewed.
The correct answer is B.
3 Turning the image 90 degrees and viewing the shape of the
distribution confirms symmetrical.
The correct answer is E.
0
5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5
Pocket money ($)
The data is approximately symmetric.
b The majority of students receive about $8.00, or within the
range $7 to $9 per week.
11 a The histogram has a slight positively skewed distribution.
b i Adding up the last 5 columns of the histogram gives:
5 + 4 + 3 + 2 + 1 = 15
(3 + 5 + 4 + 3 + 2)
× 100 = 85%
ii
20
12 a Most of the data is listed on the lower stems. This suggests
that the data is positively skewed.
b Since most of the data is linked to the lower stems, this
suggests that some students do little exercise, but those
students who exercise, do quite a bit each week. This could
represent the students in teams or training squads.
13 a Club A: Has most of the data at the higher stems; therefore,
the data is negatively skewed.
Club B: Has most of the data at the lower stems; thus, the
data is positively skewed.
b Since Club A has more members of its bowling team at the
higher stems as compared to Club B, you could say Club A
has the older team.
c i Club A: 11 members over 70 years of age
ii Club B: 4 members over 70 years of age
14 a
Cars sold
Score
30
25
20
15
10
5
0
Apr May Jun
1.6 Summary statistics
1.6 Exercise
1 The median is the middle data point. There are 30 data points.
The middle is between the 15th and the 16th score.
Score 15 = 33 and score 16 = 33. Therefore, the median
is 33.
2 Order the data first and highlight the two centre scores.
21, 24, 24, 28, 29, 32, 33, 36, 37, 37, 41, 45, 46, 47, 49, 52
36 + 37
Median =
2
= 36.5
3 Order the data.
10, 12, 13, 15, 16, 19, 21, 21, 21, 23, 24, 26, 27, 28, 28, 29,
30, 31, 33, 33, 39, 39, 39, 39
Median = 27
Q1 = 19
Q3 = 33
IQR = Q3 − Q1
= 33 − 19
= 14
4 Order the data.
15, 16, 18, 18, 19, 19, 19, 20, 21, 21, 24, 24, 24, 25, 27, 27,
29, 31, 32, 33
21 + 24
Median =
2
= 22.5
19 + 19
Q1 =
2
= 19
Jul Aug Sep Oct Nov
Month
P df_Fol i o: 9
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
10
TOPIC 1 Investigating data distributions • EXERCISE 1.6
27 + 27
2
= 27
Q3 =
IQR = Q3 − Q1
= 27 − 19
=8
5 Order the data.
2, 3, 3, 4, 5, 5, 7, 7, 7, 8, 8, 9, 9, 10, 11, 12, 13, 15, 15, 21
5+5
Q1 =
2
=5
11 + 12
Q3 =
2
= 11.5
IQR = Q3 − Q1
= 11.5 − 5
= 6.5
6 Order the data.
40.0, 41.9, 42.7, 42.9, 43.1, 43.8, 43.9, 44.7, 45.1, 45.1,
45.3, 45.7, 45.8, 45.9, 46.4, 47.1, 48.6, 48.9, 49.1, 49.9
43.1 + 43.8
Q1 =
2
= 43.45
46.4 + 47.1
Q3 =
2
= 46.75
IQR = Q3 − Q1
= 46.75 − 43.45
= 3.3
n+1
7 a Median =
th position
( 2 )
24 + 1
th position
( 2 )
= 12.5th position
=
12th term = 36
13th term = 38
Therefore, the median is 37.
Range = 63 − 7
= 56
Mode = 38, 49
n+1
b Median =
th position
( 2 )
35 + 1
th position
( 2 )
= 18th position
Therefore, the median is 5.
Range = 17 − 0
= 17
Mode = 5
n+1
c Median =
th position
( 2 )
=
37 + 1
th position
( 2 )
= 19th position
Therefore, the median is 11.
Range = 19 − 1
= 18
Mode = 8, 11
=
P df_Fol i o: 10
n+1
th position
( 2 )
22 + 1
th position
=
( 2 )
= 11.5th position
d Median =
11th term = 42
12th term = 43
Therefore, the median is 42.5.
Range = 49 − 31
= 18
Mode = 43
5+1
8 a Median position =
th position
( 2 )
= 3rd position
Therefore, the median is 6.
Range = 9 − 2
=7
5+1
b Median position =
th position
( 2 )
= 3rd position
Therefore, the median is 17.
Range = 21 − 12
=9
7+1
th position
c Median position =
( 2 )
= 4th position
Therefore, the median is 6.
Range = 9 − 3
=6
8+1
d Median position =
th position
( 2 )
= 4.5th position
4th term = 8
5th term = 12
Therefore, the median is 10.
Range = 16 − 3
= 13
10 + 1
e Median position =
th position
( 2 )
= 5.5th position
5th term = 18
6th term = 19
Therefore, the median is 18.5.
Range = 26 − 12
= 14
f Ordered data:
1234568
7+1
th position
Median position =
( 2 )
= 4th position
Therefore, the median is 4.
Range = 8 − 1
=7
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.6
9 a Ordered data:
8 9 12 14 14 15 18 18 21 24 24 24 25 25 25
15 + 1
th position
Median =
( 2 )
= 8th position
Therefore, the median is 18.
To calculate Q1 , consider 8 9 12 14 14 15 18.
Q1 = 14
To calculate Q3 , consider 21 24 24 24 25 25 25.
Q3 = 24
So IQR = Q3 − Q1
= 24 − 14
= 10
b Ordered data:
7 9 10 11 12 13 14 16 18 18 19 19 20 20 21
15 + 1
Median =
th position
( 2 )
= 8th position
Median = 16
To calculate Q1 , consider 7 9 10 11 12 13 14.
Q1 = 11
To calculate Q3 , consider 18 18 19 19 20 20 21.
Q3 = 19
So IQR = Q3 − Q1
= 19 − 11
=8
c The IQRs are similar for the two restaurants, but there isn’t
any indication of the actual number of cars.
10 An example of a set of data where n = 5, median is 6 and
range is 7 is:
2 3 6 8 9
Median
Range = 9 − 2
=7
There are many other sets of data with these parameters.
11 a To have a set of data where the lower quartile equals the
lowest score, the lowest score must occur several times. An
example is 5, 5, 5, 5, 6, 9, 10.
b To have a set of data where the IQR = 0, the scores in the
middle of the set of data must all be equal.
For example, 2, 5, 5, 5, 5, 5, 5, 10.
12 If Q1 = 13, median = 18 and Q3 = 25.
IQR = 25 − 13
= 12
Therefore, the answer is C.
13 a 1 Var Stats
n = 20
Min x = 6
Q1 = 13.5
Median = 21
Q3 = 31.5
Max x = 51
Median = 21
IQR = Q3 − Q1
= 31.5 − 13.5
= 18
P df_Fol i o: 11
Range = 51 − 6 = 45
Mode = 15, 23 and 32
b 1 Var Stats
n = 20
Min x = 19
Q1 = 22.5
Median = 27.5
Q3 = 30.5
Max x = 39
IQR = Q3 − Q1
= 30.5 − 22.5
=8
Range = 39 − 19 = 20
Mode = 29
c 1 Var stats
n = 19
Min x = 1.2
Q1 =2.4
Median = 3.7
Q3 =5.4
Max x = 7.1
IQR = Q3 −Q1
= 5.4 − 2.4
=3
Range = 7.1 − 1.2
= 5.9
14 a
Mode = 3.7
Stem
2
3
Leaf
3 5 5 6 7 8 9 9
0 2 2 3 4 6 6 7 8 8
4
2 2 4 5 6 6 6 7 9
5
6
7
8
9
10
11
0 3 3 5 6
2 4
5 9
2
7
Key: 4 2 = 42
1 Var Stats
n = 39
4
Min x = 23
Q1 = 32
Median = 42
Q3 = 53
Max x = 114
IQR = Q3 − Q1
= 53 − 32
= 21
Range = 114 − 23
= 91
Mode = 46
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
11
TOPIC 1 Investigating data distributions • EXERCISE 1.7
12
b
Stem
1
1∗
2
2∗
3
3∗
4
4∗
Leaf
4
1 4
5 7 8 9 9
1 2 2 2 4 4 4 4
5 5 5 6
3 4
Key: 3 2 = 32
1 Var Stats
n = 22
Min x = 14
Q1 = 28
Median = 32
Q3 = 35
Max x = 44
IQR = Q3 − Q1
= 35 − 28
=7
Range = 44 − 14
= 30
Mode = 34
15 a Range = 73 − 1
= 72
There are 22 values, so the median is calculated by
averaging the two centre scores.
37 + 38
Median =
2
= 37.5
Mode = 46
IQR = Q3 − Q1
= 46 − 24
= 22
b Range = 450 − 403
= 47
There are 27 values, so the median is 422.
Mode = 411
IQR = Q3 − Q1
= 433 − 413
= 20
16 Order the values.
2, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 7, 8, 9, 9, 9, 9, 10,
12, 13, 15
7+7
Median =
2
=7
Mode = 7
17 Using CAS: Q1 = 42.2, Q3 = 48.15
IQR = Q3 − Q1
= 48.15 − 42.2
= 5.95
Median = 45.1
P df_Fol i o: 12
18 a Using CAS: median = 93, Q1 = 91.5, Q3 = 97
IQR = Q3 − Q1
= 97 − 91.5
= 5.5
Range = 111 − 81
= 30
Mode = 93
b Since the median is 93 and par for the course is 72, then the
average handicap of the golfers should be around
93 – 72 = 21.
1.6 Exam questions
1 Divide the data in half to find the median. It will be in the
126th position.
Divide the upper half in half to find the third quartile. It will
be in the 189th position.
Count up the frequencies until you get a bar that contains the
189th data point.
The correct answer is C.
2 Range = highest value − lowest value
= 89 − 40
= 49
The correct answer is D.
3 IQR = Q3 − Q1
= 75 − 57
= 18
The correct answer is C.
1.7 The five-number summary and boxplots
1.7 Exercise
1 Range = 51 − 12
= 39
Median = 25
IQR = Q3 − Q1
= 39 − 20
= 19
2 Range = 9 − 6
=3
Median = 7.5
IQR = Q3 − Q1
= 8.2 − 6.8
= 1.4
3 a Range = Max x − Min x
= 14 − 2
= 12
IQR = Q3 − Q1
= 12 − 6
=6
Median = 8
b Range = Max x − Min x
=8−1
=7
IQR = Q3 − Q1
=6−4
=2
Median = 5
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
c Range = Max x − Min x
= 450 − 100
= 350
IQR = Q3 − Q1
= 300 − 200
= 100
Median = 250
d Range = 130 − 30
= 100
Frequency
IQR = Q3 − Q1
= 80 − 50
= 30
Median = 65
4 The histogram shows a distribution that is negatively skewed.
The boxplot shows a negatively skewed distribution as the
median is to the right of centre. Therefore, they could
represent the same data.
5 The histogram shows a distribution that is symmetrical, as
does the boxplot. Therefore, they could represent the same
data.
6 a Boxplot iii
b Boxplot iv
TOPIC 1 Investigating data distributions • EXERCISE 1.7
7 Order the data.
26, 28, 30, 32, 32, 35, 36, 37, 38, 38, 40, 40, 41, 42,
46, 48, 50, 50
38 + 38
Median =
2
= 38
Q1 = 32
Q3 = 42
24
27
30
33 36 39 42
Results out of 50
45
48
51
The data is slightly negatively skewed since the median is
slightly right of the centre. 50% of results are between
32 and 42.
8 Order the data.
4, 4, 4, 5, 6, 6, 7, 7, 8, 8, 8, 8, 10, 12, 12, 16, 20
Median = 8
5+6
Q1 =
2
= 5.5
10 + 12
Q3 =
2
= 11
4
6
8
10
12
14
16
18
20
22
The data is fairly symmetrical; however, a couple of students
work a lot more hours per week.
9 a Data:
3 5 6 8 8 9 12 14 17 18
1 Var Stats
n = 10
Min x = 3
Frequency
Q1 = 6
Median = 8.5
Q3 = 14
Max x = 18
2
Frequency
c Boxplot i
4
6
8
10 12 14 16 18 20 22 24
b Data:
3 4 4 5 5 6 7 7 7 8 8 9 9 10 10 12
1 Var Stats
n = 16
Min x = 3
Q1 = 5
Median = 7
Q3 = 9
d Boxplot ii
Max x = 12
Frequency
13
3
4
5
6
7
8
9
10 11 12
P df_Fol i o: 13
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.7
14
Q3 = 27
Max x = 32
See Image bottom of the page*
13 a Order the data.
1.30, 1.30, 1.35, 1.36, 1.38, 1.40, 1.40, 1.45, 1.45,
1.46, 1.48, 1.75
c Data:
4.3, 4.5, 4.7, 4.9, 5.1, 5.3, 5.5, 5.6
1 Var Stats
n=8
Min x = 4.3
Q1 = 4.6
Lowest = 1.30
Highest = 1.75
1.40 + 1.40
Median =
2
= 1.40
1.35 + 1.36
Q1 =
2
= 1.355
1.45 + 1.46
Q3 =
2
= 1.455
Median = 5
Q3 = 5.4
Max x = 5.6
4.2 4.4 4.6
48
5.0 5.2 5.4 5.6 5.8
d Data:
11 13 15 15 16 18 20 21 22 21 18 19 20 16 18 20
1 Var Stats
n = 16
IQR = Q3 − Q1
= 1.455 − 1.355
= 0.1
Identify any outliers.
Lower fence:
Q1 − 1.5 × IQR = 1.355 − 1.5 × 0.1 = 1.205
Upper fence:
Q3 + 1.5 × IQR = 1.455 + 1.5 × 0.1 = 1.605
Because 1.75 is greater than the upper fence, it is
considered an outlier.
Min x = 11
Q1 = 15.5
Median = 18
Q3 = 20
Max x = 22
11
12
13
14
15
10 Min x = 10, max x = 70
Median = 40
IQR = 60 − 30
= 30
The correct answer is D.
11 1 Var Stats
Min x = 2
16
17
18
19
20
21
22
1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75
Height jumped
b The data point 1.75 is an outlier. The data is symmetrical
and has one outlier.
14 Order the data.
30.3, 45.9, 46.9, 47.0, 48.5, 48.2, 49.2, 50.2, 51.6, 52.6, 55.1,
55.5, 57.3, 58.5, 60.3, 61.2
Q1 = 3
Median = 4
Lowest = 30.3
Highest = 61.2
50.2 + 51.6
Median =
2
= 50.9
47.0 + 48.5
Q1 =
2
= 47.75
55.5 + 57.3
Q3 =
2
= 56.4
Q3 = 6
Max x = 7
1
2
3
4
5
6
7
8
9
Number of clients seen in a day
10
11
12 1 Var Stats
Min x = 16
Q1 = 20
IQR = Q3 − Q1
= 56.4 − 47.75
= 8.65
Identify any outliers.
Median = 23
*12
16
P df_Fol i o: 14
17
18
19
20
21
22
23 24 25 26
Temperature (°C)
27
28
29
30
31
32
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.7
Lower Fence:
Q1 − 1.5 × IQR = 47.75 − 1.5 × 8.65 = 34.775
Upper Fence:
Q3 + 1.5 × IQR = 56.4 + 1.5 × 8.65 = 69.375
Because 30.3 is smaller than the lower fence, it is considered
an outlier.
Therefore, 30.3 is an outlier.
Lower Fence:
Q1 − 1.5 × IQR = 4.5 − 1.5 × 2.5 = 0.75
Upper Fence:
Q3 + 1.5 × IQR = 7 + 1.5 × 2.5 = 10.75
Because 15 is greater than the upper fence, it is considered an
outlier.
1.5
28
32
36
40 44 48 52
Litres of fuel
56
60
64
15 a Arrange the data in order.
0245677788889999
Minimum = 0
Maximum = 9
Median = 7.5
Q1 = 5.5
Q3 = 8.5
0
1
2
3 4 5 6 7
Number of rides
8
9 10
The data are negatively skewed with an outlier on the lower
end. The reason for the outlier may be that the person was
not at the show for long or possibly didn’t like the rides.
b Check your boxplot against that shown in the solution to
part a.
16 a The first similar property: Year 12s and Year 11s have
similar IQR values (see part b). The second similar
property: Year 12s and Year 11s have the same minimum
value, which is 25.
b Year 11s: 52 – 35.5 = 16.5
Year 12s: 55 – 39 = 16
c The value of 109 is an outlier.
Q3 + 1.5 × IQR = 55 + 1.5 × 16
= 79
Therefore, values greater than 79 would be considered
outliers. Perhaps the student did not understand how to do
the test, or stopped during the test rather than working
continuously.
17 Order data.
2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7,
7, 7, 8, 8, 9, 15
5+5
Median =
2
=5
4+5
Q1 =
2
= 4.5
7+7
Q3 =
2
=7
Max x = 15, min x = 2
IQR = Q3 − Q1
= 7 − 4.5
= 2.5
P df_Fol i o: 15
15
3 4.5 6 7.5 9 10.5 12 13.5 15 16.5
Number of times perform used per week
18 a Order data.
11, 11, 14, 16, 19, 22, 24, 25, 25, 27, 28, 28, 36,
38, 38, 39
25 + 25
Median =
2
= 25
16 + 19
Q1 =
2
= 17.5
28 + 36
Q3 =
2
= 32
Max x = 39, min x = 11
IQR = Q3 − Q1
= 32 − 17.5
= 14.5
Lower Fence:
Q1 − 1.5 × IQR = 17.5 − 1.5 × 14.5 = −4.25
Upper Fence:
Q3 + 1.5 × IQR = 32 + 1.5 × 14.5 = 53.75
None of the data is smaller than the lower fence or greater
than the upper fence, so there is no outlier.
4
8
12
16
20
24
28
32
36
40
44
b There are no outliers in this data set. There are no data
points outside the range −4.25 to 53.75.
c Check your boxplot against that in the solution to part a.
1.7 Exam questions
1 A circumference of 30 cm coincides with Q3 , so 75% of the
252 people lie below this measurement.
The correct answer is D.
2 The five-number summary consists of the smallest
measurement of 21 cm, lower quartile, Q1 = 27.4 cm, median
of 28.7 cm, upper quartile, Q3 = 30 cm and the largest
measurement of 35.9 cm.
The correct answer is B.
3 Tasman rivers are represented on the second box plot. The
five-number summary is Min, Q1 , Med, Q3 , Max.
Looking at the minimum and maximum first, the minimum is
between 30 and 40 km and the maximum is 180 km – even
though it is an outlier, it is still the maximum. That is enough
to work out that the answer is B, as that is the only option that
has the correct minimum and maximum.
The correct answer is B.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
16
TOPIC 1 Investigating data distributions • EXERCISE 1.9
8 a If the distribution is positively skewed, then the median
would be a better indicator of central tendency.
b The mean is a good indicator of central tendency if the
1.8 Exercise
distribution is symmetrical.
c
If the distribution has an outlier, then the median is a better
1
9 + 12 + 14 + 16 + 18 + 19 + 20 + 25 + 29 + 33 + 35 + 36 + 39
indicator of central tendency.
x=
13
d If the distribution is negatively skewed, then the median is a
= 23.46
better indicator of central tendency.
2
5.5 + 6.3 + 7.7 + 8.3 + 9.7 + 6.7 + 12.9 + 10.5 + 9.9 + 5.1 1.8 Exam questions
x=
10
1.53
= 8.26
1 Mean weight =
= 0.17 kg = 170 g
9
36
The correct answer is E.
3 a x=
5
2 a The wind direction with the lowest recorded wind speed
= 7.2
was south-east.
114
The wind direction with the largest range of recorded wind
b x=
16
speeds was north-east.
= 7.125
[1 mark per 2 correct responses, for a total of 1 mark]
39.9
b The minimum speed (from the boxplot) was 2, so the first
c x=
8
score is 2. The first quartile is also 2 and all scores are
= 4.9875
whole numbers, therefore the 2nd and the 3rd scores are
151
both 2 as well. The median is 3.5; therefore, the 4th score
d x=
9
is 3 and the 5th score is 4. The third quartile and the
= 16.8
maximum score are both 4; therefore, the 6th, 7th and 8th
6.47
scores are all 4.
4 a x=
Hence the scores are 2, 2, 2, 3, 4, 4, 4 and 4. [1 mark]
6
= 1.0783
3 Total shots for 15 rounds = 15 × 72.6 = 1089
Total shots needed over 17 rounds for a mean score of
Not a good indicator of central tendency due to outlier
71 = 17 × 71 = 1207
(2.3).
136
The total number of shots needed over the next two rounds
b x=
will be 1207 − 1089 = 118.
8
The correct answer is C.
= 17
Yes, the mean is a good indicator of central tendency.
1.9 Standard deviation of a sample
247
c x=
8
= 30.875
1.9 Exercise
Yes, the mean is a good indicator of central tendency.
1 Using CAS: s = 3.54 cents
109
2 Using CAS: s = 14.28 %
d x=
7
3 Using CAS: s = 9.489
= 15.57
4 Using CAS: s = 7.306
Not a good indicator of central tendency due to outlier (2).
5 a s = 1.21
243
b s = 2.36
5 x=
20
c s = 6.01
= 12.15
d s = 2.45
The mean number attending each week is approximately
6 Data:
12 people.
2, 4, 7, 7, 9, 10, 11, 14, 14, 14, 14, 15, 16, 16, 17, 18, 18, 19,
6 Data:
21, 22, 22, 23, 23, 23, 27
21, 22, 24, 25, 26, 26, 27, 28, 28, 28, 29, 30, 31, 31,
s = 6.47
32, 33, 34, 34, 36, 38
∴ The answer is C.
1 Var Stats
7 s = 0.48%
583
x=
a
2
20
x
x−x
(x − x)
= 29.15
2.3
0.6625
0.4389
Therefore, the answer is D.
0.8
–0.8375
0.7014
7 Data:
4, 7, 12, 14, 15, 15, 16, 17, 18, 21, 22, 24, 27, 27, 27
1.6
–0.0375
0.0014
1 Var Stats
2.1
0.4625
0.2139
266
x=
1.7
0.0625
0.0039
15
1.3
–0.3375
0.1139
= 17.7
1.4
–0.2375
0.0564
Therefore the answer is A.
1.9
0.2625
0.0689
1.8 The mean of a sample
P df_Fol i o: 16
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.10
∑ ( x − x)
1.5987
=
n−1
7
1.5987
s =√
= 0.48
7
8 Data:
8, 9, 13, 14, 17, 19, 20, 21, 23, 27, 27, 33, 34, 41, 45, 46, 47,
2
b
48, 53, 58
s = 15.49
9 Using CAS: s = 2.96 km/h
10 Using CAS: s = 6.067 pens
11 Using CAS: s = 2.39 °C
12 Using CAS: x = 75.7, s = 5.6
13 Using CAS: s = 3.786 players
14 Using CAS: s = 2.331
1.9 Exam questions
1 Calculate the one-variable statistics with the CAS calculator.
x = 27.77 and sX = 1.664 03...
The correct answer is D.
2 Enter the data into CAS.
x = 5.25
s = 2.38
The correct answer is E.
3 Using a calculator, the standard deviation is s = 1.37 (don’t
confuse with the population standard deviation of 1.33).
The correct answer is B.
1.10 The 68–95–99.7% rule and z-scores
1.10 Exercise
1 a 68% = x ± s
49 − 14 = 35
49 + 14 = 63
68% of group’s concentration span falls between 35 secs
and 63 secs.
b 95% = x ± 2s
49 − 14 × 2 = 21
49 + 14 × 2 = 77
95% of group’s concentration span falls between 21 secs
and 77 secs.
c 99.7% = x ± 3s
49 − 14 × 3 = 7
49 + 14 × 3 = 91
99.7% of the group’s concentration span falls between
7 secs and 91 secs.
2 a x − s = 45 − 1.7 = 43.3 mm
x + s = 45 + 1.7 = 46.7 mm
68% of rainfall is between 43.3 mm and 46.7 mm.
b x − 2s = 45 − 2 × 1.7 = 41.6 mm
x + 2s = 45 + 2 × 1.7 = 48.4 mm
95% of rainfall is between 41.6 mm and 48.4 mm.
c x − 3s = 45 − 3 × 1.7 = 39.9 mm
x + 3s = 45 + 3 × 1.7 = 50.1 mm
99.7% of rainfall is between 39.9 mm and 50.1 mm.
P df_Fol i o: 17
17
3 a 68% is all data up to 1 standard deviation from the mean.
x ± s = 68%
10 − 2 = 8
10 + 2 = 12
b 95% is all data up to 2 standard deviations from the mean.
x ± 2s = 95%
10 − 2 × 2 = 6
10 + 2 × 2 = 14
c 99.7% is all data up to 3 standard deviations from the mean.
x ± 3s = 99.7%
10 − 3 × 2 = 4
10 + 3 × 2 = 16
4 a x + s = 68%
5 − 13 = 3.7
5 + 13 = 6.3
b x + 2s = 95%
5 − 2 × 1.3 = 2.4
5 + 2 × 1.3 = 7.6
c x + 3s = 99.7%
5 − 3 × 1.3 = 1.1
5 + 3 × 1.3 = 8.9
5 a Bell shaped
b Bell shaped
c Not bell shaped
d Not bell shaped
e Not bell shaped
f Bell shaped
6 a 68% = x ± s
1.9 − 0.6 = 1.3
1.9 + 0.6 = 2.5
68% of group hair growth rate falls between 1.3 mm and
2.5 mm per week.
b 95% = x ± 2s
1.9 − 0.6 × 2 = 0.7
1.9 + 0.6 × 2 = 3.1
95% of group hair growth rate falls between 0.7 mm and
3.1 mm per week.
c 99.7% = x ± 3s
1.9 − 0.6 × 3 = 0.1
1.9 + 0.6 × 3 = 3.7
99.7% of group hair growth rate falls between 0.1 mm and
3.7 mm per week.
7 a 68% = x ± 3s
7−2=5
7+2=9
68% of seedlings lie between 5 cm and 9 cm.
b 95% = x ± 2s
7−2×2=3
7 + 2 × 2 = 11
95% of seedlings lie between 3 cm and 11 cm.
c 99.7% = x ± 3s
7−3×2=1
7 + 3 × 2 = 13
99.7% of seedlings lie between 1 cm and 13 cm.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
18
TOPIC 1 Investigating data distributions • EXERCISE 1.10
8
49.85% 47.5%
95%
21
26
As 95% = x ± 2s
26 − 21
𝜎=
2
= 2.5
The correct answer is C.
130
(–3s)
31
160
180
(+2s)
11 a More than 15 days
9
0.15%
34%
34%
9
0.15% 2.35%
13.5%
13.5%
235 240 245 250 255
a More than 214 g = 2.35% + 0.15%
= 2.50%
2.35% 0.15%
260
15
(+3s)
0.15% of employees take more than 15 days off.
b Fewer than 5 days
265
b More than 200 g = 34% + 13.5% + 2.35% + 0.15%
= 50%
c Less than 193 g = 13.5% + 2.35% + 0.15%
= 16%
d Between 193 g and 214 g = 34% + 34% + 13.5%
= 81.5%
2.5%
5
9
(–2s)
0.15% + 2.35% = 2.5% of employees take fewer than 5
days.
c More than 7 days
10 a Less than 170 cm tall
34%
34%
50%
50%
50% + 34% = 84% are less than 170 cm tall.
b Less than 140 cm tall
170
7
9
(–s)
34% + 50% = 84% of employees take more than 7 days.
d Between 3 and 11 days
34%
49.85%
3
(–3s)
2.5%
140
160
(–2s)
0.15% + 2.35% = 2.5% are less than 140 cm tall.
c Greater than 150 cm tall
9
11
(+s)
49.85% + 34% = 83.85% take between 3 and 11 days.
e Between 7 and 13 days
34%
47.5%
34%
50%
150 160
(–s)
34% + 50% = 84% are taller than 150 cm.
d Between 130 cm and 180 cm in height
49.85% + 47.5% = 97.35% are between 130 cm and
180 cm tall.
7 9
(–s)
13
(+2s)
34% + 47.5% = 81.5% take between 7 and 13 days.
12 x = 20 and s = 1
19 marbles is 1 standard deviation below the mean.
Percentage above 19 = 50% + 34%
= 84%
P df_Fol i o: 18
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • EXERCISE 1.10
Number of marbles = 84% × 500
= 420 bags
13 a More than 245 mL
iii Between 2.6 cm and 2.8 cm
0.15%
34%
34%
50%
2.5 2.6
(+s)
245 250
(–s)
34% + 50% = 84%
Number of containers = 84% × 400
= 336 containers
b Less than 240 mL
2.5%
47.5%
47.5%
240
250
260
(–2s)
(+2s)
47.5% + 47.5% = 95%
Number of containers = 95% × 400
= 380 containers
14 a i Between 2.4 cm and 2.6 cm
34%
34%
2.4 2.5 2.6
(+s)
(–s)
34% + 34% = 68%
Number of bolts = 68% × 2000
= 1360 bolts
ii Less than 2.7 cm
50%
47.5%
2.5
P df_Fol i o: 19
Shaded area = 50% − (34% + 0.15%)
= 15.85%
Number of bolts = 15.85% × 2000
= 317 bolts
b Less than 2.3 cm and greater than 2.7 cm
2.5%
240
250
(–2s)
Number of containers = 2.5% × 400
= 10 containers
c Between 240 and 260 mL
2.7
(+2s)
50% + 47.5% = 97.5%
Number of bolts = 97.5% × 2000
= 1950 bolts
2.8
(+3s)
2.5%
2.5
2.7
2.3
(–2s)
(+2s)
2.5% + 2.5% = 5%
Number of bolts rejected = 5% × 2000
= 100 bolts
15 x = 76% and s = 9%
x−x
z=
s
97 − 76
=
9
= 2.33
x−x
16 z =
s
96 − 60
=
12
36
=
12
=3
17 Specialist Maths: x = 83, x = 67, s = 9
English: x = 88, x = 58, s = 14
x−x
a zs =
s
83 − 67
=
9
= 1.78
x−x
ze =
s
88 − 58
=
14
= 2.14
b English has the higher result as it has the higher z-score.
x−x
18 a English z =
s
75 − 65
=
8
= 1.25
x−x
Maths z =
s
72 − 56
=
12
= 1.33
b The Maths mark is better as it has a higher z-score.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
19
20
TOPIC 1 Investigating data distributions • REVIEW 1.11
x−x
s
54 − 60
=
12.5
−6
=
12.5
= −0.48
x−x
b Maths z =
s
78 − 60
=
12.5
18
=
12.5
= 1.44
x−x
c Biology z =
s
61 − 60
=
12.5
1
=
12.5
= 0.08
x−x
d Geography z =
s
32 − 60
=
12.5
−28
=
12.5
= −2.24
x−x
e Art z =
s
95 − 60
=
12.5
35
=
12.5
= 2.8
19 a English z =
i See table in part d.
ii The median number of eggs produced is greatest for
cage chickens; therefore, it is possible that the more
space a chicken has, the fewer eggs it lays.
1.10 Exam questions
1 76.0 is a score 1 standard deviation above the mean.
Therefore, there would be 84% of participants below this
value.
84% × 800 = 672 unsuccessful participants.
The correct answer is D.
81.5 − 69.5
2 Amy z-score:
= 1.846 …
6.5
80.5 − 69.5
Brian z-score:
= 1.692 …
6.5
82 − 69.5
Cherie z-score:
= 1.923 …
6.5
Amy and Cherie scored over 1.8, so will be offered a leading
role, only Brian will not.
The correct answer is C.
3 Greater than 16% is at 1 standard deviation above the mean:
x + 𝜎 = 160
Less than 2.5% is at 2 standard deviations below the mean:
x + 2𝜎 = 115
Solve on your CAS: x = 145 and 𝜎 = 15
The correct answer is C.
1.11 Review
1.11 Exercise
Multiple choice
1 Recording length, which is numerical. Since the lengths are
not fixed values, it is continuous.
The correct answer is B.
2 The observations shown on the stem plot are:
20 a Using CAS:
Living conditions Cage Barn
Free range
Mean
5.2 4.4
4.1
Standard deviation 0.3 0.3
0.2
x−x
b z=
s
4.3 − 4.1
=
0.17
= 1.18
c z = 1 means 1 standard deviation above the mean. The
percentage below this is:
34% + 34% + 13.5% + 2.35% + 0.15% = 84%
d Boxplots with the following 5-number summary:
Cage
Min = 4.7
Q1 = 5
Barn
Free range
Med = 5.15
Q3 = 5.5
Min = 3.9
Q1 = 4.1
Min = 3.8
Q1 = 4
Max = 5.8
Med = 4.35
Q3 = 4.6
Max = 4.9
Med = 4.1
Q3 = 4.2
Max = 4.4
Stem
2
2
3
3
Leaf
0 0 1
6 7 8 9 9
0 1 1
5
Key: 2 1 = 21
20, 20, 21, 26, 27, 28, 29, 29, 30, 31, 31, 35
The correct answer is D.
3 The largest number of people attending was 142, not 140.
The correct answer is C.
Frequency
Score
4
1
2
3
4
5
6
7
8
9
2
3
3
4
1
3
4
3
4
The correct answer is D.
P df_Fol i o: 20
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
5
10 Range = 80
IQR = 30
Q2 = 50
Leaf
Stem
2
2∗
3
3∗
4
4∗
5
5∗
6
6∗
3 4
5 6 8
0 1 2 3 4 4
5 5 7 9 9
0 1 3 3
6 8 8
0 1
6
9
Range = 90 − 10 = 80
Q2 = 50
10
Key: 3 1 = 31
The distribution of the stem plot is positively skewed.
The correct answer is C.
6 The distribution of the data shown in the histogram is
negatively skewed.
The correct answer is A.
7 A set of data has n = 7
Median = 5
Range = 3
This could be
4 5 5 6 6 7 7
Median
Range = 7 − 4
=3
The correct answer is C.
Stem
Leaf
8
1
2 3
2
0 4 5 7
3
1 2 5 9
4
1 3 6 7
2 9 9
5
6
3
Key: 2 4 = 24
18 + 1
th
( 2 )
1
= 9 th
2
35 + 39
Median =
2
= 37
The correct answer is E.
Median position =
9
5
10
Min x = 5
Max x = 35
Range = 30
15
20
25
30
35
IQR = 25 − 10
= 15
Median = 15
The median is equal to the interquartile range.
The correct answer is E.
P df_Fol i o: 21
21
20
30
40
50
60
70
80
90
IQR = Q3 − Q1
= 65 − 35 = 30
The correct answer is B.
11 Q3 = 0.8 m
Q1 = 0.54 m
IQR = Q3 − Q1
= 0.8 − 0.54 = 0.26 m
IQR is 0.26 m, not 0.2 m.
The correct answer is D.
12 The mean of 14, 18, 20, 21, 23, 23, 24, 25, 29, 30 is
227
x=
10
= 22.7
The correct answer is C.
Stem
Leaf
13
1
1∗
2
2∗
3
3∗
5 7 7 8 8 8 8 8 8 9
0 0 0 1 1
6 8
1
5
Key: 2 0 = 20 years
Mean of stem plot:
398
x=
19
= 20.9
The correct answer is D.
14 If the distribution is positively skewed, it is likely that the
mean is greater than the median.
The correct answer is A.
15 If the data is symmetric, then the median is not necessarily the
better measure of centre.
The correct answer is B.
16 Using 1 Var Stats,
sx = $277.75 ≈ $278
The correct answer is B.
17 Using 1 Var Stats,
sx = 14.05 ≈ 14
The correct answer is C.
18 95% of batteries
95%
x − 2s = 1200 − (2 × 10)
= 1180
–2s
1200
+2s
x + 2s = 1200 + (2 × 10)
= 1220
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
22
c
So 95% of batteries sampled have a lifetime (hours) between
1180 hours and 1220 hours.
The correct answer is D.
19 68% = x ± s
45 − 5 = 40
45 + 5 = 50
40 ≤ x ≤ 50
95% = x ± 2s
45 − 2 × 5 = 35
45 + 2 × 5 = 55
35 ≤ x ≤ 55
99.7% = x ± 3s
45 − 3 × 5 = 30
45 + 3 × 5 = 60
30 ≤ x ≤ 60
95% lie between 35 and 55 not 35 and 50
The correct answer is C.
114
120
(–2s)
2.5% of broom handles are shorter than 114 cm.
The correct answer is B.
Short answer
21 a An example of categorical data is the colour of cars.
b i An example of discrete numerical data is the number of
phone calls made in one month.
ii An example of continuous numerical data is the length
of each call made in one month. Many other answers
are possible.
b
0 50
52
*24 a
P df_Fol i o: 22
Frequency
Key: 6 0 = $6
8 8 9
2 3
4 4
7
Range = 26 − 2
= 24
Mode = 14
IQR = 18 − 8.5
= 9.5
24 a See figure at bottom of the page*
Check by entering appropriate values using statistics menu
and suitable window.
X Minimum = 49
X Maximum = 79
X Scale = 2
Y Minimum = 0
Y Maximum = 12
Y Scale = 1
b The data are approximately symmetrical. More than half
the drivers exceed the speed limit. The fastest drivers were
about 15 km/h over the limit.
Many other conclusions are possible.
Key: 6 0 = $6
Stem
Leaf
0 1
5 7
1 1 3
8 8 9
2 3 4 4
7
1 1
3
Max x = 26
Leaf
0 1 5 7
1 1 3 8 8 9
2 3 4 4 7
6
6∗
7
7∗
8
8∗
5
7
Stem plot b is probably an appropriate display. No real
need for stems in fifths.
23 2, 5, 6, 6, 8, 9, 10, 12, 12, 14, 14, 14, 15,
15, 17, 18, 18, 19, 21, 23, 26
Input data into L1 .
1 Var Stats:
Min x = 2
Q1 = 8.5
Median = 14
Q3 = 18
2.5%
Stem
6
7
8
Leaf
0 1
Key: 6 0 = $6
20
22 a
Stem
6
6
6
6
6
7
7
7
7
7
8
8
8
8
8
10
8
6
4
2
54
56
58
60 64 66 68
Speed (km/h)
70
72
74
76
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
25 a The distribution is positively skewed.
b There would need to be a shift of some of the amounts in
the 20s to the 30s and 40s.
26 a Using CAS input data into L1 .
1 Var Stats
Min x = 1
Q1 = 3.5
Median = 5
Q3 = 7
Max x = 9
290 060
5.46
59 070
4.77
8450
3.93
log10 (income)
Frequency
3−<4
4−<5
5−<6
6−<7
2
6
3
1
23
Extended response
1
2
3
4
5
6
7
8
9
b The distribution is approximately symmetrical.
27 a Min value = 150
Max value = 650
10
Range = 650 − 150
= 500
7
× 100 = 58.3%
b
12
28 Data:
21, 22, 22, 24, 24, 24, 25, 26, 26, 28
28, 29, 29, 30, 32, 32, 34, 35, 36, 38
Input data into L1 .
1 Var Stats:
sx = 0.05 mL
29 a i 68% = x ± s
12 + 1.2 = 13.2
12 − 1.2 = 10.8
68% of the dogs would be between 10.8 and 13.2 years.
ii 95% = x ± 2s
12 + 2 × 1.2 = 14.4
12 − 2 × 1.2 = 9.6
95% of the dogs would be between 9.6 and 14.4 years.
iii 99.7% = x ± 3s
12 + 3 × 1.2 = 15.6
12 − 3 × 1.2 = 8.4
99.7% of the dogs would be between 8.4 and 15.6
years.
b There is a large range of life spans for these dogs. The
oldest dog is almost twice as old as the youngest.
30 Kory, as he has a greater z-score (1.5 compared with 0.875).
31 8.2 − 6.6 = 1.6
101.6 = 39.81
Therefore, the first measurement is 40 times greater than the
second.
32
Income
log10 (income)
P df_Fol i o: 23
45 000
4.65
20 000
4.30
360 000
5.56
750 000
5.88
3200
3.51
1 048 500
34 590
6.02
4.54
37 250
4.57
65 710
4.82
33 a
i A stem plot is more appropriate since there are only 25
observations in each set.
10E
10C
Leaf
Stem
Stem
Leaf
0
0
0
0
0
0
4
0
7 7
0
0
8 9
0
9 9
0 1 1
1
1
0 0 1
2 2 2 3 3 3 3 3
1
2 3
1
1
4 4 4 4 4 4 5 5 5
1
4 4 5 5 5
1
6 6
1
6 7 7 7 7
9
1
1
8 8 8 9 9
Key: 1 3 = 13
The distribution of 10C is negatively skewed with no
outliers.
The distribution of 10E is symmetric with no outliers.
ii Class 10C:
Input data into L1 .
From 1 Var Stats:
Min x = 4
Q1 = 10
Median = 15
Q3 = 17
Max x = 19
Therefore, the median is 15.
IQR = 17 − 10 = 7
Range = 19 − 4 = 15
Mode = 17
iii Class 10E:
Input data into L2 .
From 1 Var Stats:
Min x = 8
Q1 = 12
Median = 13
Q3 = 14.5
Max x = 19
IQR = 14.5 − 12 = 2.5
Range = 19 − 8 = 11
Mode = 14
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
24
4 5
10E
8
8+1
1
th = 4 th position
( 2 )
2
Q1 = 108.5, Q3 = 128
IQR = 128 − 108.5 = 19.5 beats/ min
Mode = 130 beats/ min
c Office workers
Q3 /Q1 =
10C
9
6
7
8
9
10 11 12 13 14 15 16 17 18 19
10 11 12 13 14 15 16 17 18 19
70
iv Class 10C:
341
Mean =
25
x = 13.64
sx = 4.24
Class 10E:
330
Mean =
25
x = 13.2
sx = 2.35
v For 10C the median provides a better indication,
whereas for 10E the mean and the median are close.
vi The 68 − 95 − 99.7% rule can be applied to the
distribution of 10E′ s data since it is approximately
bell-shaped.
b We use the median of the 10C scores to give us an
indication of the centre of the distribution. The median of
the 10C scores is 15. For 10E, the distribution is symmetric
and hence we use the mean to give us an indication of the
centre of the distribution. The mean of 10E scores is 13.2.
The range of the 10C scores is 15, whereas for 10E the
range is 11. Also, the standard deviation for 10C is 4.24
and for 10E it is 2.35. This means that the scores in 10C are
more spread out than those in 10E, which are relatively
bunched. So, while 10C has more students with higher
marks than in 10E, the range of marks in 10C is greater and
this would make it a more challenging class to teach.
34 Office workers:
Stem
7
8
9
10
11
Leaf
6
5
6 7
0 2
12
0 1 2 4 6 7 9
13
0 0 4
Key: 12 4 = 124 beats/ min
a Office workers distribution is negatively skewed with one
outlier (76).
b
Min x = 76, max x = 134
Range = 134 − 76 = 58 beats/ min
18 + 1
1
th = 8 th position
Median =
( 2 )
2
= 121.5 beats/ min
80
90
100
110
120
130
140
d Mean =
Beats per minute
1869
16
x = 116.8125 beats/ min
sx = 15.3 beats/ min
Sports instructors:
Stem
6
7
8
9
10
Leaf
2 4 8 8 9
2 2 3 5 7 9
2 8
6
8
Key: 6 2 = 62 beats/ min
a Sports instructors distribution is positively skewed with one
outlier (108).
b Min x = 62, max x = 108
Range = 108 − 62 = 46
Median = 73 beats/ min
Q3 = 82, Q1 = 68
IQR = 82 − 68 = 14 beats/ min
Mode∶ 68, 72
c Sports instructors
60 65 70 75 80 85 90 95 100 105 110
Beats per minute
1153
d Mean =
15
x = 76.87 beats/ min
sx = 12.43 beats/ min
e Since distributions are not symmetric, the 68−95−99.7%
rule cannot be applied.
f Office workers: Pulse rates are generally very high,
clustered around 120−130 beats/min. Also, there is one
person whose rate was much lower than the rest. This
outlier (76) produces a large range and makes the mean
slightly lower than the median. As a result, the median is a
more appropriate measure of the centre of the data rather
than the mean.
Sports instructors: Pulse rates are generally low, clustered
around 60−70 beats/min, although there are a few people
with rates much higher, which makes the mean slightly
higher than the median and also produces quite a large
range. As a result of the skewed distribution, the median is
the more appropriate measure of the centre of the data
rather than the mean, although there is little difference
between these values.
P df_Fol i o: 24
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
35 a Since rounding to the nearest cm is used, this is discrete
numerical data.
b Lowest score = 13
Highest score = 19
Use 1 stem, split into fifths.
Stem
1
1
1
Leaf
6
1
8 8 9 9
6
6
6
6
6 7 7 7
Key: 1 3 = 13 cm
The data looks slightly negatively skewed.
c There are 20 values, so the median lies between the 10th
and 11th value.
10th value = 16; 11th value = 16
Median = 16 cm
Q1 will be between the 5th and 6th values.
5th value = 15; 6th value = 15
Q1 = 15 cm
Q3 will lie between the 15th and 16th values.
5th value = 17; 6th value = 17
Q3 = 17 cm
Lowest score = 13
Highest score = 19
14
15 16 17 18
Fish length (cm)
19
20
d From the boxplot, the distribution seems to be symmetrical.
e Given the data itself, the boxplot gives a better indication of
the distribution’s shape.
f Mean = median = 16 cm
Since the mean and median are the same, this suggests that
this is a symmetric distribution.
g From part f:
s = 1.75 cm
x + s = 17.75
16%
16
17.75
Since 68% of data lies between x ± s, 16% lies beyond x + s.
The proportion of salmon with lengths greater than
17.75 cm is 16%.
16% of 10 000 = 1600
h Use the statistics menu and appropriate functions to draw a
boxplot on CAS. Determine values:
Median = 18
Q1 = 16.5
Q3 = 19
From the boxplot, the mean is lower than the median, this
suggests that the data is negatively skewed, with a lower end
outlier (12).
x = 17.3
i
s = 1.95
x + s = 19.25
Since 68% of data lies between x ± s, 16% lies beyond x + s.
3 3
4 4 5 5 5
1
13
25
16%
17.3
19.25
The proportion of salmon, in this population, with lengths
greater than 19.25 cm is 16%.
16% of 10 000 = 1600
j The river fish seem to be larger overall. Only 1600 of the
coastal fish lie above 17.75 cm, whereas 1600 of the river
fish lie above 19.25 cm. All of the quartiles of the river fish
are higher than those for the coastal ones. It would seem
that the river fish have grown more than the coastal fish.
1.11 Exam questions
1. Q1 = 148, Q3 = 159, therefore IQR = 159 − 148 = 11
UF = 159 + 1.5 × 11 = 175.5
The correct answer is E.
2. a. There are 3 numerical variables.
[1 mark]
b. 1.81 (use your CAS)
[1 mark]
14.50 − 13.74
c. z =
= 0.5 [1 mark – note that rounding
1.43
applies here]
d. A z-score of −1 means 1 standard deviation below the
mean. Therefore, the percentage of athletes who would be
expected to jump higher than Chara is
34% + 50% = 84%
[1 mark]
e. Q3 and the maximum are the same value of 1.87 [1 mark]
f. Calculate the upper fence:
UF = 42.88 + 1.5 × (42.88 − 40.88)
[1 mark]
= 45.88
[1 mark]
3. The data is bunched down the lower end of the histogram with
a tail going to the right, therefore it is positively skewed.
Given the large range of the data, it is likely that the data at
the upper end is an outlier.
The correct answer is B.
4. log10 (100 000) = log10 (105 ) = 5
Therefore, less than 5 on the horizontal scale is 7 + 1 = 8
countries.
The correct answer is D.
5. a There are three categorical variables: City, Congestion
level and Size.
[1 mark]
b There are two categories that are ordinal: Congestion level
and Size.
[1 mark]
c The large cities with medium traffic congestion levels are
Newcastle-Sunderland and Liverpool.
[1 mark]
P df_Fol i o: 25
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 1 Investigating data distributions • REVIEW 1.11
26
d
City size
Congestion level
Small
Large
high
4
2
medium
low
Total
4
8
16
2
3
7
[1 mark per 3 correct responses, for a total of 2 marks]
e The percentage of small cities with high traffic congestion
4
× 100 = 25%.
[1 mark]
is
16
f The distribution of the increase in travel time is positively
skewed.
[1 mark]
g Q1 is at 30 minutes and Q3 is at 39 minutes. The IQR is
Q3 − Q1 = 39 − 30 = 9.
The upper fence is 1.5 × IQR + Q3 = 1.5 × 9 + 39
= 52.5
[1 mark]
P df_Fol i o: 26
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.3
27
Topic 2 — Investigating associations between two
variables
2.2 Exercise
1 a Since the number of air conditioners sold could reasonably
be a response to the daily temperature:
daily temperature = explanatory variable
air conditioners sold = response variable
b You wouldn’t think that a person’s favourite colour could
be affected by their age.
Therefore, it is not appropriate to label one variable
explanatory and one variable response.
2 a Since the size of the crowd could be a response to the teams
playing:
size of the crowd = response variable
teams playing = explanatory variable
b The net score is dependent upon the golfer’s handicap:
net score of a round of golf = response variable
golfer’s handicap = explanatory variable
3 See table at bottom of the page*
4 See table at bottom of the page*
5 It is reasonable to expect that a mother’s reaction time to a
certain stimulus would depend on how well rested she is (i.e.
the amount of sleep she had). The answer is C.
6 Since it is possible to identify the explanatory variable as
being ‘vitamin D level’, the answer is True.
7 It is reasonable to expect the number of training sessions to be
explanatory, and the weight loss to be the response variable.
The statement is False.
8 a The number of minutes on a basketball court: Explanatory
variable
b The number of points scored: Response variable
9 a Response variable: electricity bill
b Explanatory variable: number of Christmas lights
10 a Explanatory variable: hours spent playing
b Response variable: vocabulary
*3
2.2 Exam questions
1 Both variables define categorical variables.
The correct answer is B.
2 The data represents a numerical variable and a categorical
variable. Parallel boxplots graph this data.
A histogram requires two numerical variables.
A scatterplot requires two sets of numerical data.
A time series plot requires two sets of numerical data.
A back-to-back stem plot requires categorical sets (only two
options) and numerical data.
The correct answer is E.
3 Training sessions would generate or create data linked to
weight loss. Therefore, training sessions are the explanatory
(horizontal) variable and weight loss is the response (vertical)
variable.
The correct answer is D.
2.3 Contingency (two-way) frequency tables and
segmented bar charts
2.3 Exercise
1 Note: Black data are given in the question; grey data are the
answers.
Explanatory
Response
a
The age of an AFL footballer
His annual salary
b
The amount of fertiliser a plant
receives
The growth of a plant
c
e
The number of members in a
household
a
The month of the year
The size of the electricity bill
b
Hours spent on preparation
Marks obtained for the test
d
15 to 30
23
Over 30
15
Total
38
Samsung
Total
15
7
22
38
22
60
Marginal distribution: Apple =
38
22
Samsung =
60
60
= 0.63
= 0.37
Not appropriate
The size of the house
c
P df_Fol i o: 27
Apple
The number of books read and the eye colour of
the reader
The voting intentions of a woman and her weekly
consumption of red meat
d
*4
Age group
Phone
2.2 Response and explanatory variables
Mark obtained on English test and the mark
obtained on a Maths test
The season
The cost of grapes
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.3
2 Note: Black data are given in the question; grey data are the
answers.
i 47 − 25 = 22
ii 51 − 25 = 26
iii iv − ii = 45 − 26
= 19
iv 92 − 47 = 45
v 92 − 51 = 41
b i 100 − 53 = 47%
ii 100 − 42 = 58%
iii 100%
iv 100%
7 a
Teenagers
Adults
Total
Coke
Pepsi
221
166
155
108
376
274
Total
387
263
650
Conditional distribution:
221
Teenagers who prefer Coke =
387
= 0.57
166
Teenagers who prefer Pepsi =
387
= 0.43
37 (given)
Attitude
Inner city
Against 102 (139 − 37)
For
Total
139 (given)
Residents
Outer suburban
79 (given)
23 (102 − 79)
102 (given)
Lesson
length
116
Marginal distribution∶ For =
= 0.48,
241
125
Against =
= 0.52
241
4
Student type
45 min
1 hour
Total
86 (given)
93 (given)
125 (102 + 23)
241 (139 + 102)
179 (86 + 93)
Conditional distribution:
33
Senior students prefer 45 mins =
= 0.35,
93
60
= 0.65
Senior students prefer an hour =
93
5
116 (37 + 79)
Total
Junior
Senior
Total
50 (86 − 36) 33 (93 − 60) 83 (50 + 33)
36 (given)
60 (given) 96 (36 + 60)
Age group
Teenagers
Phone
Apple
Samsung
Total
6
Adults
23
× 100 = 60.5%
38
15
× 100 = 39.5%
38
38
× 100 = 100%
38
15
× 100 = 68.2%
22
7
× 100 = 31.8%
22
22
× 100 = 100%
22
Age group
Teenagers
Drink
Coke
Pepsi
Total
221
× 100 = 57.1%
387
166
× 100 = 42.9%
387
387
× 100 = 100%
387
Adults
155
× 100 = 58.9%
263
108
× 100 = 41.1%
263
263
× 100 = 100%
263
Age group
Preference
3
8
Rent by
themselves
18 to 30
31 to 40
9
× 100 = 43%
21
16
× 100 = 41%
39
100%
100%
12
23
× 100 = 57%
× 100 = 59%
( 21
)
( 39
)
Share with
friends
Total
The information in the following two-way frequency table
relates to questions 9 and 10.
Attitude
Drink
Age group
Technical staff
Total
For
Against
Administrative
staff
53
37
98
31
151
68
Total
90
129
219
9 From the above table, we can conclude that 59% of
administrative staff were for the upgrade.
53
× 100% = 58.88%
( 90
)
≃ 59%
The correct answer is D.
10 From the above table, we can conclude that 76% of the
technical staff were for the upgrade.
98
× 100% = 75.968%
)
( 129
≈ 76%
The correct answer is C.
11
Age group
Choice
28
Entrée
Main
Dessert
Total
Entrée (children) =
Children
8
31
19
58
8
× 100%
58
= 13.79%
≈ 14%
31
Main (children) =
× 100%
58
= 53.45%
≈ 53%
Adults
18
16
27
61
P df_Fol i o: 28
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Total
26
47
46
118
TOPIC 2 Investigating associations between two variables • EXERCISE 2.3
Dessert (children) =
Children
14
53
33
100
Choice
Adults
30
26
44
100
Dessert
Main
Entrée
100
90
80
70
60
50
40
30
20
10
0
Superannuation
32
50
Investment
Total
23
100
32
100
Investment
Superannuation
Mortgage
100
90
80
70
60
50
40
30
20
10
0
40+
x
Age
The under 40s have a focus on their mortgage, whereas the 40
and overs prioritise their superannuation.
13 a
Adults
Delegates
x
Age
Children enjoy the main meal the most, while adults enjoy the
dessert the most.
Mortgage
<40
42
Age group
40+
18
Total
60
Superannuation
30
51
81
Investment
Total
21
93
33
102
54
195
42
Mortgage < 40 =
× 100%
93
= 45.16%
≈ 45%
30
Superannuation < 40 =
× 100%
93
= 32.26%
≈ 32%
21
Investment < 40 =
× 100%
93
= 22.58%
≈ 23%
18
Mortgage 40+ =
× 100%
102
= 17.65%
≈ 18%
For
Against
Total
Liberal
22 (62 − 40)
40 (given)
62 (given)
Labor
28 (71 − 43)
43 (given)
71 (given)
Total
50
83
133
Delegates
Attitude
12
Choice
40+
18
<40
Children
P df_Fol i o: 29
Mortgage
<40
45
Age group
Percentage
Entrée
Main
Dessert
Total
51
× 100%
102
= 50%
33
× 100%
Investment 40+ =
102
= 32.35%
≈ 32%
Superannuation 40+ =
Attitude
Percentage
Choice
19
× 100%
58
= 32.76%
≈ 33%
18
× 100%
Entrée (adults) =
61
= 29.51%
≈ 30%
16
Main (adults) =
× 100%
61
= 26.23%
≈ 26%
27
Dessert (adults) =
× 100%
61
= 44.26%
≈ 44%
29
For
Against
Total
Liberal
22 100
×
= 35.5%
62
1
40 100
×
= 64.5%
62
1
100%
Labor
28 100
×
= 39.4%
71
1
43 100
×
= 60.6%
71
1
100%
b There is not a lot of difference in the reactions.
c To construct the segmented bar chart:
• Rule out and label the axes.
• Draw 2 columns of the same width (one for Liberal, one
for Labor) and height to 100%.
Break each column into 2 segments: bottom segment
corresponds to the 1st cell in each column. So draw the
line at 35.5% level for the ‘Liberal’ column and at 39.4%
level for the ‘Labor’ column.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
30
TOPIC 2 Investigating associations between two variables • EXERCISE 2.4
Percentage
• Add legend.
100
90
80
70
60
50
40
30
20
10
0
2.4 Back-to-back stem plots
Attitude
Against
For
Liberal
Labor
2.4 Exercise
1 Lowest score = 30
Highest score = 39
Use a stem of 3, divided into fifths.
4B
Stem 4A
3
0 0 1
1
3 2
3
2 3
3
5 5
4
7 6
3
9 9
3
8
x
Key 3 ∣ 1 = 31
Delegates
There is not a lot of difference in the reactions.
14 Answer C: The number of mid-sized towns is
7 + 31 + 16 = 54 (column total)
16
× 100 ≈ 30%
54
15 Conditional distribution:
4
Large town and no waste =
= 0.15
27
5
Large town and medium waste =
= 0.19
27
18
Large town and high waste =
= 0.67
27
Note: Rounding causes the total to be greater than 100%.
5
16 Percentage small town and high waste =
× 100
6+8+5
5
× 100
=
19
= 26.32%
2 Lowest score = 21
Highest score = 49
Use a stem of 2, 3 and 4 divided into halves.
Stem
Chemistry
Physics
4
9 8
4 2 2 0
6
4 1
9 8 5 5
Key 2 ∣ 4 = 24
1 Select the option that shows an association between two
variables, not just a fact about the table (eliminate A and B).
The supportive statement containing correct values is
option E.
20% of sugar traps contained more than 500 moths, while
10% of light traps contained more than 500 moths.
The correct answer is E.
2 There are 11 adults with high blood pressure who are also
under 50 years of age. There are a total of 58 adults who are
under 50 years of age.
*4
P df_Fol i o: 30
Lion cubs
Tiger cubs
38
25
3.4
3.0
45
36
21
46
5.0
2.7
30
44
39
39
4.2
3.7
41
38
3.7
3.3
22
24
5 5 6 7 9
4 See table at the bottom of the page*
Highest − 5.0 kg
Lowest − 2.6 kg
Use a stem of 1, divided into halves.
Key: 2 ∣ 7 = 2.7 kg
11
× 100 = 18.9%.
58
= 19%
The correct answer is B.
3 30% of 300 = 90.
The correct answer is B.
20
23
1
8 8
0 1 2 3 4
8
3 See table at bottom of the page*
Highest − 48
Lowest − 20
Use a stem of 1, divided into halves.
Key: 2 ∣ 3 = 23
Stem French
Chinese
2110
2
34
7655
2*
558
3 2 100
3
0144
9 877
3*
56889
4
2344
2 1
5
4*
68
2.3 Exam questions
*3 Chinese
French
2
2*
3
3*
4
4*
27
25
4.9
4.0
33
42
30
38
3.4
3.1
21
34
3.8
2.6
25
28
32
31
4.8
3.2
37
44
42
30
3.6
3.6
26
35
31
48
4.3
3.1
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
25
43
37
34
TOPIC 2 Investigating associations between two variables • EXERCISE 2.4
Stem
Tiger cubs
Use a stem of 1, divided into halves.
2*
3
3*
4
4*
5
6 7
0 1 1 2 3
6 7
0
a
5 Company A
Stem
Company B
4 4 2 2 2
8
4 2 1
9 8
13
13*
14
14*
15
15*
16
16*
0 1 3 4
8 8
Lion cubs
4 4
8 7 6
3 2
9 8
0
6
0
8
Key 13 ∣ 0 = 130
Mean
Median
IQR
Supermarket A
Mean = 19
Supermarket B
s = 4.88
Q1 = 16
s = 7.24
Q1 = 20
*7
Company A
Company B
143.57
141.5
149 − 134 = 15
144.71
145.5
153 − 134 = 19
11.42
10.87
Stem
English
2
9
7 4
8 8 7 5
4 2
8 2
4
5
6
7
8
9
10
7 8
2
1 4 5 8 9
0 7
6
0
English
Mean
Median
IQR
74.67
77.5
83 − 65.5 = 17.5
76.42
76.5
83.5 − 66.5 = 17
Standard
deviation
15.07
13.60
11
10
Median = 25.5
15
15
20
20
25
25
Q2 = 30
Max x = 27
IQR = 7
Max x = 35
IQR = 10
b For supermarket A, the mean is 19, the median is 18.5, the
standard deviation is 4.9 and the interquartile range is 7.
The distribution is approximately symmetric.
For supermarket B, the mean is 24.4, the median is 25.5, the
standard deviation is 7.2 and the interquartile range is 10.
The distribution is approximately symmetric. The centre
and spread of the distribution of supermarket B is higher
than that of supermarket A. There is greater variation in the
number of trucks arriving at supermarket B.
8 10A
10B
12
10
a
10A
10B
0
2 3
4 4 5
7
9
Mean = 41.5
s = 1.60
Min x = 12
Q1 = 13.5
Median = 14.5
Q3 = 15.5
Max x = 17
IQR = 2
Mean = 14.25
s = 2.81
Min x = 10
Q1 = 12.5
Median = 14
Q2 = 16
Max x = 19
IQR = 3.5
13
12
14
13
14
14
15
14
Highest − 19
Lowest − 10
Use a stem of 1, divided into fifths.
History
A
B
Min x = 10
Q3 = 23
Stem
1
3 2
1
1
5 5 4 4
7 6
1
1
Key 1 ∣ 0 = 10 marks
Statistical analysis:
10A
10B
History has a slightly higher median; however, English has
a slightly higher mean. Their standard deviations are
similar, so overall the results are quite similar.
7 See table at bottom of the page*
Highest − 35
Lowest − 10
P df_Fol i o: 31
Mean = 24.36
Median = 18.5
History
b
B
0
5 6
0 1 3
5 6 8 9
0 1 2
5
Min x = 11
They are both positively skewed. The median is a better
indicator of the centre of the distribution than the mean. This
shows that Company B handed out more pamphlets, taking
into account that the IQR and standard deviations are quite
similar.
Key 5 ∣ 7 = 57
Stem
1
1*
2
2*
3
3*
Key 1 ∣ 0 = 10 trucks
Statistical analysis:
5 6 8
3 3
5
0 2
Standard deviation
6 a
A
2 1
7 7 6 6 5
4 3 2 1 0
7 5
12
30
16
35
31
21
16
27
31
16
32
17
21
17
23
22
26
15
15
16
17
23
28
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
17
19
24
29
TOPIC 2 Investigating associations between two variables • EXERCISE 2.4
32
b For the 10A marks, the mean is 14.5, the median is 14.5,
the standard deviation is 1.6 and the interquartile range is 2.
The distribution is approximately symmetric. For the 10B
marks, the mean is 14.25, the median is 14, the standard
deviation is 2.8 and the interquartile range is 3.5. The
distribution is symmetric.
The centre of each distribution is about the same. The
spread of marks for 10B is greater, however. This means
that there is a wider variation in the marks for 10B
compared to the marks for 10A.
9
2021
2022
30
22
31
26
35
27
37
28
39
30
Highest − 46
Lowest − 22
Use stems of 1, divided into halves.
a
Leaf
2021
1 0
9 7 5
3 2 1 1
6
Stem
41
31
41
31
42
33
43
34
46
36
Mean = 29.8
sx = 4.16
Min x = 22
Max x = 46
IQR = 7
Max x = 36
IQR = 6
Q1 = 35
Median = 40
Q3 = 42
2
5
0 1
6 7
2
6
Mean = 26.75
Mean = 33.63
s = 8.19
s = 2.82
Min x = 23
Min x = 22
Q1 = 27.5
Median = 33.5
Q3 = 39.5
Max x = 31
Max x = 46
IQR = 4.5
IQR = 12
b For the distribution of Species 1, the mean is 26.75, the
median is 26.5, the standard deviation is 2.8 and the
interquartile range is 4.5.
For the distribution of Species 2, the mean is 33.6, the
median is 33.5, the standard deviation is 8.2 and the
interquartile range is 12.
The centre of the distributions is very different: it is much
higher for Species 2. The spread of the ages of Species 1 is
very small but very large for Species 2.
11 Kindergarten 3 13 14 25 28 32 36 41 47 50
Q1 = 27
Median = 30.5
Q3 = 33
b The distribution of marks for 2021 and for 2022 are each
approximately symmetric.
For the 2021 marks, the mean is 38.5, the median is 40, the
standard deviation is 5.2 and the interquartile range is 7.
The distribution is approximately symmetric.
For the 2022 marks, the mean is 29.8, the median is 30.5,
the standard deviation is 4.2 and the interquartile range is 6.
The spread of each of the distributions is much the same,
but the centre of each distribution is quite different with the
centre of the 2022 distribution quite a lot lower. The work
may have become a lot harder.
10 Species 1 23
24
25
26
27
28
30
31
Species 2 22
25
30
31
36
37
42
46
Highest − 46
Lowest − 22
Use stems of 1, divided into halves.
Species 2
Key 2 ∣ 2 = 22 years old
Statistical analysis:
Species 1
Species 2
Q3 = 29
2022
Mean = 38.5
sx = 5.2
Min x = 30
2
2*
3
3*
4
4*
Median = 26.5
Key 2 ∣ 2 = 22 marks
Statistical analysis:
2021
Stem
4 3
8 6 8 7
1 0
Q1 = 24.5
Leaf
2022
2
6 7 8
0 1 1 3
6
2
2*
3
3*
4
4*
a Species 1
Prep. school
Highest − 52
Lowest − 3
Use stems of 1.
5 12 17 25 27 32 35 44 46 52
a Kindergarten
Stem
Prep. school
0
1
2
3
4
5
5
2 7
5 7
2 5
4 6
2
3
4 3
8 5
6 2
7 1
0
Key 1 ∣ 2 = 12 points
Statistical analysis:
Kindergarten
Mean = 28.9
Prep. school
Q1 = 14
Q1 = 17
s = 15.4
Min x = 3
Median = 30
Q3 = 41
Max x = 50
IQR = 27
Mean = 29.5
s = 15.3
Min x = 5
Median = 29.5
Q3 = 44
Max x = 52
IQR = 27
P df_Fol i o: 32
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.4
b For the distribution of scores of the kindergarten children,
the mean is 28.9, the median is 30, the standard deviation is
15.4 and the interquartile range is 27.
For the distribution of scores for the prep. children, the
mean is 29.5, the median is 29.5, the standard deviation is
15.3 and the interquartile range is 27. The distributions are
very similar. There is not a lot of difference between the
way the kindergarten children and the prep. children
scored.
12 The pair of variables that could be displayed on a
back-to-back stem plot is: the time put into completing an
assignment and a pass or fail score on the assignment.
The correct answer is B.
Why: bivariate data, involving a numerical variable and a
categorical variable with 2 categories.
13 A back-to-back stem plot is a useful way of displaying the
relationship between the age and attitude to gambling (for or
against).
The correct answer is C.
Why: Age is a numerical variable; attitude is categorical with
two categories (for or against).
14 Key: 7 ∣ 2 = 72
Stem
Year 11
Year 12
2
7
0 2
7*
6
2 0
8
0 1 3 3 4
8*
8
8 6
4 4 2 0 0
9*
0 2 4
9*
6
0
10
15 A back-to-back stem plot is used to display bivariate data,
involving numerical and categorical variables.
The correct answer is D.
16 Key: 3 ∣ 1 = 31
Mathematical
General
Methods
Stem
Mathematics
98
2∗
8
431
3
0 1
3∗
5 7 7 8
9966
44210
4
1 3 3 4
4∗
6 8 8
Mean
Median
IQR
Standard
deviation
Mathematical
Methods
36.86
37.5
44 − 33 = 11
General
Mathematics
39.21
39.5
44 − 35 = 9
5.29
6.58
Mathematical Methods has a higher IQR, but a slightly lower
standard deviation. General Mathematics had a greater mean
(39.21) compared to Mathematical Methods (39.5), as well as
a greater median: 39.5 compared to 36.86. This suggests that
students do better in General Mathematics compared to
Mathematical Methods by an average of two points.
33
2.4 Exam questions
1 Back-to-back stem plots are used when comparing categorical
data with numerical data. The car’s speed is the numerical
data. Options B and E both contain categorical data, but for a
back-to-back stem plot, the categorical data can only contain
two categories, that is, M or F.
The correct answer is E.
2 The male IQR = 43 − 23 = 21.
Adding 43 or 45 to the female data would find Q1 = 52 and
Q3 = 74. Does not give IQR = 20
Adding 74 to the female data would find Q1 = 55 and
Q3 = 74. Does not give IQR = 20
Adding 75 to the female data would find Q1 = 55 and
Q3 = 75. Give IQR = 20
Adding 84 to the female data would find Q1 = 55 and
Q3 = 76. Does not give the IQR = 20
The correct answer is D.
3 a Using CAS: mean (x) = 20, median = 18, mode = 21,
standard deviation = 4.7
Award 1 mark for each correct value.
b Min X = 17
Q1 = 17.5
Med = 18
Q3 = 21
Max X = 37
Range = 37 − 17
= 20
[1 mark]
IQR = 21 − 17.5
= 3.5
[1 mark]
c Using CAS: mean (x) = 25.4, median = 21, mode = 18,
standard deviation = 10.5
Award 1 mark for each correct value.
d Min X = 18
Q1 = 18
Med = 21
Q3 = 33
Max X = 45
Range = 45 − 18
= 27
[1 mark]
[1 mark]
IQR = 33 − 18
= 15
[1 mark]
e Answers will vary. Students should include at least 3 of the
following.
There are many more male offenders than female
offenders.
[1 mark]
The ages of the male offenders include many younger
males, even below legal driving age.
[1 mark]
The mean age of male offenders is 20, while the mean age
for female offenders is 25.4 (much higher).
[1 mark]
There is a greater deviation in the female results as
compared to the male results, as seen by the different
standard deviation values.
[1 mark]
Most of the offenders were below 25 (only 3 out of
26 were older).
[1 mark]
P df_Fol i o: 33
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
[1 mark]
TOPIC 2 Investigating associations between two variables • EXERCISE 2.4
34
b Clearly, the median height increases from Year 9 to
Year 11. There is greater variation in 9A’s distribution than
in 10A’s. There is a wide range of heights in the lower 25%
of 9A’s distribution. There is a greater variation in 11A’s
distribution than in 10A’s, with a wide range of heights in
the top 25% of the 11A distribution.
2.5 Parallel boxplots and dot plots
2.5 Exercise
1
Mean
n
Min
s
Q1
6A 13 16.2 … 1.52 … 13.8 15
6B 13 17.7 … 1.73 … 14.8 16.4
Median
Q3
Max
16.1
17.6 18.8
17.3
19.6 20.3
4 Statistical analysis for boxplots:
6A
Min x
Q1
20–29
2000
5000
30–39
4000
6000
40–49
10 000
12 000
6B
Median
Q3
6350
7000
6900
9000
13 600
14 000
Max x
10 000
12 000
14 16 18 20
Time (seconds)
a
From the boxplots, 6A has a significantly lower median. The
6A median is lower than Q1 of the 6B time; that is, the lowest
25% of times for 6B is greater than the lowest 50% of times
for 6A.
2
10A 14 71.4 … 17.2 … 42
10D 14 72.6 … 16.4 … 40
n
Mean
s
30–39 age
group
20–29 age
group
Min Q1 Median Q3 Max
59
62
70.5
70.5
0
2
4
6
8 10 12 14 16
Annual superannuation contribution (× $1000)
87 98
84 100
b Clearly, there is a great jump in contributions to
superannuation for people in their 40s. The spread of
contributions for that age group is smaller than for people
in their 20s or 30s, suggesting that a high proportion of
people in their 40s are conscious of superannuation. For
people in their 20s and 30s, the range is greater, indicating
a range of interest in contributing to superannuation.
See image bottom of the page*
10A
10D
30
40
50 60 70 80
Exam mark (%)
90 100
From the boxplots the medians are the same but 10D has a
higher mean. 10D has the highest score of 100% but 10D also
has the lowest score. Since Q1 and Q3 are closer together for
10D, their results are more consistent around the median. The
parallel dot plot confirms this but doesn’t give any further
information.
3 Statistical analysis for boxplots:
9A
Min x = 120
Q1 = 140
10A
Min x = 140
Q1 = 149
Median = 153
Q3 = 160
11A
Min x = 151
Q1 = 160
Median = 163
Q3 = 170
Min x = 170
Min x = 199
a
11A
10A
9A
120 130 140 150 160 170 180 190 200
Height (cm)
*2
5 a For Company A, the median = 80, Q1 = 75 and
Q3 = 85(5 units either side of the median).
Min x = 65 and max x = 95 (15 units either side of the
median). So the distribution is fully symmetrical.
Answer: True
b Company A: Max x = 95; Company B: Q3 = 95. So the
highest 25% of the values for Company B were equal to or
higher than the highest share price for Company A.
Answer: True
c Company A: range = 95 − 65 IQR = 85 − 75
= 30
= 10
Company B: range = 100 − 70 IQR = 95 − 80
= 30
= 15
Although the range for both companies was the same, the
IQR was bigger for Company B. So the spread of the share
prices was not the same.
Answer: False
d Company A: median = 80
Company B: Q1 = 80
So 75% of the data for Company B were higher (or equal
to) in value than the median for Company A.
Answer: True
6 a All five values from the five-figure summary for The
Pearlfishers are higher than corresponding numbers for
Orlando:
Median = 167
Q3 = 180
Min x = 180
15 000
40–49 age
group
10A
40
45
50
55
60
65
70
75
80
85
90
95
100
P df_Fol i o: 34
10D
Exam mark (%)
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.6
The Pearlfishers
Orlando
Min x
Q1
550
700
400
600
Median
Q3
750
800
650
750
Max x
900
850
So The Pearlfishers was definitely more popular with
the public.
The answer is The Pearlfishers.
b
Range
The
Pearlfishers
Orlando
4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0
Day 2
Diving score
The dives on day 1 were more consistent than the dives on
day 2; with most of the day 1 dives between 5.4 and 6.0
(inclusive), despite two lower dives. Day 2 was more spread
with dives from 4.9 to 6.0 (inclusive); it must be noted that
there were no very low scoring dives on the second day.
10 B
11 B
12 C
13 Statistical analysis for boxplots:
Vitamin A Vitamin B Vitamin C Multi
5
10
8
12
7
11
9
13
Median
Q3
8
11
14.5
15
9.5
12
16
19
Max x
14
19
13
20
A
B
C
Multi-vitamin
6
Min.
Max.
Range
Median
Thursday
70
90
20
81
Friday
77
89
12
83
Saturday
81
89
8
86
Sunday
70
94
24
89
Sunday
Saturday
850 − 400 = 450 750 − 600 = 150
Day 1
4
Day
900 − 550 = 350 800 − 700 = 100
7 Answer D: 87 is the value of Q3 , which is the 75th percentile,
hence 25% of values must be above this.
8 Answer C: The boxplot for Year 8s is to the left of the Year
10s, plus the IQR and range are both greater for Year 8s,
hence their results must be lower and more variable.
9
Min x
Q1
14 For all four days, the median is the 6th score.
For all four days, Q1 is the 3rd score. For all four days, Q3 is
the 9th score.
IQR
Both range and IQR were higher for Orlando, so this opera
had larger variability in the number of A-reserve tickets
sold.
The answer is Orlando.
8 10 12 14 16
Number of jars sold
18
20
Overall, the biggest sales were of multi-vitamins, followed by
vitamin B, then vitamin C and finally vitamin A.
35
Friday
Thursday
70
75 80 85 90 95 100
Golf scores on 4 days
in a tournament
2.5 Exam questions
1 Parallel boxplots must be plotted on the same scale; therefore,
they must have the same units.
Annual rainfall has the same units (millimetres) as monthly
median rainfall.
The correct answer is E.
2 a Negatively skewed
[1 mark]
b The median life expectancy increases as the years increase
(in 1953 the median is 51; in 1973 the median is 63; in 1993
the median is 69). There is a positive correlation. [2 marks]
3 Only the upper quartile of jellyfish found at location A have a
diameter greater than 14 mm. That is 25%.
The correct answer is C.
2.6 Scatterplots
2.6 Exercise
1 Since the data is a reasonable fit to a straight line and goes
from bottom left to top right, it can be said that there is a
moderate positive linear relationship between ‘weekly
shooting practice’ and ‘shooting percentage’.
2 There are points at the top where the average speed tapers off
due to reading near maximum speed of the freeway. Despite
this, the data is a reasonable fit to a straight line. There is a
moderate positive linear relationship between ‘hours after
5 pm’ and ‘average speed’.
3 a Time spent in a supermarket and money spent —
relationship exists — positive association
b Income and value of car driven — relationship exists —
positive association
c Number of children living in a house and time spent
cleaning the house — relationship exists — positive
association
d Age and number of hours of competitive sport played per
week — relationship exists — negative association
e Amount spent on petrol each week and distance travelled by
car each week — relationship exists — positive association
P df_Fol i o: 35
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.6
20
10
16
14
12
10
8
6
4
2
0
15
10
5
0
0
20
0
19
0
18
17
16
0
0
Height (cm)
TV per week (h)
As would be expected, there is no relationship between ‘hair
length’ and ‘height’.
7
4 8 12 16 20 24 28 32
Time of booking
(number of days before
the performance)
Strong negative association of linear form, no outliers
11 Since random points, there is no relationship (correlation).
The correct answer is B.
12 a, b
Cost ($)
Hair length (cm)
f Number of hours spent in front of a computer each week
and time spent playing the piano each week — relationship
exists — negative association
g Amount spent on weekly groceries and time spent
gardening each week — no relationship exists so therefore
there is no association
4 a relationship exists, negative, weak, linear form
b relationship exists, negative, moderate, linear form
c relationship exists, positive, moderate, linear form
d relationship exists, positive, strong, linear form
e no relationship exists, therefore there is no association
f due to shape non-linear association
5 From this scatterplot, it would be reasonable to observe that
‘as the value of x increases, the value of y decreases’.
The correct answer is B.
6
Row number in A-Reserve
36
30
26
24
22
20
18
16
14
12
10
8
6
4
2
0
20
1 2 3 4 5 6 7
Number of dry
cleaning items
10
0
10 20 30 40 50 60
Age (years)
13 a, b
Number of drinks sold
As the points follow no pattern, there is no relationship
between ‘TV watched per week’ and ‘age’.
Number of
primary schools
8
10
8
6
4
10
0
12
0
14
0
16
0
18
0
20
0
0
190
180
170
160
150
140
130
120
110
100
90
80
0
Population (× 1000)
Moderate positive association of linear form, no outliers
14 a There is a strong positive correlation between the number
of dry cleaning items and the cost.
b There is a strong positive correlation between the
maximum daily temperature and the number of drinks sold.
15 There appears to be no correlation between population and
area of the various states and territories.
4000
3500
3000
2500
2000
1500
1000
0
5 10 15 20 25 30
Hours of paving
Strong positive association of linear form, no outliers
Area of land (km2)
Cost ($)
9
2 800 000
2 400 000
2 000 000
1 600 000
1 200 000
800 000
400 000
0
P df_Fol i o: 36
15 20 25 30 35 40
Maximum daily
temperature (°C)
1 2 3 4 5 6 7 8
Population (millions)
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.7
Finished length of
recording (min)
16 a
50
45
40
35
30
25
20
15
10
5
0
2 4 6 8 10 12 14 16 18 20
Hours spent re-engineering
in studio
b There is a relationship. It is strong, positive and linear.
2.6 Exam questions
1 Points are tightly grouped in a good linear format. Best
description is strong association between the points in a linear
form with a positive gradient.
The correct answer is D.
2 The graph shows that as the number of cigarettes smoked
increases, the level of fitness decreases.
The correct answer is B.
3 The graph shows that as the number of hours of work
increases, the recreation hours decreases. Therefore, there is a
moderate, linear, negative relationship between the number of
hours worked and the number of hours for recreation.
The correct answer is D.
37
d i r ≃ −0.2
ii No linear association
6 a i r=1
ii Perfect positive linear association
b i r ≃ 0.8
ii Strong positive linear association
c i r≃0
ii No association
d i r ≃ −0.7
ii Moderate, negative linear association
7 A set of data relating the variables x and y is found to have an
r value of 0.62, and is represented by a scatterplot.
The correct answer is B.
8 A set of data relating the variables x and y is found to have an
r value of −0.45. This indicates ‘There is a weak linear
relationship between x and y and when the x-values increase,
the y-values tend to decrease’.
The correct answer is E.
9 a r = 0.1: no linear relationship
b r = 0.2: no linear relationship
c r = 0.95: strong and positive
10 A weak, negative, linear association between two variables
would have an r value closest to r = −0.45.
The correct answer is D.
11 A value of Pearson product–moment correlation must be
between −1 and 1.
The correct answer is C.
12 y
2.7 Estimating and interpreting Pearson’s
product–moment correlation coefficient
2.7 Exercise
1 a i r = −0.9
ii The relationship can be described as a strong, negative,
linear relationship.
b i r = 0.7
ii The relationship can be described as a moderate,
positive, linear relationship.
2 a The relationship can be described as a strong, positive,
linear relationship.
b The relationship can be described as a weak, negative,
linear relationship.
3 a No association
b Moderate positive association
c Strong negative association
d Strong negative association
4 a Strong positive association
b Strong positive association
c Weak negative association
d No association
5 a i r ≃ −0.8
ii Strong negative linear association
b i r ≃ 0.6
ii Moderate positive linear association
c i r ≃ 0.2
ii No linear association
0
x
13 If two variables have an r value of 1, then they are said to
have a perfect positive linear relationship. Note that the
distractor option C does not contain the word linear.
The correct answer is E.
14 The correct ascending order of positive values of r: No linear
association, Weak, Moderate, Strong.
The correct answer is C.
2.7 Exam questions
1 A positive association means that as the explanatory variable
(HDI) increases, so too does the response variable (carbon
dioxide emissions).
The correct answer is D.
2 A scatterplot shows the relationship between two numerical
variables. Because gender is nominal rather than numerical,
gender cannot be shown on a scatterplot.
The correct answer is A.
3 The data is tightly grouped in a very linear formation. This
indicates r is positive and would be close to 1. Option A is the
most representative of this data set.
The correct answer is A.
P df_Fol i o: 37
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • EXERCISE 2.8
Points scored
1 a
25
20
15
10
5
19
5
20
20 0
5
21
21 0
5
22
0
0
Height (cm)
Birth mass (kg)
b The data shows that there is no linear relationship. We
might expect r to be a value close to 0.3.
c Using CAS, r ≈ 0.36. This indicates that there is a weak,
positive, linear relationship between the ‘height of
basketball players’ and the ‘number of points scored in a
game’.
2 a
3.5
3
2.5
2
1.5
1
0.5
0
30 32 34 36 38 40
Gestation time
(weeks)
Number of votes
b The data shows that there is a strong, positive, linear
relationship. We might expect r to have a value close to
0.95.
c Using CAS, r ≈ 0.99. This confirms our original thought
that there is a very strong, linear relationship between
‘gestation time’ and ‘birth mass’.
3 a
35
30
25
20
15
10
5
28
0
30
0
32
0
34
0
36
0
38
0
40
420
0
44
0
46
0
48
0
50
0
0
Yearly salary (× $1000)
b There is moderate, negative, linear association; r is
approximately −0.6.
c r = −0.66. There is a moderate, negative, linear association
between the yearly salary and the number of votes. That is,
the larger the yearly salary of the player, the fewer the
number of votes we might expect to see.
P df_Fol i o: 38
4 Coefficient of determination = r2
= (0.77)2
= 0.59
We can conclude that 59% of the variation in the number of
people found to have a blood alcohol reading over 0.05 can be
explained by the variation in the number of booze buses in
use. Therefore, the number of booze buses in use is a factor in
predicting the number of drivers with a reading over 0.05.
Yearly profit increase (%)
2.8 Exercise
5 Coefficient of determination = r2
= (0.87)2
= 0.7569
Comment: 75.7% of the variation in visits to the doctor is due
to the number of cigarettes smoked per day and 24.3% is due
to other factors such as age, health and so on.
6 a
10
9
8
7
6
5
4
3
2
1
0
10 12 14 16 18 20 22 24 26 28 30
Annual advertising budget (× $1000)
b There is strong, positive, linear association; r is
approximately 0.8.
c r = 0.98
d Coefficient of determination = (0.98)2 = 0.96
e 96% of the variation in yearly profit is due to the annual
advertising budget and 4% is due to other factors, for
example, economic climate.
7 a
Weekly grocery costs ($)
2.8 Calculating r and the coefficient of
determination r2
225
200
175
150
125
100
75
50
25
0
2 3 4 5 6
Number of people
in household
b There is strong, positive association of a linear form and r is
approximately 0.9.
c r = 0.98
d Coefficient of determination = (0.98)2 = 0.96
e 96% of variation in the weekly grocery bill is due to the
number of people in the household and 4% is due to other
factors, for example, age, health and pets.
8 a
Total weight loss (kg)
38
10
9
8
7
6
5
4
3
2
1
0
1 2 3 4 5 6 7 8 9 10
Number of weeks on the diet
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • REVIEW 2.10
b The scatterplot shows strong, positive association of linear
form.
c It is appropriate to use Pearson’s correlation coefficient
since the scatterplot indicates a linear association and there
are no outliers.
d The estimated value of r is 0.9.
e r = 0.96
f We cannot say whether total weight loss is affected by the
number of weeks people stayed on the Certain Slim diet.
We can only note the degree of correlation of the two
variables.
g The coefficient of determination, r2 = 0.962
= 0.92
h The coefficient of determination tells us that 92% of the
variation in total weight loss can be explained by the
variation in the number of weeks on the Certain Slim diet.
2.8 Exam questions
1 Using a CAS calculator and entering the rate of pay for 1990
as the independent variable and the rate of pay for 2010 as the
dependent variable gives r = 0.9622, which is closest to 0.96.
The correct answer is E.
2 The value of r = −0.563 indicates moderate (negative)
correlation between the population density and distance from
the centre of the city, while the value of r = 0.357 indicates
weak (positive) correlation between the house size and
distance from the centre of the city.
Therefore, population density is more strongly associated
with distance from the centre of the city than is house size.
The correct answer is E.
3 We don’t know how fit the students were or whether they
played computer games very much. There is a negative
correlation, which tells us that greater fitness levels tended to
occur with lower time playing computer games and vice versa.
The correct answer is D.
2.9 Cause and effect
1 a
Annual funds raised (× $1000)
2.9 Exercise
13
12
11
10
9
8
7
6
5
4
0
39
2 There is a strong positive association between the number of
books read and the spelling levels for this group of students.
3 Correlation does not imply causation.
The correct answer is E.
4 It cannot be stated for sure that the number of hours worked is
the major factor in the cost of child care; other factors such as
income and means testing could also be major factors.
The correct answer is B.
5 Breeding season attracts larger numbers of visitors and more
visitors means more drivers on the roads, which leads to more
wildlife being killed. The association between the number of
penguins born and the number of wildlife killed can be
explained by the common response variable, the number of
visitors to the island.
The correct answer is C.
6 It cannot be concluded that diabetes causes deterioration of
the hip joint. The common cause could be age.
7 It cannot be concluded that holidays cause weight gain. The
common cause could be any or a combination of the
following: amount of exercise, alcohol consumption, diet.
8 Confounding variables could be time away from family,
amount of exercise, diet.
2.9 Exam questions
1 Three variables have been listed in this question:
the number of stray cats, the number of stray dogs and the
population of the city (which grew in size over the time period
data was collected).
Correlation does not imply causality. Option D is the only
logical conclusion.
The correct answer is D.
2 There is no causal relationship between the two variables,
which makes the correlation value unfounded. There would
be other factors affecting the variables. For example, the
weather could be one possibility. If the investigation involved
data taken during the colder months of the year, a rise in
tissue sales would be more than likely as would a rise in hot
chocolate sales.
The correct answer is C.
3 Association is not causation, making option B the incorrect
response.
The correct answer is B.
2.10 Review
2.10 Exercise
Multiple choice
3 4 5 6 7 8
Number of people
on committee
b There is almost perfect positive correlation of a linear form
and r is nearly 1.
c Use a CAS calculator to find r = 0.99.
d Causation cannot be established on a strong association
alone.
1 A back-to-back stem plot could be used to display the number
of children at a day-care centre and whether the centre had
federal funding.
The correct answer is A.
2 The company with the largest interquartile range is
Company A (95 − 32 ≃ 63).
The correct answer is A.
3 Company E has the lowest median (41).
The correct answer is E.
P df_Fol i o: 39
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • REVIEW 2.10
Short answer
13 a Highest − 15
Lowest − 1
Use stems of 1, divided into fifths.
Full-time
1
2 2
4 4 3 3
6 5
Stem
0
0
0
0
0
1
1
1
1
1
Volunteer
8
0 1 1
2 3 3
4 5
b Both distributions are symmetric with the same spread. The
centre of the volunteers’ distribution is much higher than
that of the full-time firefighters’ distribution. Clearly, the
volunteers needed more counselling.
14 Statistical analysis for boxplots:
Min x
Q1
Team A
98
103.5
Team B
95
101
Team C
114
115.5
Median
Q3
110.5
125
105
109.5
120.5
126.5
Max x
140
120
145
Team C
Team B
Team A
90 100 110 120 130 140 150
IQ
15
Attitude
Party preference
For
Against
Liberal
30 (45 − 15)
15 (given)
Labor
10 (53 − 43)
43 (given)
Total
40
58
Total
45 (given)
53 (given)
98
a
Party preference
Liberal
For
Attitude
4 From the table, we can conclude that 42.6% of junior staff
were for relocation.
23 100
×
% = 42.6%
54
1
The correct answer is B.
5 From the table we can conclude that 74.5% of senior staff
were against the relocation.
41 100
×
% = 74.5%
55
1
The correct answer is E.
6 The relationship between the variables x and y in the
scatterplot could be described as a strong negative
association, although data appears non-linear.
The correct answer is D.
7 A set of data relating the variables x and y is found to have an
r value of −0.83. The scatterplot that could best represent this
data set is D.
The correct answer is D.
8 If Pearson’s correlation coefficient = 0.86
Coefficient of determination = (0.86)2
= 0.7396
≈ 0.74
The correct answer is E.
9 The line of best fit follows the trend of the data and lies within
the data.
The correct answer is A.
10 The correlation between two variables x and y is −0.88. As x
increases, y decreases.
The correct answer is E.
11 When calculating a least-squares regression line, a correlation
coefficient of −1 indicates that all the data lie on the same
straight line.
The correct answer is D.
12 Cities that have large numbers of tourists tend to have low
satisfaction levels of local residents.
The correct answer is C.
Against
Total
66.7%
30 100%
×
( 45
1 )
15 100%
×
( 45
1 )
= 33.3%
100%
Labor
18.9%
10 100%
×
( 53
1 )
43 100%
×
( 53
1 )
= 81.1%
100%
To construct the segmented bar chart from the table:
• Rule out and label the axes (percentage is on the vertical
axis).
• Draw two columns of equal width (one for Liberal, one
for Labor) and equal height — to 100%.
• Divide each column into segments, corresponding to the
cells in the table. That is, draw a line to the level of
66.7% in the 1st column and to 18.9% in the 2nd column.
• Add a legend to the graph.
Percentage
40
100
90
80
70
60
50
40
30
20
10
0
Attitude
Against
For
Liberal
Labor
x
Delegates
b Comment: Clearly the reaction to uranium mining is
affected by political affiliation with 66.1% of Liberals for
uranium mining compared to 18.9% of Labor delegates.
P df_Fol i o: 40
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • REVIEW 2.10
16 Age
b
Summary
Symphony
Jazz
i
ii
Min x
Q1
20
40.5
16
21.5
15 16 17 18 19
iii
iv
v
vi
vii
Median
Q3
Max x
Mean
IQR
Age
viii
Standard
deviation
Pulse rate
15
17
18
16
19
19
17
15
17
79
74
75
85
82
76
77
72
70
Pulse rate
a
85
80
75
70
0
b There appears to be an extremely weak association or no
association between the variables.
Exam result (%)
17 r is approximately equal to −0.7. There is a moderate,
negative, linear association between the variables x and y.
18 a Lectures attended — explanatory
Exam result — response
b
0
10
90
80
70
60
0
Lectures attended (%)
c There is strong, positive correlation of a linear form
between the variables, and r is approximately equal to 0.8.
d r = 0.96
e The coefficient of determination is 0.93.
f The proportion of the variation in the exam results that can
be explained by the variation in the number of lectures
attended is 93%.
19 a Key∶1 ∣ 6 = 16
9 9 8 6
9 7 4 3 0
9 8 4 3 0
6 5 3 2 0
2
Symphony
1
2
3
4
5
6
0 3
0 5 9
2 5 5 7 8 8 9 9
0 3 4 6 8 8
0
11.20
Min x
Q1
12
34
Median
Q3
47.5
56.5
Max x
Mean
IQR
68
44.95
56.5 − 34 = 22.5
Sx
Extended response
Stem
48
31.5
53.5
41
60
62
45.45
32.35
53.5 − 40.5 = 13 41 − 21.5 = 19.5
15.55
f See figure at bottom of the page*
g The distributions for symphony and opera are similar: both
are skewed to the left and have approximately the same
centre (means: 45.45 and 44.95; medians: 48 and 47.5,
respectively), but opera has a much bigger spread. For jazz,
the distribution is positively skewed; the centre is much
lower than for both symphony and opera, while the spread
is in the middle (slightly bigger than for the symphony, but
smaller than the opera). Overall, it appears that people who
*19 f
Opera
Jazz
Symphony
15
P df_Fol i o: 41
20
25
30
35
40 45
Age
50
55
12.04
c All values (except for maximum) in the five-figure
summary are higher for symphony goers than for jazz. The
median (48) for symphony is much higher than for jazz
(31.5); the mean is also much higher for symphony (45.45
compared to 32.35). So the centre is much higher for
symphony goers. The spread is higher for the jazz goers:
IQR is 19.5 (as opposed to 13 for symphony) and standard
deviation is also slightly higher for jazz. Overall, it appears
that people who went to the symphony concert were older
than those who went to jazz. The spread of ages was nearly
the same, but slightly higher for jazz goers.
d Back-to-back stem plots can only be used for numerical
data with 2 categories. To compare distributions
for 3 different events (i.e. 3 categories) we need to use
boxplots.
e
Opera
100
95
90
85
80
75
70
65
60
55
Jazz
41
60
65
70
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • REVIEW 2.10
42
Percentage
attend symphony concerts and opera tend to be
approximately the same age and older than those who
attend jazz. The most spread out distribution is for the
opera, followed by jazz and then by the symphony goers.
So the widest range of ages is among the opera goers.
20 a 6 + 52 + 130 = 188 students reported high level of stress.
b These are people who reported ‘regularly’ or ‘sometimes’.
So the total of the 1st and 2nd columns:
16 + 12 + 6 + 32 + 40 + 52 = 158
c See table at bottom of the page*
d To construct the chart:
• Rule out and label the axes.
• Draw 3 columns of equal width (labelled ‘Regularly’,
‘Sometimes’ and ‘Never’) to the height of 100%.
• Segment each column so that height of each segment
corresponds to % in the cell of the table.
• Add the legend.
100
90
80
70
60
50
40
30
20
10
0
Based on the parallel boxplots, we can clearly see that the
median of Pond B is greater than the maximum of Pond A; it
indicates that at least 50% of the fish caught in Pond B is
greater than all the fish caught in Pond A.
The correct answer is B.
2 This question is asking for the gradient connecting the two
variables. It is investigating the effect that drinking coffee has
on sleep, so sleep is the response variable.
b=r
ssleep
sy
1.12
=r
= −0.770 ×
= −0.55282
sx
scoffee
1.56
VCAA Examination Report note:
Students needed to recognise that sleep was the response
variable and coffee the explanatory variable, and be familiar
with the rules connecting a least squares regression line
equation to summary statistics.
The correct answer is A.
3 a Place of capture
[1 mark]
b The most frequently occurring value in the forest section
of the stem plot is 20, so the modal wingspan
is 20 mm.
[1 mark]
c The minimum wingspan in the forest is 16 mm.
[1 mark]
The upper quartile (Q3 ) in the grassland is 36 mm. [1 mark]
d IQR = Q3 − Q1
= 32 − 20
= 12
The wingspan of 52 mm is at the upper end of the forest
values. Students need to show that this value is greater than
the upper fence, that is, greater than Q3 + 1.5 × IQR.
Q3 + 1.5 × IQR
= 32 + 1.5 × 12
= 50
[1 mark]
As 52 mm is greater than this upper-fence value of 50 mm,
it is an outlier.
[1 mark]
e Possible solution:
The wingspan is associated with the place of
capture.
[1 mark]
Those captured in the grassland had a median wingspan of
30 mm, which is greater than the median wingspan of
21 mm of the moths captured in the forest.
[1 mark]
4 A positive association means that as the explanatory variable
(HDI) increases, so too does the response variable (carbon
dioxide emissions).
The correct answer is D.
Stress level
High
Medium
Low
Regularly
Sometimes
Never
x
Amount of exercise
e It appears that low level of stress is the highest for students
who exercise regularly (47.06%) and smallest for those who
do not exercise (16%). With high level of stress it is vice
versa: it is largest for those who do not exercise (58.55%),
second largest for those who exercise sometimes (41.93%)
and lowest for those who exercise regularly (17.65%).
Overall it appears that for this group of students’ stress
levels are related to the amount of physical activities they
had.
2.10 Exam questions
1 As we know, 50% of the difference is between the minimum
and the median, 50% of the difference is between Q1 and Q3 ,
and the other 50% could also be between the median and the
maximum.
*20 c
Sport engagement
Level of stress
Sometimes
Never
16
× 100 = 47.06%
34
32
× 100 = 25.81%
124
36
× 100 = 16.22%
222
High
6
× 100 = 17.65%
34
52
× 100 = 41.93%
124
130
× 100 = 58.55%
222
Total
100%
Low
Medium
P df_Fol i o: 42
Regularly
12
× 100 = 35.29%
34
40
× 100 = 32.26%
124
100%
56
× 100 = 25.23%
222
100%
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 2 Investigating associations between two variables • REVIEW 2.10
5 a Reading off Australia’s column, the black section ends at
19. Therefore 19% of people were aged 0–14 years old in
2010.
[1 mark]
b Reading off Japan’s column, the grey section starts at 77, so
the percentage of people in Japan 65 years and over is
100 − 77 = 23%.
23
× 128 000 000 = 29 440 000
[1 mark]
100
c 15–64 age group:
Australia’s population percentage = 86 − 19 = 67%
India’s population percentage = 95 − 31 = 64%
Japan’s population percentage = 77 − 13 = 64%
As the percentage of people aged 15–64 is almost the same
for all three countries, there is no association between these
percentages and the country in which they live.
[1 mark]
P df_Fol i o: 43
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
43
44
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.2
Topic 3 — Investigating and modelling linear associations
b b=r
3.2 Least squares line of best fit
3.2 Exercise
1 Using CAS: y = −37.57 + 1.87x
Air conditioner sales = −37.57 + 1.87 × (temperature)
Coefficient of determination (r2 ) = 0.97
Therefore, a strong linear association exists between
temperature and air conditioner sales.
2 Using CAS: y = 2.11 + 2.64x
Number of dialysis patients = 2.11 + 2.64 × (month)
Coefficient of determination (r2 ) = 0.95
Therefore, a strong linear association exists between y and x.
3 x
4
6
7
9
5
10 12 15 17
y
10
8
13
15
19
14
18
19
23
Using CAS:
The least squares line of best fit is y = 4.57 + 1.04x.
r2 = 0.91; therefore, a strong linear association exists
between y and x.
4
5
x
y
1
35
2
28
3
22
4
16
5
19
6
14
7
9
8
7
9
2
Using CAS:
The least squares line of best fit is y = −1.691 − 3.72x.
x
y
−4
6
−2
7
−1
3
0
10
1
16
2
9
4
12
5
16
5
11
Using CAS:
The least squares line of best fit is y = 9.06 + 1.20x.
6 a The explanatory variable is ‘height of netball player’.
sy
b b=r×
sx
6.1
= 0.82 ×
5.2
= 0.96
c a = y − bx
= 182 − 0.96 × 166
= 22.64
sy
7 b=r×
sx
1.4
= −0.67 ×
1.2
= −0.781 67
a = y − bx
= 10. 5 − (−0.78 167 × 4.4)
= 13.94
∴ y = 13.94 − 0.78x (rounded to 2 decimal places)
sy
8 a b = r with r = 0.7, sy = 5.7 and sx = 1.2
sx
b = 3.325
a = y − bx, with y = 110.4, x = 5.6 and b = 3.325
a = 91.78
The equation of the least squares line of best fit is
y = 91.78 + 3.33x.
7
21
sy
with r = −0.7, sy = 1.2 and sx = 5.7
sx
b = −0.1474
a = y − bx with y = 5.6, x = 110.4 and b = −0.1474
a = 21.873
The equation of the least squares line of best fit is
y = 21.87 − 0.15x.
sy
c b = r with r = 0.88, sy = 250 and sx = 4.2
sx
b = 52.381
a = y − bx with b = 52.381, y = 10 200 and x = 25
a = 8890.476
The equation of the least squares line of best fit is
y = 8890.48 + 52.38x.
sy
d b = r with r = −0.5, sy = 2 and sx = 1
sx
b = −1
a = y − bx with y = 20, b = −1 and x = 10
a = 30
The equation of the least squares line of best fit is
y = 30 − x.
9 a The response variable is the subject of the equation in the
form y = a + bx (that is, y).
Therefore, the response variable is the Mathematics exam
mark.
sy
b b = r , with r = 0.64, sy = 14.5 and sx = 9.8
sx
b = 0.9469
c a = y − bx, with y = 64, x = 74 and b = 0.9469
a = −6.07
d y = −6.07 + 0.95 × 85
y = 75%
10 (3) x = 10, sx = 4.472, y = 15, sy = 4.899, r = 0.952
sy
b=r
sx
4.899
= 0.952 ×
4.472
b = 1.043
a = y − bx
= 15 − 1.043 × 10 = 4.571
The least squares equation is y = 4.57 + 1.04x.
(4) x = 5, sx = 2.739, y = 16.889, sy = 10.446, r = −0.974
sy
b=r
sx
10.446
= −0.974 ×
2.739
b = −3.716
a = y − bx
= 16.889 − 3.716 × 5
= −1.691
The least squares equation is
y = −1.691 − 3.72x.
P df_Fol i o: 44
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.2
(5) x = 1.7, sx = 3.529, y = 11.1, sy = 5.384, r = 0.7853
sy
b=r
sx
5.384
= 0.785 ×
3.529
b = 1.198
a = y − bx
= 11.1 − 1.198 × 1.7
= 9.063
The least squares equation is y = 9.06 + 1.20x.
In each case, the least squares lines are the same as
calculated in questions 3, 4 and 5.
11
x
y
1
20
2
18
3
16
4
14
5
12
6
10
7
8
8
6
9
4
Cost ($)
c
50
10 0
1500
0
20 0
2500
0
30 0
3500
0
40 0
4500
0
50 0
0
55 0
0
60 0
0
65 0
0
70 0
0
75 0
00
The line does not fit closely for all data points. The
equation is not reliable due to the outlier, which was the
two-hour (7200 sec) call. This is supported by an r-value
of 0.8564. If the last two long calls are eliminated, then
there is a direct relationship.
10
2
16
3
14
4
12
5
10
6
8
7
6
8
4
9
2
10
Cost ($)
18
2
c The least squares line of best fit is y = 11 − 0.5x.
The two lines are the inverses of each other.
Using CAS:
12 The x-intercept is 2, so the best estimate of the line is
1
y = −1 + x. The answer is E.
2
13 See the table at the foot of the page.*
50
10 0
1500
0
20 0
2500
0
30 0
3500
0
40 0
4500
0
50 0
0
55 0
0
60 0
0
65 0
0
70 0
0
75 0
00
d r = 0.73; therefore, the linear association between cell
duration and cost of cell is only moderate.
sy
15 b = r
sx
5
= 0.9 ×
3
= 1.5
The correct answer is D.
16
x
1
2
3
4
5
6
Duration (s)
The least squares line of best fit is y = −164.7 + 0.119x.
80
75
70
Age
y = −164.7 + 0.119x
65
60
y
55
12
16
17
21
25
29
Using CAS:
50
0
1800
1840
1880
1920
Year
1960
2000
2020
c The data definitely are not linear; there are big increases
between 1900 and 1940 and between 1960 and 2020.
14 a Cost of call is the response variable. Duration is
explanatory.
b Using CAS:
y = 4.27 + 0.002 57x
Cost of call ($) = $4.27 + $0.002 57 × duration of call (sec)
P df_Fol i o: 45
y = 4.27 + 0.003x
2
85
*13
18
16
14
12
10
8
6
4
2
0
a Using CAS:
b
y = 4.27 + 0.003x
Duration (s)
a The least squares line of best fit is y = 22 − 2x.
b The least squares line of best fit is a perfect fit, r2 = 1.
20
1
18
16
14
12
10
8
6
4
2
0
Using CAS:
x
y
45
Year
Lifespan (years)
1780
51.2
1800
52.4
1820
51.7
1840
53.2
1860
53.1
a The least squares line of best fit for the first two points is
y = 8 + 4x, which is a perfect fit, but meaningless.
b The least squares line of best fit for the first three points is
y = 10 + 2.5x, a good fit, but almost meaningless.
c Including point 4:
The least squares line of best fit for the first four points is
y = 9.5 + 2.8x, a good fit.
Including point 5:
The least squares line of best fit for the first five points is
y = 8.9 + 3.1x, a good fit.
1880
54.7
1900
59.9
1920
62.7
1940
63.2
1960
66.8
1980
72.7
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
2000
79.2
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.3
46
a a = 157.3, b = 14, y = 157.3 + 14x or red blood cell
count = 157.3 + 4x day of experiment
b The red blood cell count is changing at a rate of 14 cells
per day.
c The red cell blood count at day 0 ≃ 157.
Including point 6:
The least squares line of best fit for the first six points is
y = 8.4 + 3.3x, a good fit.
d From point 4 onwards, the line appears to be converging to
the correct line.
Using CAS:
4
3.2 Exam questions
1 This is the definition of how a least squares line is calculated.
The correct answer is E.
2 42.9842 will become 43.0 (Normal rounding still applies, and
the 0 after the decimal place is significant.).
–0.877447 will become –0.877 (The 0 before the decimal
place is a place holder. Significant figures start from the first
non-zero number.).
The correct answer is E.
3 Note that the scales do not start at 0, so the vertical axis
intercept is not 67.2.
Choose any two points along the line, say (1, 67.2) and (4, 64).
y2 − y1
64 − 67.2
−3.3
Gradient: m =
=
=
= 1.10
x2 − x1
4−1
3
Equation:
resting pulse rate = a + b × time spent exercising
resting pulse rate = a − 1.10 × time spent exercising
Passes through (4, 64):
64 = a − 1.1 × 4
68.4 = a
So the closest equation is
resting pulse rate = 68.3 − 1.10 × time spent exercising
The correct answer is D.
*5
P df_Fol i o: 46
Sales volume (×1000)
Selling price ($)
230
260
260
8
5
7
6
311
220
413
280
379
334
5 See the table at the foot of the page.*
Using CAS:
The least squares line of best fit is:
sales volume = 464 − 1.72x (selling price)
Interpretation:
For every $1 increase in the selling price, sales volume
decreases by 1720.
If the selling price was $0, the sales volume would be 464 000.
This is a case where extrapolation of the line makes no sense.
6 a Using CAS: y = 56.03 + 9.35x
Height (cm) = 56.03 + 9.35 × (age)
When age is 10 years∶ height (cm) = 56.03 + 9.35 × (age)
= 56.03 + 9.35 × (10)
= 149.53 cm
b From part a: height (cm) = 56.03 + 9.35 × (age)
When age is 16 years∶ height (cm) = 56.03 + 9.35 × (age)
= 56.03 + 9.35 × (16)
= 205.63 cm
c Since we are extrapolating the result (predicting outside the
data set) we cannot claim that the prediction is reliable.
7 a Using CAS: y = 40.10 + 6.54x
Length (cm) = 40.10 + 6.54 × (months)
When age is 15 months ∶ length (cm) = 40.10 + 6.54 × (months)
= 40.10 + 6.54 × (15)
= 138.20 cm
b From part a: length (cm) = 40.10 + 6.54 × (months)
When age is 2 years (24 months) ∶
1 a Using CAS: y = 12.13 + 3.49x
Monkey height (cm) = 12.13 + 3.49 × (month from birth)
b The rate at which the monkey is growing is related to the
gradient; therefore, 3.49 cm/month.
c At birth (y-intercept) when the month from
birth = 0, height at birth = 12.13 cm.
2 a Using CAS: y = 13.56 + 2.83x
Temperature (°C) = 13.56 + 2.83 × (time after 6 am.)
b The rate (gradient) is 2.83°C/hr.
c Temperature at 6 am (y -intercept) is 13.56°C.
3 Day of
5
6
8
7
9
4
experiment
240
4
a The rate of increase of visitors as the number of live
animals is increased is ≃ 49.
b The predicted number of visitors if there are no live
animals is ≃ 32.
3.3 Exercise
210
6
Using CAS:
3.3 Interpretation, interpolation and extrapolation
Red blood cell
count
Number of
animals
Number of
visitors
length (cm) = 40.10 + 6.54 × (months)
= 40.10 + 6.54 × (24)
= 197.06 cm
c Since we are extrapolating the result (predicting outside the
data set), we cannot claim that the prediction is reliable.
290
60
80
100
120
140
160
200
220
240
260
400
300
275
250
210
190
150
100
50
0
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.3
c When n = 10 000, C = $78 200
d The fixed costs in production is $600.
e Interpolation is used for a and b as 180 and 200 are in
[100, 1000].
8 See the table at the foot of the page.*
Using CAS:
a If the weekly income is $680, number of restaurant visits
= 1.2 + 0.0085 × 680 ≃ 7 visits
b If the weekly income is $2000,
numberofrestaurantvisits = 1.2 + 0.0085 × 2000
≃ 18 visits
9
x
0
1
2
4
5
6
8
10
y
2
3
7
12
17
21
27
35
12 See the table at the foot of the page.*
Using the data supplied, a calculator and the least squares line
of best fit:
Test result (%) = −4.7262 + 0.73 96IQ
The completed table is shown at the foot of the page.**
13 s = 37 000 + 1800n
a When n = 10, salary = $55 000
b When n = 12, salary = $58 600
c When n = 15, salary = $64 000
d Solve 60 800 = 37 000 + 1800n. So number of years
schooling is 13 years.
e Number of years of schooling is a good indicator of salary,
although secondary to tertiary would be a better indicator.
a The least squares equation is y = 0.286 + 3.381x.
b When x = 3, y = 10.4 (interpolation).
c When x = 12, y = 40.9 (extrapolation).
d When y = 7,
7 = 0.286 + 3.381x
x = 1.99 (interpolation)
Using CAS:
14 a Using CAS: r = 11.73 + 5.35q
b When q = 4, r = 11.73 + 5.35q
= 11.73 + 5.35 (4)
= 33.13
e When y = 25,
25 = 0.286 + 3.381x
x = 7.31 (interpolation)
f Part c is extrapolation.
c When q = 18, r = 11.73 + 5.35q
= 11.73 + 5.35 (18)
= 108.03
d When r = 100∶
r = 11.73 + 5.35q
100 = 11.73 + 5.35q
100 − 11.73 = 5.35q
5.35q = 88.27
88.27
q=
5.35
= 16.50
e Both c and d are extrapolation, since they are both outside
the data range.
15 a C = 180 + 80n, when n = 3
C = 180 + 80 (3)
= $420
10 See the table at the foot of the page.*
a Using CAS:
The least squares line of best fit for Factory 1 is:
cost = 43.21 + 1.51 × distance
The least squares line of best fit for Factory 2 is:
cost = 56.61 + 0.964 × distance
b Factory 1 is cheaper at $43.21 (compared to Factory 2 at
$56.61).
c When distance (x = 115 km), Factory 1 cost is $216.86 and
Factory 2 cost is $167.47, so Factory 2 is cheaper by
$49.39.
d Factory 2 has the most linear rates as r2 = 0.9763
compared to Factory 1 with r2 = 0.9332.
11 C = 600 + 7.76n for [100, 1000]
a When n = 200, C = $2152.
b When C = 2000,
2000 = 600 + 7.76n
So n = 180 calculators
*8
*10
*12
**12
P df_Fol i o: 47
47
Weekly income ($)
100
200
300
400
500
600
700
800
900
1000
Number of restaurant
visits per year
5.8
2.6
1.4
1.2
6
4.8
11.6
4.4
12.2
9
Distance from Melbourne (km)
10
20
30
40
50
60
70
80
Factory 1 cost ($)
70
70
90
100
110
120
150
180
Factory 2 cost ($)
70
75
80
100
100
115
125
135
115
IQ
80
92
102
Test result %
56
60
68
65
IQ
80
87
92
102
Test result %
56
60
68
65
107
111
74
71
73
105
106
107
111
115
121
73
74
71
73
80
92
105
121
92
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.4
b
C = 180 + 80n, when C = 1250
1250 = 180 + 80n
80n = 1250 − 180
80n = 1070
1070
n=
80
n = 13.38 hours
c C = 180 + 80n, when n = 9 hours and 30 minutes (n = 9.5)
C = 180 + 80 (9.5)
= $940
d C = 180 + 80n, when n = 0
C = 180 + 80 (0)
= $180
Therefore, the callout fee is $180.
e Since the function is only for an 8-hour day, b and c are
extrapolation.
16 a Using CAS:
y = 42 956 + 14 795 × (millions spent on advertising)
Members = 42 956 + 14 795 ×
(millions spent on advertising)
2 a The male life expectancy increases by 0.88 years with each
year of increase in the female life expectancy.
[1 mark]
b Male = 6.3 + 88.0 × 35 = 34 4. years
[1 mark]
c About 95% of the variability in the male life expectancy is
explained by the variability of the female life expectancy
using the linear least squares model.
[1 mark]
3 (width)2 = 1.8 + 0.8 × 120
= 1.8 + 96
= 97.8
width =√97.8
= 9.9
The correct answer is B.
3.4 Residual analysis
3.4 Exercise
b Members = 42 956 + 14 795 × (2)
= 72 546
As this is within the data range, it is interpolation.
c When members = 70 000∶
70 000 = 42 956 + 14 795 × (millions spent on advertising)
70 000 − 42 956 = 14 795 × (millions spent on advertising)
14 795 × (millions spent on advertising) = 27 004
27 004
millions spent on advertising =
14 795
millions spent on advertising = 1.83 million dollars
d r2 = 0.90
Therefore, this data shows a strong linear relationship
between the amount of money spent on advertising and the
number of members.
1 a. Using CAS: y = −0.03 + 10.85x
See the table at the foot of the page.*
b
Residual plot
Residual
48
8
6
4
2
0
–2
–4
–6
–8
2
4
6
8
10
x
c Since the points are randomly scattered above and below
the x-axis, the original data probably has a linear
relationship. r2 = 0.98, which is consistent with a linear
relationship.
2 a Using CAS: y = 0.69 + 9.82x
See the table at the foot of the page.*
3.3 Exam questions
1 The gradient is 0.422, which means that as the HDI increases
by 10 units, life expectancy increases by 4.22 years,
NOT43.0 years as suggested by option C.
The correct answer is C.
*1a
*2a
P df_Fol i o: 48
x
y
1
12
2
20
3
35
4
40
5
50
6
67
7
83
8
88
9
93
ypredicted
Residual (y − ypredicted )
10.82
21.67
43.37
−4.22
65.07
75.92
86.77
97.62
2.48
−3.37
54.22
1.18
−1.67
32.52
1.93
7.08
1.23
−4.62
x
y
5
45
7
61
10
89
12
122
15
161
18
177
25
243
30
333
40
366
ypredicted
49.79
69.43
98.89
118.53
147.99
177.45
246.19
295.29
393.49
3.47
13.01
Residual (y − ypredicted )
−4.79
−8.43
−9.39
−0.45
−3.19
37.71
−27.49
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.4
b
4 The following data set (ii) appears to be linear; therefore, the
answer is D.
Residual
Residual plot
40
30
20
10
0
–10
–20
–30
–40
15 20 25 30 35 40 45
y
90
80
70
60
50
40
30
20
10
x
c r2 = 0.98, consistent with a linear relationship
3 a
ypred
y
x
Residual
−4.1
5.1
1
1
9.7
2.24
7.46
2
9.82
3
2.88
12.7
13.7
12.18
1.52
4
−0.14
14.4
5
14.54
16.9
−2.4
6
14.5
0
Residual
–1
–2
–3
–4
–5
x
a Using the least squares line of best fit, the defective rate
% = 16.34 − 1.19 × day, substitute day number into
equation.
See the table at the foot of the page.*
b See the table at the foot of the page.*
c
Residual
3
2
1
4
3
2
1
1 2 3 4 5 6
10 20 30 40 50 60 70 80
5 See the table at the foot of the page.*
b = −1.19, a = 16.34, r = −0.87
Using CAS:
Since the points are randomly scattered above and below
the x-axis, the original data probably has a linear
relationship. r2 = 0.98, which is consistent with a linear
relationship.
b Residual graph for data in question 5.
0
0
–1
–2
–3
–4
–5
12
2
x
No apparent pattern in the residuals — likely to be linear
d After 1 day, the rate = 16.34 − 1.19 × 1
= 15.15
e 0 = 16.34 − 1.19 × day
1.19 × day = 16.34
16.34
day =
1.19
= 13.7 days
Unlikely that extrapolation that far from data points is
accurate. It is unlikely that there would be 0% defectives.
x
From the residual plot, as a pattern exists, the data is clearly
non-linear.
6 See the table at the foot of the page.*
*5
Defective rate %
Day
2
4
5
7
8
9
10
11
15
10
12
4
9
7
3
4
*5 a
ypred
13.96
11.58
10.39
8.01
6.82
5.63
4.44
3.25
*5 b
Residual = y−ypred
1.04
1.61
−4.01
2.18
1.37
−1.44
0.75
*6
Day
Mon (1)
−1.58
Bookings in hotel
49
158
Tue (2)
Wed (3)
Thurs (4)
Fri (5)
Sat (6)
Sun (7)
124
74
56
31
35
22
P df_Fol i o: 49
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
50
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.4
c Coefficient of determination r2 = 0.9177. Therefore, 91.8%
of the variation in y can be explained by the variation in x.
d
x
Residual
a Using CAS:
Using the least squares line of best fit: bookings in
hotel = 161.3 − 2.25 × day, substitute day number into
equation.
See the table at the foot of the page.*
b See the table at the foot of the page.*
c
0
1
2
3
4
5
6
7
8
9
10
Residual
25
20
15
10
5
–5
–10
–15
–20
–25
0
M T W T F S S
x
e
Residual
Slight pattern in the residual plot, so data may not be linear.
d As this is the first week of drought (an atypical event), a
decline in occupancy would be expected.
7 Looking at the placement of the data above and below the
regression line, and comparing this to the residual plot, the
answer is C.
8 a Non-linear as it goes from negative values to positive and
back to negative values.
b Linear as there is no pattern and as the values change sign
randomly.
c Non-linear as it goes from positive values to negative and
back to positive values.
9 a
y
Residual
1 2 3 4 5 6 7 8 9 10
–15
–30
–45
0
1 2 3 4 5 6 7 8 9 10
x
D
20
15
10
5
–5
–10
–15
–20
0
2 4 6 8 10 12 14 16 18 k
x
Since the data randomly jumps from above to below the
k-axis, the data probably has a linear relationship.
b Coefficient of correlation r = 0.958, which indicates a
strong positive relationship.
Using CAS:
P df_Fol i o: 50
60
45
30
15
The points of the data show a curved pattern, indicating that
the data is non-linear.
10 a Using CAS: D = 13.72 + 7.22k
See the table at the bottom of the page.*
b
Residual plot
400
360
320
280
240
200
160
120
80
40
0
58.7
23.8
−3.1
−23.1
−34.0
−37.9
−35.8
−27.8
−5.7
16.4
68.5
*6 a
ypred
138.8
116.3
93.8
71.3
48.8
26.3
3.8
*6 b
Residual = y−ypred
19.2
7.7
−19.8
−15.3
−17.8
8.7
18.2
*10a
k
D
Dpredicted
7.7
66.9
69.31
8.1
82.5
72.20
9.7
88.7
83.75
10.3
91.6
88.09
10.30
4.95
3.51
Residual (D − Dpredicted )
1.6
22.5
25.27
−2.77
2.5
37.8
31.77
6.03
5.9
41.5
56.32
−14.82
−2.41
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
15.4
120.4
124.91
−4.51
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.5
c Coefficient of determination: r2 = 0.94
This also tends to suggest that there is a linear association
between D and k.
b See table at the bottom of the page.*
y = 11.14 + 2.07x2 with r = 0.999 76
This transformation improved the correlation coefficient
from 0.97 to 0.09976; thus the transformed equation is a
better fit of the data.
2 a
y
3.4 Exam questions
1 Residual = actual − predicted
= 63 − 64
= −1
The residual is −1.0 beats per minute.
The correct answer is B.
2
LE = 43.0 + 0.422 × 92.9
= 82.2038
Residual = actual − predicted
= 81.8 − 82.2
= −0.4
The correct answer is B.
3 a Morning median 52%, Evening median 56%
[2 marks]
b Evening congestion level
[1 mark]
c Evening congestion level = 8.48 + 0.922 × 60 = 63.8, i.e.
63.8%
[1 mark]
d Residual = 50 – (8.48 + 0.922 × 47) = −1.8, i.e. −1.8%
[2 marks]
e 0.922 = 0.85 = 85%
[1 mark]
600
500
400
300
200
100
0
3
L1
2
3
4
5
7
9
_________
L3 (1) = 4
x
2
Apply an x transformation to stretch the x-axis.
*1 b
x2
1
4
9
16
25
36
49
64
81
y
12
19
29
47
63
85
114
144
178
*2 b
x2
y
9
5
25
12
81
38
144
75
256
132
441
209
576
291
1089
578
*4
Time (sec)
1
2
3
4
5
6
7
8
1.95
1.85
1.74
1.65
1.59
1.54
1.51
1.48
*5
P df_Fol i o: 51
L2
96
95
92
90
14
−100
_________
log10 (speed) (m s )
x
log10 (y)
−1
5
3
10
2.70
15
2.35
20
2.17
L3
4
9
16
25
49
81
_________
Original data
L3 has been transformed using x2 .
The least squares line of best fit is y = 128.15 − 2.62x2 .
The r2 value has increased from 0.8346 to 0.9425, so the
data has become more linear.
4 See the table at the bottom of the page.*
log10 (speed) = 1.97 − 0.07 × time with r = −0.97
Therefore, the correlation coefficient improved after applying
a logarithmic transformation.
5 See the table at the bottom of the page.*
log10 (y) = 3.01 − 0.04x with r = −0.96
Therefore, the correlation coefficient significantly improved
after applying a logarithmic transformation.
y
200
180
160
140
120
100
80
60
40
20
1 2 3 4 5 6 7 8 9
x
Apply an x transformation to stretch the x-axis.
b See the table at the bottom of the page.*
y = −4.86 + 0.53x2 with r = 0.9990
This transformation improved the correlation coefficient
from 0.96 to 0.9990, thus the transformed equation is a
better fit of the data.
3.5 Exercise
0
5 10 15 20 25 30 35
2
3.5 Transforming to linearity
1 a
51
25
2.00
30
1.85
35
1.77
40
1.75
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
52
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.5
6
Age group (years)
Average height (cm)
9
10
11
12
13
14
15
16
17
18
128
144
148
154
158
161
165
164
166
167
7 a The least squares line of best fit for the transformed data is
y = 24.21 + 117.2 log10 (x).
i When x = 7, y = 24.21 + 117.2 log10 (7) = 123.3 cm
ii When x = 10.5, y = 24.21 + 117.2 log10 (10.5)
= 143.9 cm
iii When x = 20, y = 24.21 + 117.2 log10 (20.5)
= 176.7 cm
b As the data ranges from 9 to 18, answer (2) is interpolated.
8 Normal growth is linear only within a given range; eventually
students stop growing. The logarithmic transformation is a
big improvement over the original regression.
9 a See the table at the bottom of the page.*
1
y = 3.85 + 29.87xT , where xT = or
x
29.87
temperature = 3.85 +
time after 6 p.m.
b 10.30 pm is 4.5 hours after 6 pm.
29.87
Temperature = 3.85 +
time after 6 pm.
29.87
= 3.85 +
4.5
= 10.49 °C
10 a See the table at the bottom of the page.*
1
y = −13.51 + 273.78xT , where xT = or
x
273.78
y = −13.51 +
x
273.78
b When x = 12. y = −13.51 +
x
273.78
= −13.51 +
12
= 9.31
273.78
c Wheny = 20, y = −13.51 +
x
273.78
20 = −13.51 +
x
273.78
20 + 13.51 =
x
273.78
33.51 =
x
33.51x = 273.78
273.78
x=
33.51
x = 8.17
11 See the table at the bottom of the page.*
Average height (cm)
a Using a calculator:
y = 104.69 + 3.76x
r2 = 0.84r = 0.917
180
170
160
150
140
130
120
0
b
9 10 11 12 13 14 15 16 17 18
Age group (years)
log (age group)
Average height (cm)
0.954
1
1.041
1.079
1.114
1.146
1.176
1.204
1.230
1.255
128
144
148
154
158
161
165
164
166
167
y = a + bx
b = 117.2
a = 24.21
r2 = .901
r = .949
c As the r2 value has increased from 0.841 to 0.901, the data
has become more linear.
*9 a
1
Time after 6 pm
Temperature (°C)
P df_Fol i o: 52
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
32
22
16
11
9
8
7
7
1
10
9
1
13
5
1
15
2
1
18
1
*10 a
1
x
y
*11
Distance from light source (metres)
1
2
3
4
5
10
Intensity (candlepower)
90
60
28
22
20
12
1
2
120
1
5
50
1
7
33
1
9
15
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • EXERCISE 3.5
16 a x = 2.5
1
= −12.5 + 0.2x
y
1
= −12.5 + 0.2 × 2.5 = −12
y
1
y=
= −0.08333 ≈ −0.08
−12
a Original data
y
b x = −2.5
1
= −12.5 + 0.2x
y
1
= −12.5 + 0.2 × −2.5 = −13
y
1
y=
= −0.07692 ≈ −0.08
−13
17 a
x
0
Using CAS, the regression equation is:
y = 68.557 − 7.174x
r2 = 0.576
r = −0.759
Use a reciprocal transformation:
Seeds
y
x
0
Using CAS:
The transformed regression equation is
90.687
.
y = 2.572 +
x
90.867
b When x = 20, y = 2.572 +
= 7.1 candlepower.
20
12 a Compress the y- or x-values using logarithms or reciprocals.
b Stretch the y-values using y2 or compress the x-values using
logarithms or reciprocals.
c Compress the y- or x-values using logarithms or reciprocals.
13 a x = 2.5
y = −12.5 + 0.2x2
y = −12.5 + 0.2 × 2.52 = −11.25
x = −2.5
log10 (y) = 0.03 + 0.2x
log10 (y) = 0.03 + 0.2 × −2.5 = −0.47
y = 10−0.47 = 0.33884 ≈ 0.34
*17 d
P df_Fol i o: 53
Circle
Seeds
Residuals
1
3
40.33
2
5
20.59
y = –59.07 + 21.74x
1 2 3 4 5 6 7 8 9 10
Circle
b r =√0.761
r = 0.872
The relationship is strong and positive.
c Circle = 11 seeds = 21.74 circle − 59.07
Seeds = 180.07
Therefore, the number of seeds will be approximately 180.
d See the table at the bottom of the page.*
e
75
60
45
30
15
Residual
14 a x = 2.5
y = −25 + 1.12 log10 (x)
y = −25 + 1.12 log10 (2.5) = −24.55
b x = −2.5: cannot do this as you cannot take the log of a
negative number.
15 a
x = 2.5
log10 (y) = 0.03 + 0.2x
log10 (y) = 0.03 + 0.2 × 2.5 = 0.53
y = 100.53 = 3.3884 ≅ 3.39
240
220
200
180
160
140
120
100
80
60
40
20
0
b x = −2.5
y = −12.5 + 0.2x2
y = −12.5 + 0.2 × (−2.5)2 = −11.25
b
53
–15
–30
–45
0
1 2 3
5 6 7 8 9 10
Circle
f i Stretch using x2 .
ii Compress using log10 (y) or
1
.
y
18 Apply the log10 y transformation to the data used in
question 17.
a Fit a least squares line of best fit to the transformed data
and plot it with the data.
Using CAS, regression line:
log (y) = 0.2721 + 0.2097x
3
8
1.85
4
13
−14.89
5
21
−28.63
6
34
−37.37
7
55
−38.12
8
89
−25.58
9
144
7.41
10
233
74.67
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • REVIEW 3.6
Log seeds
54
2.4
2.1
1.8
1.5
1.2
0.9
0.6
3.6 Review
3.6 Exercise
Multiple choice
0
1 Using CAS, regression line y = −381.97 + 14.417x
The correct answer is C.
sy
2 b=
sx
7.4
= 0.9 ×
5.8
= 1.15
1 2 3 4 5 6 7 8 9 10
Circle
b The correlation coefficient r = 0.9999, this is an almost
perfect relation and is better than the use of a linear
relationship.
c The least squares line of best fit for the transformation:
log10 (y) = 0.2721 + 0.2097x
d The coefficient of determination = 0.9999 ∴ we can say
that 99.99% (100.0%) of variation in number of seeds is due
to number of circles. This is a perfect relation, often found
in nature (see the Golden Ratio).
e log10 (y) = 11
log10 (number of seeds) = 2.578
number of seeds = 102.578 = 378
f This is a much better prediction as it follows a steep
upward trend.
a = y − bx
= 172.5 − 1.15 × 154.4
= −5.06
The correct answer is B.
3 Points on line (5, 25) and (4, 20)
y2 − y1
25 − 20
=
=5
m=
x2 − x1
5−4
y − 25 = 5 (x − 5)
y = 5x
The correct answer is D.
4 Actual value 33
Predicted values 30
Residual 33 − 30 = 3
The correct answer is E.
5 The residual graph is identical to the original graph apart from
a different scale and the least squares line of best fit becomes
the horizontal line at 0.
The correct answer is C.
6 y = 0.4 × 2.52 + 12.1
= 14.6
The correct answer is C.
3.5 Exam questions
log10 N = 1.160 + 0.03617 × 3
log10 N = 1.126851
N = 101.126851
N = 18.557
= 19
The correct answer is C.
2 Enter the x-data values into List1 and the y-data values into
List2 of a CAS calculator. Then let List3 = List12 . Then draw
a scatterplot with List3 on the x-axis and List2 on the y-axis.
Fit a least squares line of best fit and its equation will be:
y = 7.1 + 2.9x2
The correct answer is A.
3 Substitute x = 1.1 into the equation.
log10 (y) = 3.1 − 2.3 (1.1)
log10 (y) = 0.57
y ≈ 3.7
The correct answer is E.
1
*8 a
x
y
*8 b
x
ypred
*8 c
x
1
23
2
21
3
20.6
5
18.1
1
Residual
4
20
−0.1
8
14
7
15.6
2
−0.85
4
0.65
Short answer
7 a −0.45
b 16.75
8 a See the table at the bottom of the page.*
Using CAS, the least squares line of best fit is
y = 24.35 − 1.25x and the correlation coefficient is
r = −0.96.
b See the table at the bottom of the page.*
c Use table of residuals in CAS:
See the table at the bottom of the page.*
9
16
10
9
11
12
9
13.1
11
10.6
13
8.1
8
−0.35
9
2.9
15
5
15
5.6
10
−2.85
17
3.2
11
1.4
20
−0.65
15
−0.6
P df_Fol i o: 54
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 3 Investigating and modelling linear associations • REVIEW 3.6
g Square of residual error is reduced, correlation (0.98) is
closer to 1, graph looks more linear.
h Prediction of 18.49 against 22.87 using untransformed data.
Given the nature of the data, likely to be more accurate.
12 a Gradient = 8.6. It shows that on average, the test mark
increases by 8.6% with every extra hour spent on studying
for the test.
The y-intercept is 21.6. It indicates the student who did not
study for the test can expect to get 21.6%.
b Number of hours spent studying = 6
Test mark (%) = 21.6 + 8.6 × 6
= 73.2
= 73%
Therefore, the equation predicts that Nathan will get 73%
on the test.
This prediction is an example of interpolation, as 6 hours is
within the original set of data.
c Number of hours spent studying = 6
Test mark (%) = 21.6 + 8.6 × 9
= 99
Therefore, the equation predicts that Rachel will get 99%
on the test.
This prediction is an example of extrapolation, as 9 hours is
outside the original set of data. Therefore, it is probably not
very reliable.
Extended response
Speed (wpm)
9 a Common sense suggests that the speed of typing would
depend on number of hours spent practising and not the
other way around. So time is the explanatory variable and
speed is the response variable.
b
65
60
55
50
45
40
35
30
0
20 25 30 35 40 45 50
Time (h)
c The scatterplot suggests a strong, positive, linear
relationship between the time spent practising and the speed
of typing.
10 a Likely to be a y versus x2 relationship
b A poor predictor for most values of x
c 128. Predictions outside the data set are not as reliable as
predictions within the data set. As the salesperson gains
experience, the sales per day will no longer increase at the
same rate.
d See the table at the bottom of the page.*
e y = −13.86 + 1.96xT , where xT = x2
f 182. The regression line is a better fit for the transformed
data and hence is likely to be a better predictor of the sales
figures.
11 a y = 4.27 + 1.55x
b 22.87
c Simplicity of eye fitting versus accuracy in this case is quite
good. Little difference in the sum of squared errors. Least
squares regression gives quite a different answer from the
other 2 methods, with consequent change in errors. (The
3-median method is subject to errors due to outliers and
computational errors.)
d Not very linear, logarithmic transformation suggested.
e
Log (month)
Production (tonnes)
55
3.6 Exam questions
1 a i Median age is 24 years.
[1 mark]
body density
= 1.065
[1 mark]
ii ∑
12
b i Body density is being predicted from weight; therefore,
weight is the explanatory variable.
[1 mark]
ii Using your CAS calculator, perform a least squares line
of best fit analysis.
10
0
0.301
0.477
0.602
0.699
0.778
0.845
0.903
0.954
3
8
10.8
12
11.6
14
15.5
15
18.1
Slope = −0.001 12 correct to 3 significant
figures.
[1 mark]
c This question is referring to the coefficient of
determination(r2 ), which is found on the least squares line
of best fit screen on CAS.
r2 = 0.28995...
Therefore, 29% of the variation in body density may be
explained by the variation in weight.
[1 mark]
f y = 3.30 + 14.08xT , where xT = log10 (x)
*10 d
P df_Fol i o: 55
Day
1
4
9
16
25
36
49
64
Units sold
1
2
4
9
20
44
84
124
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
56
TOPIC 3 Investigating and modelling linear associations • REVIEW 3.6
2 a The line needs to go through approximately (60, 1.104) and
(130, 0.998).
[1 mark]
See the graph at the bottom of the page.*
b body density = 1.195 − 0.001512 × 65 = 1.09672 =
1.10 kg/litre.
[1 mark]
c A waist measurement of 65 cm is outside of the original
data set, therefore the prediction in part b is extrapolated.
[1 mark]
d For every 1 cm increase in waist measurement, body
density will decrease by 0.001512 kg/litre.
[1 mark]
e predicted body density = 1.195 − 0.001512 × 122 =
1.010536
residual = actual − predicted
= 0.995 − 1.010536
= −0.015536
= −0.02
[1 mark]
f If r2 = 0.6783 and the data is decreasing (i.e. a negative
[1 mark]
gradient), then r = −√06783 = −0.824.
g Yes, because there is no pattern to the residual plot. The
residuals are randomly scattered above and below the
horizontal axis.
[1 mark]
3 a Difference in the mean heights = 167.1 − 156.7 = 10.4 cm.
[1 mark]
b The association is strong and negative.
[1 mark]
157 − 168
c slope =
≈ −0.169
85 − 20
intercept = 168 − (−0.169 × 20) = 171.38 = 171
Therefore, the equation will be
mean height = 171 − 0.169 × mean age.
[1 mark]
d The most appropriate transformation of the explanatory
variable would be x2 .
Use a CAS calculator to perform the transformation and the
association linear regression.
mean height = 167.9 − 0.001621 × (mean age)2 [2 marks]
4 a Relative humidity at 9 am
[1 mark]
b Perform a least squares calculation on your CAS calculator:
humidity 3 pm = −1.26 + 0.765 × humidity 9 am
[1 mark]
c From CAS: r = 0.871
[1 mark]
5 a On average, for each 1 hPa increase in pressure at 9 am, the
pressure at 3 pm increases by 0.8894 hPa.
[1 mark]
b Pressure 3 pm = 111.4 + 0.8894 × 1025 = 1023.035
= 1023 hPa
[1 mark]
c Interpolation
The value 1025 hPa is within the data range for pressure at
9 am.
[1 mark]
d When pressure 9 am = 1013 hPa, Predicted pressure 3 pm
= 111.4 + 0.8894 × 1013 = 1012.3622 hPa
According to the graph, actual pressure at 3 pm is
1015 hPa.
Residual = actual – predicted = 1015 – 1012 = 3 hPa.
[1 mark]
sy
sx
e i b = r or rearrange to be r = b
sx
sy
4.5477
r = 0.8894 ×
= 0.96569678 = 0.966 [1 mark]
4.1884
2
2
ii r = 0.966 = 0.933156
[1 mark]
93.3% of the variation in pressure at 3 pm may be
explained by the variation in pressure at 9 am.
f i That the data follows a linear relationship between the
two variables.
[1 mark]
ii There is a curved pattern in the residual plot. It is
not random enough. A transformation may be
required.
[1 mark]
y
*2a
1.15
Body density
(kg/litre)
1.10
1.05
1.00
0.95
P df_Fol i o: 56
60
70
80
90
100
110
120
130
x
Waist measurement (cm)
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.2
57
Topic 4 — Investigating and modelling time series data
Number moved
4.2 Time series plots and trends
4.2 Exercise
Sales ($)
1 a The time series plot displays seasonality only.
b The time series plot displays seasonality with an upward
trend.
c The time-series plots displays irregular fluctuations with no
obvious trend.
2 Any time-series plot may have irregular fluctuations while
still showing an upward or downward trend.
The correct answer is C.
3
1500
1450
1400
1350
1300
0
t
1 2 3 4 5 6 7 8
Year
Sales ($)
The sales can be classified as having an upward trend with a
possible outlier at the second data point.
a
Amount of rainfall per month
Seasonal
b Number of soldiers in US army,
measured annually
Irregular
c
Upwards trend
e
0
t
1 2 3 4 5 6 7 8
Year
The linear trend line is y = 1342.96 + 16.45x, where y
represents sales in dollars and x corresponds to one year after
the new business opening.
6
Temperature (°C)
Number moved
4
180
170
160
150
140
130
120
110
100
Irregular or cyclical
with slight upwards
trend
Cyclical
Number of seats held by
the Liberal Party in Federal
Parliament
40
38
36
34
32
30
28
26
24
22
20
18
16
Temp = 40 − 2.33t
t
1 2 3 4 5 6 7 8 9 10
Year
0
See the table at the foot of the page.*
The number of families who moved from Melbourne to
Ballarat can be classified as random with a secular upward
trend.
*4
Number of people living in
Australia, measured annually
d Share price in BHP Billiton,
measured annually
1500
1450
1400
1350
1300
0
t
1 2 3 4 5 6 7 8 9 10
Year
The linear trend line is y = 97.8 + 7.22x, where y is the number of
families who moved from Melbourne to Ballarat and x = 1
corresponds to the year beginning 2013.
5
0
180
170
160
150
140
130
120
110
100
Year
Number moved
Time code
2013
97
1
2014
118
2
2015
125
3
1 2 3 4 5 6 7 8 9 10 11 12
Days
t
There is a definite downward trend.
2016
106
4
2017
144
5
2018
155
6
2019
162
7
2020
140
8
2021
158
9
2022
170
10
P df_Fol i o: 57
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.2
58
7 See the table at the bottom of the page.*
The graph of umbrella sales appears symmetrical (bell
shaped) and it is almost impossible to fit a trend line due to
the cyclical nature of data.
10 a
y
y = 7.09 + 0.83t
0
2 4 6 8 10 12
t
Price (cents)
20
15
10
5
Price ($)
Although there are some random variations, the trend could
also be cyclical and upward.
8 See the table at the bottom of the page.*
Using CAS:
y = 2.257 + 0.243t
6
5
4
3
2
1
2 4 6 8 10 12 14 t
Month
Sales (× $1000)
100
90
80
70
60
50
40
30
20
10
240
220
200
180
160
140
120
100
t
0
0
t
1 2 3 4 5 6 7 8 9 10 11 12
*7
t
y
*8
1
6
2
9
3
13
Month
Jan.
Price $
Time code
2.50
1
2020
2021
Quarter
2022
10
11
11
15
Using CAS, the trend line equation is:
y = 111.515 + 4.510x
Sales = 111.52 + 4.51 × quarter
It is difficult to fit a trend line due to the likely cyclical nature
of the software sales business.
Using CAS to find the least squares line:
y = 35.606 + 1.573t
P df_Fol i o: 58
2 4 6 8 10 12 14 16 18 20 22 24 26
Week
Using CAS:
y = 39.42 + 2.21t
b When t = 25, price = 39.42 + 2.21 × 25
= 95 cents
11 See the table at the bottom of the page.*
Share prices are irregular and increasing. In January
(13th month) the estimated share price is $5.42, but this
is an estimated price only as the value has been extrapolated
(i.e. the value is outside the plotted values).
9 See the table at the bottom of the page.*
*11
t
0
0
*9
100
95
90
85
80
75
70
65
60
55
50
45
40
4
8
5
9
6
14
7
15
8
17
9
14
Feb.
Mar.
Apr.
May
June
July
Aug.
2.70
2
3.00
3
3.20
4
3.60
5
3.70
6
3.90
7
4.20
8
12
19
Month
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sep.
Oct.
Nov.
Dec.
Sales
Time
code
5
10
15
40
70
95
100
90
60
35
20
10
1
2
3
4
5
6
7
8
9
10
11
12
Quarter Q1 − 20 Q2 − 20 Q3 − 20 Q4 − 20 Q1 − 21 Q2 − 21 Q3 − 21 Q4 − 21 Q1 − 22 Q2 − 22 Q3 − 22 Q3 − 22
Sales
Time
code
120
135
150
145
140
120
100
110
120
140
190
220
1
2
3
4
5
6
7
8
9
10
11
12
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.3
b Monday in week 4 = time code 22
Millions of viewers = 1.18 − 0.02 × time code
= 1.18 − 0.02 × 22
= 0.74
Number of viewers = 0.74 × 1 000 000
= 740 000
Number of employees
12 See the table at the bottom of the page.*
7000
6000
5000
4000
3000
2000
1000
0
1 2 3 4 5 6 7 8 9 10
Month
59
2 Using CAS,
y = 128.1 − 0.77x
or driver fatalities = 128.1 − 0.77 × time code
t
a At June year 7, time code = 78
Driver fatalities = 128.1 − 0.77 × time code
= 128.1 − 0.77 × 78
= 128.1 − 60.06
= 68.04
The predicted number of driver fatalities in June year 7
is 68.
b At April year 10, time code = 112
Driver fatalities = 128.1 − 0.77 × time code
= 128.1 − 0.77 × 112
= 128.1 − 86.24
= 41.86
The predicted number of driver fatalities in April year 10
is 42.
Using CAS, the trend line equation is:
y = 6333.33 − 300.61x
Number of employees = 6333.33 − 300.61 × months
There is a decreasing trend of approximately 300 employees
per month. Eventually the bank will have no employees if this
trend continues. Although not likely, there is a clear
downward trend.
4.2 Exam questions
1 There is evidence of a decreasing trend and irregular
fluctuations.
However, there is no evidence of a seasonality, as the peaks
and troughs do not occur at regular intervals.
The correct answer is D.
2 Find the median of the vertical axis (‘passengers’). There are
12 data values, so the median will be in the 6th spot counting
from the bottom (or the top).
The correct answer is A.
3 Temperature is seasonal, which we can see from the time
series plot. However, there are irregular fluctuations in the
data.
The correct answer is C.
3
y
40
30
20
10
The equation is y = 33.72 − 2.9t (found using CAS).
0
1 2 3 4 5 6 7 8 9 10
t
a y = 33.72 − 2.9 × 11
= 1.82
The number of cars remaining after 11 hours is 2.
b The rate is 2.9 cars per hour.
c Solve y = 33.72 − 2.9t, where y = 0.
0 = 33.72 − 2.9t
−33.72
t=
−2.9
= 11.6276
There will be no cars left after 11.63 hours.
4.3 Fitting the least squares line and forecasting
4.3 Exercise
1 See the table at the bottom of the page.*
Using CAS: y = 1.18 − 0.02x or
millions of viewers = 1.18 − 0.02 × time code
a Wednesday in week 3 = time code 17
Millions of viewers = 1.18 − 0.02 × time code
= 1.18 − 0.02 × 17
= 0.84
Number of viewers = 0.84 × 1 000 000
= 840 000
*12
*1
P df_Fol i o: 59
Month
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
Dec.
Employees
6100
5700
5400
5200
4800
4400
4200
4000
3700
3300
Time code
1
2
3
4
5
6
7
8
9
10
Viewers (1 000 000s)
M
T
W
Th
F
S
Su
M
T
W
Th
F
S
Su
1.20
1.18
1.16
1.18
0.9
0.75
1.0
1.21
1.23
1.19
1.16
0.95
0.68
0.98
Time code
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Day
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.3
60
4 y = 2.68 + 1.16x, where y = numbers of staff and x = months,
where x = 1 is the first month of business
a
Number of staff
20
18
16
14
12
10
8
6
4
2
8 a Share price ($) = $18.57 − $0.1 × time code
Time code of t = 1 represents 2020.
b The y-intercept of $18.57 represents the approximate value
of the shares in 2019. The gradient of −$0.1 means that on
average the share price will decline by $0.1 (10 cents) each
year.
c At 2029, time code = 10∶
share price ($) = $18.57 − $0.1 × time code
= $18.57 − $0.1 × 10
= $18.57 − $1.0
= $17.57
9 a y = share price, t = time code
So the equation is:
Share price = $2.56 + $0.72 × time code
b The y-intercept of $2.56 represents the approximate value
of the shares in 2019. The gradient of +$0.72 means that
the share value will grow by $0.72 each year.
c Time code for year 2030 is 11.
So share price = $2.56 + $0.72 × 11
= $10.48
t
0
1 2 3 4 5 6 7 8 9 10 11 12
Month
b A further 12 months: t = 24
y = 2.68 + 1.16x
= 2.68 + 1.16(24)
= 30.52
Therefore, an approximate number of 31 staff will be
required in a further 12 months.
5 See the table at the bottom of the page.*
a The least squares line is y = 2.48 + 0.044t.
y
5
4
3
2
1
b The 3-median least squares line is y = 2.79 + 0.018t.
t
0
large values well. Least squares line of best fit equation is
y = 0.283 + 0.83x.
7 a Share price ($) = $22.74 + $0.28 × time code
Time code of t = 1 represents 2020.
b The y-intercept of $22.74 represents the approximate value
of the shares in 2019. The gradient of +$0.28 means that on
average the share price will grow by $0.28 each year.
c At 2030, time code = 11∶
share price ($) = $22.74 + $0.28 × time code
= $22.74 + $0.28 × 11
= $22.74 + $3.08
= $25.82
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
y
5
4
3
2
1
10 a, b
t
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Neither line is very effective as a predictor. Note the small
gradients (0.044 and 0.018). This is either a random or
cyclical trend.
10
9
8
7
6
5
4
3
2
1
Sales
Visitors (millions)
6
0
1 2 3 4 5 6 7 8 9 10
Month
P df_Fol i o: 60
Jan.
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sales
Time
code
65
95
130
115
145
170
190
220
1
2
3
4
5
6
7
8
300
275
250
225
200
175
150
125
100
75
50
25
t
0
Not a good predictor as data is curved (appears exponential)
and the least squares line of best fit will not predict small and
*5
Month
1 2 3 4 5 6 7 8 9 10 11 12
Month
t
The least squares line of best fit is y = 49.64 + 20.36t,
where y = sales and t = months.
Day
1
2
3
4
5
6
7
8
Price ($)
2.75
3.30
3.15
2.25
2.10
1.80
1.50
2.70
Day
9
10
11
12
13
14
15
16
Price ($)
4.10
4.20
3.55
1.65
2.60
2.95
3.25
3.70
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.3
c The predicted value for December is 293.93, approximately
294 sales.
a, b
Y1
293.93
314.29
334.64
355
375.36
395.71
416.07
Price ($)
X
12
13
14
15
16
17
18
X = 12
c
Sales (× 1000)
60
55
50
45
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Week
t
X
13
14
15
16
17
18
19
X = 13
Mark (%)
a, b
P df_Fol i o: 61
85
80
75
70
65
60
55
50
0
1 2 3 4 5 6 7 8 9 10
Month
t
The least squares line of best fit is y = 56.9 + 1.93t
where y = marks and t = months.
c
X
11
12
13
14
12 See the table at the bottom of the page.*
*13
Y1
398.38
404.73
411.09
417.44
423.79
430.14
436.5
13 See the table at the bottom of the page.*
Y1
35.833
43.367
50.9
58.433
65.967
73.5
81.033
Sales in week 10 = $35.83, sales in week 12 = $43.37 and
sales in week 14 = $50.90. There may be slight differences
in your answers due to rounding.
d Even though the r2 value is close to 1, due to the shape of
the graph, this trend line is not a good fit of the data.
*12
t
The predicted price for the next quarter is
approximately $398.
d The first two years seemed seasonal, the last year an
upward trend, so the overall trend line would appear to be a
poor predictor.
The least squares line of best fit is y = −1.83 + 3.77t,
where y = sales and t = weeks.
X
10
12
14
16
18
20
22
X = 10
1 2 3 4 5 6 7 8 9 10 11 12
Quarter
The least squares line of best fit is y = 315.8 + 6.35t
where y = price and t = quarters.
11 a, b
c
460
440
420
400
380
360
340
320
300
0
d As r2 is close to 1, this is a very good prediction of sales.
0
61
Y1
78.133
80.067
82
83.933
Quarter
1
2
3
4
5
6
Price ($)
358
323
316
336
369
333
Quarter
7
8
9
10
11
12
Price ($)
328
351
389
387
393
402
Test
Mark (%)
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
57
63
62
67
65
68
70
72
74
77
Time code
1
2
3
4
5
6
7
8
9
10
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.3
62
15
85.867
16
87.8
17
89.733
X = 11
The results for the last exam in December would be
predicted to be around 78%.
d The trend line appears to be an excellent predictor of scores
(increasing trend, r2 close to 1).
14 a
Time
(yrs)
1
2
3
4
5
6
7
Stock price ($)
300
100
1
2
3
4
5
6
7
8
9
Hotel
price ($)
250
240
235
237
239
230
228
237
332
t
3200
3000
2800
2600
2400
2200
2000
0
400
375
350
325
300
275
250
225
200
0
1 2 3 4 5 6 7 8
Time (years)
t
1 2 3 4 5 6 7 8 9 10
Month (2019)
b The outlier appears to be the September value.
t
Hotel price ($)
c y = 1777.14 + 192.14x, where y = All Ordinaries
and x = time
d The gradient increased and the y-intercept is lower when
the outlier is taken out.
15 See the table at the bottom of the page.*
a y = 199.71 + 8.62 x, where y = stock price ($) and
x = month, where x = 1 is January 2022
400
375
350
325
300
275
250
225
200
0
1 2 3 4 5 6 7 8 9 10
Month (2019)
t
c y = 245.79 − 1.95x
d The gradient has changed significantly, going from a
positive value to a negative value. The y-intercept
increased.
300
Stock price ($)
t
Month
(2019)
Hotel price ($)
1 2 3 4 5 6 7 8
Time (years)
b Taking out the 2000 data value:
250
200
150
100
50
4.3 Exam questions
0
2
1 On your CAS, enter the data and find the LSR equation:
time = 44 − day
When day = 10, time = 44 − 10 = 34, which is the same as
day 4.
The correct answer is B.
4 6 8 10 t
Month (2022)
b Difficult to see any obvious outliers; from the graph, the
April value is a possible outlier.
*15
4 6 8 10
Month (2022)
16 a y = 222.72 + 4.97x, where y = hotel price ($) and
x = months, where x = 1 is January 2019
0
P df_Fol i o: 62
2
c y = 193.72 + 9.13x
d The gradient increases and the y-intercept is lower when
the April value is taken out.
y = 1848.57 + 173.1x, where y = All Ordinaries value and
x = time, where x = 1 is 1995.
All Ordinaries
150
0
All
1925 2140 2435 2500 2820 3200 2950 3050
Ords
All Ordinaries
200
50
8
3200
3000
2800
2600
2400
2200
2000
250
Month 2022
Stock prices ($)
1
2
3
4
5
6
7
8
9
192.06
204.62
235.00
261.09
256.88
251.53
257.25
243.10
283.75
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.4
c The last smoothed data point is 24.83, so the temperature at
9 hours is predicted to be 24.83 °C.
2 On your CAS calculator, enter the data, generate the
scatterplot and add the LSR line. 6 values lie below the
regression line.
2 a
y
x
The correct answer is D.
3 Enter the original data into the CAS calculator and perform a
least squares line of best fit.
y = 28.5 + 1.767x
The correct answer is E.
4.4 Smoothing using the moving mean with an
odd number of points
Time (h) Temp (°C)
1
20.3
2
21.8
3
20.9
4
22.0
5
23.4
6
24.9
7
24.3
8
25.3
Smoothed temperature (°C)
2015
38 587
2016
42 498
2017
45 972
2018
57 408
2019
71 271
2020
72 688
2021
78 427
2022
72 170
Smoothed members (m)
1
(38 587 + 42 498 + 45 972) = 42 352
3
1
(42 498 + 45 972 + 57 408) = 48 626
3
1
(45 972 + 57 408 + 71 271) = 58 217
3
1
(57 408 + 71 271 + 72 688) = 67 122
3
1
(71 271 + 72 688 + 78 427) = 74 129
3
1
(72 688 + 78 427 + 72 170) = 74 428
3
b
1
(20.3 + 21.8 + 20.9) = 21.00
3
1
(21.8 + 20.9 + 22.0) = 21.57
3
1
(20.9 + 22.0 + 23.4) = 22.10
3
1
(22.0 + 23.4 + 24.9) = 23.43
3
1
(23.4 + 24.9 + 24.3) = 24.20
3
1
(24.9 + 24.3 + 25.3) = 24.83
3
Original
Smoothed
80 000
75 000
70 000
65 000
60 000
55 000
50 000
45 000
40 000
35 000
0
t
Year
c The last smoothed data point is 74 428, so the number of
members in 2024 is predicted to be 74 428.
3 a
Year (t)
Sales (y)
3-point moving mean
2015
2016
2017
2018
2019
2020
2021
2022
b
Temperature (°C)
Members
(m)
20
1
20 5
1
20 6
1
20 7
1
20 8
1
20 9
2
20 0
2
20 1
22
1 a
Year
(t)
Members
4.4 Exercise
63
26
25.5
25
24.5
24
23.5
23
22.5
22
21.5
21
20.5
20
2250
2600
2400
2750
2900
2450
3100
3400
2416.7
2583.3
2683.3
2700.0
2816.7
2983.3
The 3-point moving mean for 2016 is used as an example.
2250 + 2600 + 2400
3-point moving mean for 2014 =
3
7250
=
3
= 2416.7
0
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8
Time (hours)
Original
t
Smoothed
P df_Fol i o: 63
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.4
64
b
d
Original
Smoothed
3400
3300
3200
3100
3000
2900
2800
2700
2600
2500
2400
2300
2200
2100
t
20
2015
1
20 6
1
20 7
1
20 8
1
20 9
2
20 0
2
20 1
2
20 2
23
20
24
0
Jan.
Feb.
Mar.
Apr.
120
70
100
110
96.7
93.3
100.0
May
90
93.3
June
July
80
70
80.0
80.0
Aug.
90
80.0
Sep.
80
90.0
Oct.
Nov.
Dec.
100
60
60
80.0
73.3
Sales
130
120
110
100
90
80
70
60
50
40
1 2 3 4 5 6 7 8 9 10 11 12
Month
Original
Smoothed
t
Sales
5-point moving mean
Jan.
Feb.
Mar.
Apr.
120
70
100
110
98
90
May
90
90
June
July
80
70
88
82
Aug.
90
84
Sep.
80
80
Oct.
Nov.
Dec.
100
60
60
78
130
120
110
100
90
80
70
60
50
40
0
1 2 3 4 5 6 7 8 9 10 11 12
t
Month
Smoothed
Original
Smoothed data: A definite downward trend is now
apparent.
5
Smoothed data: There is a possible downward trend, but
there are still fluctuations.
0
Sales
c The predicted value for 2023 is 3350 books.
4 a
Month
Sales
3-point moving mean
Month
Quarter
Rainfall (mm)
3-point moving mean
1
2
3
4
5
6
7
8
9
10
11
12
100
50
65
120
90
50
60
110
85
40
50
100
71.7
78.3
91.7
86.7
66.7
73.3
85.0
78.3
58.3
63.3
b Price = 99.78 − 2.39t
c For March next year, t = 15.
Price = 99.78 − 2.39 × 15
= 63.93
The data are cyclical and the prediction is based on
smoothed data that have removed this trend.
P df_Fol i o: 64
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Rainfall (mm)
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.4
120
110
100
90
80
70
60
50
40
30
20
10
0
1 2 3 4 5 6 7 8 9 10 11 12
Quarter
Original
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
t
Smoothed
Spring 2018–Winter 2021
Smoothed data: Did not remove seasonal fluctuations.
From the figure, there may be a slight trend downward.
Attendance
6 a
Year
Attendance
3-point moving mean
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
75
72
69
74
66
72
61
64
69
65
72.0
71.7
69.7
70.7
66.3
65.7
64.7
66.0
Attendance
Smoothed
Linear
80
75
70
65
60
55
0
b
t
1 2 3 4 5 6 7 8 9 10
Year 2013–2022
The smoothed data shows a clear downward trend.
b Using CAS:
Attendance = 74.47 − 1.11x
c For 2024, t = 12
Attendance = 74.47 − 1.11 × 12
= 61.15
Attendance in 2024 will be 61. This is a reasonable
prediction as long as the trend continues to decline as given
by the negative gradient.
7 a The formula for cell C3 has been given as an example.
Copy the formula into cells C4 to C21.
= sum (B2 ∶ B4) /3
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
A
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
Sales
34
27
31
37
41
29
32
37
47
38
41
44
47
49
41
52
48
44
49
56
54
Sales
34
27
31
37
41
29
32
37
47
38
41
44
47
49
41
52
48
44
49
56
54
65
C
Smoothed data
30.67
31.67
36.33
35.67
34
32.67
38.67
40.67
42
41
44
46.67
45.67
47.33
47
48
47
49.67
53
Smoothed data
34.00
33.00
34.00
35.20
37.20
36.60
39.00
41.40
43.40
43.80
44.40
46.60
47.40
46.80
46.80
49.80
50.20
P df_Fol i o: 65
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.4
Sales
66
Q2 − 21
55
50
45
40
35
30
0
2 4 6 8 10 12 14 16 18 20
The 5-point moving mean smoothing has made it more
evident that there is an upward trend in the data.
Sales
5-point moving mean
Feb.
Mar.
Apr.
141
270
234
253.8
May
357
303
June
July
267
387
306.6
320.4
Aug.
288
322.4
Sep.
303
362
Oct.
Nov.
Dec.
367
465
398
364.2
Price ($)
Month
Sales
Q1 − 22
351
356
Q2 − 22
389
375.67
Q3 − 22
387
389.67
393
394
402
Original
Smoothed
420
410
400
390
380
370
360
350
340
330
320
310
300
1 2 3 4 5 6 7 8 9 10 11 12
t
The smoothed data has some but not all the seasonal
fluctuations removed.
b
Original
Smoothed
600
550
500
450
400
350
300
250
200
150
100
1 2 3 4 5 6 7 8 9 10 11 12
t
Quarter
Q1 − 20
Price ($)
Q2 − 20
358
Q3 − 20
323
Q4 − 20
316
325
Q1 − 21
336
340.33
369
346
3-point moving mean
332.33
Price ($)
Month
P df_Fol i o: 66
Q4 − 21
337.33
Quarter
Most random variation smoothed, slight upward trend.
Using the last smoothed value, the expected sales for the next
month are 364.
9 a
343.33
328
0
Calculation for the 5-point moving mean (April’s figure has
been used as an example):
141 + 270 + 234 + 357 + 267
For April, sales =
5
1269
=
5
= 253.8
0
333
Q4 − 22
t
Week
8
Q3 − 21
Quarter
Sales
Q1-20
358
Q2-20
323
Q3-20
316
340.40
Q4-20
336
335.40
Q1-21
369
336.40
Q2-21
333
343.40
Q3-21
328
354.00
Q4-21
351
357.60
Q1-22
389
369.60
Q2-22
387
384.40
Q3-22
393
Q4-22
402
385
380
375
370
365
360
355
350
345
340
335
0
(continued)
Smoothed data
Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
20 20 20 20 21 21 21 21 22 22 22 22
Quarters
t
It has further smoothed out the seasonal fluctuations.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.5
10
Week
Employees
3-point moving mean
1
2
3
4
5
6
7
67
78
54
82
69
88
94
66.33
71.33
68.33
79.67
83.67
The predicted number of people for week 8 is
approximately 84.
5
63
6
52
7
47
8
61
(50 + 63 + 52 + 47)
= 53
4
(63 + 52 + 47 + 61)
=
4
55.75
(52 + 53)
= 52.5
2
(53 + 55.75)
= 54.38
2
4.4 Exam questions
186 + x + y + z + 346
= 206
5
x + y + z = 205 × 5 − 186 − 346
The three-mean smoothed value on Thursday will be:
x + y + z 205 × 5 − 186 − 346
=
3
3
= 166
The correct answer is B.
2160 000
= 432 000
2
5
The correct answer is D.
3. For seven-mean smoothing, you would lose three data points
at the beginning and another three at the end. Six data points
would be lost in total, which means that six smoothed data
points would be left.
The correct answer is C.
4.5 Smoothing using the moving mean with an
even number of points
Sales (× $1000)
1 The five-point moving mean smoothed value on Thursday is:
67
Original
Smoothed
64
62
60
58
56
54
52
50
48
46
44
42
40
0
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8
t
Time
b The data shows a steady increasing trend (even with four
smoothed points). This was not obvious from the original
data.
2 a
4-point centred
Time Sale 4-point moving mean
moving mean
1
554
2
503
3
467
4.5 Exercise
1 a
Time Sale
1
59
2
48
3
43
4
50
4-point moving mean
(59 + 48 + 43 + 50)
= 50
4
(48 + 43 + 50 + 63)
= 51
4
(43 + 50 + 63 + 52)
= 52
4
4-point centred
moving mean
(50 + 51)
= 50.5
2
(51 + 52)
= 51.5
2
(554 + 503 + 467 + 587)
4
= 527.75
(503 + 467 + 587 + 636)
4
= 548.25
4
587
(467 + 587 + 636 + 533)
4
= 555.75
(527.75 + 548.25)
2
= 538
(548.25 + 555.75)
2
= 552
P df_Fol i o: 67
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.5
68
5
636
(587 + 636 + 533 + 493)
4
= 562.25
6
533
Cost ($)
(636 + 533 + 493 + 684)
4
= 586.50
7
493
8
684
9
10
62
64
Calculations:
4-point moving mean:
75 + 54 + 62 + 60
= 62.75
4
4-point centred mean:
62.75 + 61.5
= 62.125
2
80
75
70
65
60
55
50
45
40
0
1 2 3 4 5 6 7 8 9 10 11 12
Original
Smoothed
4
t
t
Price
1
2
45
67
3
51
4-point
moving mean
4-point mean
after centring
51.75
1 2 3 4 5 6 7 8
Time
t
y
1
2
75
54
4-point
moving mean
62
4
4-point mean
after centring
60
5
70
62.125
60.375
59.25
58.25
57.25
45
57.125
57
54
44
54.63
55.75
5
52
6
76
57.25
58.75
59.25
59.75
7
63
8
48
60.50
61.25
61.5
4
52.625
53.5
t
62.75
7
57.375
b Smoothed data indicated a general downwards trend,
possibly with a cyclic trend in the original data.
3 a
6
59
59.75
(562.25 + 586.50)
2
= 574.38
b This data shows a steady increasing trend. This is not
obvious with the original data.
3
8
Original
Smoothed
700
680
660
640
620
600
580
560
540
520
500
480
460
0
(555.75 + 562.25)
2
= 559
61.75
62.25
9
58
62.63
63
10
80
11
12
66
52
63.50
64
56
55
P df_Fol i o: 68
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
4
80
75
70
65
60
55
50
45
40
73
69
78.375
77.5
5
62
6
85
76.625
75.75
75.375
75
7
83
74.875
74.75
0
1 2 3 4 5 6 7 8 9 10 11 12
Season
Original
Smoothed
t
Smoothed data indicates a strong upward trend of almost
11 cents over 3 years.
5 a
4-point
4-point centred
Day Temperature moving mean moving mean
1
36.6
2
36.4
36.75
36.7875
3
36.8
36.825
36.8375
4
37.2
36.85
36.9
5
36.9
36.95
36.975
6
36.5
37
37.025
7
37.2
37.05
37.1625
8
37.4
37.275
37.325
9
37.1
37.375
37.3125
10
37.4
37.25
37.2625
11
37.6
37.275
37.3
12
36.9
37.325
37.2375
13
37.2
37.15
14
37.6
15
36.9
Calculations:
4-point centred moving mean:
For day 3, the formula used in cell D4 is C3
= sum (C3 ∶ C4) /2
Copy cell D4 down into cells D5 to D14.
b Using CAS:
Temperature = 36.766 + 0.056d
c Temperature = 36.766 + 0.056d
On day 16, d = 16
Temperature = 36.766 + 0.056 × 16
= 37.7 °C
6 You leave the first two and the last two points.
The correct answer is A.
7 With a 6-point centred moving mean, you lose 6 points.
April has 30 days, thus 30 − 6 = 24.
The correct answer is D.
8 a
4-point
4-point mean
Season
Sales
moving mean
after centring
1
2
78
92
90
70
9
61
73.875
71.875
70.75
10
78
69.375
68
11
12
74
59
b
100
95
90
85
80
75
70
65
60
55
50
0
1 2 3 4 5 6 7 8 9 10 11 12
t
Season
Original
Smoothed
c The smoothed data indicates a clear downward trend in
sales over three years.
d
6-point
6-point moving
t
Temperature moving mean
mean after centring
1
2
3
36.6
36.4
36.8
4
37.2
5
36.9
36.73
36.78
36.83
36.92
37.00
6
36.5
7
37.2
37.03
37.05
37.07
37.08
8
37.4
37.14
37.20
83.25
3
8
73
Sales (× $1000)
Price ($)
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.5
81.25
9
37.1
37.24
37.27
79.25
P df_Fol i o: 69
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.5
70
10
37.4
11
37.6
37.27
c There was significant improvement in her race times over
the 10-day period.
37.29
d
37.27
37.30
12
36.9
37.29
37.27
13
14
15
37.2
37.6
36.9
6-point moving mean for day 4:
36.73 + 36.83
= 36.78
2
9 a
4-point
Day Time (s) moving mean
188
179
3
183
171
171
166
171.75
Time (s)
5
173
178.5
177.09
171
175.17
182
173
168
9
171
10
166
Calculations: first 6-point moving average
(188 + 179 + 183 + 180 + 173 + 171)
= 179.00
6
First 6-point moving average
(179 + 178)
= 178.50
2
10 a
4-point
4-point moving
mean after centring
Day Index moving mean
177.75
678
726
176.625
3
175
4
714
173.25
5
689
172.375
6
702.5
692
703.875
705.25
700.375
695.5
705.5
715.5
687
711.875
708.25
7
772
708.125
708
8
b
685
711.125
714.25
190
185
180
175
170
165
160
155
150
0
180
1
2
173
9
10
4
8
180.625
182
168
183
171.83
173.5
8
3
7
176.5
7
179
174.17
176.75
6
2
6
4-point moving
mean after centring
180
173
188
176.17
178.75
5
1
178
182.5
4
Time (s)
6-point
centred moving
average
179
Calculations:
Point between day 3 and 4:
36.6 + 36.4 + 36.8 + 37.2 + 36.9 + 36.5
=
6
220.4
=
6
= 36.73
1
2
Day
6-point moving
average
9
10
1 2 3 4 5 6 7 8 9 10
Day
Original
688
712
t
Smoothed
P df_Fol i o: 70
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Index
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.6
Original
Smoothed
780
770
760
750
740
730
720
710
700
690
680
670
660
650
0
2456 + 4651 + 3456 + 2823
= $3346.5
4
Left of May + Right of May
Centred on May:
=
2
3241.25 + 3346.5
= $3293.875 ≈ $3294
2
The correct answer is B.
390 + 126 + 85 + 130
3 The first 4-point mean =
= 182.75.
4
(This value corresponds to the time between autumn and
126 + 85 + 130 + 460
= 200.25.
The second 4-point mean =
4
(This value corresponds to the time between winter and spring
2007.)
The centred value for winter
182.75 + 200.25
2007 =
= 191.5 ≅ 192.
2
The correct answer is D.
winter 2007.)
1 2 3 4 5 6 7 8 9 10
t
Day
Smoothing indicates a very flat trend.
Index
b
720
715
710
705
700
695
690
685
0
4.6 Median smoothing from a graph
4.6 Exercise
1
1 2 3 4 5 6 7 8
Day
t
y = 700.2 + 1.88x, where y = price index and
x = day number
c y = 700.2 + 1.88x, sub in x = 15
= 700.2 + 1.88(15)
= 700.2 + 28.2
= 728.4
Price index is approximately 728 after 15 days.
4.5 Exam questions
1 Six-mean smoothed value to the left of August:
92.6 + 77.2 + 80.0 + 86.8 + 93.8 + 55.2
= 80.9333...
6
Six-mean smoothed value to the right of August:
77.2 + 80.0 + 86.8 + 93.8 + 55.2 + 97.3
= 81.71666...
6
80.93333 + 81.71667
Centred on August:
= 81.325
2
The correct answer is C.
2 For the 4-mean smoothing centring on May, the profits from
March, April, May and June must be used in one calculation,
and the profits from April, May, June and July must be used in
the second calculation.
March + April + May + June
Left of May:
=
4
2402 + 2456 + 4651 + 3456
= $3241.25
4
April + May + June + July
=
Right of May:
4
71
y
20
18
16
14
12
10
0
2
Original
Smoothed
1 2 3 4 5 6 7 8 9 10 11 12
y
Original
Smoothed
50
45
40
35
30
25
0
x
1 2 3 4 5 6 7 8 9
x
3 a In a 3-point median smoothing you lose 2 data points, so
would be left with 10 data points.
b In a 5-point median smoothing you lose 4 data points, so
you would be left with 7 data points.
4 y
Original
80
75
70
65
60
55
50
45
40
35
0
Smoothed
1 2 3 4 5 6 7 8 9 10
x
P df_Fol i o: 71
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.6
5
Original
Smoothed
Temperature (°C)
32
30
28
26
24
22
20
0
1 2 3 4 5 6 7 8 9 10 11 12
b The median smoothing smoothed out much of the variation
and indicated an upward trend.
8
Share price ($)
72
t
Month
The median smoothing had virtually no effect on the data,
only minor variations smoothed.
6 a y
Original
20
16
12
8
4
Smoothed
1.9
1.85
1.8
1.75
1.7
1.65
1.6
1.55
1.5
1.45
1.4
1.35
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Day
Original
0
1 2 3 4 5 6 7 8 9 10 11 12
Smoothed
9 See the graph at the bottom of the page.*
x
b The median smoothing was effective at smoothing out the
random fluctuations.
7 a
y
100
90
80
70
60
50
40
30
Original
Smoothed
0
Share price ($)
*9
1 2 3 4 5 6 7 8 9 10 11 12
x
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0.9
0.8
0.7
0.6
0
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
t
Day
P df_Fol i o: 72
Original
Smoothed
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
t
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.7
10 a
y
56
54
52
50
48
46
44
42
40
38
36
34
32
30
28
26
24
0
b
Original
Smoothed
Order these to find the middle value: 19, 19, 26, 30, 37, 46
and 48.
The middle value (centred on day 4) is 30 (km/h).
The correct answer is D.
4.7 Seasonal adjustment
4.7 Exercise
Note: Your answers may vary slightly, depending on rounding.
Try to round to 4 decimal places for all intermediate calculations.
1 2 3 4 5 6 7 8 9 10 11 12 13 14
x
y
56
54
52
50
48
46
44
42
40
38
36
34
32
30
28
26
24
0
73
Original
Smoothed
1 a Find the yearly averages:
(1.55 + 1.50 + 1.65 + 1.70)
= 1.60
2018:
4
(1.60 + 1.55 + 1.75 + 1.80)
2019:
= 1.68
4
(1.50 + 1.50 + 1.70 + 1.75)
= 1.61
2020:
4
(1.60 + 1.45 + 1.70 + 1.80)
2021:
= 1.64
4
(1.45 + 1.40 + 1.60 + 1.65)
= 1.53
2022:
4
Year
Average
2018
1.60
2019
1.68
2020
1.61
2021
1.64
2022
1.53
Divide each term in the original time series by the yearly
average.
1.55
For example: summer 2018 =
= 0.97
1.60
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Season
Summer
Autumn
Winter
Spring
x
The 3-point median smoothing in part a reduces the
fluctuations more than the 5-point median smoothing in
part b.
4.6 Exam questions
1 Centre the median on game 10. A nine-median smoothing
would be 4 values either side of game 10.
The nine-median smoothed value will be the points scored at
game number 9, which is 110.
The correct answer is C.
2 Quickly perform three and five-median smoothing.
Day
2
3
4
5
6
7
8
Three-median
Five-median
28
40
28
28
40
51
33
33
37
37
37
37
Day number 7 is the same value.
The correct answer is E.
3 The seven values centred on day 4 are: 30, 19, 19, 46, 37, 26
and 48.
2018
0.97
0.94
1.03
1.06
2019
0.95
0.92
1.04
1.07
2020
0.93
0.93
1.06
1.09
2021
0.98
0.88
1.04
1.10
2022
0.95
0.92
1.05
1.08
Find the seasonal averages:
(0.97 + 0.95 + 0.93 + 0.98 + 0.95)
Summer:
= 0.96
5
(0.94 + 0.92 + 0.93 + 0.88 + 0.92)
Autumn:
= 0.92
5
(1.03 + 1.04 + 1.06 + 1.04 + 1.05)
Winter:
= 1.04
5
(1.06 + 1.07 + 1.09 + 1.10 + 1.08)
Spring:
= 1.08
5
Season
Summer Autumn Winter
Seasonal index
0.96
0.92
1.04
Spring
1.08
b Divide each term in the original series by its seasonal index.
1.55
For example: summer 2018 =
= 1.61
0.96
Season
Summer
Autumn
Winter
Spring
2018
1.61
1.63
1.59
1.57
2019
1.67
1.68
1.68
1.67
2020
1.56
1.63
1.63
1.62
2021
1.67
1.58
1.63
1.67
P df_Fol i o: 73
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
2022
1.51
1.52
1.54
1.53
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.7
74
c
y
0
Season
Summer
Autumn
Winter
Spring
Original
Deseasonalised
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
2 4 6 8 10 12 14 16 18 20 22
x
Year
2018 2019 2020 2021 2022
Average 1.5325 1.5125 1.5125 1.4775 1.4075
Summer 2018∶1.60 ÷ 1.5325 = 1.0440
Autumn 2018∶1.48 ÷ 1.5325 = 0.9657
Winter 2018∶1.40 ÷ 1.5325 = 0.9135
Spring 2018∶1.65 ÷ 1.5325 = 1.0767
Summer 2019∶1.60 ÷ 1.5125 = 1.0579
Spring 2022∶1.59 ÷ 1.4075 = 1.1297
Sales
P df_Fol i o: 74
Original
2021
1.1032
0.8866
0.8663
1.1438
2022
1.1226
0.9023
0.8455
1.1297
Season
Summer
Autumn
Winter
Spring
Seasonal index
1.0731
0.9423
0.8874
1.0972
Season
Summer
Autumn
Winter
Spring
2019
1.4910
1.5918
1.5213
1.4583
2020
1.4630
1.5494
1.5664
1.4856
2021
1.5190
1.3902
1.4424
1.5402
2022
1.4724
1.3478
1.3410
1.4491
2018
1.4910
1.5706
1.5776
1.5038
c See the graph at the bottom of the page.*
d There appears to be a slight downward trend in the price of
1 litre of petrol.
e Using CAS:
Deseasonalised 1 L of petrol cost =
1.5659 − 0.007 38 × time code
Time code 1 represents summer 2018.
f Summer 2023: t = 21
Deseasonalised 1 L of petrol cost = 1.5659 − 0.007 38 × 21
= 1.41092
≈ $1.41
2 a 2018∶(1.60 + 1.48 + 1.40 + 1.65) ÷ 4 = 1.5325
2019∶(1.60 + 1.50 + 1.35 + 1.60) ÷ 4 = 1.5125
2020∶(1.57 + 1.46 + 1.39 + 1.63) ÷ 4 = 1.5125
2021∶(1.63 + 1.31 + 1.28 + 1.69) ÷ 4 = 1.4775
2022∶(1.58 + 1.27 + 1.19 + 1.59) ÷ 4 = 1.4075
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
2020
1.0380
0.9653
0.9190
1.0777
b Summer 2018∶1.6 ÷ 1.0731 = 1.4910
Autumn 2018∶1.48 ÷ 0.9423 = 1.5706
⋯
Spring 2022∶1.59 ÷ 1.0972 = 1.4491
Seasonalised value (reseasonalised predicted value)
= deseasonalised value × seasonal index
= 1.5633 × 0.96
= 1.5008
≈ $1.50
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
2019
1.0579
0.9917
0.8926
1.0579
Summer∶(1.0440 + 1.0579 + 1.0380 + 1.1032 + 1.1226) ÷ 5 = 1.0731
Autumn∶(0.9657 + 0.9917 + 0.9653 + 0.8866 + 0.9023) ÷ 5 = 0.9423
Winter∶(0.9135 + 0.8926 + 0.9190 + 0.8663 + 0.8455) ÷ 5 = 0.8874
Spring∶(1.0767 + 1.0579 + 1.0777 + 1.1438 + 1.1297) ÷ 5 = 1.0972
d There seems to be a very slight downward trend in milk
prices.
e Using CAS:
Deseasonalised 1 L of milk cost
= 1.6557 − 0.004 40 × time code
Time code 1 represents summer 2018.
f Summer 2023: t = 21
Deseasonalised 1 L of milk cost = 1.6557 − 0.004 40 × 21
= 1.5633
≈ $1.56
*2c
2018
1.0440
0.9657
0.9135
1.0767
Seasonalised value (reseasonalised predicted value)
= deseasonalised value × seasonal index
= 1.410 92 × 1.0731
= 1.514 06
≈ $1.51
t
Month
Deseasonalised
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.7
3 a Season
Summer
Autumn
Winter
Spring
Yearly average
2020
1.03
1.26
1.36
1.14
2021
0.98
1.25
1.34
1.07
2022
0.95
1.21
1.29
1.04
1.1975
1.16
1.1225
Season
1
2
3
4
5
6
7
8
9
10
11
12
Yearly average calculation example:
1.03 + 1.26 + 1.36 + 1.14
2020 average =
4
4.7900
=
4
= 1.1975
Divide each term by the yearly average.
Example for 2020:
1.03
Summer:
= 0.8601
1.1975
1.26
= 1.0522
Autumn:
1.1975
1.36
Winter:
= 1.1357
1.1975
1.14
Spring:
= 0.9520
1.1975
Now calculate the seasonal indices.
Example: seasonal index for summer:
(0.8601 + 0.8448 + 0.8463) ÷ 3 = 0.8504
Summer
Autumn
Winter
Spring
2020
0.8601
1.0522
1.1357
0.9520
2021
0.8448
1.0776
1.1552
0.9224
2022
0.8463
1.0780
1.1492
0.9265
c
2020
1.21
1.18
1.19
1.22
2021
1.15
1.17
1.17
1.15
2017
0.8918
0.8052
1.2381
1.0649
2018
0.9372
0.8116
1.1981
1.0531
2019
0.9383
0.8099
1.1951
1.0568
*4a
Summer
Autumn
Winter
Spring
Price ($)
y
1 2 3 4 5 6 7 8 9 10 11 12
t
Season
Seasonal
indices
0.8504
1.0692
1.1467
0.9336
Original
Summer
Autumn
Winter
Spring
103
93
143
123
97
84
124
109
95
82
121
107
117
100
156
125
118
99
155
122
Yearly
average
115.5 103.5 101.25 124.5 123.5 123.25
See table at the bottom of the page.*
2022
1.12
1.13
1.12
1.11
2020
0.9398
0.8032
1.2530
1.0040
Deseasonalised
d The deseasonalised data shows that there is a slight
downward trend in sugar prices over the three years.
4 a Season 2017 2018 2019 2020 2021 2022
Deseasonalised data example:
Summer 2020: 1.03 ÷ 0.8504 = 1.2111
Autumn 2020: 1.26 + 1.0692 = 1.1785
Deseasonalised data
Summer
Autumn
Winter
Spring
Deseasonalised
1.21
1.18
1.19
1.22
1.15
1.17
1.17
1.15
1.12
1.13
1.12
1.11
1.5
1.4
1.3
1.2
1.1
1
0.9
0
4.0000
b
Actual
1.03
1.26
1.36
1.14
0.98
1.25
1.34
1.07
0.95
1.21
1.29
1.04
2021
0.9555
0.8016
1.2551
0.9879
2022
0.9736
0.7951
1.2252
1.0061
Seasonal index
0.9394
0.8044
1.2274
1.0288
4.0000
P df_Fol i o: 75
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
120
98
151
124
75
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.7
76
Rainfall (mm)
b See table at the bottom of the page.*
c
d The smoothed data indicates that youth unemployment
increases in all seasons except summer.
6 a Begin by rewriting the table as follows:
160
140
120
100
80
0
2 4 6 8 10 12 14 16 18 20 22 24
t
Season
Original
Youth unemployment
2020
2021
2022
1
2
3
4
Yearly average
5.8
4.9
3.5
6.7
5.225
6.1
5.1
3.2
6.5
5.225
5.7
4.5
4.1
7.1
5.35
Deseasonalised
d Rainfall in 2018 and 2019 was much lower than other
years, so there may have been a drought.
5 a See the table at the bottom of the page.*
See the table at the bottom of the page**.
b See the table at the bottom of the page.*
c
2021
1.1675
0.9761
0.6124
1.2440
2022
1.0654
0.8411
0.7664
1.3271
Deseasonalised data
1
2
3
4
2020
5.205
5.336
5.125
5.216
1
2
3
4
14
13
12
11
10
9
8
7
b
0
Quarter
2020
1.1100
0.9378
0.6699
1.2823
Seasonal index
1.1143
0.9183
0.6829
1.2845
4.0000
2021
5.474
5.554
4.686
5.060
t
2 4 6 8 10 12 14 16 18 20
Season
Original
Deseasonalised
*4b
Season
Summer
Autumn
Winter
Spring
*5a
Season
Summer
Autumn
Winter
Spring
Yearly average
**5a
Summer
Autumn
Winter
Spring
2017
109.644
115.614
116.506
119.557
2018
0.7581
1.0873
1.1671
0.9875
2018
103.257
104.426
101.027
105.949
2019
101.128
101.939
98.582
104.005
2020
124.548
124.316
127.098
121.501
2021
125.612
123.073
126.283
118.585
2022
127.741
121.830
123.024
120.529
2018
7.6
10.9
11.7
9.9
2019
7.7
11.3
12.4
10.5
2020
7.8
11.9
12.8
10.8
2021
7.7
12.6
13.5
11.4
2022
7.9
13.1
13.9
11.9
10.025
10.475
10.825
11.3
11.7
2019
0.7351
1.0788
1.1838
1.0024
2020
0.7206
1.0993
1.1824
0.9977
2021
0.6814
1.1150
1.1947
1.0088
2022
0.6752
1.1197
1.1880
1.0171
Seasonal index
0.7141
1.1000
1.1832
1.0027
4.0000
*5b
P df_Fol i o: 76
Deseasonalised data
Summer
Autumn
Winter
Spring
2018
10.6428
9.9091
9.8884
9.8733
2019
10.7828
10.2727
10.4801
10.4717
2020
10.9228
10.8182
10.8181
10.7709
2021
10.7828
11.4545
11.4097
11.3693
2022
11.0629
11.9091
11.7478
11.8680
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
2022
5.115
4.900
6.004
5.527
TOPIC 4 Investigating and modelling time series data • EXERCISE 4.7
8
7
6
5
4
3
2
1
0
1 2 3 4 5 6 7 8 9 10 11 12
Original
t
Time period
Deseasonalised
Revenue
Unemployment
c
d Using CAS, the deseasonalised unemployment rate
= 5.1449 + 0.0188t
e i Quarter 1 in 2023:
The t-value for this quarter would be 13.
Unemployment rate = 5.1449 + 0.0188 × 13
= 5.3893
P df_Fol i o: 77
1734
1790
1804
1789
2078
2204
2215
2184
2167
Saturday
2467
2478
2504
2526
2589
Sunday
1895
1786
1824
1784
1755
Yearly
1668
1666
1710
average
See the table at the bottom of the page.*
See the table at the bottom of the page.**
1717
1701
2800
2200
1600
1000
400
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
t
Time period
Deseasonalised
From the original data, the restaurant should close on Monday
and Tuesday. Otherwise, the smoothed data shows that sales
remain fairly constant over the five-week period.
8 June 2022, t = 1
July 2022, t = 2 …
June 2023, t = 13
Deseasonalised monthly sales = 10 000 + 1500 × time code
= 10 000 + 1500 × 13
= 29 500
Seasonalised value = 29 500 × 0.8
= $23 600
The correct answer is A.
9 Since there are 12 ‘seasons’, their sum is 12.
Missing index (Feb.) = 12 − (1.05 + 1.0 + 1.0 + 0.95 + 0.85+
0.8 + 0.9 + 0.95 + 1.05 + 1.10 + 1.15)
= 1.2
The correct answer is E.
Week 1 Week 2 Week 3 Week 4 Week 5
1036
1089
1064
1134
1042
Tuesday
1103
1046
1085
1207
1156
Wednesday
1450
1324
1487
1378
1408
**7
1645
Friday
Original
Seasonalised value = 5.7277 × 0.6829
= 3.9
(Seasonal index for quarter 3 is 0.6829.)
*7
Thursday
0
Seasonalised value = 5.3893 × 1.1143
= 6.0
(Seasonal index for quarter 1 is 1.1143.)
ii Quarter 3 in 2027:
The t-value for this quarter would be 31.
Unemployment rate = 5.1449 + 0.0188 × 31
= 5.7277
7 Season
Monday
77
Season
Monday
Week 1
0.6211
Week 2
0.6537
Week 3
0.6222
Week 4
0.6605
Week 5
0.6126
Seasonal index
0.6340
Tuesday
0.6613
0.6279
0.6345
0.7030
0.6796
0.6613
Wednesday
0.8693
0.7947
0.8696
0.8026
0.8277
0.8328
Thursday
0.9862
1.0408
1.0468
1.0507
1.0517
1.0353
Friday
1.2458
1.3229
1.2953
1.2720
1.2740
1.2820
Saturday
1.4790
1.4874
1.4643
1.4712
1.5220
1.4848
Sunday
1.1361
1.0720
1.0667
1.0390
1.0317
1.0691
Deseasonalised data
Monday
Week 1
1634.07
Week 2
1717.67
Week 3
1678.23
Week 4
1788.64
Week 5
1643.53
Tuesday
1667.93
1581.73
1640.71
1825.19
1748.07
Wednesday
1741.11
1589.82
1785.54
1654.66
1690.68
Thursday
1588.91
1674.88
1728.97
1742.49
1728.00
Friday
1620.90
1719.19
1727.77
1703.59
1690.33
Saturday
1661.50
1668.91
1686.42
1701.24
1743.67
Sunday
1772.52
1670.56
1706.11
1668.69
1641.57
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
78
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
10 Deseasonalised figure = actual figure ÷ seasonal index
June index = 0.85
Deseasonalised figure = $102 000 ÷ 0.85 = $120 000
The correct answer is D.
11 Since there are 4 seasons, the sum of the indices is 4.
Missing index (winter) = 4 − (1.23 + 0.89 + 1.45)
= 0.43
12 a Deseasonalised data (for e.g., January):
24
= 20.8696
=
1.15
Using CAS, the deseasonalised umbrella sales
= 21.8788 + 0.9161 × time code
b Time code for the following January is 13.
Deseasonalised umbrella sales = 21.8788 + 0.9161 × 13
= 33.7881
The expected deseasonalised sale for January is 33.79
umbrellas.
Seasonalised umbrella sales
= deseasonalised value × seasonal index
= 33.7881 × 1.15
= 38.8563
= 39 (rounded to a whole number of umbrellas).
The expected sales for January the following year would be
39 umbrellas.
actual figure
13 a Deseasonalised figure =
seasonal index
4345
=
1.2
= $3620.83
actual figure
b Deseasonalised figure =
seasonal index
950
=
0.6
= $1583.33
c Seasonalised value (reseasonalised predicted value)
= deseasonalised value × seasonal index
= 5890 × 1.4
= $8246
The forecast sales figure for the first quarter in 2023 is
$8246.
actual figure
14 a Deseasonalised figure =
seasonal index
8945
=
1.3
= $6880.77
actual figure
b Deseasonalised figure =
seasonal index
3250
=
0.4
= $8125
c Seasonalised value (reseasonalised predicted value)
= deseasonalised value × seasonal index
= 7950 × 1.7
= $13 515
The forecast sales figure for the first quarter in 2023 is
$13 515.
15 Deseasonalised figure = actual figure + seasonal index
a Deseasonalised figure = $3000 ÷ 1.0 = $3000
b Deseasonalised figure = $800 ÷ 0.25 = $3200
c Predicted value = deseasonalised figure × seasonal index
Predicted value = $3200 × 1.50
= $4800
16 There are 7 ‘seasons’ (Monday to Sunday); therefore, the sum
of the indices is 7.
Friday index = 7 − (sum of all other indices)
= 7 − (0.4 + 0.3 + 0.6 + 0.8 + 1.8 + 1.3)
= 7 − 5.2
= 1.8
17 There are 7 ‘seasons’ (Monday to Sunday); therefore, the sum
of the indices is 7.
Saturday index = 7 − (sum of all other indices)
= 7 − (0.4 + 1.1 + 0.5 + 0.6 + 1.3 + 1.6)
= 7 − 5.5
= 1.5
18 Since there are 7 ‘seasons’, the sum of the seasonal
indices is 7.
The missing index = 7 − sum of the other six
Wednesday’s index = 7 − (0.5 + 0.2 + 0.6 + 1.5 + 2.2 + 1.1)
= 0.9
4.7 Exam questions
1
29 685
25 420
31 496
+
+
÷ 3 = 1.088
( 27 194.0 23 183.5 29 243.0 )
2 February 2020 is when n = 2
Deseasonalised number of visitors
= 2349 − 198.5 × 2 = 1952
Actual number of visitors = deseasonalised value × SI =
1952 × 1.25 = 2440
The correct answer is E.
92.6
3
= 75.777
1.222
The correct answer is B.
The correct answer is C.
4.8 Review
4.8 Exercise
Multiple choice
1 Prices of oranges over a 16-month period appear random.
The correct answer is C.
2 Price = 0.415 × month + 8.45
= 0.415 × 18 + 8.45
= 15.92
The correct answer is D.
3 Gradient is positive and y-intercept is 10, so most appropriate
line of best fit is y = 8t + 10.
The correct answer is E.
P df_Fol i o: 78
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
4
Time
y-value
2016
2017
12
13
4-point
moving
average
4-point moving
average after
centring
16
520
= 667
0.78
14.875
The correct answer is B.
9 Since p = length of cycle, quarterly: length of cycle = 4.
The correct answer is B.
10 Increasing trend with seasonal variation due to the peaks
occurring at similar time intervals.
The correct answer is E.
16.25
Short answer
14.25
2018
b Deseasonalised figure =
79
15.5
2019
16
2020
17
11
17
Month
1
2
3
4
5
6
7
8
9
10
11
12
17.75
18.5
2021
2022
19
22
2015
2016
2017
2018
2019
2020
2021
2022
350
320
300
310
270
240
200
160
323
310
293
273
237
200
The first 2 points are 323 and 310.
The correct answer is C.
b The number of points obtained from the smoothed trend in
part a is 6.
The correct answer is C.
6
t
y
3-point median smoothing
1
2
3
4
5
6
7
8
9
10
30
20
25
20
15
25
30
20
15
10
25
20
20
20
25
25
20
15
1.12
0.78
130
110
90
70
50
30
10
0
2 4 6 8 10 12
Month
t
Uniform sales are high at the start of the school year and at
the beginning of winter and decrease throughout the other
parts of the year. This data is seasonal, so it is very difficult to
fit a trend line.
12 Using CAS, the line is y = 2057.14 + 184.52x.
The gradient is the b value and the y-intercept is the a value.
Gradient = 184.52
y-intercept = 2057.14
13 Day
1 2 3 4 5 6 7 8 9 10 11
Rooms
12 18 15 20 22 20 25 24 26 28 30
y
The last smoothed value is 15.
The correct answer is D.
7 Seasonal indices and adjustment can be used when there are
seasonal variations along with secular trends.
The correct answer is B.
8 a
Season Spring Summer Autumn Winter
Index
Number of
uniforms sold
The value after a 4-point smoothing with centring plotted
against 2019 is 16.25.
The correct answer is A.
5 a
Year
Number
3-point moving average
Sales
118
92
53
20
47
102
90
42
35
26
12
58
x
0.92
Winter index = 4 − (1.12 + 0.78 + 0.92) = 1.18
The correct answer is A.
P df_Fol i o: 79
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
80
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
X
12
13
14
15
16
17
18
X = 12
Y1
31.309
32.891
34.473
36.055
37.636
39.218
40.8
Using CAS least squares line of best fit, the equation is
y = 12.32 + 1.58x.
5-point moving average
1
2
3
4
5
6
7
8
690
500
400
720
780
660
550
440
618
612
622
630
Day
Rain (mm)
3-point moving average
1
2
3
4
5
6
7
8
2
5
4
6
3
7
6
9
3.7
5.0
4.3
5.3
5.3
7.3
Sales ($)
5-pt mov av
Linear (5-pt mov av)
850
800
750
700
650
600
550
500
450
400
0
14 Calculation example: day 2
(2 + 5 + 4)
Day 2’s value =
3
11
=
3
= 3.7
Rainfall (mm)
Sales ($)
Coat sales
Sales
a Gradient = 1.58
y-intercept = 12.33
b Number of rooms booked on day 12:
R = 1.58 × 12 + 12.33
= 31.28
≃ 31 rooms on day 12
On day 13:
R = 1.58 × 13 + 12.32
≃ 33 rooms
Season
1 2 3 4 5 6 7 8 9 10
Seasons
b Calculation example: column 3
690 + 500 + 400 + 720
Between day 2 & 3 =
4
= 577.5
577.5 + 600
Column 4: day 3 =
2
1177.5
=
2
= 588.75
Season
Sales ($)
1
2
690
500
3
400
4-point
moving
average
4-point moving
average after
centring
577.5
588.75
600
10
8
6
4
2
4
720
620
640
0
t
1 2 3 4 5 6 7 8
Day
Smoothed
Original
15 a Calculation example: day 3
690 + 500 + 400 + 720 + 780
Day 3′ s value =
5
3090
=
5
= 618
5
780
6
660
658.75
677.5
642.5
607.5
7
8
550
440
There is the same number of smoothed points in a 4-point
moving average with centring as a 5-point moving average.
P df_Fol i o: 80
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
t
y
1
2
3
4
5
6
7
8
9
10
30
20
15
20
15
10
15
10
15
5
3-point median smoothing
20
20
15
15
15
10
15
10
Season
Seasonalised
Share price deseasonalised
Spring
150
250
Summer
Autumn
Winter
100
300
400
200
250
235
Extended response
18 a The time series is seasonal. There are peaks and troughs
occurring on the same days of the week. There is also an
upward secular trend. This can also be seen from the table
as each day, week after week, more video games are sold.
Number of video
games sold
16
35
30
25
20
15
10
5
60
50
40
30
20
10
0
2 4 6 8 10 12 14 16 18
t
Week 1 Week 2 Week 3
0
17
81
1 2 3 4 5 6 7 8 9 10
Cycle
t
Season
Winter
Spring
Summer
Autumn
Index
1.7
0.6
0.5
1.2
b There are 6 seasons — each of the days of the week the
store opened.
c See the table at the bottom of the page.*
From the table above (last column):
Week 1∶ 19
Week 2∶ 22
Week 3∶ 25.5
d See the table at the bottom of the page.*
e See the table at the bottom of the page.*
Calculation example for spring:
Deseasonalised = seasonalised value ÷ index
= 150 ÷ 0.6
= 250
*18c
Jazza’s online store daily sales figures — number of video games sold
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Week 1
10
8
12
15
24
45
Week 2
12
9
14
18
26
53
Week 3
15
10
16
21
33
58
Monday
Tuesday
*18d
1
2
3
10
= 0.5263
19
12
= 0.5455
22
0.5882
*18e
0.4211
0.4091
0.3922
Monday
Seasonal indices
0.5533
Wednesday
Thursday
Friday
0.7895
1.2632
12
= 0.6316
19
14
= 0.6364
22
16
= 0.6275
25.5
21
= 0.8235
25.5
Tuesday
Wednesday
1.2224
= 0.4075
3
0.8182
0.6318
26
= 1.1818
22
1.2941
Thursday
2.4312
= 0.8104
3
Average daily sales for the week
114
= 19
6
132
= 22
6
153
= 25.5
6
Saturday
45
= 2.3684
19
2.4091
2.2745
Friday
3.7391
= 1.2464
3
P df_Fol i o: 81
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Saturday
7.052
= 2.3507
3
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
f An index of 2.3507 means that Saturdays have
approximately 2.35 times the sales compared to the average
daily sales.
19 a y
20 a
Rainfall (mm)
82
100
80
60
40
20
0
ii Saturday week 4, t = 24,
deseasonalised video game sales
= 0.4993t + 17.43
= 0.4993 × 24 + 17.43
= 29.41
iii Saturday week 6, t = 36,
deseasonalised video game sales
= 0.4993t + 17.43
= 0.4993 × 36 + 17.43
= 35.40
c i Monday week 4,
deseasonalised video games sales = 26.92
Seasonalised value = deseasonalised value × seasonal index
= 26.92 × 0.5533
= 14.89 = 15 video games approximately
ii Saturday week 4,
deseasonalised video games sales = 29.41
Seasonalised value = deseasonalised value × seasonal index
= 29.41 × 2.3507
= 69.13 = 69 video games approximately
iii Saturday week 6,
deseasonalised video game sales = 35.40
Seasonalised value = deseasonalised value × seasonal index
= 35.40 × 2.3507
= 83.21 = 83 video games approximately
The first two predictions are reliable as they are only
one week into the future. The Saturday week 6
prediction of 83 video games is not so reliable as it is
far into the future and the trend may change in the
meantime due to events such as holidays etc.
Rainfall (mm)
Using CAS, deseasonalised video games sales
= 17.43 + 0.4993t.
The y-intercept of 17.43 means that these sales were
expected the day before the data was calculated. The
gradient of 0.4993 (0.5) means that as each day goes by
Jazza can expect an increase of half a video game each day
(or more logically 1 video game extra each two days).
b i Monday week 4, t = 19, deseasonalised video game sales
= 0.4993t + 17.43
= 0.4993 × 19 + 17.43
= 26.92
t
Due to the nature of the data it is very difficult to fit a trend
line. However, there appears to be an upward trend.
b The least squares line of best fit equation is
y = 48.47 − 0.021t, found using CAS.
100
80
60
40
20
0
2 4 6 8 10 12
Season
t
This line is a poor fit for the data.
c Season Rainfall
3-point moving average
1
2
3
4
5
6
7
8
9
10
11
12
43
75
41
13
47
78
50
19
51
83
55
25
53
43
33.7
46
58.3
49
40
51
63
54.3
3-point smooth:
Rainfall (mm)
x
2 4 6 8 10 12
Season
100
80
60
40
20
0
2 4 6 8 10 12
Season
Season Rainfall (mm)
1
2
3
4
43
75
41
13
t
5-point moving average
43.8
50.8
P df_Fol i o: 82
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
47
78
50
19
51
83
55
25
45.8
41.4
49.0
56.2
51.6
46.6
Rainfall (mm)
5
6
7
8
9
10
11
12
t
Rainfall (mm)
The 3-point moving average has only slightly reduced the
variation, while the 5-point moving average smoothing is
more effective.
d Using CAS, the trend line for the 5-point moving average is
y = 44.81 + 0.74x. The gradient is still fairly weak at
b = 0.74.
e The correlation is also still weak.
Season
Rainfall
1
2
43
75
3
41
4
13
4-point
moving
average
Quarter
Year 1
Year 2
Year 3
Summer
Autumn
Winter
Spring
43
75
41
13
47
78
50
19
51
83
55
25
Yearly average
43
48.5
53.5
43.5
44
Summer
Autumn
Winter
Spring
44.4
44.75
47
t
Seasonal index calculation example:
1.0000 + 0.9691 + 0.9533
Summer index =
3
2.9224
=
3
= 0.9741
4-point moving
average after
centring
43
5
2 4 6 8 10 12
Season
Least squares: y = 42.01 + 1.36t
A stronger increasing trend is shown with the 4-point
centring method compared to the 5-point moving average.
This could suggest seasonal data.
f Calculation example:
43 + 75 + 41 + 13
Year1average =
4
172
=
4
= 43
100
80
60
40
20
2 4 6 8 10 12
Season
100
80
60
40
20
0
5-point smooth:
0
45.9
Year 1
1.0000
1.7442
0.9535
0.3023
Year 2
0.9691
1.6082
1.0309
0.3918
Year 3
0.9533
1.5514
1.0280
0.4673
47
6
78
50
8
19
9
51
4.0000
Deseasonalised data
Summer
Autumn
Winter
Spring
49
49.5
50.1
50.75
83
11
12
55
25
Year 1
44.1
45.9
40.8
33.6
Year 2
48.2
47.7
49.8
49.1
51.4
52
10
Seasonal
index
0.9741
1.6346
1.0041
0.3871
47.8
48.5
7
83
g
52.8
53.5
Season
1
2
3
Actual
43
75
41
Deseasonalised
44.1
45.9
40.8
P df_Fol i o: 83
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Year 3
52.4
50.8
54.8
64.6
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
4
5
6
7
8
9
10
11
12
13
47
78
50
19
51
83
55
25
33.6
48.2
47.7
49.8
49.1
52.4
50.8
54.8
64.6
80
Rainfall
(1908, 75.26)
75
Rainfall
Actual
Deseas
Linear (Deseas)
90
80
70
60
50
40
30
20
10
0
Key
Women
Men
85
Winning time (seconds)
84
70
65
60
55
50
(2020, 48.04)
45
1908
M
1 2 3 4 5 6 7 8 9 10 11
Season
t
The least squares line of best fit equation is
y = 37.70 + 1.66t.
y
This has given the strongest trend yet (b = 1.66). Also, all
12 points of data are used, rather than 8 or 10 as in the
moving average smoothing.
x
4.8 Exam questions
1 a The end point at 1908 needs to be between 75 ≤ M ≤ 76
and the endpoint at 2020 needs to be between 47 ≤ N ≤ 49.
Award 1 mark for the line correctly drawn.
b Winning time men = 356.9 − 0.1544 × 2024 = 44.3944
Winning time women = 538.9 − 0.2430 × 2024 = 47.068}
Award 1 mark if one of these is correct.
Difference = 47.068 − 44.3944 ≃ 2.7 seconds
Award 1 mark — note that rounding applies here
1924
1940
1956
1972
Year
1988
2004
c Solve on your CAS: winning time women < winning time
men
538.9 − 0.2430x < 356.9 − 0.1544x
Therefore, year =2054.176.
Award 1 mark for an answer rounding to 2054.
So the next Olympics will be in 2056. [1 mark]
2 a First, calculate the yearly averages:
142 + 156 + 222 + 120
= 160
2015∶
4
135 + 153 + 216 + 96
2016∶
= 150
4
Divide each row by its yearly average:
Year
2015
2016
Summer
0.8875
0.9
Autumn
0.975
1.02
Winter
1.3875
1.44
Spring
0.75
0.64
Lastly, find the average of the season to get the seasonal
index:
0.8875 + 0.9
Summer:
= 0.89375 = 0.89
2
0.975 + 1.02
Autumn:
= 0.9975 = 1.00
2
0.75 + 0.64
Spring:
= 0.695 = 0.7
2
Award 2 marks for all correct; 1 mark for 2 correct,
0 marks for 1 or none correct.
262
b
= 185.8156 = 186 mm
[1 mark]
1.41
P df_Fol i o: 84
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
2020
N
TOPIC 4 Investigating and modelling time series data • REVIEW 4.8
19 + 22 + 43 + 37
= 30.25
4
26.5 + 30.25
Smoothed value for day 11 =
2
= 28.375 (km/h)
This is closest to 28 km/h.
The correct answer is D.
5 a
=
Minimum
3 a See the graph at the bottom of the page*
[1 mark]
b i See the graph at the bottom of the page*
[1 mark]
ii The average rate of increase in percentage congestion
level from 2008 to 2016 in Sydney is equal to the
graph’s gradient.
This is 1.15% per year on average. [1 mark]
iii Substitute 43 for congestion level in the equation and
solve for year. The predicted year that Sydney will
have a 43% congestion level is 2020. [1 mark]
c –1514.755 56 rounded to 4 significant figures
is –1515. [1 mark]
d Using the CAS calculator to find the line of least squares,
we get:
congestion level = −1515 + 0.7667 × year
Award 1 mark for –1515, 1 mark for 0.7667.
e Sydney’s traffic congestion in 2008 (28%) is greater than
Melbourne’s traffic congestion in 2008 (25%), and the
gradient for Sydney’s least squares line is 1.15, which is
greater than Melbourne’s least squares gradient of 0.7669.
Therefore, Sydney’s traffic congestion will increase faster
than Melbourne’s, meaning it will always exceed future
traffic congestion levels in Melbourne.
Award 1 mark for appropriate statistics, 1 mark for proving
why Sydney’s congestion will be greater.
4 The first four values are 22, 19, 22 and 43. They will be
centred between days 10 and 11. Mean
22 + 19 + 22 + 43
= 26.5
=
4
The second four values are 19, 22, 43 and 37. They will be
centred between days 11 and 12. Mean
*3a
Unsmoothed
Five-median
smoothed
100
90
80
70
60
50
40
30
20
10
0
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Months
Award 2 marks for correctly drawing the five-median
smoothing by marking smoothed values with crosses (×).
124 + 140
= 132
b Mean of September and October =
2
140 + 225
Mean of October and November =
= 182.5
2
132 + 182.5
Centred on October =
= 157.25
2
Award 1 mark for calculating the two means correctly and
1 mark for calculating the centred on October 157.25 mm.
45
Key
Sydney
Melbourne
40
35
congestion
level (%)
30
25
20
2008
*3bi
2009
2010
2011
2012
year
2013
2014
2015
2016
45
Key
Sydney
Melbourne
40
35
congestion
level (%)
30
25
20
2008
P df_Fol i o: 85
2009
2010
2011
2012
year
2013
2014
85
2015
2016
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
86
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.2
Topic 5 — Modelling depreciation of assets using
recursion
5.2 A first-order linear recurrence relation
5.2 Exercise
1 u0 = −2
u1 = −2 − 3 = −5
u2 = −5 − 3 = −8
u3 = −8 − 3 = −11
u4 = −11 − 3 = −14
−2, −5, −8, −11, −14
2 u0 = 3
u1 = 3 × 2 = 6
u2 = 6 × 2 = 12
u3 = 12 × 2 = 24
u4 = 24 × 2 = 48
3, 6, 12, 24, 48
3 un = 4un − 1 + 3, u0 = 5
u1 = 4u0 + 3
=4×5+3
= 23
u2 = 4u1 + 3
= 4 × 23 + 3
= 95
u3 = 4u2 + 3
= 4 × 95 + 3
= 383
P df_Fol i o: 86
u4 = 4u3 + 3
= 4 × 383 + 3
= 1535
The sequence is 5, 23, 95, 383, 1535.
4 fn+1 = 5fn − 6, f0 = −2
f1 = 5f0 − 6
= 5 × −2 − 6
= −16
f2 = 5f1 − 6
= 5 × −16 − 6
= −86
f3 = 5f2 − 6
= 5 × −86 − 6
= −436
f4 = 5f3 − 6
= 5 × −436 − 6
= −2186
The sequence is −2, −16, −86, −436, −2186.
5 a There is no description of two consecutive terms.
This is not a first-order recurrence relation.
b There is a description of two consecutive terms and a
starting term (u0 ).
This is a first-order recurrence relation.
c There is a description of two consecutive terms and a
starting term (u0 ).
This is a first-order recurrence relation.
d By rearranging the equation: there is a description of two
consecutive terms, but there is no starting term.
un = 4un−1 + 5
This is not a first-order recurrence relation.
e There is a description of two consecutive terms, but there is
no starting term.
This is not a first-order recurrence relation.
f There is no description of two consecutive terms.
This is not a first-order recurrence relation.
g There is a description of two consecutive terms and a
starting term (u0 ).
This is a first-order recurrence relation.
h There is no description of two consecutive terms.
This is not a first-order recurrence relation.
i There is a description of two consecutive terms, but there is
no starting term.
This is not a first-order recurrence relation.
j There is a description of two consecutive terms and a
starting term (p0 ).
This is a first-order recurrence relation.
6 a u0 = 6
u1 = u0 + 2
=6+2
=8
u2 = u1 + 2
= 10
u3 = u2 + 2
= 12
u4 = u3 + 2
= 14
The sequence is 6, 8, 10, 12, 14.
b u0 = 5
u1 = u0 − 3
=5−3
=2
u2 = u1 − 3
= −1
u3 = u2 − 3
= −4
u4 = u3 − 3
= −7
The sequence is 5, 2, −1, −4, −7.
c u0 = 23
u1 = 1 + u0
= 1 + 23
= 24
u2 = 1 + u1
= 25
u3 = 1 + u2
= 26
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.2
P df_Fol i o: 87
u4 = 1 + u3
= 27
The sequence is 23, 24, 25, 26, 27.
d u0 = 7
u1 = u0 − 10
= 7 − 10
= −3
u2 = u1 − 10
= −13
u3 = u2 − 10
= −23
u4 = u3 − 10
= −33
The sequence is 7, −3, −13, −23, −33.
7 a u0 = 1
u1 = 3u0
=3×1
=3
u2 = 3u1
=9
u3 = 3u2
= 27
u4 = 3u3
= 81
The sequence is 1, 3, 9, 27, 81.
b u0 = −2
u1 = 5u0
= 5 × −2
= −10
u2 = 5u1
= −50
u3 = 5u2
= −250
u4 = 5u3
= −1250
The sequence is −2, −10, −50, −250, −1250.
c u0 = 1
u1 = −4u0
= −4 × 1
= −4
u2 = −4u1
= 16
u3 = −4u2
= −64
u4 = −4u3
= 256
The sequence is 1, −4, 16, −64, 256.
d u0 = −1
u1 = 2u0
= 2 × −1
= −2
u2 = 2u1
= −4
u3 = 2u2
= −8
u4 = 2u3
= −16
The sequence is −1, −2, −4, −8, −16.
8 a u0 = 1
u1 = 2u0 + 1
=2×1+1
=3
u2 = 2u1 + 1
=7
u3 = 2u2 + 1
= 15
u4 = 2u3 + 1
= 31
The sequence is 1, 3, 7, 15, 31.
b u0 = 5
u1 = 3u0 − 2
=3×5−2
= 13
u2 = 3u1 − 2
= 37
u3 = 3u2 − 2
= 109
u4 = 3u3 − 2
= 325
The sequence is 5, 13, 37, 109, 325.
9 a u0 = 6
u1 = −u0 + 1
= −6 + 1
= −5
u2 = −u1 + 1
=6
u3 = −u2 + 1
= −5
u4 = −u3 + 1
=6
u5 = −u4 + 1
= −6 + 1
= −5
u6 = −un − 1 + 1
= −u5 + 1
=−−5+1
=6
The first seven terms are 6, −5, 6, −5, 6, −5, 6.
b u0 = 1
u1 = 5u0
=5×1
=5
u2 = 5u1
=5×5
= 25
u3 = 5u2
= 5 × 25
= 125
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
87
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.3
u4 = 5u3
= 5 × 125
= 625
u5 = 5u4
= 5 × 625
= 3125
u6 = 5u5
= 5 × 3125
= 15 625
The first seven terms are 1, 5, 25, 125, 625, 3125, 15 625.
10 u0 = 2
u1 = 3u0 + 4
=3×2+4
= 10
The correct answer is C.
11 u0 = −3
u1 = 2u0 − 1
= 2 × −3 − 1
= −7
The correct answer is E.
1
12 a un+1 = 3un ; u0 =
4
3
u1 = 3u0 =
4
9
1
u2 = 3u1 = = 2
2
4
3
27
=6
u3 = 3u2 =
4
4
1
81
= 20
u4 = 3u3 =
4
4
b un+1 = 2000 + un ; u0 = 200 000
u1 = 2000 + u0 = 202 000
u2 = 2000 + u1 = 204 000
u3 = 2000 + u2 = 206 000
u4 = 2000 + u3 = 208 000
c un+1 = un − 7; u0 = 100
u1 = u0 − 7 = 93
u2 = u1 − 7 = 86
u3 = u2 − 7 = 79
u4 = u3 − 7 = 72
d un+1 = 2un − 50; u0 = 200
u1 = 2u0 − 50 = 350
u2 = 2u1 − 50 = 650
u3 = 2u2 − 50 = 1250
u4 = 2u3 − 50 = 2450
5.2 Exam questions
1. A0 =3
A1 =2 × A0 + 4 = 2 × 3 + 4 = 10
A2 =2 × A1 + 4 = 2 × 10 + 4 = 24
A3 =2 × A2 + 4 = 2 × 24 + 4
The correct answer is D.
2. Note that the sequence does not increase by a constant
amount so it is not linear. Eliminate clear distractors and test
the remaining options with the given sequence.
Option D:
2,
4 × (2) − 2 = 6,
4 × (6) − 2 = 22,
4 × (22) − 2 = 86,
4 × (86) − 2 = 342
The correct answer is D.
3 A0 = 2
A1 = 3 (2) + 1 = 7
A2 = 3 (7) + 1 = 22
A3 = 3 (22) + 1 = 67
The correct answer is E.
5.3 Modelling flat rate depreciation with a
recurrence relation
5.3 Exercise
1 a V0 = 25 000
d = V0 ×
r
100
= 25 000 ×
15
100
= 3750
Annual depreciation is $3750.
b Vn + 1 = Vn − d
Vn + 1 = Vn − 3750
c Time, n(years) Depreciation, d($)
0
1
2
3
4
5
Future value
(× $1000)
88
Vn
25
20
15
10
5
0
Future value, Vn ($)
3750
3750
3750
3750
3750
25 000
21 250
17 500
13 750
10 000
6250
(0, 25 000)
(5, 6250)
1 2 3 4 5
Time (years)
n
d Since its useful life is five years, from the depreciation
schedule, the scrap value is $6250.
2 a V0 = 30 000, Vn+1 = Vn − 6000
b V0 = 2000, Vn+1 = Vn − 200
c V0 = 6000, Vn+1 = Vn − 500
50 000 − 25 000
3 a i Annual depreciation =
5
25 000
=
5
= $5000
P df_Fol i o: 88
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.3
d = $5000
V0 = $50 000
∴ Vn = 50 000 − 5000n
0 = 50 000 − 5000n
5000n = 50 000
n = 10
The time taken to be written off is 10 years.
850 − 150
b i Annual depreciation =
7
700
=
7
Future value ($)
ii
Vn
10 000
8000
6000
4000
2000
0
6 a
= $100
0
1
2
3
4
5
0
1500
1500
1500
1500
1500
10 000
8500
7000
5500
4000
2500
Future value
(× $1000)
Vn
50
1 2 3 4 5 n
Time (years)
Future value, Vn ($)
0
4000
4000
4000
4000
4000
4000
4000
4000
4000
4000
45 000
41 000
37 000
33 000
29 000
25 000
21 000
17 000
13 000
9000
5000
(0, 45 000)
40
30
20
10
(10, 5000)
0
2 4 6 8 10
Time (years)
n
b As its useful life is 10 years, using the depreciation
schedule, the scrap value is $5000.
7 a
Time, n(years) Depreciation, d($) Future value, Vn ($)
0
1
2
3
4
5
6
Future value ($)
d = $95
V0 = $1235
∴ Vn = 1235 − 95n
0 = 1235 − 95n
95n = 1235
n = 13
∴ The time taken to be written off is 13 years.
4 a V0 = 22 500
Vn+1 = Vn − 3200
b When n = 5:
V5 = 22 500 − 3200(5)
= 22 500 − 16 000
= 6500
The expected value will be $6500.
5 a Annual depreciation = 15% of $10 000
= 0.15 × $10 000
= $1500
b
Time, n(years) Depreciation, d($) Future value, Vn ($)
ii
(5, 2500)
0
1
2
3
4
5
6
7
8
9
10
d = $100
V0 = $850
∴ Vn = 850 − 100n
0 = 850 − 100n
100n = 850
n = 8.5
∴ The time taken to be written off is 8.5 years.
1235 − 285
c i Annual depreciation =
10
950
=
10
= $95
(0, 10 000)
Time, n(years) Depreciation, d($)
ii
89
0
2000
2000
2000
2000
2000
2000
Vn
14 000 (0, 13 500)
12 000
10 000
8000
6000
4000
2000
0
1 2 3 4 5 6
Time (years)
13 500
11 500
9500
7500
5500
3500
1500
(6, 1500)
n
b From the depreciation schedule in part a, the scrap value is
$1500.
P df_Fol i o: 89
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
90
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.3
8 a Annual depreciation = 20% of $7750
= 0.20 × $7750
= $1550
b Vn + 1 = Vn − 1550, Vo = 7750
c
Time, n(years) Depreciation, d($)
Future value
(× $1000)
0
1
2
3
4
0
1550
1550
1550
1550
Vn
8
6
4
2
b
Future value, Vn ($)
7750
6200
4650
3100
1550
(4, 1550)
1 2 3 4 n
Time (years)
d Using the depreciation schedule in part c, it will take
4 years to reach a scrap value of $1550.
9 a Annual depreciation = 15% of $92 000
= 0.15 × $92 000
= $13 800
b Vn+1 = Vn − 13 800, Vo = 92 000
c
Time, n(years) Depreciation, d($) Future value, Vn ($)
Future value ($)
0
1
2
3
4
5
6
0
13 800
13 800
13 800
13 800
13 800
13 800
Vn
100 000
80 000
60 000
40 000
20 000
0
i Cost price = $800
iii The time taken to reach the scrap value is 3
(0, 7750)
0
i Cost price = $1750
ii Annual depreciation = $500
92 000
78 200
64 400
50 600
36 800
23 000
9200
1
years.
2
800 − 100
5
700
=
5
= $140
iii The time taken to reach its scrap value is 5 years.
d i Cost price = $18 000
ii Annual depreciation = $2400
1
iii The time taken to reach its scrap value is 7 years.
2
11 V0 = $31 000
V5 = $5000
n = 5 years
c
ii Annual depreciation =
5000 = 31 000 − 5d
5d = 31 000 − 5000
5d = 26 000
d = 5200
∴ Annual depreciation = $5200
The correct answer is C.
12 a 400 = 2200 − 225n
225n = 2200 − 400
225n = 1800
n = 8 years
500 = 3600 − 310n
310n = 3600 − 500
310n = 3100
n = 10 years
The cheaper machine costing $2200 needs to be replaced
first.
b 10 − 8 = 2 years later, the other machine is replaced.
(0, 92 000)
5.3 Exam questions
(6, 9200)
1 2 3 4 5 6 n
Time (years)
d Using the depreciation schedule in part c, it will take
6 years to reach a scrap value of $9200.
10 a i Cost price = $3000
3000 − 1000
ii Annual depreciation =
5
2000
=
5
= $400
iii The time taken to reach scrap value is 5 years.
1a $15 000
[1 mark]
b V1 = V0 − 15 000 = 120 000 − 15 000 = $105 000
V2 = V1 − 15 000 = 105 000 − 15 000 = $90 000 [1 mark]
15000
c
× 100 = 12.5%
[1 mark]
120 000
d Vn = 120 000 − 15000n, where n = 0, 1, 2 …
[1 mark]
2 10% of the purchase price is 10% × $4500 = $450
Geoff will depreciate his computer by $450 per year, so
Vn+1 = Vn − 450
The correct answer is A.
3 The decrease in the values is not linear and is not a growth, so
some of the options can be eliminated.
The graph highlights a geometric decay with a reducing value
of 6% p.a.
Test the values with the reducing-balance recurrence relation:
V0 = 7000, Vn+1 = (1 − 0.06) × Vn .
The correct answer is B.
P df_Fol i o: 90
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.4
5.4 Modelling reducing balance depreciation with
a recurrence relation
5.4 Exercise
1 a R=1−
r
100
17
=1−
100
= 0.83
Vn+1 = RVn
Vn+1 = 0.83Vn , V0 = 1500
Time, n(years)
Future value ($)
V2 = RV1
= 0.87 × 2871
= 2497.77
V3 = RV2
= 0.87 × 2497.77
= 2173.059 …
V4 = RV3
= 0.87 × 2173.059 …
= 1890.562 …
V5 = RV4
= 0.87 × 1890.562 …
= 1644.789 …
Time, n(years)
Future value, Vn ($)
0
1500.00
1
1245.00
2
1033.35
3
857.68
4
711.87
5
590.86
b The future value of the laptop after five years will be
$590.85.
c
Vn
(0, 1500)
1500
1300
1100
900
700
500
V1 = RV0
= 0.87 × 3300
= 2871
3300
3100
2900
2700
2500
2300
2100
1900
1700
1500
0
2 a R=1−
1 2 3 4 5
Time (years)
r
100
13
=1−
100
= 0.87
Vn + 1 = RVn
Vn+1 = 0.87Vn , V0 = 3300
n
(0, 3300)
1
3 a R = 0.8, hence
Vn+1 = 0.8 Vn
V0 = 60 000
b Time,
n(years)
0
Future value, Vn ($)
0
3300.00
1
2871.00
2
2497.77
3
2173.06
4
1890.56
5
1644.79
b The future value of the road bike after five years will be
$1644.79
c
Vn
Future value ($)
V1 = RV0
= 0.83 × 1500
= 1245
V2 = RV1
= 0.83 × 1245
= 1033.35
V3 = RV2
= 0.83 × 1033.35
= 857.680 …
V4 = RV3
= 0.83 × 857.680 …
= 711.874 …
V5 = RV4
= 0.83 × 711.874 …
= 590.856 …
91
2
3
4
Time (years)
5
Depreciation, d($)
n
Future value,
Vn ($)
0
—
60 000
1
20% of 60 000 = 12 000
48 000
2
20% of 48 000 = 9600
38 400
3
20% of 38 400 = 7680
30 720
4
20% of 30 720 = 6144
24 576
5
20% of 24 576 = 4915.20
19 660.80
c The future value of the bulldozer after five years is
$19 660.80.
P df_Fol i o: 91
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.4
b
Vn
60 000 (0, 60 000)
55 000
50 000
45 000
40 000
35 000
30 000
25 000
20 000
0
1
4 a R = 0.6, hence
Vn+1 = 0.6 Vn
V0 = 4000
b
Time,
n(years)
Future value ($)
0
5 a
(4, 24 576)
2
3
4
Time (years)
5
20 000
15 000
5000
(5, 0)
0
Future value,
Vn ($)
Depreciation, d($)
1
2
3
4
Time (years)
5
0
1
2
3
4
5
—
5998
5998
5998
5998
5998
29 990
23 992
17 994
11 996
5 998
0
Depreciation ($)
Future
value ($)
27% of 29 990 = 8097.30
27% of 21 892.70 = 5911.03
27% of 15 981.67 = 4315.05
27% of 11 666.62 = 3149.99
27% of 8516.63 = 2299.49
29 990.00
21 892.70
15 981.67
11 666.62
8 516.63
6 217.14
n
value ($)
0
1
2
3
4
5
—
4600
4600
4600
4600
4600
23 000
18 400
13 800
9 200
4 600
0
Time
(years)
Depreciation ($)
Future
value ($)
28% of 23 000 = 6440
28% of 16 560 = 4636.80
28% of 11 923.20 = 3338.50
28% of 8584.70 = 2403.72
28% of 6180.99 = 1730.67
0
1
2
3
4
5
n
Future
value ($)
5
Depreciation ($)
(years)
(3, 864)
1
2
3
4
Time (years)
c The future value for the reducing balance method is greater
than the flat rate method after four years.
6 a Time
Future
(0, 4000)
Depreciation ($)
0
1
2
3
4
5
25 000
10 000
n
Time
(years)
Time
(years)
(0, 29 900)
30 000
0
—
4000
1
40% of 4000 = 1600
2400
2
40% of 2400 = 960
1440
3
40% of 1440 = 576
864
4
40% of 864 = 345.60
518.40
5
40% of 518.40 = 207.36
311.04
c The future value of the computer after five years is $311.04.
d
Vn
4000
3500
3000
2500
2000
1500
1000
500
Vn
Future value ($)
Future value ($)
d
b
Future value ($)
92
Vn
25 000
(0, 23 000)
22 500
20 000
17 500
15 000
12 500
10 000
7500
5000
2500
0
1
2
3
4
Time (years)
23 000.00
16 560.00
11 923.20
8 584.70
6 180.99
4 450.31
(5, 0)
n
5
c The future value for the reducing balance method is greater
than the flat rate method after four years.
P df_Fol i o: 92
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.4
b
Flat rate
Time,
n(years)
Depreciation, d($)
Future value,
Vn ($)
0
1
2
3
4
5
—
110
110
110
110
110
550
440
330
220
110
0
Reducing balance
Time,
n(years)
Depreciation, d($)
0
1
2
3
4
5
—
30% of 550 = 165
30% of 385 = 115.50
30% of 269.50 = 80.85
30% of 188.65 = 56.60
30% of 132.06 = 39.62
b
Future value,
Vn ($)
550
385
269.50
188.65
132.06
92.44
Future value ($)
Vn
600
550
500
450
400
350
300
250
200
150
100
50
(0, 550)
Future value (× $1000)
7 a
Vn
40
35
30 (0, 30 000)
25
20
15
10
5
0
1
(3, 0)
2
3
Time (years)
0
1
2
3
4
Time (years)
c 4 years (from graph)
8 a
22.5
10 V5 = 2675 1 −
(
100 )
= 2675(0.775)5
= $747.88
5
20
11 a V4 = 20 000 1 −
(
100 )
= 20 000(0.8)4
= $8192
12 V0 = $1200
r = 20%
n = 4 years
Flat rate
Time,
n(years)
Depreciation, d($)
Future value,
Vn ($)
0
1
2
3
—
10 000
10 000
10 000
30 000
20 000
10 000
0
Reducing balance
Time,
n(years)
Depreciation, d($)
0
1
2
3
—
50% of 30 000 = 15 000
50% of 15 000 = 7500
50% of 7500 = 3750
Future value,
Vn ($)
30 000
15 000
7500
3750
n
c 3 years (from graph)
6
15
9 V6 = 45 000 1 −
(
100 )
= 45 000(0.85)6
= $16 971.73
25
b V4 = 30 000 1 −
(
100 )
= 30 000(0.75)4
= $9492.19
(5, 0)
n
5
4
20
V4 = 1200 1 −
(
100 )
= 1200(0.8)4
= $491.52
The correct answer is B.
13 a r = 25%
n = 8 years
V0 = $1150
4
4
4
25
V8 = 1150 1 −
(
100 )
= 1150(0.75)8
= $115.13
8
Depreciation = $1150 − $115.13
= $1034.87
b r = 25%
n = 4 years
V0 = $3740
25
V4 = 3740 1 −
(
100 )
= 3740(0.75)4
= $1183.36
4
Depreciation = $3740 − $1183.36
= $2556.64
P df_Fol i o: 93
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
93
94
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.4
c r = 25%
n = 6 years
V0 = $7320
V6 = 7320
n
1−
(
= 7320(0.75)6
= $1302.80
25
100 )
6
Depreciation = $7320 − $1302.80
= $6017.20
14 a r = 30%
n = 5 years
V0 = $685
30
V5 = 685 1 −
(
100 )
= 685(0.7)5
= $115.13
5
Depreciation = $685 − $115.13
= $569.87
b r = 30%
n = 4 years
V0 = $32 500
30
V4 = 32 500 1 −
(
100 )
= 32 500(0.7)4
= $7803.25
4
Depreciation = $32 500 − $7803.25
= $24 696.75
c r = 30%
n = 3 years
V0 = $1075
30
V3 = 1075 1 −
(
100 )
= 1075(0.7)3
= $368.73
3
Depreciation = $1075 − $368.73
= $706.27
15 n = 7 years
V0 = $3000
r = 25%
25
V7 = 3000 1 −
(
100 )
= 3000(0.75)7
= $400.45
7
Depreciation = $3000 − $400.45
= $2599.55
The correct answer is D.
16 V0 = $17 500
r = 20%
n = 15 years
P df_Fol i o: 94
25
900 = 4500 1 −
(
100 )
900 = 4500(0.75)n
900
0.75n =
4500
0.75n = 0.2
Using CAS, n = 5.59 years = 6 years
n
15
18 1500 = 7600 1 −
(
100 )
1500 = 7600 (0.85)n
1500
0.85n =
7600
0.85n = 0.1974
Using CAS, n = 9.98 years = 10 years.
19 V0 = $1250
n = 3 years
r = 60%
3
60
V3 = 1250 1 −
(
100 )
= 1250 (0.4)3
= $80
The correct answer is C.
n
20
20 a 3000 = 40 000 1 −
(
100 )
3000 = 40 000 (0.8)n
3000
0.8n =
40 000
0.8n = 0.075
Using CAS, n = 11.61 years
17
20
V15 = 17 500 1 −
(
100 )
= 17 500 (0.8)15
= $615.73
The correct answer is E.
15
30
b 500 = 3000 1 −
(
100 )
500 = 3000(0.7)n
500
0.7n =
3000
0.7n = 0.1667
Using CAS, n = 5.02 years
n
5.4 Exam questions
1 Only the 2nd and the 3rd recurrence relations indicate
depreciation:
• the 2nd because it is subtracting 2500
• the 3rd because Vn is being multiplied by a number less
than zero.
That is, 2 of the given recurrence relations indicate
depreciation.
The correct answer is C.
2 a V0 = 75 000
V1 = 75 000 − 3375 = 71 625
V2 = 71 625 − 3375 = 68 250
b i The constant decrease amount from the recurrence
relation is the annual depreciation amount $3375.
ii Let x represent the annual flat rate of depreciation.
x % of $75 000 = $3375
x
× 75 000 = 3375
100
solving for x, x = 4.5% p.a.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.5
c From the recurrence relation, R = 0.943
For the reducing balance method of depreciation,
r
, where r is the annual rate of depreciation.
R=1−
100
r
0.943 = 1 −
, solving for r
100
r = 5.7% p.a.
3 If the value of a car is reduced by 8% each year, then the value
of a car each year will be 100% − 8% = 92%
∴ Cn+1 = 0.92Cn , where C0 = 26 000
The correct answer is A.
5.5 Modelling unit cost depreciation with
a recurrence relation
5.5 Exercise
1 a V0 = 800
d = 0.35
Vn + 1 = Vn − 0.35n, V0 = 800
2 a
P df_Fol i o: 95
Number of washes (n)
Future value, Vn ($)
1
2
3
4
5
799.65
799.30
798.95
798.60
798.25
V0 = 1400
d = 0.65
Vn + 1 = Vn − 0.65n, V0 = 1400
b V1 = V0 − 0.65
= 1400 − 0.65
= 1399.35
V2 = V1 − 0.65
= 1399.35 − 0.65
= 1398.70
V3 = V2 − 0.65
= 1398.70 − 0.65
= 1398.05
V4 = V3 − 0.65
= 1398.05 − 0.65
= 1397.40
Hours of use (n)
Future value, Vn ($)
1
2
3
4
5
1399.35
1398.70
1398.05
1397.40
1396.75
3 a Annual depreciation = 14 000 × 0.285
= $3990
b Total depreciation = 29 600 − 12 000
= $17 600
17 600
Useful life =
0.285
= 61 754 km
4 a Depreciation = distance × rate
= distance × $0.25/km
First year: distance = 64 000 km
Depreciation = distance × $0.25/km
= 64 000 × 0.25
= $16 000
Second year: distance = 56 000 km
Depreciation = distance × $0.25/km
= 56 000 × 0.25
= $14 000
b V1 = V0 − 0.35
= 800 − 0.35
= 799.65
V2 = V1 − 0.35
= 799.65 − 0.35
= 799.30
V3 = V2 − 0.35
= 799.30 − 0.35
= 798.95
V4 = V3 − 0.35
= 798.95 − 0.35
= 798.60
V5 = V4 − 0.35
= 798.60 − 0.35
= 798.25
V5 = V4 − 0.65
= 1397.40 − 0.65
= 1396.75
b Depreciation = $20 000
Depreciation = distance × $0.25/km
20 000 = distance × 0.25
20 000
distance =
0.25
= 80 000 km
5 a i Annual depreciation = 12 600 × 0.26
= $3276
ii Total depreciation = 25 000 − 10 000
= $15 000
15 000
∴ Useful life =
0.26
= 57 692 km
b i Annual depreciation = 13 700 × 0.216
= $2959.20
ii Total depreciation = 21 400 − 8000
= $13 400
13 400
Useful life =
0.216
= 62 037 km
6 a Annual depreciation = 15.340 × 0.23
= $3528.20
b Total depreciation = 32 000 − 9500
= $22 500
22 500
∴ Useful life =
0.23
= 97 826 km
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
95
96
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.5
7 a Annual depreciation = 28 461 × 0.272
= $7741.39
b Total depreciation = 29 500 − 8200
= $21 300
21 300
∴ Useful life =
0.272
= 78 309 km
1.50
8 a d1 = 620 000 ×
1000
= $930
∴ Depreciation in the first year is $930.
1.50
d2 = 540 000 ×
1000
= $810
∴ Depreciation in the second year is $810.
b Total depreciation = 930 + 810
= $1740
∴ Future value = 7200 − 1740
= $5460
9 a Vn+1 = Vn − 0.0025n, V0 = 11 300
b See table at the foot of the page*.
10 a Vn+1 = Vn − 0.025n, V0 = 14 750
b See table at the foot of the page*.
0.22
11 a d1 = 400 000 ×
100
= $880
∴ Depreciation in the first year is $880.
0.22
d2 = 480 000 ×
100
= $1056
∴ Depreciation in the second year is $1056.
b Total depreciation after 2 years = 880 + 1056
= $1936
*9 b
*10 b
P df_Fol i o: 96
Time (years)
Copies made per year
1
350 000
2
425 000
3
376 200
4
291 040
5
385 620
Time (years)
Bottles corked per year
1
40 000
2
42 500
3
46 700
4
38 250
5
43 060
∴ Future value = 8600 − 1936
= $6664
1.50
12 a d1 = 385 000 000 ×
1 000 000
= $577.50
∴ Depreciation in the first year is $577.50.
1.50
d2 = 496 000 000 ×
1 000 000
= $744
∴ Depreciation in the second year is $744.
b Total depreciation = 577.50 + 744
= $1321.50
∴ Future value = 38 000 − 1321.50
= $36 678.50
13 Depreciation = 13 690 × 0.216
= $2957.04
∴ Future value = 25 900 − 2957.04
= $22 942.96
The correct answer is D.
14 a d1 = 15 620 × 0.248
= $3873.76
∴ Depreciation at the end of first year is $3873.76.
d2 = 16 045 × 0.248
= $3979.16
∴ Depreciation at the end of second year is $3979.16.
b Total depreciation = 3873.76 + 3979.16
= $7852.92
∴ Future value = 32 600 − 7852.92
= $24 747.08
15 a d1 = 21 216 × 0.292
= $6195.07
∴ Depreciation at the end of first year is $6195.07.
Annual depreciation ($)
0.025
350 000 ×
= 875.00
10
0.025
= 1062.50
425 000 ×
10
0.025
376 200 ×
= 940.50
10
0.025
291 040 ×
= 727.60
10
0.025
385 620 ×
= 964.05
10
Depreciation ($)
40 000 ×
2.50
= 1000.00
100
2.50
42 500 ×
= 1062.50
100
2.50
46 700 ×
= 1167.50
100
2.50
38 250 ×
= 956.25
100
2.50
43 060 ×
= 1076.50
100
Future value at end of year, Vn ($)
11 300 − 875 = 10 425.00
10 425 − 1062.50 = 9362.50
9362.50 − 940.50 = 8422.00
8422 − 727.60 = 7694.40
7694.40 − 964.05 = 6730.35
Future value at end of year, Vn ($)
14 750 − 1000 = 13 750.00
13 750 − 1062.50 = 12 687.50
12 687.50 − 1167.50 = 11 520.00
11 520 − 956.25 = 10 563.75
10 563.75 − 1076.50 = 9487.25
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.5
d2 = 19 950 × 0.292
= $5825.40
∴ Depreciation at the end of second year is $5825.40.
b Total depreciation = 6195.07 + 5825.40
= $12 020.47
∴ Future value = 35 099 − 12 020.47
= $23 078.53
30 000
16 a Depreciation rate =
2 000 000
= $0.15
= 15 cents per km
2 000 000
b Useful life =
1600
= 125 weeks
= 2 years, 21 weeks
c Vn = 30 000 − 0.15n
13 800
d Distance travelled =
0.15
= 92 000 km
Depreciation = 160 000 × 0.15
= $24 000
∴ Future value = 30 000 − 24 000
= $6000
f For every 20 000 km:
Depreciation = 20 000 × 0.15
= $3000
e
Distance, n (km)
Future value ($)
0
20 000
40 000
60 000
80 000
100 000
120 000
140 000
160 000
180 000
200 000
30 000
27 000
24 000
21 000
18 000
15 000
12 000
9 000
6 000
3 000
0
17 See table at the foot of the page *
18 a i
d = 35 000 × 0.10
= $3500
10 000 = 35 000 − 3500n
3500n = 35 000 − 10 000
3500n = 25 000
∴ n = 7.14
∴ Time taken = 8 years
*17
P df_Fol i o: 97
Time (years)
Distance travelled (km)
1
2
3
4
5
13 290
15 650
14 175
9674
16 588
ii
r = 20%
P = $35 000
Vn = $10 000
97
20
∴10 000 = 35 000 1 −
(
100 )
10 000 = 35 000(0.8)n
10 000
0.8n =
35 000
0.8n = 0.2857
Now 0.85 = 0.3277
0.86 = 0.2621
∴ n = 6 by inspection
∴ Time taken = 6 years
iii Total depreciation = 35 000 − 10 000
= $25 000
25 000
Distance travelled =
0.25
= 100 000 km
100 000
Time taken =
10 000
= 10 years
The reducing balance method (6 years) enables the car
to reach scrap value sooner than the flat rate (8 years) or
unit cost method (10 years).
b i Flat rate:
Depreciation in the first year = 35 000 × 0.10
= $3500
∴ Tax deduction in the first year is $3500.
ii Reducing balance:
1
20
V1 = 35 000 1 −
(
100 )
= 35 000 (0.8)1
= $28 000
n
Depreciation in the first year = 35 000 − 28 000
= $7000
∴ Tax deduction in the first year is $7000.
iii Unit cost:
Depreciation in the first year = 10 000 × 0.25
= $2500
∴ Tax deduction in the first year is $2500.
c i Flat rate:
Depreciation in the fifth year = 35 000 × 0.10
= $3500
∴ Tax deduction in the fifth year is $3500.
There is no change in the tax deduction in the first and
fifth year.
13 290 × 0.236 = 3136.44
15 650 × 0.236 = 3693.40
14 175 × 0.236 = 3345.30
9674 × 0.236 = 2283.06
16 588 × 0.236 = 3914.77
Depreciation ($)
Future value at end of year, Vn ($)
28 395 − 3136.44 = 25 258.56
25 258.56 − 3693.40 = 21 565.16
21 565.16 − 3345.30 = 18 219.86
18 219.86 − 2283.06 = 15 936.80
15 936.80 − 3914.77 = 12 022.03
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
98
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.6
ii Reducing balance:
20
V4 = 35 000 1 −
(
100 )
= 35 000 (0.8)4
= $14 336
20
V5 = 35 000 1 −
(
100 )
= 35 000 (0.8)5
= $11 468.80
4
5.6 Review
5.6 Exercise
Multiple choice
5
∴ Depreciation in the fifth year = 14 336 − 11 468.80
= $2867.20
∴ Tax deduction in the fifth year is $2867.20.
The deduction is $4132.80 ($7000 − 2867.20) less than
in part b.
iii Unit cost:
Depreciation in the fifth year = 10 000 × 0.25
= $2500
∴ Tax deduction in the fifth year = $2500
There is no change in the tax deduction in the first and
fifth year.
19 Total depreciation = 8500 − 2000
= $6500
6500
∴ Usage =
0.02
= 325 000 units
The correct answer is E.
5000
20 Usage =
0.02
= 250 000 units
The correct answer is B.
5.5 Exam questions
1440
= 28 800
[1 mark]
0.05
1440
b
× 100 = 12%
[1 mark]
12000
c Mn = 12 000 − 0.05 × n
[1 mark]
2 depreciation by reducing balance – depreciation by unit cost =
26 166.24
3
k
45000 × 1 −
− (30000 × 0.15) = 26166.24
(
100 )
k = 12
The correct answer is B.
1 a
3 The rule for unit cost depreciation is Vn = V0 − nd, which
gives the value of the asset after n outputs.d is the
depreciation per output.
Here: V0 = 30000, and when n = 3, V3 = 18480
30000 − 18480
d=
3 × 24000
d = 0.16
Therefore: Vn = 30000 − 0.16n
The correct answer is E.
1 un+1 = un − 4, u0 = −6
u0 = −6
u1 = −6 − 4 = −10
u2 = −10 − 4 = −14
u3 = −14 − 4 = −18
u4 = −18 − 4 = −22
The sequence is −6, −10, −14, −18, −22.
The correct answer is B.
2 Vn+1 = 3Vn − 2, V0 = 2
V0 = 2
V1 = 3 × 2 − 2 = 4
V2 = 3 × 4 − 2 = 10
V3 = 3 × 10 − 2 = 28
The sequence is 2, 4, 10, 28.
The correct answer is C.
3 Vn+1 = 1.025Vn − 200, V0 = 10 000
On a calculator screen, type in 10 000, press ENTER/EXE.
Type × 1.025 − 200, press ENTER/EXE 4 times.
10 000
10 000 × 1.025 − 200
10 050 × 1.025 − 200
10 101.25 × 1.025 − 200
10 153.78 × 1.025 − 200
10 000
10 050
10 101.25
10 153.78
10 207.62
The fifth term is 10 208 to the nearest whole number.
The correct answer is D.
4 The first term is 4 and each successive term decreases by 2.
Written as a recurrence relation this could be:
un+1 = un − 2, u0 = 4
The correct answer is C.
5 The first term is −3 and each successive term is multiplied
by −4. Written as a recurrence relation this could be:
An+1 = −4An , A0 = −3
The correct answer is A.
6 Y0 = 54 000
Depreciation rate = 12%
r
12
R=1−
=1−
= 0.88
100
100
Yn+1 = 0.88Yn , Y0 = 54 000
The correct answer is E.
7 B0 = 4000
Depreciation = 300
Bn+1 = Bn + 300, B0 = 4000
The correct answer is B.
8 Vn+1 = Vn − 500, V0 = 9500 — this is an example of a linear
recurrence model of decay as each term decreases by the
same amount.
Pn+1 = 0.7Pn , P0 = 8000 — this is an example of a geometric
recurrence model of decay as each term decreases (the
multiplying factor is less than one) by a different amount.
The correct answer is A.
P df_Fol i o: 98
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.6
9 V0 = 8400
Depreciation rate = 12%
r
12.5
R=1−
=1−
= 0.875
100
100
Vn+1 = 0.875Vn , V0 = 8400
The correct answer is D.
10 12 500 − 6500 = 6000
6000
=6
1000
It would take 6 years to reach a book value of $6500.
The correct answer is C.
Short answer
11 a un+1 = 4un − 3 u0 = −1
On a calculator screen, type in −1, press ENTER/EXE.
Type × 4 − 3, press ENTER/EXE 4 times.
−1
−1 × 4 − 3
−7 × 4 − 3
−31 × 4 − 3
−127 × 4 − 3
−1
−7
−31
−127
−511
The terms are −1, −7, − 31, −127, −511.
b un+ 1 = 3 + 5un u0 = 0
On a calculator screen, type in 0, press ENTER/EXE.
Type × 5 + 3, press ENTER/EXE 4 times.
0
0×5+3
3×5+3
18 × 5 + 3
93 × 5 + 3
0
3
18
93
468
The terms are 0, 3, 18, 93, 468.
12 a Let Mn be the number of club members at the beginning of
the nth year, where n = 0, 1, 2, ...
r
4
R=1−
=1−
= 0.96
100
100
Mn+1 = 0.96Mn + 20, M0 = 300
b M4 will be the number of members at the end of the
4th year.
On a calculator screen, type in 300, press ENTER/EXE.
Type × 0.96 + 20, press ENTER/EXE 4 times.
300
300 × 0.96 + 20
308 × 0.96 + 20
315.68 × 0.96 + 20
323.05 × 0.96 + 20
300
308
315.68
323.053
330.131
The membership at the end of 4 years is 330 members.
13 Cn+ 1 = 0.5Cn + 100 and C0 = 400
On a calculator screen, type in 400, press ENTER/EXE.
Type × 0.5 + 100, press ENTER/EXE 4 times.
P df_Fol i o: 99
400
400 × 0.5 + 100
300 × 0.5 + 100
250 × 0.5 + 100
225 × 0.5 + 100
C4 = 212.5
400
300
250
225
212.5
99
14 a R = 1 −
r
8
=1−
= 0.92
100
100
Pn+1 = 0.92Pn , P0 = 35 000
b Pn+1 = 0.92Pn , P0 = 35 000
On a calculator screen, type in 35 000, press ENTER/EXE.
Type × 0.92, press ENTER/EXE 5 times.
35 000
35 000 × 0.92
32 200 × 0.92
29 624 × 0.92
27 254.1 × 0.92
25 073.75 × 0.92
35 000
32 200
29 624
27 254.1
25 073.75
23 067.85
Highlight the last value to round off correctly to 2 decimal
places.
The future value after 5 years is $23 067.85.
15 a 0.15 × 18 000 = 2700
The depreciation each year is $2700.
b Method 1:
On a calculator screen, type in 18 000, press ENTER/EXE.
Type − 2700, press ENTER/EXE until the amount reaches
$2000.
18 000
18 000 − 2700
15 300 − 2700
12 600 − 2700
9900 − 2700
7200 − 2700
4500 − 2700
18 000
15 300
12 600
9900
7200
4500
1800
It took 6 iterations for the future value to be 1800.
It will take 6 years for the equipment to reach its scrap
value.
Method 2:
Vn = V0 − nd
1800 = 18 000 − n × 2700
2700n = 16 200
16 200
n=
=6
2700
It will take 6 years for the equipment to reach its scrap
value.
16 a Year 1
2000
= 100
20
100 × 5.20 = 520
Depreciation in first year is $520.
Year 2
2500
= 125
20
125 × 5.20 = 650
Depreciation in second year is $650.
b 6500 − 520 − 650 = 5330
At the end of the second year, the value is $5330.
Extended response
17 a Flat rate method:
Depreciation = 0.15 × 150 000 = $22 500 per year.
On a calculator screen, type in 150 000, press
ENTER/EXE.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
100
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.6
Type − 22 500, press ENTER/EXE 5 times.
150 000
150 000 − 22 500
127 500 − 22 500
105 000 − 22 500
82 500 − 22 500
60 000 − 22 500
150 000
127 500
105 000
82 500
60 000
37 500
Reducing balance method:
On a calculator screen, type in 150 000, press
ENTER/EXE.
Type × 0.75, press ENTER/EXE 5 times.
150 000
150 000 × 0.75
112 500 × 0.75
84 375 × 0.75
63 281.25 × 0.75
47 460.94 × 0.75
150 000
112 500
84 375
63 281.25
47 460.94
35 595.70
Time
(n)
Future value
( flat rate)
Future value
(reducing balance)
0
1
2
3
4
5
6
150 000
127 500
105 000
82 500
60 000
37 500
15 000
150 000.00
112 500.00
84 375.00
63 281.25
47 460.94
35 595.70
26 696.80
b Based on the depreciation table, the reducing balance
future value becomes greater than the flat rate future value
after 6 years.
18 a V0 = 60 000
r
d = V0 ×
100
12
= 60 000 ×
100
= 7200
Vn+1 = Vn − d
Vn+1 = Vn − 7200, V0 = 60 000
b V0 = 60 000, d = 7200
Vn = V0 − d × n
Vn = 60 000 − 7200n
c n=5
Vn = 60 000 − 7200n
V5 = 60 000 − 7200 × 5
= 24 000
Therefore, book value after 5 years is $24 000.
Total depreciation after 5 years = $60 000 − $24 000
= $36 000
d Depreciation = 25 000 × 5 × 0.30 = $37 500
Therefore, book value after 5 years = 60 000 − 37 500
= $22 500
e V0 = 60 000, r = 16, n = 5
r
R=1−
100
16
=1−
100
= 0.84
Vn = V0 Rn
V5 = 60 000 × 0.845
= 25 092.72
Therefore, future value after 5 years is $25 092.72.
Total depreciation = V0 − V5
= 60 000 − 25 092.72
= 34 907.28
f Flat rate: total depreciation = $36 000
Unit cost: total depreciation = $37 500
Reducing balance: total depreciation = $34 907.28
Therefore, unit cost depreciation gives greatest
depreciation over the 5-year period.
5.6 Exam questions
1 Vn+1 = 2480 + 45(n + 1)
= 2480 + 45n + 45
Note that Vn = 2480 + 45n and V0 = 2480.
Therefore, V0 = 2480, Vn+1 = Vn + 45.
The correct answer is C.
2 a V1 = 0.9 × 60000 = 54000
[1 mark]
V2 = 0.9 × 54000 = $48600
b 1 − 0.9 = 0.1 or 10%
[1 mark]
c 0.9n × 60000 < 20000
Solve on your CAS calculator:
n = 10.427 …
Phil will replace these tools after 10 years, or sometime in
the 11th year.
[1 mark]
d 8% × 60000 = 4800
Vn+1 = Vn − 4800, where V0 = 60 000
It could also be written as Vn + 1 = Vn − 0.08 V0 , where
V0 = 60 000
[1 mark]
Responses generally used the notation well, but many
wrote a recurrence relation reflective of reducing balance
depreciation rather than flat rate.
A few wrote both a recurrence relation and a rule with Vn in
terms of n which could not be accepted.
3 In the 4 years, the printer prints 1920 × 4 = 7680 pages.
Let Vn = 680 − 7680n,
Where n represents the depreciation amount per page
Solve 125 = 680 − 7680n (use CAS)
n = 0.072265 ($ per page)
≃ 7 cents
Although this was a standard application of unit-cost
depreciation, almost half of the students were unable to
determine the depreciation per page printed as required.
The correct answer is E.
P df_Fol i o: 100
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 5 Modelling depreciation of assets using recursion • EXERCISE 5.6
4. It is linear depreciation, so it must be flat rate or unit cost
depreciation.
The gradient is in dollars per kilometre travelled and is
worked out using the coordinates (0, 35 000) and
(20 000, 30 000):
35 000 − 30 000
m=
0 − 20 000
= −$0.25/km
The correct answer is D.
5. Amount lost in four years = 18 000 − 5000
= 13 000
13 000
Therefore, each year $
was lost in value.
4
After one year:
13 000
Value = 18 000 −
4
18 000 − 5000
= 18 000 −
(
)
4
The correct answer is B.
P df_Fol i o: 101
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
101
102
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.2
Topic 6 — Modelling compound interest investments and
loans using recursion
6.2 Simple interest
6.2 Exercise
1 a This is a flat rate recurrence model.
V0 = 1020, r = 8.5
8.5
× 1020 = 86.7
100
The account grows by $86.70 per year.
Vn+1 = Vn + 86.70, V0 = 1020
b Using your calculator, type 1020, press ENTER/EXE, then
type +86.7 and press ENTER 5 times.
1020
1020
1020 + 86.7
1106.7
1106.7 + 86.7
1193.4
1193.4 + 86.7
1280.1
1280.1 + 86.7
1366.8
1366.8 + 86.7
1453.5
At the end of 5 years, the investment is $1453.50.
2 a This is a flat rate recurrence model.
B0 = 713, r = 6.75
6.75
× 713 = 48.13
100
The account grows by $48.13 per year.
Bn+1 = Bn + 48.13, B0 = 713
b Using your calculator, type 713, press ENTER/EXE, then
type + 48.13 and press ENTER 5 times.
713
713
713 + 48.13
761.13
761.13 + 48.13
809.26
809.26 + 48.13
857.39
857.39 + 48.13
905.52
905.52 + 48.13
953.65
At the end of 5 years, the investment is $953.65.
3 a This is a flat rate recurrence model.
A0 = 1500, r = 12.5
12.5
× 1500 = 187.50
100
The account grows by $187.50 per year.
An+1 = An + 187.50, A0 = 1500
b Using your calculator type 1500, press ENTER/EXE, then
type +187.50 and press ENTER twice.
1500
1500 + 187.50
1687.50 + 187.50
1500
1687.50
1875
At the end of 2 years, the investment is $1875.
c Total interest = 1875 – 1500 = $375
4 a This is a flat rate recurrence model.
A0 = 2400, r = 5.5
5.5
× 2400 = 132
100
The account grows by $132 per year.
An+1 = An + 132, A0 = 2400
b The account grows by $132 p.a., so in 1.5 years the interest
will be 2 × 132 + 66 = 330.
Total amount = 2400 + 330 = $2730
c Interest from part b = $330
5 a This is a flat rate recurrence model.
A0 = 680, r = 5
5
× 680 = 34
100
The account grows by $34 per year.
An+1 = An + 34, A0 = 680
680
680 + 34
714 + 34
748 + 34
782 + 34
680
714
748
782
816
The value of the investment at the end of 4 years is $816.
b This is a flat rate recurrence model.
A0 = 13 000, r = 7.5
7.5
× 13 000 = 975
100
The account grows by $975 per year.
An+1 = An + 975, A0 = 13 000
13 000
13 000 + 975
13975 + 975
14950 + 975
13 000
13 975
14 950
15 925
The value of the investment at the end of 3 years is $15 925.
6 a V0 = 10 500, n = 5
Vn = V0 + nd
d = 0.065 × 10 500 = 682.50
Vn = 10 500 + 5 × 682.50
= 13 912.50
The account balance after 5 years was $13 912.50.
P df_Fol i o: 102
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.2
b
n
0
1
2
3
4
5
9 A3 − A2 = 3530.50 − 3377
= 153.50
Vn
10 500
11 182.50
11 865
12 547.50
13 230
13 912.50
Amount invested
Vn
14 500
13 500
12 500
11 500
10 500
0
1.0
2.0 3.0 4.0
Time in years
5.0
n
7 a For an initial investment of $6200, the building society is
offering a flat rate of interest set at $565 per annum.
r
× 6200 = 565
b
100
565 × 100
r=
6200
= 9.11
The interest rate is 9.11% p.a.
8 a V0 = 1500, n = 5
Vn = V0 + nd
d = 0.1225 × 1500 = 183.75
Vn = 1500 + 5 × 183.75
= 2418.75
After 5 years, Silvio’s return from the bond is $2418.50.
b
n
V
n
0
1
2
3
4
5
1500
1683.75
1867.50
2051.25
2235
2418.50
Amount invested
I = 4 × 90
= 360
Interest charged is $360.
b V0 = 7500, r = 12%, n = 3
d = 0.12 × 7500
= 900
I = 3 × 900
= 2700
Interest earned is $2700.
c V0 = 250, r = 1.75 × 12 = 21% p.a., n = 2.5
d = 0.21 × 250
= 52.5
6.2 Exam questions
2200
1900
1600
0
A2 − A1 = 3377 − A1
= 153.50
A1 = 3377 − 153.50
= 3223.5
A0 = 3223.5 − 153.50
= 3070
The amount invested was $3070.
10 0.05 × 3000 = 150
$150 interest earned per annum
450
=3
150
Carol should invest her money for 3 years.
11 $800 of interest is paid.
800
= 160
5
Interest paid per year = $160
r
× 1000 = 160
100
r = 16%
The correct answer is B.
12 a V0 = 690, r = 12%, n = 4
d = 0.12 × 750
= 90
I = 2.5 × 52.5
= 131.25
Interest earned is $131.25.
Vn
2500
P df_Fol i o: 103
1.0
103
2.0 3.0 4.0
Time in years
5.0
n
1 Interest = (1.00485 × 3000) − 3000 = $72.69
The correct answer is D.
2 From the graph, the value of investment increases by the same
amount each year.
This is simple interest investment with
P = 1000, I = 200 and n = 4.
100I
100 × 200
r=
=
= 5%p.a.
pn
1000 × 4
For the other investment, r = 5, n = 8 and I = 600.
100I
100 × 600
P=
=
= $1500
rn
5×8
The correct answer is C.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
104
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.3
3 I=
V0 rn
100
where n = 5, I = 1800 and r = 4.5
4.5
1800 = 5 × V0 ×
100
V0 = $8000
The correct answer is D.
6.3 Compound interest as a geometric recurrence
relation
6.3 Exercise
1 a V0 = 7500
r=6
6
R=1+
= 1.06
100
Vn+1 = 1.06Vn , V0 = 7500, where V is the amount invested
and n is the number of years after the initial investment.
b 7500
7500
7500 × 1.06
7950 × 1.06
8427 × 1.06
8932.62 × 1.06
7950
8427
8932.62
9468.58
At the end of 4 years, the amount invested is $9468.58.
2 a V0 = 3250
r = 7.5
7.5
R=1+
= 1.075
100
Vn+1 = 1.075Vn , V0 = 3250, where V is the amount
invested and n is the number of years after the initial
investment.
b 3250
3250
3250 × 1.075
3493.75 × 1.075
3755.78 × 1.075
4037.46 × 1.075
4340.37 × 1.075
3493.75
3755.78
4037.46
4340.27
4665.80
At the end of 5 years, the amount invested is $4665.80.
3 a V0 = 600
r = 6% p.a.
6
=
12
= 0.5% per month
0.5
R=1+
= 1.005
100
Vn+1 = 1.005Vn , V0 = 600, where V is the amount invested
and n is the number of months after the initial investment.
b 600
600
600 × 1.005
603 × 1.005
606.02 × 1.005
609.05 × 1.005
612.09 × 1.005
615.15 × 1.005
603
606.02
609.05
612.09
615.15
618.23
At the end of 6 months, the amount invested is $618.23.
4 a V0 = 2500
P df_Fol i o: 104
r = 12% p.a.
12
=
12
= 1% per month
R=1+
1
= 1.01
100
Vn+1 = 1.01Vn , V0 = 2500, where V is the amount
borrowed and n is the number of months after the initial
loan.
b At the end of 6 months, the total amount of the loan is
$2550.25.
2500
2500 × 1.01
2525 × 1.01
2500
2525
2550.25
The interest owing is $50.25.
5 The initial amount is multiplied by 1.05, so the compound
interest rate per annum is 5%.
The correct answer is B.
5
6 Interest earned =
× 3000 = $150
100
The correct answer is C.
7 Balance at the end of the first year = 3000 + 150 = $3150
The correct answer is D.
8 Vn+1 = Vn
(
1+
V1 = 25 000 ×
4.7
2
100 )
, V0 = $25 000
1+
4.7
2
1+
4.7
2
100 )
(
= 25 587.50
First period interest is 25 587.50 − 25 000 = $587.50
V2 = 25 587.50 ×
(
100 )
= 26188.81
Second period interest is 26 188.81 − 25 587.50 = $601.31
Difference in payments is 601.31 − 587.50 = $13.81
The correct answer is A.
5
9 Interest earned =
× 3150 = $157.50
100
The correct answer is D.
10 Compound interest grows with each compounding period, so
the interest in the fourth year must be greater than the interest
in the third year.
The correct answer is E.
11
V ($)
Interest ($)
V ($)
n
5750
B
6706.80
7243.34
F
A% of 5750 = 460
8% of C = 496.80
8% of 6706.80 = 536.54
8% of 7243.34 = 579.47
8% of 7822.81 = 625.82
A = 8%
8
× 5750 = 460
100
B = 5750 + 460
= $6210
C = B = 6210
D = 6210 + 496.80
= $6706.80
E = F = 7243.34 + 579.47
= $7822.81
n+1
6210
D
7243.34
E
8448.63
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.4
12
Vn ($)
Interest ($)
12 000
12 900
C
14 907.56
E
7.5% of 12 000 = 900
7.5% of 12 900 = A
7.5% of 13 867.50 = 1040.06
7.5% of 14 907.56 = 1118.07
7.5% of 16 025.63 = 1201.92
Vn ($)
Vn+1 ($)
7.5
× 12 900 = $967.50
100
B = 12 900 + 967.50
= $13 867.50
C = B = 13 867.50
D = 14 907.56 + 1118.07
= $16 025.63
E = D = $16 025.63
F = 16 025.63 + 1201.92
= $17 227.55
6.5
13 Interest rate per month =
= 0.542%
12
A=
5500
5529.81
5559.78
5589.92
5620.21
Vn+1 ($)
12 900
B
14 907.56
D
F
1.00 542 × 5500 = 5529.81
1.00 542 × 5529.81 = 5559.78
1.00 542 × 5559.78 = 5589.92
1.00 542 × 5589.92 = 5620.21
1.00 542 × 5620.21 = 5650.67
The value of the investment at the end of 5 months is
$5650.67.
14 a V0 = 15 000
r = 1.2% per month
1.2
R=1+
= 1.012
100
Vn+1 = 1.012Vn , V0 = 15 000, where V is the amount owing
and n is the number of months after the account due date.
b 15 000
15 000
15 000 × 1.012
15 180 × 1.012
15 362.1 × 1.012
15 180
15 362.16
15 546.51
At the end of 2 months, the total amount of the account is
$15 546.51.
Interest = 15 546.51 − 15 000
= $546.51
15 Geometric sequences grow or decay by the same factor; linear
sequences grow and decay by the same amount. These
sequences change by factors of 0.95 and 1.15 respectively so
they are both geometric sequences. The first sequence has a
factor less than 1, so it is a model of decay; the second has a
factor greater than 1, so it is a model of growth.
The correct answer is D.
1.5
× 200
100
= 200 + 3
= $203
[1 mark]
VCAA Assessment Report note:
Many students did not read that the given interest rate was
per month, so $200.25 was a common incorrect answer.
r
b A0 = 428 An+1 = R × An , where R = 1 +
100
1.5
R=1+
100
= 1.015
An+1 = 1.015 × An
A0 = 428,
Award 1 mark for stating A0 = 428, 1 mark for
An+1 = 1.015 × An .
VCAA Assessment Report note:
Many students did not write a recurrence relation in the
required form.
A recurrence relation has the initial value written first.
The name of the variable needed to be consistent in the
recurrence relation.
Some students did not recognise the difference between the
recurrence relation above and the rule An = 428 × 1.015n .
c A0 = 428
A1 = 1.015 × 428
= 434.42
A2 = 1.015 × 434.42
= 440.9363
A3 = 1.015 × 440.9363
= 447.5503445
A4 = 1.015 × 447.5503445
≃ 454.2635997
After 4 months, Lily pays $454.26.
The interest charged is 454.26 − 428 = $26.26. [1 mark]
VCAA Assessment Report note:
Some students who answered parts a. and b. correctly gave
$454.26 as the answer here, which was the total amount
Lily was charged rather than the interest. An answer to the
nearest (one) cent was required, not the nearest five or
10 cents.
Payment = 200 +
3 V0 = $25000
4.8
4.8
V1 = V0 1 +
= 25000 1 +
= $26200
(
(
100 )
100 )
[1 mark]
4.8
4.8
V2 = V1 1 +
= 26200 1 +
= $27457.60
(
(
100 )
100 )
4.8
4.8
V3 = V2 1 +
= 27457.60 1 +
= $28775.56
(
(
100 )
100 )
[1 mark]
6.4 Compound interest using a rule
6.3 Exam questions
P df_Fol i o: 105
1 The interest rate is the number found in front of Vx .
1.01 means 101% = 100% + 1%, so the interest rate, per
annum, is 1%.
The correct answer is C.
2 a Marcus pays the $200 owing plus 1.5% of the amount owed
as interest.
105
6.4 Exercise
1 a Vn = V0 (1 +
r n
100 )
7.5
V5 = 2500 1 +
(
100 )
5
b V5 = $3589.07, I = 3589.07 − 2500 = $1089.07
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.4
106
2 Vn = V0 (1 +
r n
100 )
5.25
= 6750 1 +
(
100 )
= $9657.36
6
7
V0 =
8
7 a Amount Vn = 500 1 +
(
100 )
= 500(1.08)2
= $583.20
12
Balance
7.5
c Amount Vn = 3600 1 +
(
100 )
= 3600(1.075)3
= $4472.27
1 2 3 4 5 6 7 8 9 10 11 12
Quarter
x
Even though it is difficult to see on the graph,
the graph is exponential.
4 a n = 2 × 12
= 24
5.5%
b r% =
12
= 0.4583%
r n
c V n = V 0 (1 +
100 )
0.4583
= 7500 1 +
(
100 )
= $8369.92
Balance
5
2 4 6 8 10 12 14 16 18 20 22 24
Months
n
Even though it is difficult to see on the graph,
the graph is exponential.
r n
Vn = V0 (1 +
100 )
( )
5000 = V0 1 + 4
100 ]
[
7.5
5000 = V0 (1.018 75)20
P df_Fol i o: 106
24
Vn
8500
8300
8100
7900
7700
7500
V0 =
2
13
b Amount Vn = 1000 1 +
(
100 )
= 1000(1.13)4
= $1630.47
y
5200
5050
4900
4750
4600
4450
4300
4150
4000
0
7×12
6300
(1.004 583)84
V0 = $4290.73
1.75
= 4200 1 +
(
100 )
= $5172.05
d
( )
6300 = V0 1 + 12
100 ]
[
6300 = V0 (1.004 583)84
r n
c V n = V 0 (1 +
100 )
0
r n
100 )
5.5
3 a n=3×4
= 12
7%
b r% =
4
= 1.75%
d
V n = V 0 (1 +
5000
(1.018 75)20
V0 = $3448.40
5×4
4
3
5.25
d Amount Vn = 2915 1 +
(
100 )
= 2915(1.0525)5
= $3764.86
5
8 a i V0 = 2000
r = 7.5
7.5
R=1+
100
= 1.075
Hence Vn + 1 = 1.075Vn , V0 = 2000
V1 = 1.075V0
= 1.075 × 2000
= 2150
The balance after 1 year is $2150.
ii I = Vn − V0
= 2150 − 2000
= 150
The interest earned is $150.
b i V0 = 2000
r = 7.5
7.5
R=1+
100
= 1.075
Hence Vn + 1 = 1.075Vn , V0 = 2000
V1 = 1.075V0
= 1.075 × 2000
= 2150
V2 = 1.075V1
= 1.075 × 2150
= 2311.25
The balance after 2 years is $2311.25.
ii I = Vn − V0
= 2311.25 − 2000
= 311.25
The interest earned is $311.25.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.4
c i V0 = 2000
r = 7.5
7.5
R=1+
100
= 1.075
Hence Vn + 1 = 1.075Vn , V0 = 2000
V1 = 1.075V0
= 1.075 × 2000
= 2150
V2 = 1.075V1
= 1.075 × 2150
= 2311.25
V3 = 1.075V2
= 1.075 × 2311.25
= 2484.59 …
V4 = 1.075V3
= 1.075 × 2484.59 …
= 2670.93 …
V5 = 1.075V4
= 1.075 × 2670.93 …
= 2871.25 …
V6 = 1.075V5
= 1.075 × 2871.25 …
= 3086.60 (2 d.p.)
The balance after 6 years is $3086.60.
ii I = Vn − V0
= 3086.60 − 2000
= 1086.60
The interest earned is $1086.60.
9 a n=5
b n=5×4
= 20
c n=4×2
=8
d n = 6 × 12
= 72
1
e n=4 ×2
2
=9
9
f n=3 ×4
12
= 3.75 × 4
= 15
6
10 a r =
4
= 1.5% per quarter
4
b r=
2
= 2% per half-yearly
18
c r=
12
= 1.5% per month
7
d r=
4
= 1.75% per quarter
11 a V0 = 1500
r=8
n=2
R=1+
8
100
= 1.08
Vn = V0 Rn
= 1500(1.08)2
= 1749.60
b V0 = 1500
8
r=
4
=2
R=1+
2
100
= 1.02
n=8
Vn = V0 Rn
= 1500(1.02)8
= 1757.49
c V0 = 1500
8
r=
12
2
=
3
R=1+
151
=
150
2
3
100
n = 24
Vn = V0 Rn
= 1500
151
( 150 )
= 1759.33
24
d V0 = 1500
8
r=
52
R=1+
8
52
100
5 208
=
5 200
n = 52 × 2 = 104
Vn = V0 Rn
104
5 208
= 1500
( 5 200 )
= 1760.05
e The balance increases as the number of compounding
periods become more frequent.
12 a n = 3 × 12
= 36 months
7
r=
12
= 0.583 33% per month
V0 = $2600
P df_Fol i o: 107
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
107
108
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.4
Amount Vn = 2600 1 +
(
0.583 33
100 )
36
= 2600(1.005 833 3)36
= $3205.60
Amount accrued = $3205.60 − $2600
= $605.60
b n = 4 × 12
= 48 months
8
r=
12
= 0.666 67
V0 = $3500
Amount Vn = 3500 1 +
(
0.259 62
Amount Vn = 12 000 1 +
(
100 )
= 12 000(1.002 596 2)117
= $16 252.85
117
0.666 67
100 )
48
= 3500(1.006 666 7)48
= $4814.84
Amount accrued = $4814.84 − $3500
= $1314.84
1
c n = 5 × 26
2
= 143
11
r=
26
= 0.423 08
V0 = $960
0.423 08
Amount Vn = 960 1 +
(
100 )
= 960(1.004 230 8)143
= $1755.77
143
Amount accrued = $1755.77 − $960
= $795.77
d n = 5 × 52
= 260 weeks
7.3
r=
52
= 0.140 38% per week
V0 = $2370
0.140 38
Amount Vn = 2370 1 +
(
100 )
260
Amount accrued = $3413.10 − $2370
= $1043.10
0.041 78
Amount Vn = 4605 1 +
(
100 )
= 4605(1.000 417 8)730
= $6246.82
Amount accrued = $16 252.85 − $12 000
= $4252.85
The correct answer is B.
4
9
15 a 5000 = V0 1 +
(
100 )
5000 = V0 (1.09)4
5000
V0 =
(1.09)4
= $3542.13
8.2
b 2600 = V0 1 +
(
100 )
2600 = V0 (1.082)3
2600
V0 =
(1.082)3
= $2052.54
1.5
c 3550 = V0 1 +
(
100 )
3
12
3550 = V0 (1.015)12
3550
V0 =
(1.015)12
= $2969.18
= 2370(1.001 403 8)260
= $3413.10
e n = 2 × 365
= 730 days
15.25
r=
365
= 0.041 78% per day
V0 = $4605
Amount accrued = $6246.82 − $4605
= $1641.82
13 The greatest return occurs when the interest is compounded
more regularly, i.e. fortnightly.
The correct answer is E.
1
14 n = 4 × 26
2
= 117 fortnights
6.75
r=
26
= 0.259 62% per fortnight
V0 = $12 000
0.8
d 6661.15 = V0 1 +
(
100 )
36
6661.15 = V0 (1.008)36
6661.15
V0 =
(1.008)36
= $5000
16 a n = 4 × 2
=8
9.5
r=
2
= 4.75% per 6 months
Vn = $3000
730
P df_Fol i o: 108
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.5
4.75
∴ 3000 = V0 1 +
(
100 )
8
3000 = V0 (1.0475)8
3000
V0 =
(1.0475)8
= 2069.61
$2069.61 is the required principal.
b n=3×4
= 12 quarters
9
r=
4
= 2.25% per quarter
Vn = $2000
2.25
∴ 2000 = V0 1 +
(
100 )
12
2000 = V0 (1.0225)12
2000
V0 =
(1.0225)12
= 1531.33
$1531.33 is the required principal.
1
c n=5 ×4
4
= 21 quarters
8.7
r=
4
= 2.175% per quarter
Vn = $5600
2.175
∴ 5600 = V0 1 +
(
100 )
21
5600 = V0 (1.021 75)21
5600
V0 =
(1.021 75)21
$3564.10 is the required principal.
1
d n = 4 × 12
4
= 51 months
15
r=
12
= 1.25% per month
Vn = $10 000
For part c:
Interest accrued = $5600 − $3564.10
= $2035.90
For part d:
Interest accrued = $10 000 − $5307.05
= $4692.95
6.4 Exam questions
1 A = p (1 +
r n
0.279 166 7
= 1200 1 +
)
(
100 )
100
= $1283.03
2×12
I=A−P
= $1283.03 − $1200
= $83.03
The correct answer is E.
2 A after 1 quater = PRn
1
4.25/4
+ 500
= 6000 1 +
(
100 )
= 6563.75
A after 2nd quater = PRn
1
4.25/4
+ 500
= 6563.75 1 +
(
100 )
= 7133.49
A after 3rd quater = PRn
1
4.25/4
= 7133.49 1 +
+ 500
(
100 )
= 7709.28
The correct answer is D.
3 Substituting the known values, the recurrence relation
becomes:
V2 = V1
1 + 12
100 )
(
= V1 (1.0025)
⇒ 7035.04 = V1 (1.0025)
Rearrange to find the value of V1 :
7035.04
V1 =
(1.0025)
= 7017.496
Repeat the process to find V0 :
V1 = V0 (1.0025)
⇒ 7017.496 = V0 (1.0025)
7017.496
V0 =
1.0025
= $7000
Therefore the initial investment was $7000.
The correct answer is B.
1.25
∴ 10 000 = V0 1 +
(
100 )
51
10 000 = V0 (1.0125)51
10 000
V0 =
(1.0125)51
= 5307.05
$5307.05 is the required principal.
e For part a:
Interest accrued = $3000 − $2069.61
= $930.39
For part b:
Interest accrued = $2000 − $1531.33
= $468.67
109
3
6.5 Calculating rate or time for compound interest
6.5 Exercise
1
Vn = V0 Rn
5000 = 4000R2×4
5000
= R8
4000
P df_Fol i o: 109
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
110
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.5
R=
1+
∴ r = 6.99% per 6 months
∴ Annual rate = 6.99 × 2
= 13.98%
1
5 8
(4)
5 8
r
=
100 ( 4 )
1
r
5
=
−1
100 ( 4 )
r
= 0.028 285 594
100
r = 2.828 559 4% per quarter
r ≈ 11.31% per annum
1
8
2
Vn = V0 Rn
10 000 = 7500R3×12
10 000
= R36
7500
R=
1+
V0 = $12 000
Vn = $15 000
∴ 15 000 = 12 000 (1 +
r 16 15 000
(1 + 100 ) = 12 000
1+
1
r
15 16
=
(
100
12 )
1
r
15 16
=
−1
(
100
12 )
r
= 0.014044
100
r = 1.404% per quarter
∴ Annual rate = 1.404 × 4
= 5.62%
1
5 a n = 2 × 12
2
= 30 months
1
1
36
1
n=6×4
= 24 quarters
15.5
r=
4
= 3.875% per quarter
Vn = $24 000
V0 = $25 000
Vn = $40 000
∴ 40 000 = 25 000 (1 +
3.875
∴ 24 000 = V0 1 +
(
100 )
24 000 = V0 (1.038 75)24
24 000
V0 =
(1.038 75)24
= 9637.10
The correct answer is B.
r 30 40 000
(1 + 100 ) = 25 000
r 30
100 )
r 30 40
(1 + 100 ) = 25
24
4 a n=3×2
=6
V0 = $2000
Vn = $3000
r 30 × 30
40 30
=
(1 + 100 )
( 25 )
1
1+
1
r
40 30
=
100 ( 25 )
1
r
40 30
=
−1
100 ( 25 )
r
= 0.015 79
100
r = 1.58% per month
Annual rate = 0.015 79 × 12
= 18.95%
1
b n = 4 × 26
2
= 117 fortnights
1
∴ 3000 = 2000 (1 +
r 6 3000
(1 + 100 ) = 2000
r 6×6
3 6
=
(1 + 100 )
(2)
1
1+
1
1
4 36
(3)
4
r
=
100 ( 3 )
r 16
100 )
r 16 × 16
15 16
=
(1 + 100 )
( 12 )
4 36
r
=
−1
100 ( 3 )
r
= 0.008 023 183
100
r = 0.802 318 3% per month
r ≈ 9.63% per annum
3
b n=4×4
= 16 quarters
1
r
3 6
=
100 ( 2 )
1
r
3 6
=
−1
100 ( 2 )
r
∴
= 0.0699
100
1
r 6
100 )
V0 = $43 000
Vn = $60 000
∴ (1 +
60 000 = 43 000 (1 +
r 117 60 000
=
100 )
43 000
r 117
100 )
P df_Fol i o: 110
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.5
r 117 60
(1 + 100 ) = 43
r 117× 117
60 117
=
(1 + 100 )
( 43 )
1
1
1+
r
60 117
=
100 ( 43 )
1
r
60 117
=
−1
100 ( 43 )
1
r
= 0.002 851
100
r = 0.2851% per fortnight
Annual rate = 0.2851 × 26
= 7.41%
c n = 2 × 52
= 104 weeks
V0 = $1400
Vn = $1950
∴ 1950 = 1400 (1 +
r 104 1950
(1 + 100 ) = 1400
r 104
100 )
r 104 195
(1 + 100 ) = 140
(1 +
r 104× 104
195 104
=
( 140 )
100 )
1
1
r
195 104
1+
=
100 ( 140 )
1
r
195 104
=
−1
100 ( 140 )
1
r
= 0.003 191
100
r = 0.3191% per week
Annual rate = 0.3191 × 52
= 16.59%
6 n=4×4
= 16 quarters
V0 = $2300
Vn = $3200
r 16
3200 = 2300 (1 +
100 )
r 16 3200
(1 + 100 ) = 2300
r 16 × 16
32 16
=
(1 + 100 )
( 23 )
1
1+
1
r
32 16
=
(
100
23 )
1
r
32 16
=
−1
100 ( 23 )
r
= 0.020 85
100
r = 2.085% per quarter
Annual rate = 2.085 × 4
= 8.34%
The correct answer is C.
1
P df_Fol i o: 111
111
7 Using CAS:
N∶ unknown
I%∶ 7
PV∶ − 3000
PMT∶ 0
FV∶ − 4500
P/Y∶ 1
C/Y∶ 1
n = 5.992 805 314 years
n ≈ 6 years
8 Using CAS:
N∶ unknown
I%∶ 7.5
PV∶ − 7300
PMT∶ 0
FV∶ 10 000
P/Y∶ 1
C/Y∶ 1
n = 4.351 602 128 years
n ≈ 5 years
9 Using CAS:
N∶ unknown
I%∶ 9
PV∶ − 4700
PMT∶ 0
FV∶ 6100
P/Y∶ 4
C/Y∶ 4
n = 11.717 713 58 quarters
n ≈ 12 quarters
n ≈ 3 years
10 Using CAS:
N∶ unknown
I%∶ 15
PV∶ − 3800
PMT∶ 0
FV∶ 6300
P/Y∶ 4
C/Y∶ 4
n = 13.732 518 49 quarters
n ≈ 14 quarters
1
n ≈ 3 years
2
11 a Using CAS:
N∶ unknown
I%∶ 8
PV∶ − 2000
FV∶ 3173.75
P/Y∶ 1
C/Y∶ 1
n = 6.000 000 554 years
n ≃ 6 years
Note: As the required time is almost exactly 6 years, in this
instance we round down instead of up.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
112
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.5
b Using CAS:
N∶ unknown
I%∶ 6
PV∶ − 9250
FV∶ 16 565.34
P/Y∶ 1
C/Y∶ 1
n = 9.999 998 764
n ≈ 10 years
12 a Using CAS:
N∶ unknown
I%∶ 7
PV∶ − 850
FV∶ 1000
P/Y∶ 1
C/Y∶ 1
n = 2.402 042 209 years
n ≈ 3 years
b Using CAS:
N∶ unknown
I%∶ 13.25
PV∶ − 12 000
FV∶ 20 500
P/Y∶ 1
C/Y∶ 1
n = 4.303 854 847 years
n ≈ 5 years
n
3
13 a 2100 = 1200 1 +
(
100 )
2100 = 1200(1.03)n
2100
1.03n =
1200
1.03n = 1.75
log10 (1.03)n = log10 (1.75)
n log10 (1.03) = log10 (1.75)
log10 (1.75)
n=
log10 (1.03)
n = 18.9 half years
Required time = 19 half years
1
= 9 years
2
n
2.5
b 13 500 = 8300 1 +
(
100 )
13 500 = 8300(1.025)n
13 500
1.025n =
8300
1.025n = 1.6265
log10 (1.025)n = log10 (1.6265)
n log10 (1.025)n = log10 (1.6265)
log10 (1.6265)
n=
log10 (1.025)
n = 19.75 quarters
Required time = 20 quarters
= 5 years
1
c 16 900 = 9600 1 +
(
100 )
16 900 = 9600(1.01)n
16 900
1.01n =
9600
1.01n = 1.7604
n
log10 (1.01)n = log10 (1.7604)
n log10 (1.01) = log10 (1.7604)
log10 (1.7604)
n=
log10 (1.01)
n = 56.84 months
Required time = 57 months
3
= 4 years
4
n
0.25
14 a 24 000 = 16 750 1 +
(
100 )
24 000 = 16 750(1.0025)n
24 000
1.0025n =
16 750
1.0025n = 1.4328
log10 (1.0025)n = log10 (1.4328)
n log10 (1.0025) = log10 (1.4328)
log10 (1.4328)
n=
log10 (1.0025)
n = 144.03 fortnights
Required time = 145 fortnights
= 5 years 15 fortnights
8
b r=
4
= 2% per quarter
V0 = $7800
Vn = $10 000
2
10 000 = 7800 1 +
(
100 )
10 000 = 7800(1.02)n
10 000
1.02n =
7800
1.02n = 1.2821
log10 (1.02)n = log10 (1.2821)
n log10 (1.02) = log10 (1.2821)
log10 (1.2821)
n=
log10 (1.02)
n = 12.55
Required time = 13 quarters
1
= 3 years
4
11
c r=
4
= 2.75% per quarter
n
V0 = $800
Vn = $1900
P df_Fol i o: 112
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.5
2.75
1900 = 800 1 +
(
100 )
1900 = 800(1.0275)n
1900
1.0275n =
800
1.0275n = 2.375
log10 (1.0275)n = log10 (2.375)
n log10 (1.0275) = log10 (2.375)
log10 (2.375)
n=
log10 (1.0275)
0.6667
1700 = 1400 1 +
(
100 )
1700 = 1400(1.006 667)n
1700
1.006 667n =
1400
1.006 667n = 1.2143
log10 (1.006 667)n = log10 (1.2143)
n log10 (1.006 667) = log10 (1.2143)
log10 (1.2143)
n=
log10 (1.006 667)
n = 29.22 months
n
n = 31.88 quarters
Required time = 32 quarters
= 8 years
10.4
15 r =
4
= 2.6% per quarter
V0 = $1600
Vn = $2200
2.6
2200 = 1600 1 +
(
100 )
2200 = 1600(1.026)n
2200 = 1600(1.026)n
2200
1.026n =
1600
1.026n = 1.375
log10 (1.026)n = log10 (1.375)
n log10 (1.026) = log10 (1.375)
log10 (1.375)
n=
log10 (1.026)
n = 12.41 quarters
∴ Required time = 13 quarters
1
= 3 years
4
6.5
16 r =
2
= 3.25% per half–yearly
V0 = $6470
Vn = $9000
Required time = 30 months
1
= 2 years
2
9.6
b r=
26
= 0.3692% per fortnight
V0 = $8000
Vn = $8000 + $4400
= $12 400
n
0.3692
∴ 12 400 = 8000 1 +
(
100 )
12 400
n
1.003 692 =
8000
1.003 692n = 1.55
log10 (1.003 692)n = log10 (1.55)
n log10 (1.003 692) = log10 (1.55)
log10 (1.55)
n=
log10 (1.003 692)
n = 118.92 fortnights
Required time = 119 fortnights
= 4 years 15 fortnights
8
18 a r =
4
= 2% per quarter
3.25
∴ 9000 = 6470 1 +
(
100 )
9000 = 6470(1.0325)n
9000
1.0325n =
6470
1.0325n = 1.3910
log10 (1.0325)n = log10 (1.3910)
n log10 (1.0325) = log10 (1.3910)
log10 (1.3910)
n=
log10 (1.0325)
n = 10.32 half years
∴ Required periods = 11 half years
The correct answer is B.
17 a r =
8
12
= 0.6667% per month
P df_Fol i o: 113
V0 = $1400
Vn = $1400 + $300
= $1700
n
n
V0 = $11 000
Vn = $15 000
2
15 000 = 11 000 1 +
(
100 )
15 000 = 11 000(1.02)n
15 000
1.02n =
11 000
1.02n = 1.3636
log10 (1.02)n = log10 (1.3636)
n log10 (1.02) = log10 (1.3636)
log10 (1.3636)
n=
log10 (1.02)
n = 15.66 quarters
Required time = 16 quarters
= 4 years
11
b
r=
4
= 2.75% per quarter
n = 16 quarters
Vn = $15 000
16
2.75
15 000 = V0 1 +
(
100 )
n
n
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
113
114
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.6
15 000 = V0 (1.0275)16
15 000
V0 =
1.027516
V0 = $9718.11
Dawn needs to invest $9718.11.
4 a n = 12
i = 0.08
reff =
=
b n = 12
i = 0.1
1 Solve on CAS
10000 × 1.055n = 20000
n = 12.946
= 13
The correct answer is B.
2 a B1 = 1.003 × B0 = 1.003 × 5000 = 5015
[1 mark]
B2 = 1.003 × B1 = 1.003 × 5015 = 5030.045
B3 = 1.003 × 5030.045 = 5045.14
B4 = 1.003 × 5045.14 = $5060.27
b Monthly interest rate is 0.3%
[1 mark]
c Use the rule Bn+1 = 1.003 × Bn + 50 to get B12 = $5793.
Therefore, Samuel would have $5793 − $5183 = $610
extra money.
[1 mark]
3 The interest rate is the number found in front of Vx .
1.01 means 101% = 100% + 1%, so the interest rate, per
annum, is 1%.
The correct answer is C.
reff =
(
i
−1
n)
reff =
=
1+
2 n=4
i = 0.14
reff =
=
(
1+
1+
4
i
−1
n)
3 n = 12
i = 0.115
reff =
=
(
1+
1+
4
i
−1
n)
n
0.115
−1
(
12 )
= 0.121 25...
= 12.13% (2 d.p.)
1+
(
1+
1+
d n = 12
i = 0.0675
reff =
=
i
−1
n)
n
12
i
−1
n)
n
(
1+
5
1+
12
i
−1
n)
n
0.0675
−1
(
12 )
= 0.069 62...
= 6.96%
12
n = 18
i = 0.1841
n
i
reff = 1 +
−1
(
n)
=
0.1841
−1
(
12 )
= 0.200 45...
= 20.05%
n
0.14
−1
(
4 )
= 0.1475 …
= 14.8% (1 d.p.)
12
0.075
−1
(
12 )
= 0.077 63 …
= 7.76%
n
0.11
−1
(
4 )
= 0.1146 …
= 11.5% (1 d.p.)
=
1+
c n = 12
i = 0.075
6.6 Exercise
1+
n
0.1
−1
(
12 )
= 0.104 71 …
= 10.47%
=
6.6 Nominal and effective annual interest rate
reff =
(
1+
i
−1
n)
0.08
−1
(
12 )
= 0.082 99 …
= 8.30%
6.5 Exam questions
1 n=4
i = 0.11
(
1+
6
1+
n = 52
i = 0.125
reff =
=
(
1+
1+
12
i
−1
n)
n
0.125
−1
(
52 )
= 0.132 97 …
= 13.30%
52
12
P df_Fol i o: 114
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.7
7 Number of compounding periods per year = 12
i = 0.0485
n
i
−1
reff = 1 +
(
n)
0.0485
= 1+
−1
(
12 )
= 0.049 59 …
= 4.96%
The correct answer is A.
8 Number of compounding periods per year = 4
i = 0.096
4
i
reff = 1 +
−1
(
n)
12
0.096
−1
= 1+
(
4 )
= 0.099 512...
= 9.95%
The correct answer is E.
9 Number of compounding periods per year = 12
i = 0.09
n
i
reff = 1 +
−1
(
n)
4
0.09
−1
1+
(
12 )
= 0.093 81
= 9.40%
The effective annual interest rate of 9.3% is the better option
(the compound interest rate has an effective annual interest
rate of 9.38%).
10 Number of compounding periods per year = 12
i = 0.085
n
i
reff = 1 +
−1
(
n)
=
12
=
0.085
1+
−1
(
12 )
= 0.088 391 …
= 8.84%
12
6.6 Exam questions
1 reffective =
1+
8
− 1 × 100%
[(
]
100n )
Go through each option to find which one is not correct.
When interest is charged fortnightly:
26
8
reffectire =
1+
− 1 × 100% = 8.3154 =
100 × 26 )
[(
]
8.32%, which does not match the percentage given.
The correct answer is C.
2 Loan A: Nominal rate: 8%. Effective rate is
n
1 + 365
− 1 = 0.08328 = 8.33%
100 )
(
Loan B: Nominal rate: 8.1%. Effective rate is
8
1+
P df_Fol i o: 115
365
− 1 = 0.08264 = 8.26%
100 )
(
Loan A has the greater effective interest rate, 8.33%, by a
margin of 0.07%.
The correct answer is D.
8.1
2
2
3 Effective rate =
=
(
1+
1+
i
−1
n)
n
0.0585
−1
72 )
(
= 0.060
⇒ Rate = 6.0%
The correct answer is A.
72
6.7 Review
6.7 Exercise
Multiple choice
1 A flat rate of interest
r = 6%
V0 = 5000
6
× 5000 = 300
100
Vn+1 = Vn + 300
The correct answer is E.
V0 × r × n
2 I=
100
∴ V0 = $360
r = 8% p.a.
n = 3 years
360 × 8 × 3
∴ I=
100
= $86.40
Vn = V0 + I
= $360 + $86.40
= $446.40
The total amount received was $446.40.
The correct answer is E.
100 × I
3 r=
V0 × n
∴ I = $1125
V0 = $5000
n = 5 years
100 × 1125
∴ r=
(5000 × 5)
= 4.5% p.a.
The simple interest rate was 4.5% p.a.
The correct answer is B.
100 × I
4 r=
P×T
I = $2000
V0 = $10 000
n = 10 years
100 × 2000
∴r =
(10 000 × 10)
= 2% p.a.
The simple interest rate was 2% p.a.
The correct answer is C.
5 V0 = 2000, Vn+1 = 1.05 Vn
5
1.05 = 1 +
100
∴ 5% is the annual rate of interest.
The correct answer is B.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
115
116
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.7
6 V0 = 100 000
12
= 1% per month
12% p.a. =
12
r
1
R=1+
=1+
= 1.01
100
100
V0 = 100, 000 Vn+1 = 1.01 Vn
The correct answer is A.
7 V0 = $4500
6.4
r=
4
= 1.6
n=5×4
= 20
1.6
∴ Amount Vn = 4500 1 +
(
100 )
= 4500(1.016)20
= $6181.40
The correct answer is A.
1
8 n = 4 × 12
2
= 54
V0 = $1200
Vn = $1750
∴ 1750 = 1200 (1 +
r 54 1750
(1 + 100 ) = 1200
∴ Required time = 113 fortnights
= 4 years 9 fortnights
The correct answer is D.
n
i
−1
10 Effective rate of interest = 1 +
(
n)
=
0.12
(
365 )
= 12.75%
The correct answer is B.
20
r 54
100 )
r 54
(1 + 100 ) = 1.4583
1
r 54
1
54
(1 + 100 ) = 54 (1.4583)
r
1+
= 1.0069
100
r
= 0.0069
100
∴ r = 0.69% per month
= 0.69 × 12
= 8.3% per annum
The correct answer is C.
9 V0 = $850
8
r=
26
= 0.3077
Vn = $1200
0.3077
∴ 1200 = 850 1 +
(
100 )
∴
1+
T
0.3077
1200
=
(
100 )
850
n
0.3077
1+
= 1.4118
(
100 )
n
0.3077
log 1 +
= log 1.4118
(
100 )
0.3077
n log 1 +
= log 1.4118
(
100 )
n
n=
log (1 + 0.3077
100 )
n = 112.25
log 1.4118
1+
365
−1
Short answer
V0 × r × n
11 I =
100
V0 = $270
r = 8% p.a.
n = 4 years
270 × 8 × 4
∴I =
100
= $86.40
Vn = V0 + I
= $270 + $86.40
= $356.40
The amount received at the end of 4 years was $356.40.
100 × I
12 r =
V0 × n
I = $206.65
V0 = $725
n = 3 years
100 × 206.65
∴r =
(725 × 3)
= 9.5% p.a.
The yearly interest rate was 9.5%.
100 × I
13 V0 =
r×n
I = $67.50
3
r = 3 % p.a.
4
1
n = 4 years
2
100 × 67.50
∴ V0 =
3
1
(3 4 × 4 2 )
= $400
The amount invested was $400.
V0 × r × n
14 I =
100
V0 = $1800
1
r = 12 % p.a.
2
n = 4 years
1800 × 12 12 × 4
∴I =
100
= $900
Vn = V0 + I
= $1800 + $900
= $2700
The total return from the bond was $2700.
P df_Fol i o: 116
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.7
15 r =
9.25
2
= 4.625
20
Vn = $5000
n=4×2
=8
4.625
∴ 5000 = V0 1 +
(
100 )
8
5000 = V0 (1.046 25)8
5000
∴ V0 =
1.046 258
V0 = $3482.46
The amount to be invested is $3482.46.
12
16 r =
365
V0 = $950
n = 3 × 365
= 1095
∴ Amount Vn = 950 1 + 365
100 )
(
= $1361.58
12
1095
Interest = $1361.58 − $950
= $411.58
Extended response
V0 × r × n
17 I =
100
V0 = $4000
r = 10% p.a.
1
n = 2 years
2
4000 × 10 × 2 12
∴I =
100
= $1000
Vn = V0 + I
= 4000 + 1000
= $5000
The total amount received was $5000.
18 a V0 = $5400
n=5
r = 12
5400 × 12 × 5
I=
100
= $3240
The total amount at the end of
5 years = $5400 + $3240
= $8640
b
11.8
r=
4
= 2.95
V0 = $5400
n=5×4
= 20
2.95
∴ Amount A = 5400 1 +
(
100 )
= 5400(1.0295)20
= $9658.75
The total amount at the end of 5 years is $9658.75.
11.7
c r=
12
= 0.975
V0 = $5400
n = 5 × 12
= 60
Vn = 5400
1+
0.975
(
100 )
= 5400(1.009 75)60
= $9665.53
The total amount at the end of 5 years is $9665.53.
The option in c is the most productive since it gives the
biggest yield.
V0 rn
19 a I =
100
V0 = $1983.50 from question 1
r = 5.6% p.a.
n = 2 years
1983.50 × 5.6 × 2
∴I =
100
= $222.15
60
Vn = V0 + I
= $1983.50 + $222.15
= $2205.65
The total value of the investment was $2205.65.
b Extra amount needed = $3995.00 − $2205.65
= $1789.35
Time = 2 years
= 52 fortnights
∴ Fortnightly savings needed =
c
Vn = V0 × (1 +
3995 = 1983.50 ×
1789.35
52
= $34.41
r n
100 )
(
1+
5.4
12
100 )
n
Using CAS, n = 155.946 months = 13 years
r n
d Vn = V0 × (1 +
100 )
V8 = 1983.50 ×
= $2056.04
e
(
1+
8
5.4
12
100 )
1
Vn = V0 × 1 +
(
100 )
V0 = 1000 + 2056.04 = 3056.04
3995 = 3056.04 ×
n
1 + 1.2
100 )
(
5.4
n
Using CAS, n = 59.6724 months = 4.9727 years
Therefore, Geoff will have to wait 5 more years.
P df_Fol i o: 117
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
117
TOPIC 6 Modelling compound interest investments and loans using recursion • EXERCISE 6.7
118
20 a R = 1 +
r
= 1.005
100
r
= 0.005
100
r = 0.5 per month
interest rate per annum = 0.5 × 12
= 6% p.a.
b A0 = 6500
Bridie had saved $6500.
r n
c A = A0 (1 +
100 )
= 6500(1.005)20
= $7326.54
6.7 Exam questions
1 a. S1 = 1.001 × 570 000 − 1193 = $569 377
[1 mark]
b. 0.001 × 26 × 100 = 2.6%
[1 mark]
(note that 0.1 × 26 and 0.001 × 2600 were accepted)
3.6
2 The interest rate of 3.6% p.a. is equivalent to
or 0.3% p.m.
12
Shirley has an interest-only loan, so her balance will remain at
$225 000 after she pays each monthly interest charge.
The correct answer is D.
3 a $15 000
[1 mark]
b V0 = 15 000
V1 = 1.04 × 15 000 = $15 600
[1 mark]
V2 = 1.04 × 15 600 = $16 224
VCAA Assessment Report note:
Some students used only the explicit rule Vn = Rn V0 to
write V2 = 1.042 × 15 600 = 16 224, but this was not
using recursion as required.
c There is an increase of 4% each compounding
period.
[1 mark]
d i Vn = 1.04n × 15 000
[1 mark]
ii V10 = 1.0410 × 15 000 = $22 203.66
[1 mark]
VCAA Assessment report note:
Many students only wrote an answer of $22 203.7, which
might arise from a calculator setting that restricts the
number of displayed digits.
Some students appeared to round all currency answers to
the nearest five or ten cents. A common incorrect answer
was $22 203.70, which had clearly been rounded from a
written $22 203.664… This was assessed as a rounding
error.
4 a I=A−P
= 60 000 − 45 500
= $14 450
[1 mark]
b
A = PRn
= P (1 +
r n
100 )
60 000 = 45 500 1 + 4
(
100 )
r
P df_Fol i o: 118
4×4
+ 885
c i Account balance = 60 000 1 + 12
100 )
(
= 60 000(1 + 0.006) + 885
[1 mark]
VCAA Assessment Report note:
Answers equivalent to 0.006 were accepted in the first
7.2
7
1
7
box, commonly 100 or 12 . However, many students
12
100
7
7
wrote only
or .
100 12
A number of students added a power to complete a
formula to find the account balance at the end of
12 months, rather than the first month as required.
ii Using the previous equation and working out each
month’s balance, after 12 months the balance
is $75 443.
[1 mark]
7500 × 8 × 2
= $1200
5 a Interest =
100
Total to be repaid = 1200 + 7500 = $8700
Monthly instalment = 8700 ÷ 24 = $362.50
[1 mark]
VCAA Assessment Report note:
Students often confuse simple (flat) interest with compound
interest calculations. Many students inappropriately treated
this as a reducing balance problem. Students should
understand that the term ‘flat interest rate’ refers to simple
interest.
b Effective rate of interest:
2n
100I
×
re =
Pt
n+1
100 × 1200 2 × 24
re =
×
7500 × 2
24 + 1
re = 15.36%
[1 mark]
c A flat rate is charged on the initial borrowed amount (for
the whole duration of the hire-purchase agreement).
An effective rate is charged on the reduced monthly
balance. Therefore, an effective rate must be higher than
the flat interest rate to result in the same amount
of interest.
[1 mark]
d First, find the rate of depreciation. For reducing balance
depreciation:
r t
V = P × (1 −
100 )
r 1
6375 = 7500 × (1 −
100 )
r = 15% p.a.
After 5 years:
r t
V = P × (1 −
100 )
15
V = 7500 × 1 −
(
100 )
V = 3327.79
[1 mark]
Therefore, the value of the bike after 5 years is $3328.
5
solve on CAS
r = 6.9%
[1 mark]
VCAA Assessment Report note:
A common error was to treat this as a simple interest
question, despite the question stating that there were ‘four
years of compounding interest’.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.2
119
Topic 7 — Modelling reducing balance loans, annuities
and perpetuities using recursion
b 15 000
7.2 Modelling reducing balance loans with
recurrence relations
7.2 Exercise
1 a V0 = 565
r = 0.015
R = 1 + 0.015 = 1.015
b
V0 = a, Vn+1 = RVn ± d
V0 = 565, Vn+1 = 1.015Vn + 50
565
565
565 × 1.015 + 50
623.475
623.475 × 1.015 + 50
682.827
682.827 × 1.015 + 50
743.07
At the beginning of 2021, the number of enrolments was
743 students to the nearest whole number.
c Enrolments at the end of 2019 are 682.827 and at the end of
2018 are 623.475.
682.827 − 623.475 = 59.352
59 new students enrol in 2019.
2 a A0 = 3.5 m3
b
3.5
3.5
3.5 × 0.65 + 1.5
3.954
3.954 × 0.65 + 1.5
4.0699
4.0699
4.1455
After 4 years, there will be 4.15 m3 of sand, correct to
2 decimal places.
c After 3 years, the amount of sand will exceed 4 m3 .
3 The sequence described by the recurrence relation
P0 = 10 000, Pn+1 = 1.03Pn − 250 is an example of
geometric growth and linear decay.
The correct answer is D.
4 a V0 = 15 000
12.5
r=
= 6.25
2
6.25
R=1+
= 1.0625
100
d = 3074.44
V0 = a, Vn+1 = RVn − d
V0 = 15 000, Vn+1 = 1.0625Vn − 3074.44,
where Vn is the balance of the loan, after n half-yearly
payments.
*8
P df_Fol i o: 119
12 863.06
12 863.06 × 1.0625 – 3074.44
10 592.56
10 592.56 × 1.0625 – 3074.44
8180.16
8180.16 × 1.0625 – 3074.44
5616.98 × 1.0625 – 3074.44
2893.60 × 1.0625 – 3074.44
5616.98
2893.60
0.00
After 6 half-yearly payments (3 years) the loan will be
paid off.
5 a V0 = 1500
5.5
r=
= 0.4583...
12
0.4583
= 1.004 583
R=1+
100
d = 200
V0 = a, Vn+1 = RVn + d
V0 = 1500, Vn+1 = 1.004 583Vn + 200
b 1500
1500
1500 × 1.004 583 + 200
1706.88 × 1.004 583 + 200
1914.7 × 1.004 583 + 200
2123.47 × 1.004 583 + 200
2333.21 × 1.004 583 + 200
1706.88
1914.70
2123.47
2333.21
2543.90
After 5 months, the account balance is $2543.90.
9
= 0.75%
12
b Interest for first month = 2400 × 0.0075 = $18
c From the table, the balance of the loan after 3 repayments
= 1453.18 − 478.74 = $974.44
d From the table, the principal reduction of the 4th payment
= 489.64 − 7.31 = $482.33
e Interest = 18 + 14.46 + 10.90 + 7.31 + 3.69 = $54.36
f Final balance at the end of 5 months
= 492.11 − 485.95 = $6.16
5.5
= 0.458 33...%
7 Interest rate per month =
12
Interest for the fourth payment = 24 527.62 × 0.004 583 3
= 112.418 = $112.42
6 a Interest rate per month =
3.775
3.775 × 0.65 + 1.5
15 000
15 000 × 1.0625 – 3074.44
Principal reduction = 271.32 − 112.42 = $158.90
Balance of loan = 24 527.62 − 158.90 = $24 368.72
8 See table at the foot of the page*
8
Interest rate per quarter = = 2%
4
Total interest = 200 + 193.43 + 186.72 = $580.15
Balance of the loan after 3 payments is $8994.02.
Payment number (n)
Payment
Interest
0
1
2
3
0.00
528.71
528.71
528.71
0.00
10 000 × 0.02 = 200
9671.29 × 0.02 = 193.43
9336.01 × 0.02 = 186.72
Principal reduction
Balance of loan
0.00
10 000.00
528.71 − 200 = 328.71
10 000 − 328.71 = 9671.29
528.71 − 193.43 = 335.28 9671.29 − 335.28 = 9336.01
528.71 − 186.72 = 341.99 9336.01 − 341.99 = 8994.02
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
120
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.3
9 a Let Vn be the balance of the loan after n months.
6.5
Interest rate per month =
= 0.542%
12
V0 = 23 000, Vn+1 = 1.005 42Vn − 341.54
b 23 000
23 000
23 000 × 1.00542 – 341.54
22 783.12
22 783.12 × 1.00542 – 341.54
22 565.07
22 565.07 × 1.00542 – 341.54
22 345.83
After her third payment is made, Kali owes $22 345.83.
10 a Let An be the balance of the loan after n months.
12
Interest rate per month =
= 1%
12
A0 = 2000, An+1 = 1.01An − 261
b 2000
2000
2000 × 1.01 – 261
1759
1759 × 1.01 – 261
1515.59
1515.59 × 1.01 – 261
1269.75
1269.75 × 1.01 – 261
1021.44
After the 4 payment, Noah will owe $1021.44.
c 2000
2000
th
2000 × 1.01 – 261
1759 × 1.01 – 261
1515.59 × 1.01 – 261
1269.75 × 1.01 – 261
1021.44 × 1.01 – 261
770.66 × 1.01 – 261
517.36 × 1.01 – 261
261.54 × 1.01 – 261
1759
1515.59
1269.75
1021.44
770.66
517.36
261.54
3.15
After 8 months, Noah will need to pay $3.15 to discharge
his loan.
6
11 a Interest rate per month =
= 0.5%
12
A0 = 18 000, An+1 = 1.005An − 532.90
Using the iterative process on your calculator:
A5 = $15 763.24
9
= 0.75%
b Interest rate per month =
12
A0 = 18 000, An+1 = 1.0075An − 608.04
Using the iterative process on your calculator:
A5 = $15 599.06
12
c Interest rate per month =
= 1%
12
A0 = 18 000, An+1 = 1.01An − 688.66
Using the iterative process on your calculator:
A5 = $15 405.32
13
12 Interest rate per fortnight =
= 0.5%
26
A0 = 8000, An+1 = 1.005An − 124.11
Using the iterative process on your calculator: A4 = $7661.03
The correct answer is A.
16.5
13 Interest rate per month =
= 1.375%
12
A0 = 20 000, An+1 = 1.013 75An − 421.02
Using the iterative process on your calculator:
A8 = $18 774.05
The correct answer is E.
7.2 Exam questions
1 Use the initial value of $300 000 to calculate the interest
of $900.
3.6
× 300 000 = 900
100 × n
3.6
= 0.003
100 × n
3.6 = 0.3n
n = 12
The correct answer is B.
2 Calculate with financial solver on a CAS calculator using
N = 60, I = 3.72, PV = 175 260.56, PMT = −3200 and
PpY = 12. The final value will be –368.116 …
Since the final value is negative, Adam must still
pay this amount.
Therefore, Adam’s final payment will be
3200 + 368.116 ≈ $3568.12
The correct answer is E.
VCAA Assessment Report note:
Students who chose option D did not add the interest that was
required to be paid with the final payment.
3 V1 = 1.003 × (26 000) − 400
= 25 678
V2 = 1.003 × (25 678) − 400
= 25 355.034
V3 = 1.003 × (25 355.034) − 400
= 25 031.0991
V4 = 1.003 × (25 031.0991) − 400
= 24 706.1924
V5 = 1.003 × (24 706.1924) − 400
= 24 380.3110
The correct answer is A.
7.3 Solving reducing balance loan problems using
finance solver
7.3 Exercise
1 a N∶ 60
I%∶ 12
PV∶ 65 000
PMT∶ −715.71
FV∶ unknown
PY∶ 12
CY∶ 12
After 5 years the amount owing is $59 633.49.
b N∶ 120
I%∶ 12
PV∶ 65 000
PMT∶ −715.71
FV∶ unknown
PY∶ 12
CY∶ 12
After 10 years the amount owing is $49 884.16.
P df_Fol i o: 120
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.3
c N∶ 180
I%∶ 12
PV∶ 65 000
PMT∶ −715.71
FV∶ unknown
PY∶ 12
CY∶ 12
After 15 years the amount owing is $32 172.59.
2 a N∶ 104
I%∶ 13
PV∶ 52 000
PMT∶ −303.57
FV∶ unknown
PY∶ 26
CY∶ 26
After 4 years the amount owing is $46 102.98.
b N∶ 208
I%∶ 13
PV∶ 52 000
PMT∶ −303.57
FV∶ unknown
PY∶ 26
CY∶ 26
After 8 years the amount owing is $36 196.88.
c N∶ 312
I%∶ 13
PV∶ 52 000
PMT∶ −303.57
FV∶ unknown
PY∶ 26
CY∶ 26
After 12 years the amount owing is $19 556.12.
3 a N∶ 24
I%∶ 6.8
PV∶ 3500
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The monthly repayments are $156.39.
b Total interest = $156.39 × 24 − $3500
= $253.36
4 a N∶ 24
I%∶ 6.3
PV∶ 2 400
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The monthly repayments are $106.69.
b Total interest = $106.69 × 24 − $2400
= $160.56
P df_Fol i o: 121
5 a N∶ 60
I%∶ 8.1
PV∶ 18 000
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The monthly repayments are $365.84.
b i The 10th repayment
V9 = $15 740.6207
V10 = $15 481.02989
Principal repaid = V9 − V10
= $259.59
Interest paid = 365.84 − 259.59
= $106.25
ii The 50th repayment
V49 = $3865.89
V50 = $3526.14
Principal repaid = V49 − V50
= 3865.89 − 3526.14
= $339.75
Interest paid = 365.84 − 339.75
= $26.09
6 a N∶ 48
I%∶ 6.8
PV∶ 8000
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The monthly repayments are $190.83.
b i V9 = $6660.45
V10 = $6507.36
Principal repaid = V9 − V10
= $6660.45 − $6507.36
= $153.09
Interest = $190.83 − $153.09
= $37.74
ii V39 = $1669.72
V40 = $1488.36
Principal repaid = V39 − V40
= $1669.72 − $1488.36
= $181.36
Interest = $190.83 − $181.36
= $9.47
7 a N∶ unknown
I%∶ 7.6
PV∶ 3500
PMT∶ −206.35
FV∶ 0
PY∶ 12
CY∶ 12
1
∴ To repay loan in full will take 18 repayments or 1
2
years.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
121
122
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.3
b Total interest = 206.35 × 18 − 3500
= $214.30
8 Using CAS:
N∶ unknown
I%∶ 8.7
PV∶ 8000
PMT∶ −368.45
FV∶ −5489.56
PY∶ 12
CY∶ 12
n = 7.886 741 159
n ≈ 8 months
9 Using CAS:
N∶ unknown
I%∶ 6.5
PV∶ 11 000
PMT∶ −409.50
FV∶ −5565.48
PY∶ 12
CY∶ 12
n = 14.952 446 42
n ≈ 15 months
10 N∶ unknown
I%∶ 4.5
PV∶ 2400
PMT∶ −154.82
FV∶ 0
PY∶ 12
CY∶ 12
N = 16 instalments
It will take 16 months to fully pay off the loan of $2400.
11 a Using CAS:
N∶ unknown
I%∶ 11.5
PV∶ 8357.65
PMT∶ −423.82
FV∶ −2450.15
PY∶ 12
CY∶ 12
n = 15.985 845 77
n ≈ 16 months
b Put FV = 0, n = 21.961 723 9, n ≈ 22 months
12 a Using CAS:
N∶ unknown
I%∶ 8.8
PV∶ 6900.86
PMT∶ −273.56
FV∶ −1670.48
PY∶ 12
CY∶ 12
n = 21.725 9970 6
n ≈ 22 months
b Put FV = 0, n = 27.996 627 383, n ≈ 28 months
13 a N∶ 60
I%∶ 7.8
PV∶ 18 000
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The repayment is $363.25.
b Using CAS:
N∶ unknown
I%∶ 7.8
PV∶ 18 000
PMT∶ −390.50
FV∶ 0
PY∶ 12
CY∶ 12
n = 54.966 539 73
n ≈ 55 months
c Interest = $390.50 × 55 − $18 000
= $3477.50
d Interest = $363.25 × 60 − $18 000
= $3795.00
Interest difference = $3795.00 − $3477.50
= $317.50 is saved
14 a N∶ 130
I%∶ 6.5
PV∶ 25 000
PMT∶ unknown
FV∶ 0
PY∶ 26
CY∶ 26
The repayments is $225.49.
b Using CAS:
N∶ unknown
I%∶ 6.5
PV∶ 25 000
PMT∶ −245
FV∶ 0
PY∶ 26
CY∶ 26
n = 117.950 407 73
n ≈ 118 fortnights
c Interest = $245 × 118 − $25 000
= $3910
d Interest = $225.49 × 130 − $25 000
= $4313.70
Interest difference = $4313.70 − $3910
= $403.70 is saved
P df_Fol i o: 122
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.3
15 a After 6 years how much is still owing?
N∶ 24
I%∶ 6.8
PV∶ 21 000
PMT∶ −899.41
FV∶ unknown
PY∶ 4
CY∶ 4
After 6 years there is $5 089.55 still owing.
N∶ unknown
I%∶ 6.8
PV∶ 5089.55
PMT∶ −685.05
FV∶ 0
PY∶ 4
CY∶ 4
n = 8.0096
n ≈ 8 quarters (2 years)
Total term = 6 + 2
= 8 years
b Interest = $899.41 × 24 + $685.05 × 8 − $21 000
= $6066.24
c Interest = $899.41 × 30 − $21 000
= $5982.30
Interest difference = $6066.24 − $5982.30
= $83.94
This will be how much more interest will be paid.
16 a N∶ 24
I%∶ 5.9
PV∶ 17 000
PMT∶ −670.29
FV∶ unknown
PY∶ 4
CY∶ 4
After 6 years the balance of the loan is $5023.16.
Using CAS:
N∶ unknown
I%∶ 5.9
PV∶ 5023.16
PMT∶ −1724.02
FV∶ 0
PY∶ 4
CY∶ 4
n = 3.000 003 8
n ≈ 3 quarters
3
year
(4
)
Total term = 6 +
3
4
3
= 6 years
4
b Interest = $670.29 × 24 + $1724.02 × 3 − $17 000
= $4259.02
P df_Fol i o: 123
c Interest = $670.29 × 32 − $17 000
= $4449.28
123
Interest difference = $4449.28 − $4259.02
= $190.26
This will be saved by increasing payments.
17 a N∶ unknown
I%∶ 9.4
PV∶ 5500
PMT∶ −861.29
FV∶ 0
PY∶ 4
CY∶ 4
N = 6.999 ≈ 7
∴ It will take Aimee
3
7 repayments or 1 years to pay off the loan.
4
b Total interest = 861.29 × 7 − 5500
= $529.03
18 a N∶ unknown
I%∶ 7.8
PV∶ 22 000
PMT∶ −443.98
FV∶ 14 209.88
PY∶ 12
CY∶ 12
N = 24
It took 24 monthly repayments or 2 years to reach current
balance of the loan.
b N∶ unknown
I%∶ 7.8
PV∶ 22 000
PMT∶ −443.98
FV∶ 0
PY∶ 12
CY∶ 12
N = 59.999 ≈ 6
It took 5 years to pay off the loan.
19 a N∶ unknown
I%∶ 10.5
PV∶ 25 000
PMT∶ −537.35
FV∶ −11 586.64
PY∶ 12
CY∶ 12
It will take 3 years for the loan to reach $11 586.64.
b To change FV to 0.
It will take 5 years to fully repay the loan.
20 N∶ unknown
I%∶ 7.9
PV∶ 80 000
PMT∶ −639.84
FV∶ 0
PY∶ 12
CY∶ 12
N = 263.999
≈ 264
264 ÷ 12 = 22
The loan will be paid in full in 22 years.
The correct answer is D.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
124
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.3
21 a i N∶ unknown
I%∶ 10.5
PV∶ 8069.78
PMT∶ −230.43
FV∶ −3822.20
N = 24
It will take 24 monthly repayments or 2 years to reach
$3822.21.
ii N∶ unknown
I%∶ 10.5
PV∶ 3822.20
PMT∶ −230.43
FV∶ 0
PY∶ 12
CY∶ 12
N = 18
1
It will take 3 years to pay the loan in full.
2
b i N∶ unknown
I%∶ 10.5
PV∶ 3226.06
PMT∶ −230.43
FV∶ −1341.23
3
It will take 9 monthly repayments or of a year.
4
ii Change FV to 0.
1
In a further 15 monthly repayments or 1 years, the
4
loan will be repaid in full.
22 a N∶ 240
I%∶ 6.9
PV∶ 50 000
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
PMT = −384.65
The monthly repayment is $384.65.
b Using the Finance Solver:
(FV) Repayment
($) amount
Term of loan
‘N’ value
i 577.97
10 years
120
ii 486.33
13 years
156
iii 361.85
23 years
276
iv 352.90
1
24 years
2
294
c Sample calculation: Total interest paid = d × n − V0
i.e. (i) d = $577.97, N = 120, V0 = $50 000
I = 577.97 × 120 − 50 000
= $19 356.40
Sample calculation: Difference in interest from original
offer
Original interest = 384.65 × 240 − 50 000
= $42 316
Difference = 42 316 − 19 356.4
= $22 959.60
$19 356.40
i
c
d
23 a
$22 959.60
less
$25 867.48
ii
$16 448.52
less
$49 870.60
iii
$7554.60
more
i To calculate the amount owing after 5 years:
N∶ 20
I%∶ 8.6
PV∶ 20 000
PMT∶ −750.48
FV∶ unknown
PY∶ 4
CY∶ 4
FV = −12 095.80
After 5 years Jack owed $12 095.80.
Change PV to 12 095.80, FV to 0 and PMT to
−901.48, then N = 16
20 + 16 = 36 quarters or 9 years.
ii Total interest paid
= d × n − V0
= 750.48 × 20 + 901.48 × 16 − 20 000
= $9433.28
iii Difference = original loan interest − total interest
Original loan = 750.48 × 40 − 20 000
= $10 019.20
Difference = 10 019.20 − 9433.28
= $585.92
b
i N∶ unknown
I%∶ 8.6
PV∶ 12 095.80
PMT∶ −1154.34
FV∶ 0
PY∶ 4
CY∶ 4
N = 12
20 + 12 = 32
8 years.
ii Total interest paid
= d × n − V0
= 750.48 × 20 + 1154.34 × 12 − 20 000
= $8861.68
iii Difference = original loan interest − total interest
= 10 019.20 − 8861.68
= $1157.52
P df_Fol i o: 124
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
$53 752.60
iv
$11 436.60
more
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
24 N∶ 16
I%∶ 9.5
PV∶ 25 000
PMT∶ −975.06
FV∶ unknown
PY∶ 4
CY∶ 4
After 4 years balance of loan is $17 682.01. Change PV to
17 682.01, PMT to −1167.17 and FV to 0.
N = 19
16 + 19 = 35
35
= 8.75 ≈ 9 years.
4
The correct answer is D.
25 Total interest = 975.06 × 16 + 1167.17 × 19 − 25 000
= $12 777.19
The correct answer is E.
26 a N∶ 24
I%∶ 9.2
PV∶ 26 000
PMT∶ −383.61
FV∶ unknown
PY∶ 12
CY∶ 12
The balance of the loan is $21 164.60.
b N∶ 60
I%∶ 9.2
PV∶ 21 164.60
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
The monthly repayment is $441.40.
7.3 Exam questions
1 Use Finance Solver:
N∶ 60
I(%)∶ 2.8
PV∶ −12 000
PMT∶ ??
FV∶ 25 000
PpY∶ 12
CpY∶ 12
Therefore, the minimum value of the payment that Joanna
needs to make is $174.11.
The correct answer is B.
9%
2 P = $8400, n = 24 quarters, r =
= 2.25%,
4
2.25
R=1+
= 1.0225
100
Substitute into the annuities formula to find the regular
quarterly repayments:
Q=
PRn (R − 1)
Rn − 1
125
(8400)(1.0225)24 (1.0225 − 1)
1.022524 − 1
= $456.793 9228
3 years means 12 quarterly payments, so
balance remaining after 3 years (12 quarterly payments)
= $4757.41 (using the finance solver).
Balance reduced = 8400 − 4757.41
= 3642.59
=
Percentage reduced by = (3642.59/8400) × 100
= 43.4%
The correct answer is C.
3 Via TVM solver:
N = 240
I = 6.95
PV = 90 000
PMT = ?
FV = 0
PpY = 12
CpY = 12
Therefore, monthly payment = $695.09 ≈ $695.
The correct answer is C.
7.4 The effect of rate and repayment changes on
reducing balance loans
7.4 Exercise
1 a i Using CAS:
N∶ unknown
I%∶ 8.3
PV∶ 18 000
PMT∶ −369.60
FV∶ 0
PY∶ 12
CY∶ 12
n = 59.591 413 76
n ≈ 60 months
Term of loan = 5 years
ii Using CAS:
N∶ 59
I%∶ 8.3
PV∶ 18 000
PMT∶ −309.48
FV∶ unknown
PY∶ 12
CY∶ 12
FV = −217.37
Prior to the last payment owe∶ $217.37
P df_Fol i o: 125
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
126
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
b i Using CAS:
N∶ unknown
I%∶ 8.3
PV∶ 18 000
PMT∶ −170.58
FV∶ 0
PY∶ 26
CY∶ 26
n = 128.879 837
n ≈ 129 fortnights
Term of loan = 4 years 25 fortnights
ii Using CAS:
N∶ 128
I%∶ 8.3
PV∶ 18 000
PMT∶ −170.58
FV∶ unknown
PY∶ 26
CY∶ 26
FV = −149.63
Prior to the last payment owe∶ $149.63
c i Quarterly: interest = $1108.80 × 20 − $18 000
= $4176
8.3
ii Monthly: r =
12
= 0.6917%
V59 = $217.37
Interest on V59 = 0.00 6917 × 217.37
= $1.50
Final payment = $217.37 + $1.50
= $218.87
Total interest = 369.60 × 59 + 218.87 − 18 000
= $4025.27
8.3
iii Fortnightly: r =
26
= 0.3912%
V128 = $149.63
Interest on V128 = 0.00 3192 × 149.63
= $0.48
Final payment = $149.63 + $0.48
= $150.11
Total interest = 170.58 × 128 + 150.11 − 18 000
= $3984.35
2 a i Using CAS:
N∶ unknown
I%∶ 7.2
PV∶ 25 000
PMT∶ −430.67
FV∶ 0
PY∶ 12
CY∶ 12
n = 71.574 290 09
n ≈ 72 months
Term of loan = 6 years
ii Using CAS:
N∶ 71
I%∶ 7.2
PV∶ 25 000
PMT∶ −430.67
FV∶ unknown
PY∶ 12
CY∶ 12
FV = −246.167 390 3
Prior to the last payment owe∶ $246.17
b i Using CAS:
N∶ unknown
I%∶ 7.2
PV∶ 25 000
PMT∶ −198.77
FV∶ 0
PY∶ 26
CY∶ 26
n = 154.828 978 7
n ≈ 155 fortnights
Term of loan = 5 years 25 fortnights
ii Using CAS:
N∶ 154
I%∶ 7.2
PV∶ 25 000
PMT∶ −198.77
FV∶ unknown
PY∶ 26
CY∶ 26
FV = −164.359 906 3
Prior to the last payment owe∶ $164.36
c i Quarterly: interest = $1292.02 × 24 − $25 000
= $6008.48
7.2
ii Monthly: r =
12
= 0.6%
V71 = $246.17
Interest on V71 = 0.006 × 246.17
= $1.48
Final repayment = $246.17 + $1.48
= $247.65
Total interest = 430.67 × 71 + 247.65 − 25 000
= $5825.22
7.2
iii Fortnightly: r =
26
= 0.2769%
V154 = $164.36
Interest on V154 = 0.002 769 × 164.36
= $0.46
Final repayment = $164.36 + $0.46
= $164.82
Total interest = 198.77 × 154 + 164.82 − 25 000
= $5775.40
P df_Fol i o: 126
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
3 a Total interest = 424.94 × 60 − 20 000
= $5496.40
b i Using CAS:
N∶ unknown
I%∶ 11
PV∶ 20 000
PMT∶ −424.94
FV∶ 0
PY∶ 12
CY∶ 12
n = 61.878 469 35
n ≈ 62 months
Term = 5 years and 2 months
ii Using CAS:
N∶ 61
I%∶ 11
PV∶ 20 000
PMT∶ −424.94
FV∶ unknown
PY∶ 12
CY∶ 12
FV = −370.18
Amount owing after 61 payments is∶ $370.18
11
r=
12
= 0.917%
Interest on final payment = 0.00 917 × 370.18
= $3.39
Final repayment = 370.18 + 3.39
= $373.57
Total interest paid = 424.94 × 61 + 373.57 − 20 000
= $6294.91
4 a Total interest = 402.81 × 48 − 16 500
= $2834.88
b i Using CAS:
N∶ unknown
I%∶ 9
PV∶ 16 500
PMT∶ −402.81
FV∶ 0
PY∶ 12
CY∶ 12
n = 49.121 713 78
n ≈ 50 months
Term = 4 years and 2 months
ii Using CAS:
N∶ 49
I%∶ 9
PV∶ 16 500
P df_Fol i o: 127
PMT∶ −402.81
FV∶ unknown
PY∶ 12
CY∶ 12
FV = −48.82
127
Amount owing after 49 payments is∶ $48.82
9
r=
12
= 0.75%
Interest on final payment = 0.0075 × 48.82
= $0.37
Final payment = 48.82 + 0.37 = 49.19
49 × 402.81 = 19 737.69
19 737.69 + 49.19 − 16 500
= $3286.88
Total interest = $3286.88
5 a Using CAS:
N∶ 16
I%∶ 8
PV∶ 32 000
PMT∶ −1169.78
FV∶ unknown
PY∶ 4
CY∶ 4
FV = −22 125.28
Amount owing is∶ $22 125.28
b New rate = 9%
Using CAS:
N∶ unknown
I%∶ 9
PV∶ 22 125.28
PMT∶ −1169.78
FV∶ 0
PY∶ 4
CY∶ 4
n = 24.914 840 61
n ≈ 25 quarters
1
Term = 4 years + 6 years
4
1
= 10 years
4
c Changing N = 24, find FV, FV = −1047.60
9
r=
4
= 2.25%
Interest on final payment = 0.0225 × 1047.60
= $23.57
Final payment = 1047.60 + 23.57
= $1071.17
For the rate change:
Total interest = 1169.78 × 40 + 1071.17 − 32 000
= $15 862.37
For the rate of 8% only, total interest = 1169.78 × 40 − 32 000
= $14 792
Interest difference = $15 862.37 − $14 792
= $1070.37
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
128
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
6 a Using CAS:
N∶ 36
I%∶ 6.5
PV∶ 28 000
PMT∶ −374.81
FV∶ unknown
PY∶ 12
CY∶ 12
FV = −19 156.45
Amount owing after 3 years is∶ $ 19 156.45
b New rate = 7.5%
Using CAS:
N∶ unknown
I%∶ 7.5
PV∶ 19 156.45
PMT∶ −374.81
FV∶ 0
PY∶ 12
CY∶ 12
n = 61.765 566 31
n ≈ 62 months
Total term = 3 years + 62 months
= 8 years, 2 months
c Changing N = 60, find FV, FV = −656.08
7.5
r=
12
= 0.625%
Interest on final payment = 0.00625 × 656.08
= $4.10
Final payment = 656.08 + 4.10
= $660.18
For the rate change:
Total interest = 374.81 × 96 + 660.18 − 28 000
= $8641.94
For the rate of 6.5% only, total interest
= 374.81 × 96 − 28 000
= $7981.76
Interest difference = $8641.94 − $7981.76
= $660.18
7 a V0 = $30 000, r = 5.8%, n = 1 month =
I=
V0 r n
100
1
30 000 × 5.8 × 12
100
= $145
=
1
year
12
b Capital gain = selling price − purchase price
= 35 000 − 30 000
= $5000
Interest charged = 145 × 36
= $5220
Profit = $5000 − $5220
= −$220
Jodie made a loss of $220.
8 a V0 = $50 000, r = 7.5%, n = 1 month =
I=
V0 rn
100
1
year
12
1
50 000 × 7.5 × 12
100
= $312.50
=
b Capital gain = selling price − purchase price
= 63 000 − 50 000
= $13 000
Interest charged = 312.50 × 36
= $11 250
Profit = $13 000 − $11 250
= $1750
9 a Using CAS:
N∶ 4
I%∶ 7.5
PV∶ 2000
PMT∶ unknown
FV∶ 0
PY∶ 4
CY∶ 4
Quarterly payment = $523.66
b Complete the table as shown:
Column 3 is the quarterly interest due, which is calculated
by multiplying column 2 by 0.018 75.
Column 4 is the quarterly payment (found above).
Column 5 is calculated by calculating
column 2 + column 3 − column 4.
Principal Interest
Loan
Payment outstanding
due
Payment outstanding
no.
($)
($)
($)
($)
1
2
3
4
2000
1513.84
1018.56
514.00
37.50
28.38
19.10
9.64
523.66
523.66
523.66
523.66
1513.84
1018.56
514.00
−0.02
c $514.00 is remaining after the third payment has been
made.
d Total payment = $523.66 + $523.66 + $523.66
+ $523.66 − $0.02
= $2094.62
e Total interest = $2094.62 − $2000
= $94.62
10 a Using CAS:
N∶ 6
I%∶ 6
PV∶ 3750
PMT∶ unknown
FV∶ 0
PY∶ 6
CY∶ 6
Bi-monthly payment = $647.06
P df_Fol i o: 128
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
b Complete the table as shown:
Column 3 is the bi-monthly interest due, which is calculated
by multiplying column 2 by 0.01.
Column 4 is the bi-monthly payment (found above).
Column 5 is calculated by calculating
column 2 + column 3 − column 4.
Principal Interest
Loan
Payment outstanding
Payment outstanding
due
no.
($)
($)
($)
($)
1
2
3
4
5
6
3750
3140.44
2524.78
1902.97
1274.94
640.63
37.50
31.40
25.25
19.03
12.75
6.41
647.06
647.06
647.06
647.06
647.06
647.06
3140.44
2524.78
1902.97
1274.94
640.63
−0.02
c Using CAS:
N∶ 12
I%∶ 6
PV∶ 3750
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
Quarterly payment = $322.75
d Complete the table as shown:
Column 3 is the monthly interest due, which is calculated
by multiplying column 2 by 0.005.
Column 4 is the monthly payment (found above).
Column 5 is calculated by calculating
column 2 + column 3 − column 4.
Principal Interest
Loan
Payment outstanding
due
Payment outstanding
no.
($)
($)
($)
($)
1
2
3
4
5
6
7
8
9
10
11
12
P df_Fol i o: 129
3750
3446.00
3140.48
2833.43
2524.85
2214.71
1903.05
1589.81
1275.01
958.64
640.68
321.13
18.75
17.23
15.70
14.17
12.62
11.07
9.52
7.95
6.38
4.79
3.20
1.61
322.75
322.75
322.75
322.75
322.75
322.75
322.75
322.75
322.75
322.75
322.75
322.75
3446.00
3140.48
2833.43
2524.85
2214.71
1903.05
1589.81
1275.01
958.64
640.68
321.13
−0.01
e Sum column 3 in both tables.
Monthly: $122.99, bi-monthly: $132.34
The interest paid on the monthly payments is almost $10
less than the interest paid on the bi-monthly payments.
11 a Monthly:
Using CAS:
Trial N = 119
Amount owing = $102.61
Term of the loan = 10 years.
129
b Fortnightly:
Using CAS:
Trial N = 257
Amount owing = $185.59
258
Term of the loan =
26
= 9.923076923 years
= 9 years 24 fortnights
12
i Interest half-yearly = 3101.48 × 10 − 25 000
= $6014.8
ii Interest quarterly
= 1550.74 × 19 +
= $5707.11
iii Interest monthly
= 516.91 × 58 +
(
(
= $5496.63
iv Interest fortnightly
= 238.58 × 127 +
= $5440.38
1+
1+
(
8.25
4
100
8.25
1+
12
100
)
8.25
26
100
)
× 1217.93 − 25 000
× 512.33 − 25 000
)
× 140.27 − 25 000
Frequency
Total interest ($)
a
Half-yearly
6014.80
b
Quarterly
5707.11
c
Monthly
5496.63
d
Fortnightly
5440.38
13 R = 1 +
Saving ($)
$307.69
saving on
half-yearly
—
$518.17
saving on
half-yearly
$574.42
saving on
half-yearly
9.23
26
100
= 1.0035769
N∶ unknown
I%∶ 9.3
PV∶ 65 000
PMT∶ −309.66
FV∶ 0
PY∶ 26
CY∶ 26
N = 389.182
389.182
Term of loan =
= 14.969
26
14 years and 11.63 months or 15 years.
The correct answer is C.
14 The method for question 14 is the same as that for
question 13.
Fortnightly interest:
I = 389 × 309.66 + 1.003576923 × 56.53 − 65 000
= $55 514.47
Interest monthly, R = 1.00775
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
130
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.4
If Betty paid monthly:
N∶ unknown
I%∶ 9.3
PV∶ 65 000
PMT∶ −670.92
FV∶ 0
PY∶ 12
CY∶ 12
N = 180.002
When N = 180, the balance is $1.43.
I = 1.43 × 1.00775 + 670.92 × 180 − 5000
= 557 67.04108
Savings = $252.57
The correct answer is B.
15 a Total interest paid = d × n − V0
= 506.91 × 5 × 12 − 25 000
= $5414.60
The method for parts b and c is the same. Using part b as
example:
b i R=1+
P df_Fol i o: 130
9
12
100
= 1.0075
N∶ unknown
I%∶ 9
PV∶ 25 000
PMT∶ −506.91
FV∶ 0
PY∶ 12
CY∶ 12
N = 61.8
61.8
= 5.15097
Term of loan =
12
Term of loan is 5 years and 2 months.
ii Total interest paid = 506.91 × 61 + 408.68
× 1.0075 − 25 000
= $6333.26
c i 5 years, 4 months
ii $7331.64
16 Term of loan
N∶ 50
I%∶ 12
PV∶ 28 000
PMT∶ −401.72
FV∶ unknown
PY∶ 12
CY∶ 12
After 50 payments, the loan was $20 153.54.
N∶ unknown
I%∶ 10.75
PV∶ 20 153.54
PMT∶ −401.72
FV∶ 0
PY∶ 12
CY∶ 12
N = 67
117
Term = 50 + 67 =
12
= 9 years 9 months
The correct answer is A.
17 Total interest would be = 401.72 × (50 + 66) + 364.91
× 1.008 958 333 − 28 000
= $18 967.70
The correct answer is C.
V0 r n
18 a I =
100
140 000 × 10.8 × 14
100
= $3780
The quarterly repayment amount is $3780.
b Capital gain = selling price − purchase price
= $152 000 − $140 000
= $12 000
=
Total interest charged = repayment × number of payments
= $3780 × 4
= $15 120
Profit = capital gain − loan cost
= $12 000 − $15 120
= $3 120
In 1 year, they will make a loss of $3120.
19 Using CAS:
N∶ 26
I%∶ 6.79
PV∶ −210 000
PMT∶ unknown
FV∶ 210 000
PY∶ 26
CY∶ 26
Payment = $548.42
The correct answer is A.
7.4 Exam questions
1 First, use finance solver to find the value of the payment if it
is paid off over 20 years:
N∶ 240
I(%)∶ 3.14
PV∶ 400 000
PMT∶ ??
FV∶ 0
PpY∶ 12
CpY∶ 12
The monthly payment will be $2246.53.
However, Bob makes only interest-only payments for 2 years,
which means that at the end of 2 years he still owes $400 000.
Use finance solver a second time to calculate the new interest
rate for the final 18 years of the loan.
N∶ 216
I(%)∶ ??
PV∶ 400 000
PMT∶ −2246.53
FV∶ o
PpY∶ 12
CpY∶ 12
The interest rate will be 2.21%.
The correct answer is B.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.5
2 Use Finance Solver on CAS:
N∶ 10 years 12 months = 120
I (%) ∶ 4.35 p.a.
PV∶ 245 000
PMT∶ −1800
FV ∶ −108 219.1611158
PpY/CpY∶ 12
After 10 years, Xavier still owes $108 219.16
N∶ 5 years 12 months = 60
I(%) ∶ 4.142758985... p.a.
PV∶ 108 219.16
PMT∶ −2000
FV∶ 0
PpY/CpY∶ 12
To be paid off in 5 more years, the annual interest rate is
closest to 4.1%.
VCAA Assessment Report note:
Understanding of the sign convention for TVM use is very
important, as is the careful tracking of values used in
subsequent calculations.
The correct answer is B.
3 Using Finance Solver in CAS, first find the interest rate per
annum.
Enter N = 1, PV = 300 000, PMT = 2500, FV = −299 000,
PpY = 12, CpY = 12.
Solving for I gives I = 6%.
Check whether option A is correct:
Enter N = 2, PV = 300 000, PMT = 2500, PpY = 12, .
CpY = 12
Solving for FV gives FV = −297 995.
Therefore, option A (‘After 2 months, $297 995 is still owing
on the loan’) is correct.
VCAA Assessment Report note:
In this question, students needed to test the truth of five
statements relating to the repayment of a reducing balance
loan Many students struggled with this question The key to
answering this question was to use a TVM to work out the
interest rate applying to the loan. This knowledge, with the
aid of a TVM solver, could then be used to test the truth of
each of the five statements.
The correct answer is A.
7.5 Annuities and perpetuities
7.5 Exercise
1 a V0 = 12 500
6
r = = 1.5%
4
1.5
R=1+
= 1.015
100
*1d
P df_Fol i o: 131
b
131
d = 3243
V0 = a, Vn+1 = RVn − d
V0 = 12500, Vn+1 = 1.015Vn − 3243
12 500
12 500
12 500 × 1.015 – 3243
9444.50
9444.50 × 1.015 – 3243
6343.17
6343.17 × 1.015 – 3243
3195.32
3195.32 × 1.015 – 3243
0.25
After 1 year the balance of the annuity is 25 cents.
c Abdo’s last payment should be increased by 25 cents to
$3143.25.
d See table at the foot of the page*
2
20 000
20 000 × 1.004583 – 3500
20 000
16 591.66
16 591.66 × 1.004583 – 3500
13 167.70
13 167.70 × 1.004583 – 3500
9728.05
9728.05 × 1.004583 – 3500
6272.63
After 5 months, the balance of the annuity is $2801.38.
The correct answer is D.
3 a N∶ = ?
I%∶ = 9.5
PV∶ = −160 000
PMT∶ = 6000
FV∶ = 0
PpY∶ = 12
CpY∶ = 12
PMTAt∶ = END
N = 30 months
The annuity will last 30 months.
b N∶ = 360
I%∶ = 9.5
PV∶ = −160 000
PMT∶ = ?
FV∶ = 0
PpY∶ = 12
CpY∶ = 12
PMTAt∶ = END
PMT = $1345.37
The payment should be $1345.37.
4 N∶ = ?
I%∶ = 4
PV∶ = −100 000
PMT∶ = 3000
6272.63 × 1.004583 – 3500
2801.38
Payment
number (n)
Payment
Interest
Principal reduction
Balance of loan
0
1
2
3
4
0.00
3243
3243
3243
D = 3243.25
0.00
A = 187.50
141.67
95.15
47.93
0.00
3055.50
B = 3101.33
3147.85
3195.07
12500
9444.50
6343.17
C = 3195.32
0
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
132
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.5
FV∶ = 0
PpY∶ = 4
CpY∶ = 4
PMTAt∶ = END
The annuity will last for 42 months.
The correct answer is A.
v0 r
5 d=
100
150 000 × 5
=
100
= $7500
The correct answer is B.
V0 r
6 d=
100
95 000 × 3.8
=
100
= $3610
The correct answer is D.
V0 r
7 a d=
100
75 000 × 4.8
=
100
= $3600
100 × d
b r=
V0
100 × 4800
=
75 000
= 6.4% p.a.
c Using CAS:
N∶ 1
I%∶ 4.8
PV∶ −75 000
PMT∶ unknown
FV∶ 75 000
PY∶ 1
CY∶ 12
PMT = $3680.27
d Using CAS:
N∶ 1
I%∶ unknown
PV∶ −75 000
PMT∶ 1200
P df_Fol i o: 132
FV∶ 75 000
PY∶ 4
CY∶ 12
I = 6.37%
V0 r
8 a d=
100
125 000 × 5.7
=
100
= $7125
100 × d
b r=
V0
100 × 8000
=
125 000
= 6.4% p.a.
c Using CAS:
N∶ 1
I%∶ 5.7
PV∶ −125 000
PMT∶ unknown
FV∶ 125 000
PY∶ 1
CY∶ 12
PMT = $7314.12
d Using CAS:
N∶ 1
I%∶ unknown
PV∶ −125 000
PMT∶ 2000
FV∶ 125 000
PY∶ 4
CY∶ 12
I = 6.37%
6.25
9 R=
4
= 1.5625% per term
100 × d
V0 =
r
100 × 5000
=
1.5625
= $320 000
7.5
10 R =
4
= 1.875% per term
100 × d
V0 =
r
100 × 2000
=
1.875
= $106 666.67
11 d = ?, V0 = $350 000, r = 5
V0 r
d=
100
350 000 × 5
=
100
= $17 500
The correct answer is B.
12 a d = ?, V0 = $400 000, r = 4
V0 r
d=
100
400 000 × 4
=
100
= $16 000
b d = ?, V0 = $300 000, r = 1
V0 r
d=
100
300 000 × 1
=
100
= $3000 per quarter
13 a d = ?, V0 = $100 000, r = 12 ÷ 12
V0 r
d=
100
100 000 × 1
=
100
= $1000 per month
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.5
b d = ?, V0 = $2 000 000, r = 6 ÷ 4
V0 r
d=
100
2 000 000 × 1.5
=
100
= $30 000 per quarter
FV = 2 000 000
PY = 4
CY = 4
PMT∶ END
15 a d = ?, V0 = $80 000, r = 4
V0 r
d=
100
80 000 × 4
=
100
= $3200
14 (12a) Using Finance Solver∶
N=1
I% = 4
PV = −400 000
PMT = 16 000
FV = 400 000
PY = 1
CY = 1
PMT∶ END
b d = $5000, V0 = $80 000, r = ?
V0 r
d=
100
80 000 × r
5000 =
100
r = 6.25% p.a.
b N=1
I% = 4
PV = −300 000
PMT = 3000
FV = 300 000
PY = 4
CY = 4
PMT∶ END
c The perpetuity formula cannot be used as the frequency of
the payment is not the same as the compounding period.
Use Financial Solver.
N=1
I% = 4
PV = −80 000
PMT = 3248.3208
FV = 80 000
PY = 1
CY = 4
PMT∶ END
The total payment is $3248.32 per year, which is $48.32
greater.
d d = $400, V0 = $80 000, r = ?
V0 r
d=
100
80 000 × r
400 =
100
r = 0.5% per month or 6% p.a.
(13a) N = 1
I% = 12
PV = −100 000
PMT = 1000
FV = 100 000
PY = 12
CY = 12
PMT∶ END
b N=1
I% = 6
PV = −2 000 000
PMT = 30 000
*16
P df_Fol i o: 133
133
16 See table at the foot of the page*
The balance at the end of the sixth month is $17 088.13.
Time
period
Principal ($)
1
15 000.00
15 000
2
15 131.25 + 250 = 15 381.25
15 381.25
3
15 515.84 + 250 = 15 765.84
15 765.84
4
15 903.79 + 250 = 16 153.79
16 153.79
5
16 295.14 + 250 = 16 545.14
16 545.14
6
16 689.91 + 250 = 16 939.91
16 939.91
Interest earned ($)
0.105
= 131.25
( 12 )
0.105
= 134.59
( 12 )
0.105
= 137.95
( 12 )
0.105
= 141.35
( 12 )
0.105
= 144.77
( 12 )
0.105
= 148.22
( 12 )
Balance ($)
15 131.25
15 515.84
15 903.79
16 295.14
16 689.91
17 088.13
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.5
134
17 a
i d = $5000, V0 = $400 000, r = ?
V0 r
d=
100
400 000 × r
5000 =
100
r = 1.25% p.a.
ii d = $1000, V0 = $500 000, r = ?
V0 r
d=
100
500 000 × r
1000 =
100
r = 0.2% per month
= 2.4% p.a.
iii d = $30 000, V0 = $800 000, r = ?
V0 r
d=
100
800 000 × r
30 000 =
100
r = 3.75% per 6 months
= 7.5% p.a.
iv d = $200, V0 = $100 000, r = ?
V0 r
d=
100
100 000 × r
200 =
100
r = 0.2% per fortnight
= 5.2% p.a.
b
i N=1
I% = 1.25
PV = −400 000
PMT = 5000
FV = 400 000
PpY = 1
CpY = 1
PMT∶ END
ii N = 1
I% = 2.4
PV = −500 000
PMT = 1000
FV = 500 000
PpY = 12
CpY = 12
PMT∶ END
iii N = 1
I% = 7.5
PV = −800 000
PMT = 30 000
FV = 800 000
PpY = 2
CpY = 2
PMT∶ END
iv N = 1
I% = 5.2
PV = −100 000
PMT = 200
FV = 100 000
PpY = 26
CpY = 26
PMT∶ END
18 a Using Finance Solver:
N=1
I% = 1.242 895 218
PV = −400 000
PMT = 5000
FV = 400 000
PpY = 1
CpY = 12
PMT∶ END
r = 1.24% p.a.
b N=1
I% = 2.426 576 795
PV = −500 000
PMT = 1000
FV = 500 000
PpY = 12
CpY = 1
PMT∶ END
r = 2.43% p.a.
c N=1
I% = 7.430 975 749
PV = −800 000
PMT = 30 000
FV = 800 000
PpY = 2
CpY = 4
PMT∶ END
r = 7.43% p.a.
d N=1
I% = 5.206 067 34
PV = −100 000
PMT = 200
FV = 100 000
PpY = 26
CpY = 12
PMT∶ END
r = 5.21% p.a.
19 a d = $200 per month, V0 = ?, r = 3.6 ÷ 12
V0 r
d=
100
V0 × 0.3
200 =
100
V0 = $66 666.67
b 200 × 50 × 12 = $120 000
The correct answer is E.
P df_Fol i o: 134
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.6
20 a d =
V0 r
100
d × 100
1200 × 100
V0 =
=
r
6
= $20 000
10 000 × 100
b V0 =
4.5
= $222 222.22
300 × 100
c V0 =
0.5
= $60 000
120 × 100
d V0 =
0.25
= $48 000
135
VCAA Assessment Report note:
A common error was to choose option E, which corresponded
to both the PV and FV being incorrectly allocated the same
sign.
7.6 Annuity investments
7.6 Exercise
1 a r=
10
= 0.8333
12
0.8333
R=1+
= 1.00833...
100
V0 = 7000, Vn+1 = 1.0083Vn + 150
7.5 Exam questions
1 a After one month: A1 =1.0024 × 500 000 − 2000 = $499 200
After two months: A2 = 1.0024 × 499 200 − 2000 [1 mark]
= $498 398.08
r
, so r = 2.88%⋅
[1 mark]
b 1.0024 = 1 +
1200
c Perpetuity means that the balance will always remain the
same, i.e. An+1 = An
500 000 = k × 500 000 − 2000
[1 mark]
k = 1.004
2 The value of A is not changing, so the recurrence relation is
subtracting the same value as the interest earned.
0.025 × 200 000 = 5000
The correct answer is B.
3 Use Finance Solver on CAS:
N∶ 5 years × 12 months = 60
I (%) ∶ 5.2 p.a
PV∶ −130 784.92
FV∶ −66 992.27
PpY/CpY∶ 12
Therefore, the payment (PMT) each month is $1 500.
The correct answer is C.
7000
7000 × 1.0083 + 150
7208.10
7000
7208.10 × 1.0083 + 150
7417.93
7417.93 × 1.0083 + 150
7629.50
7629.50 × 1.0083 + 150
7842.82
7842.82 × 1.0083 + 150
8057.92
The value of the investment at the end of the fifth month is
$8057.92.
b See table at the foot of the page*
A = $58.10
B = $209.83
C = $63.92
D = $8057.83
2 a V0 = 10 000
12
r=
=1
12
1
R=1+
= 1.01
100
V0 = 10 000, Vn+1 = 1.01Vn + 200
b
10 000
10 000 × 1.01 + 200
10 000
10 300
10 300 × 1.01 + 200
10 603
10 603 × 1.01 + 200
10 909.03
10 909.03 × 1.01 + 200
11 218.12
11 218.12 × 1.01 + 200
11 530.30
11 530.30 × 1.01 + 200
11 845.61
The value of her investment at the end of the sixth month is
$11 845.60.
*1b
Time period
Payments
Interest
Principal increase
Balance of investment
0
1
2
3
4
5
0
150
150
150
150
150
0
0.0083 × 7000 = 58.10
59.83
61.57
0.0083 × 7629.50 = 63.32
65.10
0
208.10
150 + 59.83 = 209.83
211.57
213.32
215.01
7000
7208.10
7417.93
7629.50
7842.82
7842.82 + 215.01 = 8057.83
P df_Fol i o: 135
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
136
3
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.6
Time
period
Payments
Interest
Principal Balance of
increase investment
0
1
2
3
0
100
100
100
0
31.25
32.07
32.90
0
131.25
132.07
132.90
5000
5131.25
5263.32
5396.22
Interest = 31.25 + 32.07 + 32.90 = $96.22
4 See table at the foot of the page*
At the end of the 4th month the investment I is $10 875.37.
5 a N∶ 5
I%∶ 6.5
PV∶ −2500
PMT∶ −6500
FV∶ unknown
PY∶ 1
CY∶ 1
FV is 40 433.833. Sharyn will have $40 000 correct to the
nearest 1000.
b N∶ 5
I%∶ 7
PV∶ −2500
PMT∶ −8000
FV∶ unknown
FV is 49 512.29
Sharyn will have $50 000 correct to the nearest 1000.
c Sharyn will have earned an extra $10 000.
6 a N∶ 5
I%∶ 5.5
PV∶ −4000
PMT∶ −6000
FV∶ unknown
PY∶ 1
CY∶ 1
Rhonda will have $39 000 correct to the nearest 1000.
b N∶ 5
I%∶ 6.5
PV∶ −4000
PMT∶ −8000
FV∶ unknown
PY∶ 1
CY∶ 1
Rhonda will have $51 000 correct to the nearest 1000.
c Rhonda will have earned an extra $12 000.
*4
P df_Fol i o: 136
7 r=
6
12
= 0.5
Complete the table as shown:
Column 2 gives the balance at the start of each time period.
This is the same as the balance at the end of the previous time
period.
Column 3 is the additional deposit ($200).
Column 4 is calculated by: (column 2 × 1.005 + column 3)
n+1
1
2
3
4
5
$20 000
$200
$20 904.51
$200
Vn
$20 300
$20 601.50
$21 209.03
d
$20 300
Vn+1
$200
$20 601.50
$200
$21 515.08
$200
$20 904.51
$21 209.03
a Interest in month 2 is:
Vn+1 − Vn − d = 20 601.50 − 20 300 − 200
= $101.50
b Interest in month 5 is:
Vn+1 − Vn − d = 21 515.08 − 21 209.03 − 200
= $106.05
c Total interest is:
21 515, 08 − (20 000 + 5 × 200) = $515.08
8 a N∶ 20
I%∶ 6
PV∶ −60 000
PMT∶ −9000
FV∶ unknown
PY∶ 1
CY∶ 1
The amount available is $523 000 to the nearest 1000.
b N∶ 20
I%∶ 8
PV∶ −60 000
PMT∶ −10 000
FV∶ unknown
PY∶ 1
CY∶ 1
The amount available is $737 000 to the nearest 1000.
c 737 000 − 523 000 − $214 000
9 a N∶ 204
I%∶ 8
PV∶ −55 000
PMT∶ unknown
FV∶ 550 000
Time period
Payments
Interest
Principal increase
Balance of investment
0
1
2
3
4
0
150
150
150
150
0
66.67
68.04
69.58
10 654.27 × 0.00667 = 70.96
0
216.67
218.04
219.58
150 + 70.96 = 220.96
10 000
10 216.67
10 434.71
10 654.27
10 654.27 + 220.96 = 10 875.23
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.6
PY∶ 12
CY∶ 12
Monthly contribution is 779.70.
b N∶ 84
I%∶ 8
PV∶ −55 000
PMT∶ −779.70
FV∶ unknown
PY∶ 12
CY∶ 12
After 7 years, Peter has $183 523.49.
Change N to 120, PV to 183 523.49, PMT to 1559.40 and
find the final amount Peter will have $692 647.66 in his
fund.
c Use CAS and change values appropriately
816 247.11 − 692 647.66 = 123 599.66.
$124 000 to the nearest 1000.
10 a N∶ 132
I%∶ 9.5
PV∶ −115 000
PMT∶ unknown
FV∶ 600 000
PY∶ 12
CY∶ 12
Monthly payment: $1185.74
b N∶ 72
I%∶ 9.5
PV∶ −115 000
PMT∶ −1185.74
FV∶ unknown
After 6 years, there is a total of $317 370.62.
Change PV to −317 370.62 and PMT to −$2371.48 and N
to 60. Patricia will have $690 610.68 in her fund.
c 737 282.59 − 690 616.64 = 46 665.95
$47 000 to the nearest 1000.
11 Repeat the method from question 10.
a 1168.46
b $861 442.14
c $238 000
a $717.34
b $931 925.99
12 Repeat the method from question 10.
7.6 Exam questions
1 Use the first balance and second interest amount to calculate
the interest rate per period (x).
x% of 6977.50 = 27.91 solving for x,
x = 0.4% per period
The interest earned for payment number 20 is 0.4% of
$7233.83 = $28.94.
The principal addition for payment number 20 is
$7500 − $7233.83 = $266.17 .
137
So, interest + payment = principal addition
Payment + 28.94 = 266.17
Payment = 266.17 − 28.94
= $237.23
The correct answer is D.
VCAA Assessment Report note:
This question required the interpretation of an amortisation
table and the calculation of a missing payment value. Many
students seemed to assume that the payment value would be
constant at $100 (option B), despite the question stating that
the payment may vary.
5.2
2 a 5.2% p.a. is equivalent to
% per month.
12
× 360 000 = $1560 per month.
[1 mark]
100
VCAA Assessment Report note:
Some students gave the annual payment of $18 720, while
5.2
others knew the correct method but rounded
to
1000
2 decimal places before multiplying by $360 000, giving
$1548.
b Use Finance Solver on CAS:
N∶ 4 years 12 months = 48
I(%)∶ 3.8 p.a.
PV∶ −360 000
PMT∶ −500
FV ∶ 444 872.9444992
PpY/CpY∶ 12
After 4 years, Alex’s investment grows
to $444 872.94.
[1 mark]
N∶ 2 years 12 months = 24
I3.8
PV∶ −444 872.94
PMT∶ −805.65070094875
FV∶ 500 000
PpY/CpY∶ 12
To grow to $500 000 in a further two years, Alex’s new
monthly payment will be $805.65.
[1 mark]
VCAA Assessment Report note:
Some students entered the $500 payment as a positive into
their Finance Solver, giving $393 121.15 as the four-year
value.
A few students tried to use a formula but made little
progress.
P×r
3 a Q=
100
P × 3.68
460 =
[1 mark]
100
P = $12 500
b A perpetuity is an investment that provides regular
payments that continue forever. Therefore, they will be
able to provide the scholarship for an infinite number of
years.
[1 mark]
VCAA Assessment Report note:
Many students did not know that perpetuities pay out only
the interest earned, while the principal remains unchanged.
5.2
12
P df_Fol i o: 137
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
138
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.7
The most common incorrect answer was
12 500
≈ 27 years
460
7.7 Review
7.7 Exercise
Multiple choice
1 N∶ 104
I%∶ 9.75
PV∶ 14 000
PMT∶ unknown
FV∶ 0
PY∶ 26
CY∶ 26
The repayment is $162.82.
The correct answer is C.
8.2
2 r=
= 2.05%
4
Interest = 0.0205 × 22 000 = 451
The interest = $451
The correct answer is B.
3 N∶ unknown
I%∶ 10.5
PV∶ 41 000
PMT∶ −588.39
FV∶ 0
PY∶ 12
CY∶ 12
N = 108
N is closest to 110
The correct answer is E.
4 N∶ unknown
I%∶ 5
PV∶ 56 000
PMT∶ −1500
FV∶ 0
PY∶ 4
CY∶ 4
N = 50.6 ≈ 51
≈ 12.75 years
The correct answer is C.
5 N∶ 130
I%∶ 6.5
PV∶ 24 000
PMT∶ unknown
FV∶ 0
PY∶ 26
CY∶ 26
d = $216.47
The correct answer is E.
6 Enter the values into the Finance Solver and solve for FV.
N∶ = 20
I(%)∶ = 6.5
PV∶ = −24 000
PMT∶ = ?
FV∶ = 0
PpY∶ = 4
CpY∶ = 4
The payment (FV) = $1415.18
The correct answer is D.
7 N∶ 20
I%∶ 5.5
PV∶ 24 000
PMT∶ unknown
FV∶ 0
PY∶ 4
CY∶ 4
d = $1380.73
The correct answer is B.
8 V0 = a, Vn+1 = RVn + d
V0 = 2000
6
= 0.5
r=
12
0.5
R=1+
= 1.005
100
d = 360
V0 = 2000, Vn+1 = 1.005Vn + 360
The correct answer is D.
9 Interest only loans∶ PV = −FV
PpY = 12
N = any number because loan is not paid off
Using Finance Solver
PMT = $1644.458
The correct answer is D.
10 Using Finance Solver:
a N∶ 204
I%∶ 9.6
PV∶ − 92 200
PMT∶ unknown
FV∶ 800 000
PY∶ 12
CY∶ 12
PMT∶ $649.91.
Her employee is contributing $500 per month, so Claire
needs to contribute $149.91.
The correct answer is D.
Short answer
11 Interest = 542.71 × 240 − 70 000
= $60 250.40
12 After 10 years
N∶ 120
I%∶ 7
PV∶ 70 000
PMT∶ −542.71
FV∶ unknown
PY∶ 12
CY∶ 12
There is $46 741.44 still owing after 10 years.
P df_Fol i o: 138
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.7
13 N∶ unknown
I%∶ 7
PV∶ 46 741.44
PMT∶ −600
FV∶ 0
PY∶ 12
CY∶ 12
N = 104.18
104.18
Total term = 10 +
12
= 18.6667 years.
Total term is 18 years and 8 months or 19 years (to the nearest
whole year).
14 N∶ 20
I%∶ 7
PV∶ 55 000
PMT∶ −1487.93
FV∶ unknown
PY∶ 4
CY∶ 4
After 20 payments, the balance outstanding is $42 558.31.
Using CAS, it will take her 228 payments to pay out the loan.
Term of altered loan
20 228
+
=
4
26
= 5 + 8.769
= 13.76
= 13 years 20 fortnights
∴ Difference
= 15 − 13 years 20 fortnights
= 1 year 6 fortnights
15 Fixed rate
N∶ 60
1%∶ 8.25
PV∶ 8000
PMT∶ unknown
FV∶ 0
PY∶ 12
CY∶ 12
139
b Use Finance Solver because the compounding period is not
the same as the payment period.
set PpY = 1
and CpY = 12
Interest rate = 6.97% p.a.
Extended response
17 a V0 = $40 000, Vn = $400 000
n = 10 × 4r
= 7.7% p.a.
= 40
On Finance Solver:
N = 40, I = 7.7, PV = −40 000,
FV = 400 000, PpY = CpY = 4
d = PMT = $5287.59
b On TVM Solver:
N = 20 × 12 = 240
PV = −40 000
FV = 400 000
PMT = −200
PpY = CpY = 12
r = I = 9.39% p.a.
18 a V0 = $27 000 − $10 000
= $17 000
n=4
r=6
17 000 × 6 × 4
∴ Interest =
100
= $4080
Repayment∶ $163.17
∴ Interest = 163.17 × 60 − 8000
= $1790.20
Better off with variable interest rate by approximately $4.83.
Variable rate using d = $163.17
(1) After 1st V12 = 6670.98
(2) 2nd condition − V24
V24 = $3592.78
(3) Time of 3rd condition (n = ?)
N = 23.97
when n = 23, Vn = 157.25
Interest = 163.17 × (12 + 24 + 23) + 157.25 × 1.006916667 − 8000
= 9627.03 + 158.3376459 − 8000
= $1785.37
Pr
16 a d =
100
1 200 000 × 7.2
=
= $86 400
100
Total cost of the car = $27 000 + $4080
= $31 080
b i V0 = $2000
r=7
n=3
7
∴ Amount Vn = 2000 1 +
(
100 )
= 2000 (1.07) 3
= $2450.09
3
∴ Interest = 2450.09 − 2000
= $450.09
ii Total cost of the car = 12 000 + 450.09
= $12 450.09
c i V0 = $5000
n=4
r=5
5
∴ Amount Vn = 5000 1 +
(
100 )
= 5000 (1.05)4
= $6077.53
∴ The expected value of the painting after 4 years is
$6077.53.
4
P df_Fol i o: 139
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.7
140
V0 = $5000
r = 10
∴ d = 10% of 5000
= $500
n=4
∴ V4 = 5000 − 500(4)
= 5000 − 2000
= $3000
∴ Future value after 4 years is $3000.
i r = 15
n=4
V0 = $27 000
ii
d
15
∴ V4 = 27 000 1 −
(
100 )
= 27 000(0.85)4
= $14 094.17
∴ Future value of car in 4 years is $14 094.17.
ii r = 15
n=4
V0 = $12 000
4
15
∴ V4 = 12 000 1 −
(
100 )
= 12 000(0.85)4
= $6264.08
∴ Future value of the car after 4 years is $6264.08.
iii Clearly, the $27 000 car option loses the most money.
4
e
Description
Cost of loan
Depreciation
of goods
Total cost
Less benefits
Total cost
*19b
P df_Fol i o: 140
Payment
$27 000
car
$5000 car
$12 000 car and painting
$12 905.83
$5735.92
$4080
$31 080
$0
$450.09
$12 450.09
$0
$31 080
$12 450.09
Principal outstanding ($)
1
6000
2
5279.04
3
4549.97
4
3812.70
5
3067.13
6
2313.17
7
1550.74
8
779.72
19 a Using CAS, enter: N (n) = 8; I (r) = 4.5;
PV(V0 ) = −6000; PMT (d) = unknown; FV(Vn ) = 0;
PpY = 4; CpY = 4.
Solving for PMT gives 788.46424…
Therefore, quarterly payments are $788.46.
b After each payment, the amount of outstanding
loan = principal outstanding + interest due − payment.
For example, interest due prior to first repayment
0.045
= 67.5
= 6000 ×
4
Loan outstanding after the first repayment
= 6000 + 67.50 − 788.46 = 5279.04
See table at the foot of the page*
c Total amount paid on the loan = 788.46 × 8 = $6307.68
d Total interest paid on the loan = $6307.68 − $6000 =
$307.68
n
i
e r= 1+
−1
(
n)
i = 0.045; n = 4
4
0.045
−1
r= 1+
(
4 )
r = 0.045765
Therefore, the effective annual interest rate is 4.58%
(correct to 2 decimal places).
7.7 Exam questions
$0
$2000
$10 000
$1077.53
$8922.47
Interest due ($)
6000 ×
1 $449 060.08 − $422 051.93 = $270 08.15
The correct answer is C.
2 Use Finance Solver on CAS:
N∶ ??
I(%)∶ 4
PV∶ −500 000
PMT∶ 44 970.55
FV∶ 0
PpY∶ 1
CpY∶ 1
Therefore, N = 15 years .
The correct answer is B.
0.045
= 67.5
4
0.045
5279.04 ×
= 59.39
4
0.045
4549.97 ×
= 51.19
4
0.045
3812.70 ×
= 42.89
4
0.045
3067.13 ×
= 34.51
4
0.045
2313.17 ×
= 26.02
4
0.045
1550.74 ×
= 17.45
4
0.045
779.72 ×
= 8.77
4
Payment ($)
Loan outstanding ($)
788.46
5279.04
788.46
4549.07
788.46
3812.70
788.46
3067.13
788.46
2313.17
788.46
1550.74
788.46
779.72
788.46
0.03
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 7 Modelling reducing balance loans, annuities and perpetuities using recursion • EXERCISE 7.7
3 a i Principal reduction =
$318 718.08 − $318 074.23 = $643.85
[1 mark]
3.6
% × $318 074.23 = $954.22
ii Monthly interest =
12
Therefore, the balance after payment 4 will be
$318 074.23 + $954.22 − $1600
= $317 428.45
[1 mark]
3.6
= 3%
b Note that the monthly interest rate will be
12
S0 = 320 000, Sn+1 = 1.003 × Sn − 1600
[1 mark]
4 Change in rate:
I28
I27
1002.26
961.90
−
=
−
≈ +0.02% per month
V22 V26
227 785.76 229 023.86
+0.02 × 12 = +0.24% per annum.
The interest rate increased by 0.24% per annum.
The correct answer is A.
5 a i Use Finance Solver on CAS:
N∶ 12
I (%)∶ 6.9
PV∶ 70 000
PMT∶ −800
PpY/CpY∶ 12
Therefore, Ken will owe $65 076.22 after
12 months.
[1 mark]
VCAA Assessment Report note:
Rather than use a financial solver to answer the question
above, a number of students adopted a formulaic
approach, almost always unsuccessfully, based on the
compound interest formula.
ii Total interest after 12 payments = 800 × 12
− (70 000 − 65 076.22) = $4676.22
[1 mark]
VCAA Assessment Report note:
A common incorrect answer was $4923.78, which is the
reduction in the principal over the year. This failed to
take into account the $9600 total of repayments made in
the year.
b After 3 years, the value of the loan is $54 151.60.
Using Finance Solver on CAS:
N∶ 36
I (%)∶ 6.9
PMT∶ −800
FV∶ 0
PpY/CpY∶ 12
[1 mark]
The principal amount after a further 3 years would be
$25 947.58.
Ken’s lump sum payment, $L, will be:
L = 54 151.60 − 25 947.58 = $28 204.
[1 mark]
Correct tables of input values for the financial solver may
have illustrated working out to qualify for a method mark
even if the final answer was incorrect.
P df_Fol i o: 141
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
141
142
TOPIC 8 Matrices • EXERCISE 8.2
Topic 8 — Matrices
8.2 Matrix representation
8.2 Exercise
⎡−4 0 0⎤
⎢
⎥
1 A = ⎢ 0 6 0⎥ is a 3 × 3 diagonal matrix.
⎣ 0 0 3⎦
The element in the 2, 1 position is 0 and the number 3 is
represented by a33 .
2 −3
B=
is a 2 × 2 square matrix.
0]
[3
The element in the 2, 1 position is 3 and the number 3 is
represented by b21 .
1 1
C=
is a 2 × 2 binary matrix.
[0 1]
The element in the 2, 1 position is 0 and 3 is not in the matrix.
⎡ 3
1 −5⎤
⎢
⎥
D = ⎢ 1 −12
6⎥ is a 3 × 3 symmetrical matrix.
⎣−5
6 13⎦
The element in the 2, 1 position is 1 and 3 is represented
by d11 .
⎡ −2 0 0⎤
⎢
⎥
E = ⎢ 5 4 0⎥ is a 3 × 3 lower triangular matrix.
⎣−10 3 6⎦
The element in the 2, 1 position is 5 and 3 is represented
by e32 .
⎡3 −5
2⎤
⎥
⎢
2 A = ⎢0
5 −12⎥ is a 3 × 3 upper triangular matrix.
⎣0
0
15⎦
The element in the 1, 2 position is −5 and 3 is represented by
a11 .
B = [−1 −3 1] is a 1 × 3 row matrix.
The element in the 1, 2 position is −3 and 3 is not in the
matrix.
7 10
C=
is a 2 × 2 square matrix.
1]
[3
The element in the 1, 2 position is 10 and 3 is represented
by c21 .
⎡ 12
7 −17⎤
⎥
⎢
D = ⎢ 7 −9
3⎥ is a 3 × 3 symmetrical matrix.
⎣−17
3
12⎦
The element in the 1, 2 position is 7 and 3 is represented by
d32 and d23 .
3 a i This is a 4 × 2 rectangular matrix.
ii The 2, 1 element is 2.
iii a22
b i This is a 3 × 3 square matrix.
ii The 2, 1 element is 0.
iii b33
c This is not a matrix.
d i This is a 3 × 1 column matrix.
ii The 2, 1 element is 3.
iii d21
i This is a 1 × 3 row matrix.
ii The 2, 1 element does not exist.
iii e11
f i This is a 3 × 3 symmetrical matrix.
ii The 2, 1 element is −4.
iii f32 and f23
4 A is a (3 × 3) lower triangular matrix. The element in the 2, 2
position is 4 and the number 7 is represented by a33 .
B is a (1 × 3) row matrix. The element in the 2, 2 position is 7
and the number 7 is represented by b12 .
C is a (2 × 2) identity matrix. The element in the 2, 2 position
is 1 and the number 7 is not in the matrix.
D is a (3 × 3) symmetrical matrix. The element in the 2, 2
position is 0 and the number 7 is represented by d33 .
5 i a a21 = 1
b a21 = 2
c The a21 element does not exist.
d The a21 element does not exist.
e a21 = 3
f a21 = 3.5
ii The a32 element exists for matrices a and e.
6 The element in row 2, column 1 is −0.5.
The correct answer is C.
7 The number 3 is in row 3, column 2, so it can be represented
as a32 .
The correct answer is B.
8 ⎡1 0 0 0 ⎤
⎢
⎥
⎢0 1 0 0 ⎥
⎢0 0 1 0 ⎥
⎢
⎥
⎣0 0 0 1 ⎦
9 Use a 4 × 4 matrix to represent the connection between the
four vertices.
⎡0 1 1 1 ⎤
⎥
⎢
⎢1 0 1 2 ⎥
⎢1 1 0 0 ⎥
⎥
⎢
⎣1 2 0 0 ⎦
e
A
B
C
D
Vertex A is connected once to vertices B, C and D.
Vertex B is connected once to vertices A and C and twice to
vertex D.
Vertex C is connected once to vertices A and B.
Vertex D is connected once to vertex A and twice to vertex B.
⎡26
AFL
⎢
10 Cricket ⎢15
Total ⎣41
⎡26 22⎤
⎢
⎥
9⎥
⎢15
⎣41 31⎦
22⎤
⎥
9⎥
31⎦
P df_Fol i o: 142
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.3
11 Count the number of routes between each town. For example,
there are three routes between A and B, two between A and C,
and so on.
The matrix can be represented as:
⎡ 0 3 2 1 ⎤
⎢
⎥
⎢ 3 0 1 1 ⎥
⎢ 2 1 0 1 ⎥
⎢
⎥
⎣ 1 1 1 0 ⎦
12 ⎡12
⎢
⎢ 9
⎣21
13 15
[12
23
16
39
35⎤
⎥
25⎥
60⎦
10
15
100
87]
2 A=
C=
8.2 Exam questions
1 In a transpose matrix, the rows are swapped to the columns,
and vice versa.
⎡ 3 2⎤
3 8 13
⎢
⎥
If M = ⎢ 8 9⎥, then MT =
7]
[2 9
⎣13 7⎦
The correct answer is D.
2 The amount saved in week 2 is given by the element in row 2,
column 2 — 90.
The correct answer is B.
3 Select an element (e.g. 5 in row 2, column 3: m2,3 = 5) and
try all the options.
A∶ 2 + 3 − 1 ≠ 5
B∶ 2(2) − 3 + 1 ≠ 5
C∶ 2(2) + 3 − 2 = 5 This is the correct answer.
D∶ 2 + 2(3) − 2 ≠ 5
E∶ 2 + 3 + 1 ≠ 5
The correct answer is C.
8.3 Addition, subtraction and scalar operations
with matrices
b B+A
0 −3
4
+
2] [3
[5
−3
4+0
=
2] [3 + 5
7
0+4
=
−4] [5 + 3
7 + −3
4
=
−4 + 2] [8
−3 + 7
4
=
2 + −4] [8
=
4
−2]
4
−2]
−2
5
−
−8] [1
−4
[ 2
7
6
−10
6]
−7 − 5
−2
=
−5] [−5 − 1
7−7
6−6
0
0
0
−3]
=
12
−8
[
9
15
,B=
−10]
[18
8
[1
17
12
−12
[ −6
−21
,
−13]
20
10
−4
−9
,D=
−9]
[−12
17
5
−2 − −2
−8 − −5]
−12
−11]
a A+B
12
9
15
+
[−8 −10] [18
20
10
−21
−13]
−21
8
+
−13] [1
17
12
−4
15 + 8
=
−9] [18 + 1
This cannot be done because A is a (2 × 2) matrix and B is a
(2 × 3) matrix.
b B+C
15
[18
20
10
c B−C
15
[18
d B−D
15
[18
−9
[−12
17
5
=
=
=
−12
8
−
−11] [1
5
3
+
2] [5
4
[2
5
0]
−21 + −4
−13 + −9]
20 − 17
10 − 12
−21 − −4
−13 − −9]
−25
−22]
37
22
7
[17
=
17
12
20 + 17
10 + 12
−17
−4]
3
−2
−12
15 − −9
=
−11] [18 − −12
17
5
1
[−3
1+3
[−3 + 5
23
[19
−4
15 − 8
=
−9] [18 − 1
17
12
−21
−9
−
−13] [−12
20
10
e D−C
=
−21
8
−
−13] [1
20
10
3 a A+B=
8.3 Exercise
−2
cannot be done because B is a
−5]
7
6
(2 × 2) matrix and C is a (2 × 3) matrix.
d B−A
0 −3
4
7
0 − 4 −3 − 7
−
=
5
2]
3
−4]
[
[
[5 − 3 2 − −4]
e D−C
−7 7
[−5 6
14 Place the suppliers in the columns.
⎡ 5
7 15⎤
⎢
⎥
2
7⎥
⎢12
⎣ 6 15
0⎦
15 The matrix must have 4 rows and 2 columns.
⎡ 3
2⎤
⎢
⎥
−4
−1
⎢
⎥
⎢ 4 −1⎥
⎢
⎥
⎣ 4
1⎦
1 a A+B
4
7
0
+
[3 −4] [5
c B−C
5
0 −3
−
2] [1
[5
143
24
[30
3
5
−4
−9 − 8
=
−9] [−12 − 1
=
−17
[−13
0
−2]
5+0
2 + −2]
P df_Fol i o: 143
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
0
−7
20 − 17
10 − 5
−9
−2]
17 − 17
5 − 12
−8
−2]
−21 − −12
−13 − −11]
−12 − −4
−11 − −9]
TOPIC 8 Matrices • EXERCISE 8.3
144
b A+A=
=
=
5
1
+
2] [−3
1
[−3
1+1
[−3 + −3
5+5
2 + 2]
7 a A + 2B =
5
2]
=
2 10
4]
[−6
c B − C is not possible as B and C are of different orders.
3
d B−A=
[5
0
1
−
−2] [−3
3−1
=
[5 − −3
2
=
[8
4
e D−C=
[3
−5
−4]
2
−2
4−1
=
[3 − 3
3
=
[0
−3
1
0−5
−2 − 2]
5
2]
2−5
−2 − −3
− [23.50
5
−3
2−0
−1 − 3]
0
3]
2
−4]
= [26.00
169.00
45.00
87.50
81.50
−6
[ 6
−18
8
=
−4] [ 18
24
−12]
−2
12
8
=
−4] [−12
3
b −2A
c
135.00
90.00
6 a 3A
−6
[ 6
1
A
2
1 −6
2[ 6
−6
[ 6
d 0.4A
0.4
8
−3
=
−4] [ 3
299.95]
140.00]
159.95]
−16
8]
4
−2]
8
−2.4
=
−4] [ 2.4
b 2A − B = 2
3.2
−1.6]
8 a 3A = 3 ×
=
=
0
−3]
−12
−8
−
0] [ 7
12
[ 3
−12
3]
4
[1
5
2]
15
6]
4
[2
8
6]
0.1 × 4
[0.1 × 2
0.4
[0.2
c 2A + 3A = 5A
=5×
=
0.8
0.6]
4
[1
20
[ 5
5
2]
5×5
5 × 2]
25
10]
4
1
d A+ B=
4
[1
=
0.1 × 8
0.1 × 6]
5×4
[5 × 1
=
=
0
−3]
3×5
3 × 2]
b 0.1B = 0.1 ×
=
−6
−8
−
0] [ 7
−6
−6]
4
[10
12
[ 3
=
0
−6]
2
[5
3×4
[3 × 1
0
−3]
−16
−6
+
0] [ 14
−14
[ 19
=
=
⎡0.2 −0.5⎤ ⎡ 1.2 −0.5⎤
⎥
⎥ ⎢
⎢
4 F − E = ⎢2.4
5.0⎥
2.5⎥ − ⎢ 3.6
⎣ 0
2.2⎦
1.1⎦ ⎣−3.5
⎡0.2 − 1.2 −0.5 − −0.5⎤
⎥
⎢
= ⎢2.4 − 3.6
2.5 − 5.0⎥
⎣ 0 − −3.5
1.1 − 2.2⎦
⎡−1.0
0⎤
⎥
⎢
= ⎢−1.2 −2.5⎥
⎣ 3.5 −1.1⎦
The correct answer is D.
5 Profit = S − C
= [49.50
2
[5
=
2
1
−
−1] [3
−8
−6
+2
0]
[ 7
2
[5
5
4
1
+ ×
2] 4 [2
4
[1
5
1
+
2] [0.5
5
[1.5
7
3.5]
e 2(A + B) = 2 ×
=2
=
4
([1
8
([3
16
[ 6
8
6]
2
1.5]
5
4
+
2] [2
8
6])
13
8])
26
16]
P df_Fol i o: 144
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.3
f
4
1
1
(3A − B) =
3
2
2 ( [1
=
=
=
5
4
−
2] [2
15
4
−
6] [2
12
1
2 ([ 3
1 8
2 [1
7
0]
4
[0.5
3.5
0]
8
6])
8
6])
c 2C + A = 4B
2C = 4B − A
1
C = (4B − A)
2
3
4 5
4B − A = 4
−
[2 6] [−2
16
[ 8
20
3
−
24] [−2
7
2]
7
2]
13
[10
13
22]
=
=
9 Store both C and D in the graphics calculator and perform the
required calculations.
a ⎡4
⎢
⎢8
⎢4
⎢
⎢8
⎣4
12
8
0
4
4
b ⎡−0.4
⎢
⎢−0.2
⎢−0.3
⎢
⎢−0.2
⎣−0.5
c ⎡ −9
⎢
⎢ 0
⎢ −6
⎢
⎢ 0
⎣−12
d ⎡14
⎢
⎢16
⎢12
⎢
⎢16
⎣16
e ⎡−5.5
⎢
⎢ −2
⎢ −4
⎢
⎢ −2
⎣ −7
0
16
−8
12
20
−0.6
−0.3
0
−0.2
−0.1
−9
−3
0
−3
0
30
18
0
10
8
=
b D+
0
0.4
−0.2
0.3
−0.8
0
24
−12
18
−9
0
16
−8
12
46
6
[12
2
[−3
0
8
−4
6
−9.5
40
−8
10
−2
12⎤
⎥
0⎥
−6⎥
⎥
−30⎥
3⎦
8⎤
⎥
24⎥
−4⎥
⎥
20⎥
6⎦
−0.5
−1
−1
−1
−1
6
8
=
−7] [2
8
[2
11 For a:
a + −3 = 6
a=6+3
a=9
For b:
2 + b = −3
b = −3 − 2
b = −5
For c:
6 + c = −2
c = −2 − 6
c = −8
For d:
−6 + 2 = d
d = −4
0.2⎤
⎥
−0.3⎥
−0.1⎥
⎥
−1⎥
0⎦
12 a 6E =
4⎤
⎥
−3⎥
−2⎥
⎥
−15⎥
0.5⎦
E=
=
b D+
24
[6
1
[4
0
2]
6
−7]
13 ⎤
2 ⎥⎥
11⎦
42
0
−6
12]
5
6
=
2] [8
2
5]
1 24
6[6
4
[1
13
22]
−6
12]
42
0
−1
2]
7
0
D=
=
6
−5]
6
2
−
−5] [−3
1 13
2 [10
⎡ 13
= ⎢⎢ 2
⎣ 5
−9
−20]
D=
C=
−36
−80]
6
=
[5
P df_Fol i o: 145
−0.1
−0.1
−0.1
−0.1
−0.1
−36
−80]
40
−8
1 24
4 [48
3
0
0
0
0
14
8
8
8
8
−7.5
−3.5
0
−2.5
−1
24
10 a 4E =
[48
E=
8⎤
⎥
12⎥
−4⎥
⎥
0⎥
4⎦
8
4
4
4
4
6
[8
5
[4
2
1
−
5] [4
−3
3]
5
2]
c 2C + A = 4B
2C = 4B − A
1
C = 2B − A
2
4 8
1 4
=2
−
[2 6] 2 [1
8
[4
16
2
−
12] [0.5
5
2]
2.5
1 ]
6
[3.5
13.5
11 ]
=
=
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
145
146
TOPIC 8 Matrices • EXERCISE 8.3
13 D + A cannot be calculated as D and A are of different orders.
1
(A − 3B) cannot be calculated as A and B are of different
2
orders.
b and e are the only two that cannot be calculated.
14 a The elements for the new 4WDs are located at a21 = 4,
b21 = 2 and c21 = 3.
Store A has the highest number of new 4WDs (a21 = 4).
b Total inventory = A + B + C
=
=
6
[4
4
5
+
7] [2
14
[ 9
13
12]
6
3
+
5] [3
3
0]
⎡72 84⎤
⎢
⎥
15 a Semester 1 = ⎢76 68⎥
⎣81 82⎦
⎡78 74⎤
⎢
⎥
Semester 2 = ⎢76 77⎥
⎣89 85⎦
1
b C = (A + B) where C represents the average of the two
2
tests of the two subjects.
1
c C = (A + B)
2
⎛⎡72 84⎤ ⎡78 74⎤⎞
1 ⎜⎢
⎥ ⎢
⎥⎟
76 68⎥ + ⎢76 77⎥⎟
2 ⎜⎢
⎝⎣81 82⎦ ⎣89 85⎦⎠
⎡150 158⎤
1⎢
⎥
= ⎢152 145⎥
2
⎣170 167⎦
⎡75 79 ⎤
⎢
⎥
= ⎢76 72.5⎥
⎣85 83.5⎦
16 a The matrices with the larger values in the third row, second
column are A and C.
⎡7 16⎤ ⎡13 12⎤ ⎡14 7 ⎤
⎢
⎥ ⎢
⎥ ⎢
⎥
b Total stock = ⎢5 8 ⎥ + ⎢ 8 15⎥ + ⎢12 12⎥
⎣2 4 ⎦ ⎣ 3
1⎦ ⎣9
4⎦
⎡34 35⎤
⎢
⎥
= ⎢25 35⎥
⎣14 9 ⎦
c A markup of 100% is equivalent to 2 times the wholesale
matrix.
⎡20 000 13 000⎤
⎢
⎥
RRP = 2 × ⎢25 000 18 000⎥
⎣40 000 28 000⎦
⎡2 × 20 000 2 × 13 000⎤
⎢
⎥
= ⎢2 × 25 000 2 × 18 000⎥
⎣2 × 40 000 2 × 28 000⎦
⎡40 000 26 000⎤
⎢
⎥
= ⎢50 000 36 000⎥
⎣80 000 56 000⎦
=
d A discount of 10% is equivalent to 90% of the
recommended price.
⎡40 000 26 000⎤
⎥
⎢
Discount = 0.9 × ⎢50 000 36 000⎥
⎣80 000 56 000⎦
⎡36 000 23 400⎤
⎢
⎥
= ⎢45 000 32 400⎥
⎣72 000 50 400⎦
17 a B = [0.25 0.40 0.20 0.15]
b A = 800B
c A = 800 × [0.25 0.40 0.20 0.15]
= [200 320 160 120]
18 a
154
[207
214
180
b Average =
1138
1422]
1 154
11 [207
214
180
1138
1422]
⎡ 154 214 1138 ⎤
⎢ 11
11
11 ⎥
⎥
= ⎢⎢
⎥
⎢ 207 180 1422 ⎥
⎣ 11
11
11 ⎦
5
5
⎡ 14
103 ⎤⎥
19
⎢
11
11
⎥
=⎢
⎢18 9 16 4 129 3 ⎥
⎣ 11
11
11 ⎦
19 a New stock level (N) = 40% of current stock levels (matrix S)
= 0.4 × S
⎡23 150 112 5 1090⎤
⎢
⎥
⎢35 210 145 4 800 ⎥
b New stock level = 0.4 × ⎢45 230 130 8 1200⎥
⎥
⎢
⎢32 700 230 4 400 ⎥
⎣77 450 160 7 850 ⎦
⎡9
60 45 2 436⎤
⎢
⎥
14
84 58 2 320⎥
⎢
= ⎢18 92 52 3 480⎥
⎢
⎥
⎢13 280 92 2 160⎥
⎣31 168 64 3 340⎦
20 Enter the matrices into the graphics calculator and perform
the LHS and RHS calculations separately.
a True
b Not true
c Not true
d True
P df_Fol i o: 146
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.4
8.3 Exam questions
1 To add matrices together, all matrices must be of the same order. Therefore, only
The correct answer is C.
2 Start by choosing an element, say 5, which is i = 2, j = 1.
8
4
8
+
and
[12] [2]
[12
4
0
+
0] [0
A: P2,1 =4 − 1 ≠ 5
B: P2,1 = 2(2) + 1 = 5 — check with another element, P2,2 = 2(2) + 1 ≠ 4, which is the element in row 2, column 2.
C: P2,1 =2 + 1 + 1 ≠ 5
D: P2,1 =2 + 2(1) ≠ 5
E: P2,1 = 2(2) − 1 + 2 = 5 — check with another element, P2,2 = 2(2) − 2 + 2 = 4, which is correct.
⎡ 3
⎢
3 Using the given information, A = ⎢ 5
⎣ 7
The correct answer is E.
⎡3 4 5 ⎤ ⎡0
⎢
⎥ ⎢
A + B = ⎢5 6 7 ⎥ + ⎢1
⎣7 8 9 ⎦ ⎣2
⎡3 3 3 ⎤
⎢
⎥
= ⎢6 6 6 ⎥
⎣9 9 9 ⎦
The correct answer is D.
−1
0
1
−2⎤
⎥
−1⎥
0⎦
4
6
8
⎡0
5 ⎤
⎥
⎢
7 ⎥ and B = ⎢1
⎣2
9 ⎦
−1
0
1
−2⎤
⎥
−1⎥
0⎦
8.4 Multiplying matrices
8.4 Exercise
1 a AB =
=
b BA =
=
c B2 =
=
P df_Fol i o: 147
8
[−6
2 1
6] [3
14
[12
14
−18]
1
[3
2
8
−1] [−6
−4
[ 30
2
−1]
2
6]
14
0]
1
[3
2 1
−1] [3
7
[0
0
7]
2
−1]
2 a A is a 2 × 3 matrix.
B is a 3 × 2 matrix.
C is a 2 × 2 matrix.
b i AB: 2 × 3 × 3 × 2 does exist.
ii AC: 2 × 3 × 2 × 2 does not exist.
iii BA: 3 × 2 × 2 × 3 does exist.
iv CA: 2 × 2 × 2 × 3 does exist.
v BC: 3 × 2 × 2 × 2 does exist.
3 a The order of A is 2 × 2.
The order of B is 2 × 3.
The order of C is 2 × 2.
b i AB = 2 × 2 × 2 × 3. The middle two numbers are the same, so AB exists.
ii BA = 2 × 3 × 2 × 2. The middle two numbers are not the same, so BA does not exist.
iii AC = 2 × 2 × 2 × 2. The middle two numbers are the same, so AC exists.
iv CA = 2 × 2 × 2 × 2. The middle two numbers are the same, so CA exists.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
147
0
are possible.
2]
148
TOPIC 8 Matrices • EXERCISE 8.4
BC = 2 × 3 × 2 × 2. The middle two numbers are not the same, so BC does not exist.
CB = 2 × 2 × 2 × 3. The middle two numbers are the same, so CB exists.
The order of AB is 2 × 3. (The outer two numbers.)
The order of AC is 2 × 2.
The order of CA is 2 × 2.
The order of CB is 2 × 3.
2 4
2 −2 0
d i AB =
×
1 4]
[−3 3] [3
v
vi
c i
ii
iii
iv
=
=
=
ii AC =
=
=
=
iii CA =
=
=
=
iv CB =
=
=
=
4 M=
4
[3
4 + 12
[−6 + 9
−4 + 4
6+3
16
[ 3
16
12]
2
[−3
0
9
2×0+4×3
[−3 × 0 + 3 × 3
12
[ 9
0
[3
2 × −2 + 4 × 5
−3 × −2 + 3 × 5]
−4 + 20
6 + 15]
16
21]
−2
2
×
5] [−3
0 × 2 + −2 × −3
[ 3 × 2 + 5 × −3
0+6
6
[ − 15
6
[−9
0
[3
−6
27]
−2
2
×
5] [3
0−6
[6 + 15
−2
−1
4
3]
0 × 4+ − 2 × 3
3 × 4 + 5 × 3]
0−6
12 + 15]
−2
1
0 × 2 + −2 × 3
[ 3×2+5×3
−6
[ 21
0 + 16
0 + 12]
0−2
−6 + 5
−8
20]
0−8
0 + 20]
4
[3
3
−1]
=
5
1
×
2] [2
4×1+5×2
[3 × 1 + 2 × 2
4 + 10
[ 3+4
0
4]
0 × −2+ − 2 × 1
3 × −2 + 5 × 1
3
−1]
=
2×0+4×4
−3 × 0 + 3 × 4]
−2
5]
4
0
×
3] [3
0 + 12
[ 0+9
2 × −2 + 4 × 1
−3 × −2 + 3 × 1
1
5
and N =
2]
[2
a i MN =
P df_Fol i o: 148
2×2+4×3
[−3 × 2 + 3 × 3
0 × 0 + −2 × 4
3 × 0 + 5 × 4]
4 × 3 + 5 × −1
3 × 3 + 2 × −1]
14
12 − 5
=
9 − 2] [ 7
7
7]
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.4
ii NM =
=
=
=
b M=
=
=
ii NM =
=
=
⎡ 4
⎢
5 R=⎢ 1
⎣−4
5
2]
1×4+3×3
[2 × 4 + −1 × 3
4+9
[8 − 3
13
[ 5
1×5+3×2
2 × 5 + −1 × 2]
5+6
10 − 2]
11
8]
b
1
and N =
d]
[0
a
[c
i MN =
3
4
×
−1] [3
1
[2
a
[c
b
1
×
d] [0
a
[c
b
d]
1
[0
0 a
1] [ c
a
[c
b
d]
a×1+b×0
[c × 1 + d × 0
0
1]
0
1]
b
d]
1×a+0×c
[0 × a + 1 + c
a×0+b×1
c × 0 + d × 1]
1×b+0×d
0 × b + 1 × d]
⎡0
−2⎤
⎢
⎥
−7⎥ and P = ⎢0
⎣1
6⎦
c N in part b is the identity matrix.
3
5
9
1
0
0
0⎤
⎥
1⎥
0⎦
a Since it is RP and the permutation matrix P is the second matrix in the multiplication of RP, it is a column permutation of R.
b a12 , a23 , a31
Column 2 → 3, column 3 → 1, column 1 → 2, so they all change.
c Row permutation of R → PR
a12 , a23 , a31
Row 1 → 3, row 2 → 1, row 3 → 2, so row 1 is now row 3.
6 a Since the permutation matrix P is the first matrix in the multiplication of PR, it is a row permutation of R.
b a12 , a23 , a31
Row 1 → 3, row 2 → 1, row 3 → 2
Row 1 moves to row 3.
c Row 1 → 2, row 2 → 3, row 3 → 1; therefore 1s go to a13 , a21 , a32 .
⎡0 0 1 ⎤
⎥
⎢
P = ⎢1 0 0 ⎥
⎣0 1 0 ⎦
7 A=B×C
The element a31 is the result of the multiplication of the third row by the first column.
The correct answer is A.
8 a A AB = 2 × 3 × 3 × 2. The middle two numbers are the same, so AB exists.
B BC = 3 × 2 × 2 × 2. The middle two numbers are the same, so BC exists.
C CA = 2 × 2 × 2 × 3. The middle two numbers are the same, so CA exists.
D DE = 3 × 3 × 2 × 4. The middle two numbers are not the same, so DE does not exist.
E CE = 2 × 2 × 2 × 4. The middle two numbers are the same, so CE exists.
The correct answer is D.
P df_Fol i o: 149
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
149
150
TOPIC 8 Matrices • EXERCISE 8.4
b BE = 3 × 2 × 2 × 4. The order is 3 × 4 (the outer two numbers).
The correct answer is A.
c A BA = 3 × 2 × 2 × 3. The order is 3 × 3.
B BC = 3 × 2 × 2 × 2. The order is 3 × 2.
C CA = 2 × 2 × 2 × 3. The order is 2 × 3.
D AB = 2 × 3 × 3 × 2. The order is 2 × 2.
E BD = 3 × 2 × 3 × 3. This matrix product cannot be performed.
The correct answer is D.
d CE =
=
=
=
1
1
×
5] [0
3
[2
3×1+1×0
[2 × 1 + 5 × 0
3+0
[2 + 0
6+1
4+5
2
1
3
−1
4
4]
3×2+1×1
2×2+5×1
9−1
6−5
3 × 3 + 1 × −1
2 × 3 + 5 × −1
12 + 4
8 + 20]
3×4+1×4
2 × 4 + 5 × 4]
3 7 8 16
[2 9 1 28]
The correct answer is E.
e D2 = D × D
⎡1 4
1⎤
1 ⎤ ⎡1 4
⎥
⎥ ⎢
⎢
= ⎢0 5
2⎥
2 ⎥ × ⎢0 5
⎣3 6 −3⎦ ⎣3 6 −3⎦
⎡ 1×1+4×0+1×3
1×4+4×5+1×6
⎢
=⎢ 0×1+5×0+2×3
0×4+5×5+2×6
⎣3 × 1 + 6 × 0 + −3 × 3 3 × 4 + 6 × 5 + −3 × 6
⎡ 4 30
6⎤
⎥
⎢
= ⎢ 6 37
4⎥
⎣−6 24 24⎦
The correct answer is B.
9 a 2A + AB
=2×
=
=
=
2
[−3
4
[−6
=
=
=
4
4
×
3] [3
2×4+4×3
8
+
6] [−3 × 4 + 3 × 3
4
[−6
20
8
+
6] [−3
24
[−9
26
−3]
2
[−3
4
4
×
3] ([3
b A(B + C)
=
4
2
+
3] [−3
2
[−3
4
4
×
3] [6
32
[ 6
34
12]
2×4+4×6
[−3 × 4 + 3 × 6
1 × 1 + 4 × 2 + 1 × −3⎤
⎥
0 × 1 + 5 × 2 + 2 × −3⎥
3 × 1 + 6 × 2 + −3 × −3⎦
5
2]
2×5+4×2
−3 × 5 + 3 × 2]
18
−9]
5
0
+
2] [3
−2
5 ])
3
7]
2×3+4×7
−3 × 3 + 3 × 7]
P df_Fol i o: 150
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
c AB + CD
2 4
4
=
×
[−3 3] [3
=
=
=
2×4+4×3
[−3 × 4 + 3 × 3
20
[−3
=
=
4 × 2 + 5 × −3
3
[ × 2 + 2 × −3
−7
[ 0
5
0
+
2] [3
18
−4
+
−9] [ 13
16 20
[10 −5]
d BA + DC
4 5
2
=
×
[3 2] [−3
=
TOPIC 8 Matrices • EXERCISE 8.4
=2×
=
=
26
[10
1 × −2 + 3 × 5
2 × −2 + (−1) × 5]
13
−9]
1
11
−
8] [2
22
1
−
16] [2
−2
5]
4×4+5×3
1×0+3×3
+
3 × 4 + 2 × 3] [2 × 0 + −1 × 3
1×4+3×3
[2 × 4 + −1 × 3
13
[ 5
3
0
×
−1] [3
0 × 3 + −2 × −1
3 × 3 + 5 × −1]
2
4]
2 44
9]
[−3
e 2DB − D
4
1
3
×
=2×
[2 −1] [3
=2×
3
−1]
2×5+4×2
0 × 1 + −2 × 2
+
−3 × 5 + 3 × 2] [ 3 × 1 + 5 × 2
4
1
+
3] [2
31
9
+
18] [−3
−2
1
×
5] [2
1
5
−
2] [2
3
−1]
1×5+3×2
1
−
2 × 5 + −1 × 2] [2
3
−1]
3
−1]
3
−1]
25 19
[ 8 17]
10 Let matrix S have order m × n and matrix N have order p × q.
m×n×3×3×p×q=3×3
m × (n × 3) × (3 × p) × q = 3 × 3
Since the matrix S × E × N exists, n = 3 and p = 3.
m×q=3×3
Hence, order of matrix N will be 3 × 3.
The correct answer is C.
11 a A 3 × 2 matrix with columns representing washing machines and dryers respectively:
⎡550 160⎤
⎥
⎢
⎢750 220⎥
⎣990 350⎦
b
1.08
[0
0
0.92]
⎡550
⎢
c New prices = ⎢750
⎣990
160⎤
1.08
⎥
220⎥ ×
[0
350⎦
0
0.92]
⎡ 594
147.20⎤
⎥
⎢
= ⎢ 810
202.40⎥
⎣1069.20 322 ⎦
The washing machines are priced at $594, $810 and $1069, and the dryers are priced at $147, $202 and $322.
P df_Fol i o: 151
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
151
TOPIC 8 Matrices • EXERCISE 8.4
152
12 a ⎡1.50
⎢
⎢2.00
⎣2.75
2.50⎤
⎥
3.00⎥
3.25⎦
1.15
[0
0
0.88]
b
100% − 12% = 88%
100% + 15% = 115%
c ⎡1.50 2.50⎤
⎡1.73 2.20⎤
1.15 0
⎢
⎥
⎢
⎥
= 2.30 2.64⎥
⎢2.00 3.00⎥ × [0
0.88] ⎢
⎣2.75 3.25⎦
⎣3.16 2.86⎦
New price for lettuce is $1.73, $2.30 and $3.16.
New price for potatoes (per kg) is $2.20, $2.64 and $2.86.
⎡2.50 0.90⎤
⎢
⎥
13 a Price = ⎢3.50 1.90⎥
⎣4.00 2.50⎦
b A markdown of 15% is equivalent to 85% (0.85) of the price and a markup of 15% is equivalent to 115% (1.15) of the price.
The matrix is
0.85
[0
0
1.15]
⎡2.50 0.90⎤
0.85 0
⎥
⎢
c New prices = ⎢3.50 1.90⎥ ×
1.15]
[0
⎣4.00 2.50⎦
⎡2.50 × 0.85 + 0.90 × 0 2.50 × 0 + 0.90 × 1.15⎤
⎥
⎢
= ⎢3.50 × 0.85 + 1.90 × 0 3.50 × 0 + 1.90 × 1.15⎥
⎣4.00 × 0.85 + 2.50 × 0 4.00 × 0 + 2.50 × 1.15⎦
⎡2.13 1.04⎤
⎥
⎢
= ⎢2.98 2.19⎥
⎣3.40 2.88⎦
⎡ 25.00
9.90 ⎤
⎥
⎢
35.00
19.90 ⎥
14 a Price = ⎢⎢
75.00 ⎥⎥
⎢ 95.00
⎣140.00 128.00⎦
b A markdown of 20% is equivalent to 80% (0.8) of the price.
The matrix is
P df_Fol i o: 152
0.80
[0
0
0.80]
⎡ 25.00
9.90⎤
⎥
⎢
35.00
19.90⎥
0.80 0
c Sale price = ⎢⎢
⎥ × [0
95.00
75.00
0.80]
⎥
⎢
⎣140.00 128.00⎦
⎡ 25.00 × 0.80 + 9.90 × 0
25.00 × 0 + 9.90 × 0.80⎤
⎥
⎢
= ⎢ 35.00 × 0.80 + 19.90 × 0
35.00 × 0 + 19.90 × 0.80⎥
⎣140.00 × 0.80+128.00 × 0 140.00 × 0 + 128.00 × 0.80⎦
⎡ 20.00
7.92⎤
⎢
⎥
28.00
15.92⎥
⎢
=⎢
60.00⎥⎥
⎢ 76.00
⎣112.00 102.40⎦
15 a ⎡10 26⎤
⎡10 × 2.50 + 26 × 0 10 × 0 + 26 × 1.00⎤
⎢
⎥
⎢
⎥
25
45
2.50
0
⎢
⎥×
⎢25 × 2.50 + 45 × 0 25 × 0 + 45 × 1.00⎥
=
⎢22 30⎥ [0
1.00] ⎢⎢22 × 2.50 + 30 × 0 22 × 0 + 30 × 1.00⎥⎥
⎢
⎥
⎣ 5 22⎦
⎣ 5 × 2.50 + 22 × 0
5 × 0 + 22 × 1.00 ⎦
⎡25.00 26.00⎤
⎢
⎥
62.50 45.00⎥
= ⎢⎢
⎥
⎢55.00 30.00⎥
⎣12.50 22.00⎦
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
b ⎡10
⎢
⎢25
⎢22
⎢
⎣ 5
⎡10 × 2.50 + 26 × 1.00⎤
26⎤
⎢
⎥
⎥
45⎥
2.50
25 × 2.50 + 45 × 1.00⎥
×
= ⎢⎢
⎥
30⎥ [1.00] ⎢22 × 2.50 + 30 × 1.00⎥⎥
⎣ 5 × 2.50 + 22 × 1.00⎦
22⎦
TOPIC 8 Matrices • EXERCISE 8.4
⎡ 51.00⎤
⎢
⎥
107.50⎥
= ⎢⎢
⎥
⎢ 85.00⎥
⎣ 34.50⎦
c i The Year 10/11 line had the highest sales figures for pies ($62.50).
ii The Year 10/11 line had the highest total sales figures ($107.50).
16 a Since the permutation matrix is the first matrix in the multiplication PD, it is a row permutation of D.
b a13 , a21 , a32
Row: 3 → 1, 1 → 2, 2 → 3, so row 1 is moved to row 2.
c a13 , a21 , a32
Column: 1 → 3, 2 → 1, 3 → 2, so column 3 is moved to column 2.
d D is not a permutation matrix since it can only have a single 1 in each column and row.
e Row: 1 → 2, 2 → 3, 3 → 1, therefore 1s go to a21 , a32 , a13 .
⎡0 0 1 ⎤
⎢
⎥
P = ⎢1 0 0 ⎥
⎣0 1 0 ⎦
17
iPad
iPad mini
Store A
Store B
Store C
Store D
14
9
10
7
⎡14
⎢
9
a Sales matrix = ⎢⎢
10
⎢
⎣ 7
Price matrix =
550
[ 0
5
7
8
6
5⎤
⎥
7⎥
8⎥⎥
6⎦
0
320]
⎡14 5⎤
⎢
⎥
9 7⎥ 550
0
Total sales at each store = ⎢⎢
⎥ [ 0 320]
10
8
⎢
⎥
⎣ 7 6⎦
⎡7700 1600⎤
⎢
⎥
4950 2240⎥
= ⎢⎢
⎥
⎢5500 2560⎥
⎣3850 1920⎦
⎡14 5⎤
⎢
⎥
9 7⎥ 550
b Total sales at each store = ⎢⎢
⎥
⎢10 8⎥ [320]
⎣ 7 6⎦
⎡9300⎤
⎥
⎢
7190⎥
= ⎢⎢
⎥
⎢8060⎥
⎣5770⎦
Store A = $9300, store B = $7190, store C = $8060 and store D = $5770
c Store A has the highest sales for iPads at $7700.
Store A has the highest total sales at $9300.
P df_Fol i o: 153
⎡12
⎢
15
18 a Total stock = ⎢⎢
⎢10
⎣ 8
10
25
10
20
10
15
10
5
12⎤
⎥
25⎥
10⎥⎥
18⎦
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
153
154
TOPIC 8 Matrices • EXERCISE 8.4
⎡15
0
0
0⎤
⎢
⎥
0 20
0
0⎥
⎢
b Costs = ⎢
0 30
0⎥⎥
⎢ 0
⎣ 0
0
0 32⎦
c ⎡12 10 10 12⎤ ⎡15
0
0
0⎤ ⎡180 200 300 384⎤
⎢
⎥ ⎢
⎥ ⎢
⎥
15
25
15
25
0
20
0
0⎥ ⎢225 500 450 800⎥
⎢
⎥×⎢
=
⎢10 10 10 10⎥ ⎢ 0
0 30
0⎥⎥ ⎢⎢150 200 300 320⎥⎥
⎢
⎥ ⎢
⎣ 8 20
⎦
⎣
5 18
0
0
0 32⎦ ⎣120 400 150 576⎦
d ⎡12 10 10 12⎤ ⎡15⎤ ⎡12 × 15 + 10 × 20 + 10 × 30 + 12 × 32⎤
⎢
⎥ ⎢ ⎥ ⎢
⎥
⎢15 25 15 25⎥ × ⎢20⎥ = ⎢15 × 15 + 25 × 20 + 15 × 30 + 25 × 32⎥
⎢10 10 10 10⎥ ⎢30⎥ ⎢10 × 15 + 10 × 20 + 10 × 30 + 10 × 32⎥
⎢
⎥ ⎢ ⎥ ⎢
⎥
⎣ 8 20
5 18⎦ ⎣32⎦ ⎣ 8 × 15 + 20 × 20 + 5 × 30 + 18 × 32⎦
⎡1064⎤
⎢
⎥
1975⎥
= ⎢⎢
⎥
⎢ 970⎥
⎣1246⎦
⎡600
⎢
19 a February = ⎢480
⎣240
⎡1200
⎢
March = ⎢ 840
⎣1200
⎡1440
⎢
April = ⎢ 600
⎣1560
0⎤
⎥
840⎥
0⎦
500
750
1000
420⎤
⎥
1260⎥
0⎦
1000
1500
1750
1680⎤
⎥
2100⎥
420⎦
750
1500
250
⎡600
500
0⎤ ⎡1200
⎢
⎥ ⎢
b Total sales = ⎢480
750 840⎥ + ⎢ 840
⎣240 1000
0⎦ ⎣1200
⎡3240 2250 2100⎤
⎥
⎢
= ⎢1920 3750 4200⎥
⎣3000 3000
420⎦
⎡3240
⎢
c Number sold = ⎢1920
⎣3000
⎡27
⎢
= ⎢16
⎣25
2250
3750
3000
9
15
12
5⎤
⎥
10⎥
1⎦
420⎤ ⎡1440
⎥ ⎢
1260⎥ + ⎢ 600
0⎦ ⎣1560
1000
1500
1750
⎡ 1
⎢ 120
2100⎤ ⎢
⎥ ⎢
4200⎥ × ⎢ 0
420 ⎦ ⎢
⎢ 0
⎣
0
1
250
0
750
1500
250
1680⎤
⎥
2100⎥
420⎦
0 ⎤
⎥
⎥
0 ⎥⎥
⎥
1 ⎥
420 ⎦
⎡27
9
5⎤ ⎡120⎤
⎢
⎥ ⎢
⎥
d Total sales = ⎢16 15 10⎥ × ⎢250⎥
⎣25 12
1⎦ ⎣420⎦
⎡ 27 × 120 + 9 × 250 + 5 × 420⎤
⎢
⎥
= ⎢16 × 120 + 15 × 250 + 10 × 420⎥
⎣ 25 × 120 + 12 × 250 + 1 × 420⎦
⎡7590⎤
⎢
⎥
= ⎢9870⎥
⎣6420⎦
Store A earned $7590, store B earned $9870 and store C earned $6420.
P df_Fol i o: 154
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.5
20 a
i A2 =
ii A3 =
iii A4 =
iv A5 =
2
[1
1
1]
3
[2
2
1]
5
[3
3
2]
det B = (6 × −5) − (3 × 4)
= −30 − 12
= −42
Fn+1
[ Fn
c A8 =
=
F9
[F 8
F8
F7 ]
34
[21
21
13]
1
[1
30
d A30 =
=
1
[1
Fn
.
Fn−1 ]
1
0]
1
, An can be written as
0]
832 040
514 229]
F31
[F30
F30
, then the 30th Fibonacci
F29 ]
number, F30 , is 832 040.
8.4 Exam questions
1 It is a binary matrix, as the elements all only 0 or 1.
It does not fit the form of the other special binary matrices
(identity or permutation).
The correct answer is A.
2 Matrixes are multiplied by rows × columns.
So we need to multiply the 4th row of A by the 1st
column of B.
4×2+5×4
The correct answer is E.
3 Remember: Rows by Columns
7
(6 × 7) + (6 × 8) = [6 6]
[8]
The correct answer is D.
det A = (−6 × 12) − (15 × −8)
= −72 − −120
= 48
3 a det A = 9 × 5 − 8 × 7
= 45 − 56
= −11
1 1
1
b det B = × − − × 1
2 3
6
1 1
= +
6 6
1
=
3
d det D = −3 × 4 − 6 × −2
= −12 − −12
=0
4 D=
−3
[−2
8.5 Exercise
P df_Fol i o: 155
2
6
and B =
7]
[3
det A = (2 × 7) − (4 × 2)
= 14 − 8
=6
4
−5]
6
will not have an inverse as its determinant is
4]
equal to zero (D is a singular matrix).
5 det A = (5 × 8) − (6 × 6)
= 40 − 36
=4
A−1 =
8
1
4 [−6
−6
5]
6 a |C|| = ad − bc
= −5 × 4 − −1 × 10
= −20 + 10
= −10
b C−1 =
4
1
−10 [−10
7 a A−1 = −
b B
8.5 The inverse of a matrix and its determinant
2
[4
8
4]
det B = (6 × 4) − (3 × 8)
= 24 − 24
=0
−1
1 A=
6
−8
and B =
12]
[3
c det C = 5 × 8 − 9 × 3
= 40 − 27
= 13
1 346 269
[ 832 040
Given that A30 =
−6
[ 15
2 A=
8 5
[5 3]
b The numbers produced in the solution matrix are Fibonacci
numbers. If we let Fn represent the nth Fibonacci number,
then the situation can be represented as follows.
For a matrix A =
155
5
1
11 [−7
⎡ 1
⎢ 3
= 3⎢
⎢−1
⎣
1⎤
6⎥
⎥
1⎥
2⎦
1
−0.4
or
−5] [ 1
−8
9]
8 −9
1
13 [−3
5]
d The inverse matrix does not exist.
8 det C = 9 × 3 − 5 × 5
= 27 − 25
=2
c C−1 =
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
−0.1
0.5]
156
TOPIC 8 Matrices • EXERCISE 8.5
C−1 =
9 A=
11
[10
AB =
=
=
BA =
=
=
−5
9]
3
1
2 [−5
12
11
and B =
11]
[−10
11
[10
12
11
×
11] [−10
1
[0
0
1]
11 × 11 + 12 × −10
[10 × 11 + 11 × −10
11
[−10
−12
11
×
11] [10
11 × 11 + −12 × 10
−10
× 11 + 11 × 10
[
1
[0
−12
11]
−12
11]
11 × −12 + 12 × 11
10 × −12 + 11 × 11]
12
11]
11 × 12 + −12 × 11
−10 × 12 + 11 × 11]
0
1]
AB = BA = I
Therefore, A and B are the inverse of each other.
10 det T = −2 × 5 − 3 × −2
= −10 − −6
= −4
1 5 −3
4 [2 −2]
The correct answer is A.
11 det P = 12 × −6 − 4 × −12
= −72 − −48
= −24
T−1 = −
P−1 = −
1 −6
24 [ 12
−4
12]
1⎤
⎡ 1
⎢ 4
6⎥
⎥
=⎢
1
1
⎢−
⎥
−
⎣ 2
2⎦
The correct answer is E.
12 a The determinant = 0.1 × 0.45 − 0.2 × 0.25
= 0.045 − 0.05
= −0.005
−0.2 ⎤
⎡ 0.45
⎢ −0.005 −0.005 ⎥
⎥
b The inverse = ⎢
0.1 ⎥
⎢ −0.25
⎣ −0.005 −0.005 ⎦
−90 40
[ 50 −20]
13 Store each matrix onto the graphics calculator, then use the
calculator functions to find the determinant and inverse.
=
P df_Fol i o: 156
a i det A = −28
3 ⎤
⎡1
⎢ 14
⎥
28
⎥
ii A−1 = ⎢
⎢3 −5⎥
⎣ 14
28 ⎦
b i det B = 6
⎡ −1 − 2 ⎤
⎢
3⎥
−1
⎥
ii B = ⎢
⎢− 1 − 1 ⎥
⎣ 2
2⎦
c i det C = −1
−4
3
ii C−1 =
[ 7 −5]
d i det D = −0.1
−2.5 10
ii D−1 =
[ 2 −4]
e i det E = 0
ii The inverse does not exist because E is a singular matrix
(the determinant equals 0).
f i det F = 0
ii The inverse does not exist because F is a singular matrix
(the determinant equals 0).
g i det G = 96
1
1⎤
⎡− 5
⎢ 24
4
24 ⎥
⎢
⎥
1⎥
⎢ 1 −1
ii G−1 = ⎢
4
2
4⎥
⎢
⎥
1
5⎥
⎢ 1
−
⎣ 24
4
24 ⎦
h i det H = −3072
1
1
⎡ 1
−
0⎤
⎥
⎢ 16
16
16
⎥
⎢
1
1
1
⎥
⎢
0
−
⎢ 12
12
6⎥
−1
⎥
ii H = ⎢
1
1⎥
⎢ 0
0
−
⎢
4
4⎥
⎥
⎢
1
1⎥
⎢− 5 − 3
⎣ 48
16
48
3⎦
i i det I = −72
1
1
1 ⎤
⎡ −1
⎢ 3
6
6
6 ⎥
⎢
⎥
1
2
1
1 ⎥
⎢
−
⎢ 3
3
3
3 ⎥
⎥
ii I−1 = ⎢
1
1
2
1
⎢
⎥
−
⎢ 9
9
9
9 ⎥
⎢
⎥
1
1
1 ⎥
⎢ 1
−
⎣ 12
12
12
6 ⎦
j i and ii It is not possible to determine the determinant and
inverse of J, as J is not a square matrix.
14 a
AX = B
A−1 AX = A−1 B
X = A−1 B
A−1 =
X=
=
b AX =
3
[−4
3
[−4
3
[−3
−5
7]
−5 6
7] [3
−44
61]
−3
7]
13
−1]
[
X = A−1
13
[−1]
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
=
=
15 A =
9
[7
−5
13
7] [−1]
3
[−4
=
b AX =
5
−4
and B =
4]
[−5
9
[7
−1
5
4]
9
[−17
=
−4
[−5
−3
7]
=
8
7]
17 a FX =
9
=
[7
−1
5
4]
=
b GX =
12
−2]
[
=
−5
6
×
5] [3
−3
3]
−33
0.7
=
15] [0.5
1 21
30 [15
=
=
−1.1
0.5]
=
b XA = B (post-multiply both sides by A)
X = BA−1
=
=
=
6
[3
−3
1 6
×
3] 30 [0
−3
6
×
3] [0
1 6
30 [3
18
−5
5]
−5
5]
−45
1.2
=
0] [0.6
1 36
30 [18
=
=
d CX =
5
[0
6
5
1
×
6] 30 [−3
1 5
30 [0
15
1
30 [−18
−1.5
0]
0
5]
100
[120]
both
0.83
1]
sides
1 −1
13 [−3
5
−3
×
4] [7]
1 −26
13 [ 13 ]
2
[−1]
−1.0
−1
(pre-multiply both sides by G )
[ 2.9 ]
−1.0
[ 2.9 ]
1.2
−1.0
×
0.2] [ 2.9]
2.8
1
0.68 [−0.1
0.68
1
0.68 [0.68]
1
[1]
3
1
X=
6]
[−3
4
2]
2
X=
[3
3
6]
−1 ⎡
1
⎢
⎢
⎣−3
6⎤
8 ⎥⎥
−
3⎦
4⎤
⎥
⎥
2⎦
8.5 Exam questions
0
5]
25
0.5
=
30] [−0.6
100
(pre-multiply
[120]
X = C−1 ×
P df_Fol i o: 157
5
6
×
6] [−3
2
[3
5
[7]
⎡ 5
X = ⎢⎢
−3
⎣
c XC = A (post-multiply both sides by C)
X = AC−1
=
5
−1
(pre-multiply both sides by F )
[7]
X = G−1 ×
58
[−102]
1 6
30 [0
20
[10]
=−
16 a AX = B (pre-multiply both sides by A)
X = A−1 B
=
1 600
30 [300]
=−
12
[−2]
157
0
100
×
5] [120]
6
1
30 [−3
X = F−1 ×
8
7]
12
[−2]
X = A−1
=
=
44
[−59]
a AX = B
X = A−1 B
=
TOPIC 8 Matrices • EXERCISE 8.5
by
C−1 )
1 In the first position, we need an N, but that is located in the
last position in RAMON, so the first row will be
[0 0 0 0 1].
In the second position, we need an O, which is located in the
fourth position in RAMON, so the second row will be
[0 0 0 1 0].
In the third position, we need an R, which is located in the
first position in RAMON, so the third row will be
[1 0 0 0 0].
In the fourth position, we need an M, which is located in the
third position in RAMON, so the fourth row will be
[0 0 1 0 0].
And, finally, in the fifth position, we need the A, which is in
the second position in RAMON, so the fifth row will be
[0 1 0 0 0].
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.6
158
⎡0 0 0
⎢
⎢0 0 0
⎢1 0 0
⎢
⎢0 0 1
⎣0 1 0
The correct answer is E.
0
1
0
0
0
1⎤ ⎡ R ⎤ ⎡ N ⎤
⎥⎢ ⎥ ⎢ ⎥
0⎥ ⎢ A ⎥ ⎢ O ⎥
0 ⎥ ⎢M⎥ = ⎢ R ⎥
⎥⎢ ⎥ ⎢ ⎥
0 ⎥ ⎢ O ⎥ ⎢M⎥
0⎦ ⎣ N ⎦ ⎣ A ⎦
a 4 x
10
=
[18 b] [y] [ 6]
There will not be a unique solution when the determinant of
the coefficient matrix, a 4 , equals to zero.
[18 b]
2 Write as a matrix equation:
(a × b) − (4 × 18) = 0
ab = 72
Going through each of the options, the only factors of 72 are
when a = 2 and b = 36.
The correct answer is A.
3
Q = wP
QP−1 = w
P−1 = Q−1 w
P−1
= Q−1
w
1
∴ Q−1 = P−1
w
The correct answer is A.
8.6 Dominance and communication matrices
8.6 Exercise
1 a
i A–D, A–D
ii A–C–D, A–E–D
iii A–B–E–D
⎡0 1 1 2 1 ⎤
⎡0 0 0 2 1 ⎤
⎢
⎥
⎢
⎥
0
0
0
0
1
⎢
⎥
⎢0 0 0 1 0 ⎥
2
b A = ⎢0 0 0 1 0 ⎥, A = ⎢0 0 0 0 0 ⎥
⎢
⎥
⎥
⎢
⎢0 0 0 0 0 ⎥
⎢0 0 0 0 0 ⎥
⎣0 0 0 1 0 ⎦
⎣0 0 0 0 0 ⎦
2 a i A–D
ii A–B–D, A–C–D, A–C–D, A–E–D
iii A–C–E–D
⎡0 0 0 4 0 ⎤
⎡0 1 1 1 1 ⎤
⎥
⎢
⎥
⎢
0
0
0
1
0
⎢0 0 0 0 0 ⎥
⎥
⎢
b A = ⎢0 0 0 2 1⎥, A2 = ⎢0 0 0 0 0⎥
⎥
⎢
⎥
⎢
⎢0 0 0 0 0 ⎥
⎢0 0 0 0 0 ⎥
⎣0 0 0 0 0 ⎦
⎣0 0 0 1 0 ⎦
3 a i A–D
ii A–B–D, A–B–D, A–C–D, A–C–D
iii None possible
b i A–D
ii A–B–D, A–E–D, A–C–D, A–C–D
iii A–C–E–D, A–C–E–D
4 The one-stage pathways are denoted by A and the two-stage
pathways are denoted by A2 .
⎡0
⎢
0
a A = ⎢⎢
⎢0
⎣0
⎡0
⎢
⎢0
b A = ⎢0
⎢
⎢0
⎣0
1
0
0
0
2
0
0
0
⎡0
1⎤
⎥
⎢
2 ⎥ 2 ⎢0
,
A
=
⎢0
1⎥⎥
⎢
⎣0
0⎦
1
0
0
0
0
2
0
0
0
0
1
1
1
0
1
0
0
0
0
⎡0
1⎤
⎥
⎢
0⎥
⎢0
2
1⎥, A = ⎢0
⎥
⎢
0⎥
⎢0
⎣0
0⎦
0
0
0
0
4⎤
⎥
0⎥
0⎥⎥
0⎦
0
0
0
0
0
0
0
0
0
0
4
0
1
0
0
2⎤
⎥
0⎥
0⎥
⎥
0⎥
0⎦
5 One-stage routes: L1 −L2 (1)
Two-stage routes: L1 −B−L2 , L1 −A−L2 , L1 −A−L2 (3)
Three-stage routes: L1 −A–B−L2 , L1 −A–B−L2 (2)
There are 6 routes overall.
6 a The only way to reach D from A is A–B–D.
b The only way to reach A from D is D–C–A.
c There are two ways to reach B from D: D–C–B and
D–C–A–B.
7 a i O–G
ii O–R–G
iii O–C–R–G, O–C–O–G, O–S–O–G
b i None possible
ii R–C–S
iii R–G–O–S, R–C–O–S
c i None possible
ii S–O–R
iii S–O–C–R
8 P
9 E
10 a The dominant vertex is the one with the edges directed
away from it. In order, the results are:
First: D, second: C, third: B, fourth: A
b The one-stage pathways are given by the matrix:
⎡0 0 0 0 ⎤
⎢
⎥
1 0 0 0⎥
A = ⎢⎢
⎥
⎢1 1 0 0 ⎥
⎣1 1 1 0 ⎦
The two-stage pathways are given by the matrix:
⎡0 0 0 0 ⎤
⎢
⎥
0 0 0 0⎥
A2 = ⎢⎢
⎥
⎢1 0 0 0 ⎥
⎣2 1 0 0 ⎦
⎡0 0 0 0 ⎤
⎢
⎥
1 0 0 0⎥
A + A2 = ⎢⎢
⎥
⎢2 1 0 0 ⎥
⎣3 2 1 0 ⎦
The row with the highest sum represents the winner.
Sum of the first row is 0; A is placed fourth.
Sum of the second row is 1; B is placed third.
Sum of the third row is 3; C is placed second.
Sum of the fourth row is 6; D is placed first.
The finishing order is the same as that obtained in part a
using the inspection method.
11 a The dominant vertex is the one with the edges directed
away from it. In order, the results are:
first: C, second: E, third: A, fourth: B and fifth: D.
b The one-stage pathways are given by the matrix:
P df_Fol i o: 158
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
⎡0 1 0 1 0 ⎤
⎢
⎥
⎢0 0 0 1 0 ⎥
A = ⎢1 1 0 1 1 ⎥
⎢
⎥
⎢0 0 0 0 0 ⎥
⎣1 1 0 1 0 ⎦
The two-stage pathways are given by the matrix:
⎡0 0 0 1 0⎤
⎢
⎥
⎢0 0 0 0 0⎥
A2 = ⎢1 2 0 3 0⎥
⎢
⎥
⎢0 0 0 0 0⎥
⎣0 1 0 2 0⎦
⎡0 1 0 2 0 ⎤
⎢
⎥
⎢0 0 0 1 0 ⎥
2
A + A = ⎢2 3 0 4 1 ⎥
⎢
⎥
⎢0 0 0 0 0 ⎥
⎣1 2 0 3 0 ⎦
The row with the highest sum represents the winner.
Sum of the first row is 3; A is placed third.
Sum of the second row is 1; B is placed fourth.
Sum of the third row is 10; C is placed first.
Sum of the fourth row is 0; D is placed fifth.
Sum of the fifth row is 6; E is placed second.
The finishing order is the same as that obtained in part a
using the inspection method.
12 The dominant vertex can be found by inspection, looking for
the vertex with the most edges directed away from it.
Alternatively, the method of representing the one-stage and
two-stage pathways in matrix form and then adding these
matrices can be used. Both methods will reveal the same
answer.
a C
b Z
13 The vertex T has all its edges directed away from it. Nothing
leads into it, so it is the dominant vertex.
The correct answer is D.
14
B
TOPIC 8 Matrices • EXERCISE 8.6
159
a ‘Talk to’ (vertical) so C talks to A, B and D.
b ‘Receive calls’ (horizontal) so B receives calls from C and
D.
c They cannot call themselves.
d ‘Call’ or ‘talk to’ (vertical) so D cannot call C.
16
Q
P
P→R
Q → P, R and S
R → P, Q and S
S → P and Q
S
R
P ⎡0 0 1 0 ⎤
⎢
⎥
Q 1 0 1 1⎥
Communication matrix = ⎢⎢
R ⎢1 1 0 1⎥⎥
S ⎣1 1 0 0 ⎦
17 a ‘Talk to’ (vertical) so C talks to A and D.
b ‘Receive calls’ (horizontal) so B receives calls from D.
c ‘Call/talk to’ (vertical) B can only call D.
d D cannot call A.
e D can receive calls from A, B and C.
18 a All arrows associated with F lead away from F, into vertices
E and D. No arrows lead into F, so F exerts the most
influence.
b Both C and A have arrows leading into them and none
directed away. They exert no influence on the other
employees.
c This can be done by inspection or matrix methods. F is
most dominant, then E, then D (as E is dominant over D).
Fourth is B as it is dominant over C.
Last are A and C (jointly).
P
Q
R
S
8.6 Exam questions
A
A → B and C
B → C and D
C → A and D
D → A, B and C
D
C
A ⎡0
⎢
B 0
Communication matrix = ⎢⎢
C ⎢1
D ⎣1
B
C
1
0
0
1
1
1
0
1
A ⎡0
⎢
B 1
15 Communication matrix = ⎢⎢
C ⎢1
D ⎣1
B
C
0
0
1
1
1
0
0
0
A
A
P df_Fol i o: 159
0⎤
⎥
1⎥
1⎥⎥
0⎦
D
0⎤
⎥
1⎥
1⎥⎥
0⎦
D
1 There are four scenarios that need to be considered: Brie
losing to Andy or Andy losing to Brie, and Cleo losing to
Della or Della losing to Cleo.
When the sum of one-step and two-step dominances is
calculated (use CAS to do), we find that Brie will win with
the highest dominance of 9e.
The correct answer is B.
2 Use the diagram to set up a communication matrix.
receiver
S T U V
S ⎡0 1 0 1⎤
⎢
⎥
T 0 0 0 1⎥
sender ⎢⎢
U ⎢0 1 0 1⎥⎥
V ⎣1 1 1 0⎦
The correct answer is D.
3 Y cannot send directly to U or V, only directly to W or Z.
However, Z does not communicate directly with W. Therefore
the answer is 0 — Y cannot send a message to W by sending it
via one other person.
The correct answer is A.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.7
160
8.7 Transition matrices and Leslie matrices
P
8.7 Exercise
1 a
i A3 =
=
To
0.5
[0.5
0.5
0.5]
0.5
[0.5
0.5
0.5]
0.75
ii B =
[0.25
iii C 4 =
0.2
0.8]
0.56
[0.44
0.44
0.56]
=
0.2
0.8]
0.51
[0.49
0.8
[0.2
⎡1
⎢2
= ⎢
⎢1
⎣2
2 a
i A0 =
=
ii B0 =
iii C 0 =
=
A
To
0.68
[0.32
0.23
0.77]
b
From
North South
To
North 0.80
South [0.20
0.15
0.85]
0.80
[0.20
0.15
0.85]
=
From
A
B
0
To
0.25
0.75]
b S0 =
0
S2 =
0
S2 =
0
1]
0
0.58
S2 =
0.08
0.92]
0.08
0.92]
125
[125]
S2 = T2 × S0
0
1]
0.2
0.8]
A 0.9
B [0.1
T = 0.9
[0.1
0
1]
⎡0.7 0.1 0.2⎤
⎢
⎥
iv D0 = ⎢0.2 0.7 0.1⎥
⎣0.1 0.2 0.7⎦
⎡1 0 0 ⎤
⎢
⎥
= ⎢0 1 0 ⎥
⎣0 0 1 ⎦
b Each matrix produced in part a is the identity matrix.
3 a
0.42
P df_Fol i o: 160
0.23
0.77]
=
8
From
B
A 0.68
B [0.32
5 a
0.5
0.5]
0.8
[0.2
0.27
0.73]
4 a
0.49
0.51]
0.75
[0.25
1
[0
4
1⎤
2⎥
⎥
1⎥
2⎦
1
=
[0
From
WA WB
0.69
[0.31
46
1
[0
0.36
0.64]
WA
0.27
WB [0.31
]
They must vertically add to 1.
0.2
0.8]
0.5
[0.5
0.58
[0.42
To
0.8
[0.2
0.8
iv C =
[0.2
b C46 =
2
0.38
0.63]
8
0.36
0.64]
=
0.63
[0.38
=
P 0.58
Q [0.42
b
0.25
0.75]
2
=
3
From
Q
2
0.90
[0.10
0.08
125
0.92] [125]
0.818
[0.182
0.1456 125
0.8544] [125]
120.45
[129.55]
After two weeks, 120 wagons are located at point A and
130 wagons at point B.
c S6 = T6 × S0
S0 =
0.40
[0.60]
0.64
P
Q
0.36
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.7
S6 =
S6 =
0.90
[0.10
6
0.08
0.40
0.92] [0.60]
0.43
[0.57]
After 6 weeks, 43% of the wagons will be at point A and
57% of the wagons at point B.
6 a T = 0.84 0.7
[0.16 0.3]
b Find the missing values first.
0.65 + 0.15 + 0.20 = 1
0.20 + 0.68 + 0.12 = 1
0.03 + 0 + 0.97 = 1
⎡0.65 0.20 0.03⎤
⎢
⎥
T = ⎢0.15 0.68
0⎥
⎣0.20 0.12 0.97⎦
⎡0.52 0.16 0.14⎤
⎢
⎥
7 a T = ⎢0.12 0.40 0.07⎥
⎣0.36 0.44 0.79⎦
b March 2022, n = 5
S5 = T5 × S0
⎡1200⎤
⎥
⎢
S0 = ⎢ 800⎥
⎣1000⎦
0.60
[0.40
Let S0 =
0.85
0.15]
1
represent that the train is on time.
[0]
a On time the following Friday, then n = 4
S4 =
=
0.60
[0.40
1
0.85
×
0.15] [0]
4
0.681 25
[0.318 75]
There is a 68.1% chance that the train will be on time on
Friday.
b On time the following Monday, then n = 5. (No trains
operate on weekends.)
S5 =
=
0.60
[0.40
0.85
1
×
0.15] [0]
5
0.6797
[0.3203]
There is a 32.0% chance that the train will be late the
following Monday.
⎡0.52 0.16 0.14⎤ ⎡1200⎤
⎥
⎢
⎥ ⎢
S5 = ⎢0.12 0.40 0.07⎥ × ⎢ 800⎥
⎣0.36 0.44 0.79⎦ ⎣1000⎦
⎡ 694.2⎤
⎥
⎢
= ⎢ 368.8⎥
⎣1937.0⎦
Store A will have 694 customers, store B will have 369
customers and store C will have 1937 customers.
c Test n = 50, then n = 51
50
⎡1200⎤
⎡0.52 0.16 0.14⎤
⎥
⎢
⎥
⎢
S50 = ⎢0.12 0.40 0.07⎥ × ⎢ 800⎥
⎣1000⎦
⎣0.36 0.44 0.79⎦
5
⎡ 689.2⎤
⎢
⎥
= ⎢ 364.9⎥
⎣1945.9⎦
⎡1200⎤
⎡0.52 0.16 0.14⎤
⎥
⎥
⎢
⎢
S51 = ⎢0.12 0.40 0.07⎥ × ⎢ 800⎥
⎣1000⎦
⎣0.36 0.44 0.79⎦
⎡ 689.2⎤
⎥
⎢
= ⎢ 364.9⎥
⎣1945.9⎦
689.2
Store A:
× 100% = 23.0%
3000
364.9
Store B:
× 100% = 12.2%
3000
1945.9
Store C:
× 100% = 64.9%
3000
d Convert the answers from part c from decimals to fractions
on CAS.
9
24
17
Store A:
, store B:
, store C:
74
74
37
51
P df_Fol i o: 161
8 T=
161
⎡0.6
⎢
0.2
9 a T = ⎢⎢
⎢0.1
⎣0.1
0.1
0.8
0.1
0
0
0
1
0
0.05⎤
⎥
0 ⎥
0 ⎥⎥
0.95⎦
⎡0.6 0.1 0 0.05⎤
⎥
⎢
0.2 0.8 0 0 ⎥
T2 = ⎢⎢
⎥
⎢0.1 0.1 1 0 ⎥
⎣0.1 0
0 0.95⎦
⎡0.385 0.14 0 0.0775⎤
⎥
⎢
0.28
0.66 0 0.01 ⎥
= ⎢⎢
0.19 1 0.005 ⎥⎥
⎢0.18
⎣0.155 0.01 0 0.9075⎦
(From CAS.)
b T has no steady state because the matrix T 2 contains zero
elements.
c
From
A
B
C
A ⎡0.7 0.1 0.25⎤
⎢
⎥
To B ⎢0.2 0.9 0 ⎥
C ⎣0.1 0
0.75⎦
2
⎡0.7 0.1 0.25⎤
⎢
⎥
T = ⎢0.2 0.9 0 ⎥
⎣0.1 0
0.75⎦
The correct answer is A.
⎡0.6 0.1 0
0.05⎤
⎢
⎥
0.2 0.8 0.05 0 ⎥
10 a T = ⎢⎢
0 ⎥⎥
⎢0.1 0.1 0.9
⎣0.1 0
0.05 0.95⎦
b Use CAS for the calculations in part b.
i After two weeks, n = 2
⎡4500⎤
⎢
⎥
4000⎥
S0 = ⎢⎢
⎥
⎢2000⎥
⎣1500⎦
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
162
TOPIC 8 Matrices • EXERCISE 8.7
S2 = T2 × S0
⎡2423.75⎤
⎢
⎥
4127.5 ⎥
= ⎢⎢
⎥
⎢ 3122.5 ⎥
⎣2326.25⎦
Texcal has the largest customer base after 2 weeks
(4128).
ii After eight weeks, n = 8
⎡4500⎤
⎢
⎥
4000⎥
S0 = ⎢⎢
⎥
⎢2000⎥
⎣1500⎦
S8 = T8 × S0
⎡1296.6⎤
⎢
⎥
2981.0⎥
= ⎢⎢
⎥
⎢4115.9⎥
⎣3606.6⎦
Oilmart has the largest customer base after 8 weeks
(4116).
iii In the long term. Test large values of n.
S∞ = T ∞ × S0
⎡1200⎤
⎥
⎢
2000⎥
= ⎢⎢
⎥
⎢3200⎥
⎣5600⎦
CP has the largest customer base in the long term
(5600).
0.60
11 a S2 =
[0.40
=
S3 =
=
S4 =
=
0.35 106
1
+
0.65] [ 72] [2]
89.8
[91.2]
0.60
[0.40
86.8
[97.2]
0.60
[0.40
0.35 89.8
1
+
0.65] [91.2] [2]
0.35 86.8
1
+
0.65] [97.2] [2]
87.1
[99.9]
There are 100 Jersey cows expected to be milked on day 4.
b Day 1: 106 + 72 = 178 cows
Day 4: 87 + 100 = 187 cows
Difference = 187 − 178 = 9 more cows milked on day 4
than on day 1.
12
A
C
A 0.65
C [0.35
0.23
0.77]
P df_Fol i o: 162
S2 =
0.65
[0.35
=
S3 =
=
S4 =
=
1139
[1042]
0.84
[0.16
1296
[1066]
0.84
[0.16
0.22 1139
110
+
0.78] [1042] [ 71]
0.22 1296
110
+
0.78] [1066] [ 71]
1433
[1110]
Week 2: Sweet Cola: 1139, Tangy Lemon: 1042
Week 3: Sweet Cola: 1296, Tangy Lemon: 1066
Week 4: Sweet Cola: 1433, Tangy Lemon: 1110
b Week 1: 950 + 1050 = 2000
Week 4: 1433 + 1110 = 2543
2543 − 2000 = 543
0.75 0.4 34
5
15 a S2 =
−
[0.25 0.6] [26] [3]
=
30.9
[21.1]
So 21 students are expected to attend extra guitar lessons in
week 2.
5
0.75 0.4 30.9
−
b S3 =
[0.25 0.6] [21.1] [3]
=
26.615
[17.385]
Week 2: 31 + 21 = 52 students
Week 3: 27 + 17 = 44 students
Difference = 52 − 44 = 8; 8 students were not expected to
attend in week 3, but they were there in week 2.
0.8 0.3 136
3
16 a S2 =
−
[0.2 0.7] [108] [5]
=
S3 =
=
S4 =
5 A
B=
[−2] C
48 A
S1 =
[41] C
13 a 61%
b 82%
c 16 additional people go to supermarket A and 7 additional
people go to supermarket B.
0.84 0.22
950
110
14 a S2 =
+
[0.16 0.78] [1050] [ 71]
=
138.2
[ 97.8]
0.8
[0.2
3
0.3 138.2
−
0.7] [ 97.8] [5]
136.9
[ 91.1]
0.8
[0.2
3
0.3 136.9
−
0.7] [ 91.1] [5]
133.85
[ 86.15]
The number of sheep in paddock B in week 4 is 86.
0.23 48
5
+
0.77] [41] [−2]
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.7
b S5 =
0.3 133.85
5
−
0.7] [ 86.15] [3]
0.8
[0.2
129.925
=
[ 82.075]
Week 5: 130 + 82 = 212 sheep
Week 4: 134 + 86 = 220 sheep
Difference = 220 − 212 = 8 sheep
0.95 0.08 55 000
1200
17 S2 =
+
0.05
0.92]
72
000]
[
[
[ 955]
(Note: n = 51 produces the same result.)
The correct answer is B.
20
Number of
hamsters (s)
=
S4 =
=
0.95
[0.05
0.08 63 045
1200
+
0.92] [68 265] [ 955]
66 554
[66 911]
2022: city A: 59 210, city B: 69 945
2023: city A: 63 045, city B: 68 265
2024: city A: 66 554, city B: 66 911
0.95 0.55
18 a T =
[0.05 0.45]
S0 =
=
3
⎡ 69⎤
⎢
⎥
= ⎢138⎥
⎣ 23⎦
2.6
0.5
0
0.80
[0.20
0.25 256
6
+
0.75] [194] [4]
259.3
[200.7]
0.80
[0.20
0.25 259.3
6
+
0.75] [200.7] [4]
263.615
[206.385]
In week 3 there are expected to be 264 shoppers at
supermarket A.
b S4 =
=
0.80
[0.20
0.25 263.615
6
+
0.75] [206.385] [4]
268.488 25
[211.511 75]
After 4 weeks, supermarket A = 268 and supermarket
B = 212.
Difference = 268 − 212 = 56 shoppers
V
Q
0
[1]
To determine long-term probability, let n = 50.
P df_Fol i o: 163
25
0.60
=
0.55
1
×
0.45] [0]
0
0.55
×
0.45]
[1]
50
22
(Note: n = 51 produces the same result.)
11
1
The long-term probabilities are dry:
, wet:
.
12
12
c They should insure the event as there is a very good chance
that the day will be dry.
19 Test for n = 50 (and for n = 51)
50
⎡0.5 0.2 0.3⎤
⎡100⎤
⎢
⎥
⎢
⎥
S50 = ⎢0.3 0.8 0.3⎥ × ⎢ 50⎥
⎣0.2 0
⎣ 80⎦
0.4⎦
32
0.6
S3 =
S0 =
0.95
[0.05
⎡ 11 ⎤
⎢ 12 ⎥
=⎢ ⎥
⎢1⎥
⎣ 12 ⎦
55
0.50
=
0.922
[0.078]
The probability it will rain in three days’ time is 0.078.
b Initially the day is wet, then:
S50 =
2− < 3
Survival rate (r)
21 a S2 =
1
(initially the day is dry)
[0]
0.95
S3 =
[0.05
1− < 2
⎡0.6 2.6 0.5⎤
⎢
⎥
b L = ⎢0.5 0
0 ⎥
⎣0
0.6 0 ⎦
c Using CAS:
4
⎡0.6 2.6 0.5⎤ ⎡55⎤ ⎡387.45⎤
⎢
⎥ ⎢ ⎥ ⎢
⎥
S4 = ⎢0.5 0
0 ⎥ ⎢32⎥ = ⎢135.28⎥
⎣0
0.6 0 ⎦ ⎣25⎦ ⎣ 47.50⎦
387.45 + 135.28 + 47.50 = 570.23
The total number of hamsters after 4 years is approximately
570.
0.08 59 210
1200
+
0.92] [69 945] [ 955]
63 045
[68 265]
0.95
[0.05
0− < 1
Birth rate (b)
⎡55⎤
⎢ ⎥
a S0 = ⎢32⎥
⎣25⎦
59 210
=
[69 945]
S3 =
Age group of
hamsters (years)
163
T=
S0 =
S50 =
=
V 0.27
Q[0.73
0.10
0.90]
6000
[ 400]
0.27
[0.73
0.10
6000
×
0.90]
[ 400]
50
771
[5629]
In the long term there is a loss of 6000 − 771 = 5229
Victorians to Queensland (initial 6000 and steady 771).
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
164
TOPIC 8 Matrices • EXERCISE 8.8
8.7 Exam questions
1 Use the information given to construct the transition matrix, keeping in mind that the columns must add to 1. [1 mark]
D
C
The transition matrix will be: 0.65 0.45 D
[0.35 0.55] C
The correct answer is D.
68.125
2 S4 = T 4 × S0 =
[31.875]
32% of customers bought their coffee at Giorgio’s on Friday.
The correct answer is B.
3 Use the transition diagram to set up a transition matrix to identify the correct answer.
⎡0.55 0.25 0.35⎤
⎥
⎢
T = ⎢0.45 0.60 0.25⎥
⎣0
0.15 0.40⎦
The correct answer is C.
8.8 Review
8.8 Exercise
Multiple choice
1 a12 is in row 1, column 2.
So a12 = −0.5
The correct answer is C.
2 0 is in row 2, column 2.
So a22 = 0
The correct answer is C.
3 E−F
⎡ 1.2 −0.5⎤ ⎡0.2 −0.5⎤
⎢
⎥ ⎢
⎥
= ⎢ 3.6
5.0⎥ − ⎢2.4
2.5⎥
⎣−3.5
2.2⎦ ⎣0
1.1⎦
⎡ 1.0 0 ⎤
⎢
⎥
= ⎢ 1.2 2.5⎥
⎣−3.5 1.1⎦
The correct answer is A.
4 A=B×C
The element a13 is the result of multiplying the first row by the third column.
The correct answer is D.
5 AB = 3 × 3 × 2 × 2. AB does not exist because the inner two numbers are not the same.
The correct answer is A.
6 The order of matrix BD is 2 × 2 × 2 × 4.
The outer two numbers define the order of the product — 2 × 4.
The correct answer is E.
7 A BA = 2 × 2 × 3 × 3; does not exist.
B EC = 3 × 2 × 2 × 3; order is 3 × 3.
C CA = 2 × 3 × 3 × 3; order is 2 × 3.
D AC = 3 × 3 × 2 × 3; does not exist.
E CE = 2 × 3 × 3 × 2; order is 2 × 2.
The correct answer is E.
8 ED
⎡4
1⎤
1 2
3 4
⎥
⎢
= ⎢3
0⎥ ×
[0 1 −1 4]
⎣2 −1⎦
⎡ 4×1+1×0
⎢
=⎢ 3×1+0×0
⎣2 × 1 + −1 × 0
4×2+1×1
3×2+0×1
2 × 2 + −1 × 1
4 × 3 + 1 × −1
3 × 3 + 0 × −1
2 × 3 + −1 × −1
4×4+1×4 ⎤
⎥
3×4+0×4 ⎥
2 × 4 + −1 × 4⎦
P df_Fol i o: 164
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.8
⎡4 9 11 20⎤
⎢
⎥
= ⎢3 6
9 12⎥
⎣2 3
7
4⎦
The correct answer is D.
9
9
[5
8
6]
The determinant = 9 × 6 − 8 × 5
= 54 − 40
= 14
The correct answer is C.
10 det P = −1
0
1
0 −1
=
1] [1 −1]
[−1
The correct answer is B.
11 The columns of a transition matrix must add up to 1.
The missing value in the first column, t31 , is:
0.3 + 0.3 + t31 = 1
t31 = 0.4
The missing value in the third column, t13 , is:
t13 + 0.3 + 0.6 = 1
t13 = 0.1
The correct answer is D.
12 The total population is 10 000 + 12 000 = 22 000.
P−1 = −1
S4 =
=
0.5
[0.5
4
0.2
10 000
0.8] [12 000]
Rounding to the nearest figure that appears in the options, the
6300
answer appears to be
.
[15 700]
The correct answer is A.
Note that B and C are not representative of the answer as the
total number represented in the matrix is not equal to 22 000.
13 The columns of a transition matrix must add up to 1. The
missing shop B value must be 20% (0.2) and the missing shop
C value must be 5% (0.5).
⎡0.85 0.2 0.2 ⎤
⎢
⎥
The transition matrix would be ⎢0.1
0.6 0.05⎥.
⎣0.05 0.2 0.75⎦
The correct answer is C.
A L N R
A L N R
A ⎡0
⎢
L 1
14 A + A2 = ⎢⎢
N ⎢1
R ⎣1
A L N R
A ⎡0 0 0 0⎤
⎢
⎥
L ⎢2 0 0 1⎥
15 Consider the dominance matrix
N ⎢⎢3 1 0 2⎥⎥
R ⎣1 0 0 0⎦
The sum of the first row (Alex) = 0; the sum of the second
row (Lena) = 3; the sum of the third row (Nathan) = 6; the
sum of the fourth row (Rachel) = 1
Therefore, the finishing order (starting from the winner) is:
Nathan, Lena, Rachel, Alex.
The correct answer is D.
16
From
A
B
A 0.8 0.3
B [0.2 0.7]
The columns must add up to 1.
The correct answer is C.
To
17 A
B
A → B, C and D
B → A and D
C → A and B
D→B
D
C
A ⎡0
⎢
B 1
The communication matrix = ⎢⎢
C ⎢1
D ⎣0
The correct answer is B.
A
6315.8
[15 684.2]
0
0
1
0
0
0
0
0
0 ⎤ A ⎡0
⎥
⎢
1 ⎥ L ⎢1
+
1⎥⎥ N ⎢⎢2
0 ⎦ R ⎣0
A L N R
A ⎡0 0 0 0 ⎤
⎢
⎥
L 2 0 0 1⎥
= ⎢⎢
N ⎢3 1 0 2⎥⎥
R ⎣1 0 0 0 ⎦
The correct answer is D.
0
0
0
0
0
0
0
0
0⎤
⎥
0⎥
1⎥⎥
0⎦
165
B
C
1
0
1
1
1
0
0
0
1⎤
⎥
1⎥
0⎥⎥
0⎦
D
Short answer
18 a A has 3 rows and 2 columns.
It is a 3 × 2 rectangular matrix.
B has 3 rows and 3 columns.
It is a 3 × 3 square matrix.
b The a32 element is 2 and the b32 element is 0.
c In matrix A, the number 5 is in row 2, column 1.
a21 = 5
In matrix B, the number 5 is in row 3, column 1.
b31 = 5
19 a [A + B]
=
=
=
1
[3
5
3
+
2] [5
1+3
[3 + 5
5+0
2 + 2]
0
2]
4 5
[8 4]
b B−C
B and C are not of the same order. Subtraction is not
possible.
P df_Fol i o: 165
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.8
166
−5
−3
c 2C
=2×
=
=
1
[3
2×1
[2 × 3
2 × −5
2 × −3
−10
−6
2
[6
0
3]
2×0
2 × 3]
0
6]
20 a An increase of 12% is equivalent to 112% (1.12) of the
price.
A decrease of 5% is equivalent to 95% (0.95) of the price.
Price change =
b
249
[680
=
=
1.12
[0
29
1.12
×
49] [0
0
0.95]
0
0.95]
249 × 1.12 + 29 × 0
[680 × 1.12 + 49 × 0
278.88
[761.60
249 × 0 + 29 × 0.95
680 × 0 + 49 × 0.95]
27.55
46.55]
21 A × B = C
4 × 2 × 2 × 3=4 × 3
For the multiplication to be possible, A must be of order
4 × 2.
22
From
A
B
To
A 0.85
B [0.15
0.40
0.60]
0.85
[0.15
0.40
0.60]
T=
a S0 =
200
[200]
After 3 months, n = 3
S3 =
=
0.85
[0.15
0.40
200
×
0.60] [200]
3
282.625
[117.375]
After 3 months, 283 customers will be purchasing their
groceries from store A and 117 customers will be
purchasing their groceries from store B.
b At the end of 6 months, n = 6
40
S6 = T6 × S0
S0 =
[60]
0.85
=
[0.15
=
0.40
40
×
0.60] [60]
6
72.5
[27.5]
72.5% will purchase their groceries from store A, and
27.5% will purchase their groceries from store B.
P df_Fol i o: 166
⎡3 0 0 ⎤
⎢
⎥
23 A = ⎢1 4 0⎥, order (3 × 3); lower triangular matrix; 2,
⎣7 5 9 ⎦
2 = 4; 7 is in x31 .
⎡ 7⎤
⎢ ⎥
B = ⎢−2⎥, order (3 × 1); column matrix; 2, 2 doesn’t exist; 7
⎣ 5⎦
is in x11 .
1 0
C=
, order (2 × 2); identity matrix; 2, 2 = 1; 7 is
[0 1]
not in the matrix.
⎡−3
2 −1⎤
⎢
⎥
D=⎢ 2
0 −4⎥, order (3 × 3); symmetrical matrix; 2,
⎣−1 −4
7⎦
2 = 0; 7 is in x33 .
24 a PR is a row permutation of R.
b a13 , a21 , a32
Row: 3 → 1, 1 → 2, 2 → 3, so row 1 is moved to row 2.
c Row: 1 → 2, 2 → 1, 3 → 3, therefore 1s go to a12 , a21 , a33 .
⎡0 1 0 ⎤
⎢
⎥
P = ⎢1 0 0 ⎥
⎣0 0 1 ⎦
25 a C can ‘talk to’ (vertical) B and D.
b B can ‘receive calls’ (horizontal) from C and D.
c D can talk to all three, therefore D can talk to the most
people.
d C cannot call A or itself; therefore A and C.
Extended response
26 a January:
⎡450
⎢
A = ⎢310
⎣250
February:
⎡320
⎢
B = ⎢158
⎣130
March:
⎡540
⎢
C = ⎢212
⎣278
1200
1000
750
600
580
345
1300
1080
850
1500⎤
⎥
1200⎥
600⎦
900⎤
⎥
920⎥
700⎦
1400⎤
⎥
1569⎥
900⎦
b Total number = A + B + C
⎡1310 3100
⎢
= ⎢ 680 2660
⎣ 658 1945
3800⎤
⎥
3689⎥
2200⎦
⎡0.50⎤
⎢
⎥
c Profit matrix = ⎢1.50⎥
⎣4.50⎦
Total profit (for each store)
⎡1310 3100 3800⎤ ⎡0.50⎤
⎥ ⎢
⎥
⎢
= ⎢ 600 2660 3689⎥ × ⎢1.50⎥
⎣ 658 1945 2200⎦ ⎣4.50⎦
⎡22 405.00⎤
⎢
⎥
= ⎢20 890.50⎥
⎣13 146.50⎦
The profits are: store A: $22 405; store B: $20 890.50 and
store C: $13 146.50.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 8 Matrices • EXERCISE 8.8
d Total profit for the chain
= store A + store B + store C
= $22 405 + $20 890.50 + $13 146.50
= $56 442
e
To
A
A ⎡0.80
⎢
B ⎢0.15
C ⎣0.05
⎡0.80
⎢
T = ⎢0.15
⎣0.05
From
B
C
0.05 0.10⎤
⎥
0.70 0.30⎥
0.25 0.60⎦
0.05
0.70
0.25
0.10⎤
⎥
0.30⎥
0.60⎦
⎡40⎤
⎢ ⎥
f The 120 new tents are distributed evenly, so S0 = ⎢40⎥.
⎣40⎦
S2 = T2 × S0
2
⎡0.80 0.05 0.10⎤ ⎡40⎤
⎢
⎥ ⎢ ⎥
= ⎢0.15 0.70 0.30⎥ × ⎢40⎥
⎣0.05 0.25 0.60⎦ ⎣40⎦
⎡36.3⎤
⎥
⎢
= ⎢48.7⎥
⎣35 ⎦
After 2 weeks 36 tents will be at store A, 49 tents at store B
and 35 tents at store C.
g Determine the long-term trend.
Try n = 50.
S50 = T50 × S0
50
⎡40⎤
⎡0.80 0.05 0.10⎤
⎢ ⎥
⎢
⎥
= ⎢0.15 0.70 0.30⎥ × ⎢40⎥
⎣40⎦
⎣0.05 0.25 0.60⎦
⎡31.3⎤
⎢
⎥
= ⎢52.2⎥
⎣36.5⎦
Store A should get 31, store B 52 and store C 37.
27 a 100 adults, 0 eggs and 0 nymphs to start so the initial state
matrix is:
⎡ 0⎤
⎢
⎥
S0 = ⎢ 0⎥
⎣100⎦
⎡ 0
0
200⎤
⎥
⎢
b L = ⎢0.02 0
0⎥
⎣ 0
0.05
0⎦
c Sn = Ln S0
⎡ 0
0
200⎤ ⎡ 0⎤ ⎡20 000⎤
⎢
⎥⎢
⎥ ⎢
⎥
S1 = ⎢0.02 0
0⎥ ⎢ 0⎥ = ⎢ 0 ⎥
⎣ 0
0.05
0⎦ ⎣100⎦ ⎣ 0 ⎦
⎡ 0
0
200⎤ ⎡ 0⎤ ⎡ 0⎤
⎢
⎥ ⎢
⎥ ⎢
⎥
S2 = ⎢0.02 0
0⎥ ⎢ 0⎥ = ⎢400⎥
⎣ 0
0.05
0⎦ ⎣100⎦ ⎣ 0⎦
Continue this process up to n = 8
Putting this in a table:
2
167
Age
Eggs
Nymphs
Adults
0
1
2
3
4
5
6
7
8
0
20 000
0
0
4000
0
0
800
0
0
0
400
0
0
80
0
0
16
100
0
0
20
0
0
4
0
0
d The number of locusts continue to rise and fall over the
8 years.
0.60 0.30 184
3
28 a S2 =
+
0.40
0.70]
168]
[
[
[5]
=
S3 =
=
163.8
[196.2]
0.60
[0.40
0.30 163.8
3
+
0.70] [196.2] [5]
160.14
[207.86]
In week 3, 160 customers are expected to purchase their
fuel from station A.
0.60 0.30 160.14
3
b S4 =
+
0.40
0.70]
207.86]
[
[
[5]
=
161.442
[214.558]
Station A in week 2 = 164
Station A in week 4 = 161
Difference = 164 − 161 = 3 customers
Station A lost 3 customers.
8.8 Exam questions
1 a Brie and Dex – in the matrix, they both have 1’s to send to
each other.
[1 mark]
b Elena → Dex → Brie → Chai [1 mark]
c Alex → Brie → Dex and Alex → Elena → Dex
Award 1 mark for both paths.
[Note: The matrix M2 has been transcribed incorrectly. If
students calculated M2 themselves, they also found that
there was 2-step communication possible between Chai and
Brie, so the answers of C → A → B and C → D → B (must
have both) were also accepted].
2 a One row and three columns means the
order is 1 × 3.
[1 mark]
700
b
= 2800 shoppers
[1 mark]
0.25
⎡1104 1296 1056⎤
⎢
⎥
c i Q = ⎢ 621
729
594⎥
⎣ 575
675
550⎦
Award 1 mark for both red numbers.
P df_Fol i o: 167
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
168
TOPIC 8 Matrices • EXERCISE 8.8
ii 594 shoppers were in the clothing area
[1 mark]
at Westmall.
d Total amount =
⎡21.30⎤
⎥
⎢
[135 143 131] × ⎢34.00⎥ = [9663.20] [1 mark]
⎣14.70⎦
e If A2019 is a 3 × 1 matrix, then A2020 must be too.
A2020 =K × A2019
(3 × 1) = (m × n) × (3 × 1)
So K will be a 3 × 3 matrix.
⎡1.05
0
0⎤
⎢
⎥
K = ⎢ 0 0.85
[1 mark]
0⎥
⎣ 0
⎦
0 0.99
3 CT × (AT × B) = (3 × 4)T × ((2 × 4)T × (2 × 3))T
= (4 × 3) × ((4 × 2) × (2 × 3))T
= (4 × 3) × (4 × 3)T
= (4 × 3) × (3 × 4)
= (4 × 4)
The correct answer is E.
4 a If the determinant of a matrix is zero, there is no inverse.
But the determinant is 1; therefore, there is
an inverse. [1 mark]
b −7 9
[−4 5]
Award 1 mark for all three correct red numbers.
x
−7 9 7
5
c
=
=
[y] [−4 5] [6] [2]
Therefore, 2 sandwich bars are preferred in Grandmall’s
food court. [1 mark]
5 a 3×2
[1 mark]
b Add up column 2: 50 + 20 + 40 = 110
[1 mark]
⎡50⎤
⎢ ⎥
c L = ⎢20⎥
[1 mark]
⎣40⎦
T
⎡3
6 22 19⎤
⎢
⎥
d R = ⎢1 10
[1 mark]
7
2⎥
⎣1
3 10 26⎦
e The number of cars that parked for two hours
in area C.
[1 mark]
VCAA Examination Report note:
Some students recognised area C and 2 hours but did not
refer to the number of cars parked.
T
P df_Fol i o: 168
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.2
Topic 9 — Undirected graphs, networks and trees
9.2 Basic concepts of a network
4 a
A
C
9.2 Exercise
1 a
b
2 a
b
Degree A = 3
Degree B = 4
Degree C = 4
B
b
Vertices = 6; edges = 8
Vertices = 7; edges = 9
Vertices = 5; edges = 5
A
C
B
Degree A = 3
Degree B = 8
Degree C = 3
5
A
Vertex A; degree = 5
The correct answer is D.
6 a A
C
Degree A = 2
Degree B = 2
Degree C = 1
B
c
Vertices = 6; edges = 9
b
A
B
d
Vertices = 7; edges = 11
c
Degree A = 3
Degree B = 4
Degree C = 2
B
A
3
P df_Fol i o: 169
Vertices = 9; edges = 16
C
C
Degree A = 3
Degree B = 2
Degree C = 6
This figure has 7 vertices and 11 edges.
The correct answer is C.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
169
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.2
170
d B
C
9 a
2
A
7 a
b
C
E
b A
B
C
E
F
D
V = {A, B, C, D, E, F, G}
E = {(A, B), (B, C), (B, E), (B, F), (C, E),
(C, F), (E, F), (E, G), (F, G)}
G
8 a 1
3
b
5
V = {1, 2, 3, 4, 5}
E = {(1, 2), (1, 3), (2, 3), (3, 4), (4, 5)}
2
U
V
4
3
5
1
4
6
F
V = {A, B, C, D, E, F}
E = {(A, B), (A, D), (A, E), (B, D),
(B, E), (B, E), (C, D), (D, F)}
B
1
V = {1, 2, 3, 4, 5, 6}
E = {(1, 2), (1, 4), (1, 6), (2, 3), (2, 6), (3, 4), (4, 6)}
2
D
5
6
Degree A = 4
Degree B = 1
Degree C = 0
A
3
1 2 3 4 5 6
1 ⎡ 0 1 0 1 0 1 ⎤
⎢
⎥
2 ⎢ 1 0 1 0 0 1 ⎥
3 ⎢ 0 1 0 1 0 0 ⎥
⎢
⎥
4 ⎢ 1 0 1 0 0 1 ⎥
5 ⎢⎢ 0 0 0 0 0 0 ⎥⎥
6 ⎣ 1 1 0 1 0 0 ⎦
10 Starting with vertex A, there are single connections to vertices
C, B and F.
A
C
B
F
Add a connection (B, F) and vertex D, which is connected to
E (new vertex) and F.
D
4
Y
W
C
A
E
B
X
V = {U, V, W, X, Y, Z}
E = {(U, V), (U, W), (U, X), (V, W),
(V, X), (V, Z), (W, X), (W, Y),
(X, Z)}
F
Complete the network by adding (E, F) and (F, G) along with
new vertex G.
Z
c
1
5
7
3
2
4
D
C
A
E
B
F
11 a V = {1, 2, 3, 4}
E = {(1, 2), (1, 4), (2, 3), (3, 4)}
G
1
V = {1, 2, 3, 4, 5, 6, 7}
E = {(1, 2), (1, 3), (1, 5), (2, 4), (2, 6),
(2, 6), (2, 6), (2, 7), (3, 3), (3, 6),
(5, 7)}
4
6
d 6
7
2
1
5
9
P df_Fol i o: 170
3
4
3
A
C
E
V = {1, 2, 3, 4, 5, 6, 7, 8, 9}
E = {(1, 5), (1, 6), (1, 8), (2, 5), (2, 7),
(2, 9), (3, 4), (3, 5), (3, 8), (4, 5),
(4, 9), (5, 6), (5, 7), (5, 8), (5, 9),
(6, 7)}
8
b V = {A, B, C, D, E}
E = {(A, B), (A, C), (A, C), (B, B), (B, C),
(B, D), (C, D)}
2
B
D
12 a V = {1, 2, 3, 4}
E = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.2
b 1
n(V) = 5, d(V) = 4
n(E) = 5 × 4 ÷ 2
= 10
n(V) = 8, d(V) = 7
n(E) = 8 × 7 ÷ 2
= 28
a n(V) = 10, d(V) = 9
n(E) = 10 × 9 ÷ 2
= 45
b n(V) = 20, d(V) = 19
n(E) = 20 × 19 ÷ 2
= 190
c n(V) = 100, d(V) = 99
n(E) = 100 × 99 ÷ 2
= 4950
3
c Degree (1) = 4
Degree (2) = 4
Degree (3) = 4
Degree (4) = 4
2
4
13 i a V = {1, 2, 3, 4, 5}
E = {(1, 2), (1, 3), (1, 4), (1, 5)(2, 3),
(2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}
b
5
1
4
c Degree (1) = 4
Degree (2) = 4
Degree (3) = 4
Degree (4) = 4
Degree (5) = 4
ii a V = {1, 2, 3, 4, 5, 6, 7, 8}
E = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 3),
(2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (3, 4), (3, 5), (3, 6),
(3, 7), (2, 8), (3, 5), (3, 6) , (3, 7), (3, 8) , (4, 5), (4, 6),
(4, 7), (4, 8), (5, 6), (5, 7), (5, 8), (6, 7), (6, 8), (7, 8)}
b
1
3
2
8
15 V = {A, B, C, D, E, F, G, H, I, J, K, L, M, N}
• Allan (A) married Betty (B)∶ AB
• 3 children AB − C
AB − D
AB − E
• Charles (C) married Francis
(F) = CF
• 2 children CF − G
CF − H
• Doris (D) married Ian
(I) = DI
• 1 child DI − J
2
• Earl (E) married Karen
(K) = EK
• 3 children EK − L
3
7
6
EK − M
4
c Degree (1) = 7
Degree (2) = 7
Degree (3) = 7
Degree (4) = 7
Degree (5) = 7
Degree (6) = 7
Degree (7) = 7
Degree (8) = 7
14 From questions 12 and 13, the number of edges from each
vertex was one less than the total number of vertices (because
there are no loops). Therefore a network of:
5
a 10 vertices would have vertices of degree 9
b 20 vertices would have vertices of degree 19
c 100 vertices would have vertices of degree 99.
If looking at the total number of edges in a network where
each vertex is joined by exactly one edge and no loops, we
use the formula ‘degree of vertex multiplied by number of
vertices, then divided by 2’.
n(V) = 4, d(V) = 3
n(E) = 4 × 3 ÷ 2
=6
EK − N
AB
CF
16
DI
G
H
J
1
3
5
2
EK
L
M
N
4
1 2 3 4 5
1 ⎡ 0 1 1 0 0 ⎤
⎥
⎢
2 ⎢ 1 0 1 0 0 ⎥
3 ⎢ 1 1 0 1 0 ⎥
⎢
⎥
4 ⎢ 0 0 1 0 1 ⎥
5 ⎣ 0 0 0 1 0 ⎦
P df_Fol i o: 171
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
171
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.3
172
17 a
1
2
4
5
b
3
6
E = {(1, 2), (1, 4), (1, 5), (2, 3), (2, 4),
(2, 5), (4, 5), (4, 6), (5, 6)}
1
1 ⎡ 0
⎢
2 ⎢ 1
3 ⎢ 0
⎢
4 ⎢ 1
5 ⎢⎢ 1
6 ⎣ 0
b
2
1
0
1
1
1
0
3
0
1
0
0
0
0
1
2
3
4
5
6
4
1
1
0
0
1
1
5
1
1
0
1
0
1
1
1 ⎡ 0
⎢
2 ⎢ 1
3 ⎢ 0
⎢
4 ⎢ 0
5 ⎣ 0
1
6
0 ⎤
⎥
0 ⎥
0 ⎥
⎥
1 ⎥
1 ⎥⎥
0 ⎦
c
E = {(1, 2), (1, 4), (1, 5), (2, 3), (3, 5), (4, 4),
(4, 7), (5, 7), (5, 7), (5, 7), (5, 6)}
2
1
0
1
1
0
3
0
1
0
2
2
2
3
4
5
4
0
1
2
2
1
5
0 ⎤
⎥
0 ⎥
2 ⎥
⎥
1 ⎥
0 ⎦
1 2 3 4 5 6
1 ⎡ 0 1 1 0 0 0 ⎤
⎢
⎥
2 ⎢ 1 0 0 0 0 1 ⎥
⎢
3
1 0 0 2 0 2 ⎥
⎢
⎥
4 ⎢ 0 0 2 0 0 1 ⎥
5 ⎢⎢ 0 0 0 0 0 0 ⎥⎥
6 ⎣ 0 1 2 1 0 0 ⎦
1
2
3
6
7
1
1 ⎡ 0
⎢
2 ⎢ 1
3 ⎢ 0
⎢
4 ⎢ 1
5 ⎢ 1
⎢
6 ⎢ 0
7 ⎣ 0
c 1
2
1
0
1
0
0
0
0
3
0
1
0
0
1
0
0
4
1
0
0
1
0
0
1
2
3
4
5
6
5
1
0
1
0
0
1
3
6
0
0
0
0
1
0
0
7
0 ⎤
⎥
0 ⎥
0 ⎥
⎥
1 ⎥
3 ⎥
⎥
0 ⎥
0 ⎦
4
5
9.2 Exam questions
1 There are 5 vertices with a degree of 3:
7
E = {(1, 2), (2, 3), (2, 5), (2, 6), (3, 5), (3, 6),
(5, 6), (5, 7), (6, 6), (6, 7), (7, 7)}
1
1 ⎡ 0
⎢
2 ⎢ 1
3 ⎢ 0
⎢
4 ⎢ 0
5 ⎢ 0
⎢
6 ⎢ 0
7 ⎣ 0
18 a
2
1
0
1
0
1
1
0
3
0
1
0
0
1
1
0
4
0
0
0
0
0
0
0
5
0
1
1
0
0
1
1
6
0
1
1
0
1
1
1
7
0 ⎤
⎥
0 ⎥
0 ⎥
⎥
0 ⎥
1 ⎥
⎥
1 ⎥
1 ⎦
The correct answer is E.
2 The correct answer is E.
3 2 + 3 + 2 + 3 + 2 = 12
The correct answer is E.
9.3 Planar graphs and Euler’s formula
V(V − 1)
2
4(4 − 1)
E=
=6
2
There are 6 edges.
V(V − 1)
b E=
2
8(8 − 1)
E=
= 28
2
There are 28 edges.
9.3 Exercise
1 a E=
1 2 3 4
1 ⎡ 0 1 1 0 ⎤
⎢
⎥
2 ⎢ 1 0 1 0 ⎥
3 ⎢⎢ 1 1 0 1 ⎥⎥
4 ⎣ 0 0 1 0 ⎦
1
3
2
2
4
P df_Fol i o: 172
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.3
3 a Move vertex E to the left, removing the edge AE
crossing CD.
Move the edge AD to be ‘outside’ B.
6
1
2
3
B
A
Face 1—4 dots∶ degree = 4
Face 2—3 dots∶ degree = 3
Face 3—3 dots∶ degree = 3
Face 4 (outside)—6 dots∶ degree = 6
7 A complete graph with 7 vertices has vertices of degree 6, so
the total number of edges equals:
n(V) × d(V) ÷ 2 = 6 × 7 ÷ 2
= 21 edges
The correct answer is C.
8 a 3-dimensional original
4
D
C
E
b
A
B
1
2
4
D
C
3
F
5
Face 1—3 dots∶ degree = 3
Face 2—3 dots∶ degree = 3
Face 3—3 dots∶ degree = 3
Face 4—3 dots∶ degree = 3
Face 5 (outside) − 4 dots∶ degree = 4
E
E
4 a
D
C
A
B
Planar graph
C
F
D
E
b Number of vertices, V = 9
Number of edges, E = 16
Number of faces, F = 9
V=E−F+2
9 = 16 − 9 + 2?
Yes, Euler’s formula is confirmed.
A
b
B
9
C
D
B
c
E
J
A
F
H
(Note: already planar)
G
10 a Original
A
B
d 1
2
E
F
3
4
5
D
C
5
2
4
173
1
Planar
A
B
II
I
3
5 The graph is not planar because edge (E, F) crosses (A, D).
The correct answer is B.
C
E III
IV F
V
D
P df_Fol i o: 173
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.3
174
b Faces = 5
Planar graph
B
A
B
II
I
E III
IV F
C
V
D
A
c Degrees of each face:
Face I = 3
Face II = 3
Face III = 4
Face IV = 4
Face V = 4
11 a Original
1
2
I
6
4
Planar graph
5
III
2
3
IV
V
4 VI
6
b There are 6 faces.
5
II
1
I
Verify:
V=E−F+2
V=6−4+2
V=4
n(V ) = 4
n(E ) = 6
n(F ) = 4
14 Since the degree of a single node is determined by the number
of edges (E) ‘leaving’ it, and each such edge must be the
‘entering’ edge of another node, each edge is counted twice in
the sum of degrees (S). Thus the sum must be an even
number. And since each edge is counted twice, S = 2E.
15 n(E) = 5; V = E − F + 2
n(V) = 4; 4 = 5 − F + 2
n(F) = ?; F = 3
The correct answer is B.
16 Original diagram
III
2
C
4
A
5
II
3
1 2
D III
II
5 IV
3
6
1
I
C
B
D
E
3
IV
V
D
4 VI
6
12 a Degree of vertex E in figure is 4.
The correct answer is D.
A
A
B
Planar graph
B
C
D
C
E
C
E
D
H
G
b A
A
B
17
B
B
C
E
F
D
Euler’s rule: V = E − F + 2
V = 13 − 7 + 2
V=8
n(E ) = 13
n(F ) = 7
n(V ) = 8
13 Original diagram
H
A
E
D
C
G
B
The corridors/doorways leading to each room:
B
F
E
A
D
C
Note that the hall space between rooms E and D belongs to
room F and the hall space between rooms B and C belongs to
room A.
D
P df_Fol i o: 174
A
C
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.4
n(n − 1)
where n = 8
a
2
18 8 people in a room shake hands with each other.
8(8 − 1)
= 28 handshakes
2
b Let the 8 people be represented by the letters A to H.
A
B
H
C
G
D
3
175
B
A
E
G
1
C
F
D
J
H
Euler trail = A–B–C–D–E–G–F–H–J
4 Using the diagram from the question and starting with
vertex 2:
B
A
F
c
E
A B C D E F G H
A ⎡ 0 1 1 1 1 1 1 1 ⎤
⎥
⎢
B ⎢ 1 0 1 1 1 1 1 1 ⎥
C ⎢ 1 1 0 1 1 1 1 1 ⎥
⎢
⎥
D ⎢ 1 1 1 0 1 1 1 1 ⎥
E ⎢ 1 1 1 1 0 1 1 1 ⎥
⎥
⎢
F ⎢ 1 1 1 1 1 0 1 1 ⎥
⎢
⎥
G ⎢ 1 1 1 1 1 1 0 1 ⎥
H ⎣ 1 1 1 1 1 1 1 0 ⎦
E
G
1
C
F
D
J
H
Euler trail = A–J–H–E–G–F–D–C–B
5 Removing the edge E will mean that the top vertex is no
longer connected to the other vertices of the graph, so this
edge is a bridge. The correct answer is E.
6 2
3
5
9.3 Exam questions
1 There is no Eulerian trail, because it is not possible to get to
and from the vertex on the right without passing over the edge
twice. The other 4 statements are true.
The correct answer is D.
2 v+f−e=2
7+f−9=2
f=4
The correct answer is D.
3 For a Eulerian circuit to be possible, all vertices must be of
even degree. There are four vertices with odd degree vertices,
so two more edges are needed.
The correct answer is C.
Euler circuit = 1–2–5–3–4–5–1 (starting at vertex 1)
7 a
E
C
1
4
F
A
C
A
F
E
1
9.4 Exercise
B
For an Euler circuit, make all vertices of even
degree ⇒ join F to A.
The correct answer is B.
b Removing the edge FE will mean that the vertex F is no
longer connected to the other vertices of the graph, so this
edge is a bridge.
8 a
B
9.4 Walks, trails, paths, cycles and circuits
1
D
D
G
B
b
B
A
A
1
C
E
D
1
G
B
A
C
E
C
F
D
J
An Euler trail uses every edge exactly once.
A–B–C–D–E–F–G
2
E
G
F
H
Figure b only, as ‘all vertices must be of an even degree’.
The correct answer is D.
F
Euler trail = E–F–G–D–C–B–A
1
D
G
P df_Fol i o: 175
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.4
176
16
9
L
6
M
4
3
S
G
H
J
1
R
Q
A
B
K
C
D
N
17
5
2
3
4
6
18
6
1
5
Hamiltonian cycle = 2–3–4–5–6–1–2
The correct answer is A.
4
3
Hamiltonian path = 7–6–5–4–3–2–1
P
a Euler trail:
A–G–L–M–S–H–J–K–F–P–E–R–D–O–C–Q–B–N
b Euler circuit:
H–S–M–L–G–A–N–B–Q–C–O–P–F–E–D–R–K–J
10
5
2
F
E
O
C
3
2
B
F
1
2
8
7
1
A
D
1
7
• Start with vertex 1.
Smallest path = 1–3–2–1
∴ Subcircuit1 = {1, 2, 3}
• Vertex 2 would have same as vertex 1.
• Move to vertex 3.
Subcircuit2 = 3–4–5–3
∴ S = {1, 2, 3, 4, 5}
• Move to vertex 5.
Subcircuit3 = 5–6–7–8–5
∴ S = {1, 2, 3, 4, 5, 6, 7, 8}
Subcircuit1 and subcircuit2 joined at 3.
Subcircuit2 and subcircuit3 joined at 5.
1–3–4–5–6–7–8–5–3–2–1
11 For an Euler circuit to exist, all vertices must be of even
degree. Graph D is the only graph that has every vertex of
even degree.
The correct answer is D.
12 A Hamiltonian path uses every vertex exactly once.
2
3
4
5
a Euler trail = 2–1–3–4–5–2–3
b Euler circuit = cannot be done because vertices 2 and 3 are
odd.
c Hamiltonian path = 1–2–3–4–5
d Hamiltonian cycle = 1–3–4–5–2–1
E
19
5 6
1
4
3
7
8
10
2 9
a By joining 4 to 7 and 3 to 8, the network now contains a
Hamiltonian cycle and an Euler circuit because all vertices
are now even.
b Euler circuit∶ 3–2–1–3–4–1–5–4–7–6–10–7–8–10–9–8–3
Hamiltonian cycle∶ 3–2–1–5–4–7–6–10–9–8–3
20
Starcomed
Larebil
1
4
7
Hamiltonian path = 2–1–6–5–4–3–7
6
13
Yrtnuoc
Ruobal
5
As the path visits each vertex once only, but ends at a different
vertex, it is a Hamiltonian path.
The correct answer is C.
14 A Hamiltonian cycle uses every vertex once, starting and
finishing at the same vertex.
2
a Euler trail = Yrtnuoc–Ruobal–Noitaneno–
Yrtnuoc–Starcomed–Larebil–Noitaneno–Larebil
b ‘Does not travel more than once on any road’ = Euler trail
c
Starcomed
Noitaneno
Larebil
Yrtnuoc
Ruobal
3
Euler circuit = Larebil–Starcomed–Yrtnuoc–Ruobal–
Noitaneno–Yrtnuoc–Larebil–Noitaneno–Larebil
d An Euler circuit is required to travel each road only once
and start/end at Larebil.
Noitaneno
4
1
7
Hamiltonian cycle = 2–1–6–5–4–7–3–2
15 The network cannot contain a Hamiltonian cycle/path or
Euler circuit/trail because the graph is not connected.
The correct answer is E.
6
P df_Fol i o: 176
5
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.5
21
Starcomed
9.5 Trees and their applications
Larebil
Yrtnuoc
177
9.5 Exercise
Ruobal
1 Trees connecting vertices A and B without going through
vertex F:
B
Noitaneno
a and b A Hamiltonian path is required to visit each town
once only. Hamiltonian path = Ruobal–Yrtnuoc–Starcomed–
Larebil–Noitaneno
22
C
F
F
E
D
A
E
B
A–C–B
A–D–C–B
A–E–D–C–B
C
A
a Hamiltonian path = C–D–A–E–F–B
b Hamiltonian cycle = C–A–E–F–B–D–C
c Path a is a Hamiltonian path and path b is a Hamiltonian
cycle.
2 E
C
F
B
D
1 A Hamiltonian cycle passes through each vertex only once
and starts and finishes at the same vertex.
There is no edge connecting E and D. Therefore, option D is
incorrect.
The correct answer is D.
2 An Eulerian trail exists if the graph has two vertices with an
odd degree and the degrees of the other vertices are even. The
given graph has six vertices of which four are of an odd
degree and two are of an even degree.
Removing any edge between two vertices that are of an odd
degree will change the network to an Eulerian trail. There are
five different ways in which this can be done.
The correct answer is E.
VCAA Examination Report note:
Students should be familiar with scenarios involving adding
or removing an edge from a graph to enable an Eulerian trail.
This question relied on the knowledge that a graph will have
an Eulerian trail if exactly two of the vertices of that graph
have an odd degree.
3
8
2
2
3
7
3
School
4
2
Home
2
8
13
The diagram is highlighted to show the shortest path.
Time = 3 + 2 + 3 + 4 + 2 = 14 minutes
The correct answer is C.
P df_Fol i o: 177
ii and iv are trees, as i and iii start and finish at same vertex.
The correct answer is C.
3
9.4 Exam questions
6
A
D
A tree is a connected subgraph containing no loops, cycles or
parallel edges.
a
b
4 Trees connecting vertices A and B, without going through
vertex F:
B
C
F
D
E
A
A–D–B
A–C–B
A–D–C–B
A–C–D–B
A–E–D–B
A–E–D–C–B
5 Possible trees:
A–C–D–B
4 + 7 + 8 = 19
A–E–F–B
10 + 3 + 5 = 18
A–E–C–D–B 10 + 5 + 7 + 8 = 30
A–E–F–D–B 10 + 3 + 9 + 8 = 30
A–C–E–F–B 4 + 5 + 3 + 5 = 17
A–C–D–F–B 4 + 7 + 9 + 5 = 25
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.5
178
A
4
10
E
3
5
4
C
5
9
7
D
F
Shortest path: A–C–E–F–B = 17 km
6 Shortest possible tree:
1–2–5–6–8
=3+4+6+3+0
= 16 minutes
7 a
B 10
B
12
A
4 + 4 + 5 + 7 = 20
The correct answer is B.
10 a
7
4
4
8
5
7
2
1
2
3
5
5
4
8
C
13
9
11
7
8
2 1
D
17
E
2
3
15
5
16
F
4
i B–D–F
= 13 + 15
= 28 km
ii A–D–C
=9+8
= 17 km
b
D
28
14 17
16
14
4
H
3
5
9 Edge with smallest value = 4
F
4
Edge off this with smallest value = 4
Total length = 11 + 16 + 9 + 12 + 7 + 17 + 14 = 86
12
30
31
A
B
C
D
E
A
X
16
12
Start with E:
Next D:
4
Next B:
4
Edge off this tree with smallest value:
17
9
A
C
7
11
G
E
14
23
11
7
B
7
11
12
5
6
Total length = 7 + 2 + 1 + 3 + 2 + 5 + 8 + 4 = 32
16 9
b E–A–D–C
= 11 + 9 + 8
= 28 km
c Total length of minimum spanning tree = 37
The correct answer is B.
8 Minimum spanning tree:
8
7
15
18
10
0
B
❙
16
X
11
13
15
C
❩
12
11
X
8
D
❩
E
❙
13
8
X
10
15
Place 15 above B.
Place 10 above D.
B∶ 10 + 13 = 23, which is > 15; ignore.
C∶ 10 + 8 = 18; place above C.
E: Already been there.
A∶ 15 + 16 = 31; place above A.
C∶ 15 + 11 = 26, 26 > than 18, so ignore C.
D: Already been there.
E: Already been there.
A∶ 18 + 12 = 30 < 31; place above A.
B: Already been there.
Shortest distance from A to E = 30 km
Next C:
5
4
10
X
4
Next edge with smallest value:
P df_Fol i o: 178
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.5
12
41
43
A
X
18
21
A
B
C
D
E
Start with E:
Next D:
22
7
0
B
❙
18
X
16
20
23
C
❩
D
❩
E
❙
20
15
X
7
23
21
16
X
15
7
X
A∶ 23 + 18 = 41 < 43; place above A.
C: Already been there.
D: Already been there.
E: Already been there.
Shortest distance from A to E = 41 km
13 Statement E is the only true statement.
The correct answer is E.
14 a A–B–F–H
A–B–F–G–H
A–B–F–G–E–H
Next B:
12
6
H
15
E
b See the table at the foot of the page.*
Start with H:
Place 6 above G.
Place 12 above F.
Place 15 above E.
A∶ 14 + 10 = 24, which is > 22; ignore.
G: Already been there.
H: Already been there.
Next E:
A∶ 4 + 18 = 22, which = 22; ignore.
F: Already been there
Next C:
A∶ 22 + 7 = 27, which is > 22; ignore.
F: Already been there.
Shortest time for A–H = 22
15 Total number of possible trees connecting A to H is:
A–B–F–H
A–B–F–G–H
A–B–F–G–E–H
A–C–F–H
A–C–F–G–H
A–C–F–G–E–H
A–D–G–F–H
A–D–G–E–H
A–D–G–H
A–E–G–F–H
A–E–G–H
A–E–H
There are 12 possible trees.
The correct answer is C.
16 Smallest possible tree: A–C–F–G–H
=4+7+5+6
= 22 minutes
The correct answer is D.
Next B:
14 b*
P df_Fol i o: 179
A∶ 13 + 9 = 22; place above A.
G: Already been there.
Next D:
Place 23 above B.
Place 7 above D.
B∶ 7 + 20 = 27, which is > 23; ignore.
C∶7 + 15 = 22; place above C.
E: Already been there.
B
7
9
4
7 F
A
9 C
5
7
10
D 8 G
B∶ 11 + 9 = 20; place above B.
C∶ 11 + 7 = 18; place above C.
G: Already been there.
H: Already been there.
Next F:
A∶22 + 21 = 43; place above A.
C∶ 22 + 16 = 38, 36 > than 23, so ignore C.
D: Already been there.
Next C:
D∶ 6 + 7 = 13; place above D.
E∶ 6 + 8 = 14; place above E.
F∶ 6 + 5 = 11 < 12; place above F.
H: Already been there.
Next G:
23
11
A
B
C
D
E
F
G
H
22
A
X
7
4
9
10
20
B
7
X
18
C
4
13
D
9
14
E
10
12
❩
F
6
G
0
H
7
8
5
X
6
15
12
6
X
9
7
X
X
X
9
7
7
179
8
15
X
5
12
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.5
180
(5) A–C–F–H–E–G
There are 5 trees connecting A to G.
b Times:
(1) 6 + 5 + 7 = 18
(2) 6 + 7 + 9 = 22
(3) 8 + 4 + 9 = 21
(4) 8 + 4 + 7 + 5 + 7 = 31
(5) 8 + 3 + 5 + 8 + 9 = 33
Shortest path between A and G: A–B–D–G = 18
c Trees connecting D → F:
(1) D–G–E–H–F
(2) D–G–E–C–F
(3) D–G–E–B–A–C–F
(4) D–B–E–H–F
(5) D–B–E–C–F
(6) D–B–A–C–F
(7) D–B–A–C–E–H–F
There are 7 different trees connecting D to F.
d Times:
(1) 7 + 9 + 8 + 5 = 29
(2) 7 + 9 + 4 + 3 = 23
(3) 7 + 9 + 7 + 6 + 8 + 3 = 40
(4) 5 + 7 + 8 + 5 = 25
(5) 5 + 7 + 4 + 3 = 19
(6) 5 + 6 + 8 + 3 = 22
(7) 5 + 6 + 8 + 4 + 8 + 5 = 36
Shortest path between D and F:
D–B–E–C–F = 19
17 See the table at the foot of the page.*
Start with K:
Place 12 above J.
Place 12 above I.
G∶ 12 + 11 = 23; place above G.
H∶ 12 + 16 = 28; place above H.
K: Already been there.
Next J:
E∶ 12 + 14 = 26; place above E.
F∶ 12 + 18 = 30; place above F.
K: Already been there.
Next I:
C∶ 23 + 16 = 39; place above C.
D∶ 23 + 10 = 33; place above D.
J: Already been there.
Next G:
B∶ 26 + 13 = 39; place above B.
C∶ 26 + 12 = 38, 38 < 39; place above C.
I: Already been there.
Next E:
D∶ 28 + 14 = 42, 42 > 33; ignore.
J: Already been there.
Next H:
B∶ 30 + 15 = 45, 45 > 39; ignore.
I: Already been there.
Next F:
A∶ 33 + 11 = 44; place above A.
G: Already been there.
H: Already been there.
Next D:
A∶ 38 + 8 = 46, 46 > 44; ignore.
E: Already been there.
G: Already been there.
Next C:
A∶ 39 + 9 = 48, 48 > 44; ignore.
E: Already been there.
F: Already been there.
Shortest distance from A to H = 44 km
Next B:
18
19 a
5
7
B
7
6
18
E
4
C
24
Length of minimum spanning tree = 12 + 13 + 18
+ 24 = 67
32
H
3
5
a Trees connecting A → G:
(1) A–B–D–G
(2) A–B–E–G
(4) A–C–E–G
(4) A–C–E–B–D–G
F
17*
12
13
G
A
P df_Fol i o: 180
36
D
38
A
B
C
D
E
F
G
H
I
J
K
44
A
X
9
8
11
39
B
9
X
3
9
❩
❩
C
8
33
D
11
X
26
E
30
F
13
12
15
X
13
15
12
23
G
28
H
16
10
14
X
10
14
X
0
K
11
16
X
14
12
J
14
18
X
16
12
I
18
X
11
16
12
X
12
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
12
12
X
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.5
b
21 a
24
D
386
E
6
C
8
15
9
c
301
346
16
Length of minimum spanning tree = 6 + 8 + 9 + 16
+ 15 = 54
23
26
5
15
356
456
B
F
b Maximum number = 346 + 386 + 456 + 301 + 356 = 1845
22 a
C
A
80
14
120
D
10
85
26
Length of minimum spanning tree = 5 + 10 + 14 + 26 = 55
10
5
F
200
A
30
E
Entrance/exit
i 9 edges
ii 6 vertices
iii The degree of each vertex:
A–4, B–4, C–2, D–3, E–2, F–3
6
4
150
40
14
5
90
70
B
d
181
b Minimum distance is obtained from the circuit
E–A–D–F–C–B–E 30 + 70 + 90 + 80 + 120 + 40 = 430 m
c
C
7
4
6
Length of minimum spanning tree = 4 + 4 + 6 + 7 = 21
F
D
7
4
A
B
4
20 a
14
19
7
7
2
5
6
13
5
1
6
3
12
11
7
7
6
b
2
5
1
6
12
Length of minimum spanning tree = 49
3
3
3
4
3
5
3 2
4
4
4
5
3
5
4
3
3
4
3
3
E
A B C D E F
A ⎡0 1 0 1 1 1 ⎤
⎢
⎥
B ⎢1 0 1 1 1 0 ⎥
C ⎢1 1 0 0 0 1 ⎥
⎢
⎥
D ⎢1 0 1 1 0 0 ⎥
E ⎢⎢1 1 0 0 0 0⎥⎥
F ⎣1 0 1 1 0 0 ⎦
e A Hamiltonian cycle would be E–A–D–F–C–B–E.
d
9.5 Exam questions
1 A spanning tree includes all the vertices and some of the
edges of the original network, and no loops, multiple edges or
cycles.
Option B has an edge connecting vertices 3 and 5 that is not
present in the original network.
The correct answer is B.
2 Draw a minimum spanning tree that has a minimum length of
53 m.
2
4
3
Length of minimum spanning tree = 29
4
P df_Fol i o: 181
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.6
182
C
7
6 Connected graphs:
D
10
i
B
8
11
E
10
5
x
A
F
8
6
G
9
5
6
ii
7
I
10
H
x = 53 − (7 + 6 + 5 + 6 + 5 + 8 + 7)
= 53 − 44
=9
The correct answer is D.
3 A minimum spanning tree is required.
Not connected:
iii
28
A
B
35
MS
15
C
iv
28
D
E
35
30
F
32
G
Add up the edges:
15 + 28 + 30 + 32 + 35 + 35 + 28 = 203
The correct answer is B.
The correct answer is C.
7
A
B
C
H
E
D
9.6 Review
G
9.6 Exercise
Multiple choice
1 Degree (A) = 3
Degree (B) = 3
Degree (C) = 3
Degree (D) = 4
Degree (E) = 1
Total = 14
The correct answer is E.
2 E = {(A, B), (A, C), (A, D), (B, C), (B, D), (C, D), (D, E), }
The correct answer is C.
3 Degree of A = 6 (2 + 2 + 1 + 0 + 1)
The correct answer is D.
4 The isolated vertex is D.
The correct answer is D.
5 V=E−F+2
if V = 6
F=4
P df_Fol i o: 182
then E = V + F − 2
=6+4−2
=8
The correct answer is C.
F
A Hamiltonian path visits each vertex just once, starting and
finishing at a different position.
The correct answer is C.
C–G–H–A–B–D–E–F
8 An Euler trail exists because there are exactly 2 vertices
whose degree is 3.
Vertex C = 3
Vertex E = 3
The correct answer is B.
9 a
A
B
9
4
10
4
H 7
C
8
G
5
9
11
E
3 D
8
F
Shortest path from H to E:
H–G–F–E = 8 + 11 + 8 = 27
H–A–C–B–E = 10 + 4 + 4 + 9 = 27
H–A–C–B–D–E = 10 + 4 + 4 + 5 + 3 = 26
H–G–A–C–B–D–E = 8 + 7 + 4 + 4 + 5 + 3 = 31
H–G–A–C–B–E = 8 + 7 + 4 + 4 + 9 = 32
The correct answer is C.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.6
b
A
b Euler’s rule
n (E) = 6
n (F) = 3
n (V) = 5
B
4
4
H 7 C E
8
5
3D
8
8 + 7 + 4 + 4 + 5 + 3 + 8 = 39
The correct answer is A.
10 Using Euler’s formula:
V=E−F+2
V = 12 − F + 2
Hence V = 6 and F = 8
The correct answer is C.
G
F
Short answer
11 a V = {A, B, C, D, E, F, G, H}
E = {(A, B) , (A, C) , (A, G) , (A, H) , (B, C) , (B, D) ,
(B, E) , (D, E) , (E, F) , (F, G) , (G, H)}
b
A B C D
A ⎡ 0 1 1 0
⎢
B ⎢ 1 0 1 1
C ⎢ 1 1 0 0
⎢
D ⎢ 0 1 0 0
E ⎢ 0 1 0 1
⎢
F ⎢ 0 0 0 0
⎢
G ⎢ 1 0 0 0
H ⎣ 1 0 0 0
12 A
B
183
E
0
1
0
1
0
1
0
0
F
0
0
0
0
1
0
1
0
G
0
0
0
0
0
1
0
1
H
1 ⎤
⎥
0 ⎥
0 ⎥
⎥
0 ⎥
0 ⎥
⎥
0 ⎥
⎥
1 ⎥
0 ⎦
V=E−F+2
V=6−3+2
V=5
14 a Euler circuit = A–B–C–D–B–F–D–E–F–A
b Hamiltonian cycle = A–B–C–D–E–F–A or
A–F–E–D–C–B–A
15 a
Shiva Paul Matt Carlo Kevin
Shiva ⎡ 0
2
1
0
1 ⎤
⎢
⎥
Paul ⎢ 2
0
0
1
1 ⎥
Matt ⎢ 1
0
0
1
1 ⎥
⎢
⎥
Carlo ⎢ 0
1
1
0
1 ⎥
Kevin ⎣ 1
1
1
1
0 ⎦
b
Shiva
Paul
Kevin
Matt
c
Carlo
Shiva
Paul
Kevin
Matt
Carlo
d So they could start and finish at the same house.
e M–C–P–S–M–K–P–S–K–C–M
16 2
3
D
9
4
E
C
8
5
1
Degree (A) = 2
Degree (B) = 4
Degree (C) = 3
Degree (D) = 2
Degree (E) = 3
Degree (F) = 2
Total = 16
13 a Network
6
F
17
E
44
21
27 D
E
D
Planar graph
C
D
G
23
21
17
C
C
E
31
23
45
A
B
A
15
B
B
A
7
Hamiltonian paths — visit each vertex once, with different
start/finish point.
1–2–3–4–5–6–7–8–9
15 13
H
F 12 J
19
23
K 18
11
L
15 + 23 + 21 + 12 + 44 + 27 + 15 + 13 + 18 + 11 = 199 km
Extended response
18 a Loops are indicated by vertices of degree 2, between the
same vertex.
∴ There is one loop.
b Multiple edges are values greater than 1 between different
vertices.
(A, D), (A, D), (A, D)
∴ There are 3 multiple edges between vertices A and D.
c To find the degree of a vertex, add up the row or column of
the matrix.
P df_Fol i o: 183
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
184
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.6
Degree (A) = 2 + 0 + 1 + 3 + 1 + 0
=7
d Several possibilities; for example:
d Degree (D) = 3 + 1 + 1 + 0 + 0 + 1
=6
e A
C
F
B
M
E
W
G
e
F
S
C
C
S
F
B
D
f Answers will vary. A possible tree is:
B
A
M
Total distance = 510 km
This is of little use because they need to start and finish in
Melbourne, and flight paths would backtrack.
W
G
E
F
f
D
19 a Planar graph
E1
OT
K
E2
E4
F
L
PT
G
E3
b To create an Euler circuit, all vertices must be even, so join
the flowerbeds to the lake with a path.
Path∶ E2–K–L–F–K–E1–OT–E4–L–E3–E4–G–E3–PT–F–E2
c
E1 55 K
50
OT 40
50
E4
40
G
45
90
65
L
135 65
E2
35
F
40
PT
d Shortest path connecting all the vertices ⇒ minimum
spanning tree
40 + 50 + 40 + 55 + 45 + 50 + 35 + 40 + 45 = 400 m
105
OT 40
50
E4
40
G
E3
45
E1 55 K 50
45
65
90
L
135 65
110
45
E3
E2
35
F
40
PT
e E2–K–L–E4
= 50 + 45 + 90
= 185 m
20 a V = 7, E = 11, F = 6; so, 7 = 11 − 6 + 2
b
M G B C S F W
M ⎡0 1 1 0 1 0 0 ⎤
⎢
⎥
G ⎢1 0 1 0 0 0 1 ⎥
B ⎢1 1 0 1 1 0 0 ⎥
⎢
⎥
C ⎢0 0 1 0 1 0 0 ⎥
S ⎢1 0 1 1 0 1 1 ⎥
⎢
⎥
F ⎢0 0 0 0 1 0 1 ⎥
W ⎣0 1 0 0 1 1 0 ⎦
c This is a Hamiltonian cycle.
M–G–W–F–S–C–B–M and M–B–C–S–F–W–G–M
G
B
M
C
S
W
F
240
G
×
80
70
185
B
80
×
95
65
100
170
M
70
95
×
85
170
C
85
S
65
100
85
85
×
145
85
×
85
140
125
W
140
0
F
145
×
125
85
125
×
Therefore, the shortest distance between Geelong and
Mansfield is 240 km.
9.6 Exam questions
1 a Shortest distance is G − O − N − M for 86 km
[1 mark]
b George will pass through K twice
(G − H − I − K − L − K − J − O − N − M)
[1 mark]
2 a Emerson has played cricket with only ONE other player
(i.e. there is only one edge connecting Emerson). [1 mark]
b Cameron and Dale (edges joining both these players to
Alex and Bo)
[1 mark]
c
Alex
Bo
Finn
Emerson
Dale
P df_Fol i o: 184
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
Cameron
[1 mark]
TOPIC 9 Undirected graphs, networks and trees • EXERCISE 9.6
3 a Shortest distance = 0.6 + 1.2 + 0.6 + 0.8 = 3.2
kilometers.
[1 mark]
b i Eulerian trail
[1 mark]
ii Eulerian trails start and finish at vertices with an odd
degree. The training program starts at S, with a degree of
3, and will finish at P, also with a degree of 3. [1 mark]
c This track is between exercise station S and exercise
station T.
[1 mark]
1.8
M
T
S
1.5
N
1.2
1.0
0.7
0.9
U
R
0.8
0.8
1.2
0.6
1.2
Q
0.8
V
O
0.4
1.4
P
4 a The office
[1 mark]
b i Hamiltonian cycle
[1 mark]
A few responses erroneously named the route as a path
or circuit.
ii One possibility:
office
library
science
laboratories
computer
rooms
mathematics
classrooms
gymnasium
[1 mark]
5 Isomorphic means the equivalent number of vertices and
edges.
Graph 1 has four vertices and five edges, while Graph 2 has
five vertices and six edges. Therefore, Graphs 1 and 2 are not
isomorphic
The correct answer is A.
P df_Fol i o: 185
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition
185
TOPIC 10 Directed graphs and network flow • 10.2
186
Topic 10 — Directed graphs and network flow
Activity G has E and F as immediate predecessors.
10.2 Precedence tables and activity networks
D
E
G
F
10.2 Exercise
4
19
Activity letter
Immediate predecessor
M, 5
N
O
P
—
N
O, T
Q
P
R
S
T
—
N
S, Y
11
U
O, T
b Earliest completion time is 30 minutes.
c With an earliest completion time of 30 minutes:
A–B–L–M–N
A–B = 9 + 6 = 15 minutes, compared to
E–J = 5 + 4 = 9 minutes
So a difference of 15 − 9 = 6
The maximum time that path J can be delayed is by 6
minutes.
2 a The earliest completion time is 147 days.
V
W
X
Y
Z
O, T
V
Y
R
U, X
1 a
9
15
B, 6
A, 9
C, 4
D, 10
L, 6
J, 4
K, 8
E, 5
0
5
F, 6
G, 4
21
30
26
N, 4
H, 10
First edges N and R have no immediate predecessors.
N
R
64
N is a predecessor for activity O and S.
N
K, 25
F, 23
O
D, 22
B, 13
41
19
G, 20
R
114
Q, 32
P, 29
R, 30
147
L, 27
A, 6
0
N, 26
6
C, 11
H, 19
61
34
J, 21
88
S
R is a predecessor for activity Y.
N
O
R
E, 17
17
S
Y
117
M, 24
55
Y is a predecessor for activity X, S and Y are predecessors
for activity T.
N
b With an earliest completion time of 147:
A–B–D–G–L–P–R
G–L–P–R = 20 + 27 + 29 + 30 = 106 days
F–K–Q = 23 + 25 + 32 = 80 days
So a difference of 106 − 80 = 26
K can be delayed by 26 days.
3
Activity letter
Immediate predecessor
D
E
F
G
—
D
D
E, F
Activity D has no predecessor and is the first edge.
D
O
S T
R
X
Y
O and T are predecessors for activities P, U and V.
N
P
O
V
S T
R
U
X
Y
P is a predecessor for activity Q, V is a predecessor for
activity W.
N
O
S
R
P
T
U V
W Q
Y
X
Z
Activity E and F have D as an immediate predecessor.
D
P df_Fol i o: 186
E
F
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.2
b For tasks that can be delayed, identify sections of the
network where there is a choice.
5
A, 7
187
C, 12
D, 8
B, 2
F, 9
C, 5
B, 9
D, 9
E, 4
a A Activity A is an immediate predecessor of F. — False,
A is a predecessor of C.
B Activity D is an immediate predecessor of F. — True.
C Activity F must be done before activity D. — False, F
must be done after D.
D Activity F must be done before activity E. — False.
E Activity D is an immediate predecessor of E. — No, the
two can occur simultaneously.
The correct answer is B.
b Minimum time to complete all activities follows path
A–C–F = 7 + 12 + 9
= 28 minutes
The correct answer is D.
c The critical path for the network is A–C–F.
d Working backwards to find predecessors:
C and D are predecessors of F.
and
F, 4
E, 6
G, 8
B–C = 2 + 5 = 7 mins
D = 9 mins
∴ B and C can be delayed.
E–F = 6 + 4 = 10 mins
G = 8 mins
H = 11 mins
∴ E, F, and G can be delayed.
H, 11
7 a
B, 4
C, 5
F, 8
C, 5
F, 8
Select maximum value: 5 + 8 = 13
The correct answer is E.
C
F
b
D
B is a predecessor of E and D.
13 + 5 = 18
The correct answer is D.
c The earliest completion time for all tasks is:
D
B
E
A is a predecessor of C.
5
A
9
C
27
0
A and B have no predecessors.
13
18
A
5
B
Activity letter
Immediate predecessor
Time
A
B
C
D
E
F
—
—
A
B
B
C, D
7
9
12
8
4
9
6 a For the earliest completion time, fill in triangles with
maximum time to each vertex/node.
5
B, 2
A, 3
Calculate the time values for the paths to the last vertex.
A–D–G = 3 + 6 + 18 = 27
B–E–H = 4 + 5 + 8 = 17
B–E–J = 4 + 5 + 6 = 15
C–F–E–H = 5 + 8 + 5 + 8 = 26
C–F–E–J = 24
The earliest completion time is:
A–D–G = 3 + 6 + 18 = 27 minutes.
The correct answer is A.
d Working backwards to find predecessors:
E is a predecessor of H and J.
E
H
18
J
C, 5
3
E, 6
F, 4
D is a predecessor of G.
D
G, 8
D, 11
12
23
G
H, 11
The earliest completion time is 23 minutes.
B and F are predecessors of E.
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188
TOPIC 10 Directed graphs and network flow • 10.2
D
E
F
C is a predecessor of F.
C
35–G = 35 + 6
= 41
35–G–J = 41 + 11
= 52
41
F
G
A is a predecessor of D.
L
D
H
A
35–G–L = 35 + 6 + 8
= 49
*Take maximum value
35–H = 35 + 10
= 45
35
A, B and C have no predecessors.
Activity letter
Immediate predecessor
Time
A
B
C
D
E
—
—
—
A
B, F
3
4
5
6
5
F
G
H
J
C
D
E
E
8
18
8
6
e i Critical path – follows activities that cannot be delayed.
The path that takes the largest time is
A–D–G = 3 + 6 + 18
= 27 minutes
ii and iii Float time is the maximum time that an activity
can be delayed without delaying a subsequent activity
on the critical path.
Activity B can be delayed 10 minutes, activity C can be
delayed 1 minute, activity E can be delayed 1 minute,
activity F can be delayed 1 minute, activity H can be
delayed 1 minute and activity J can be delayed 3
minutes.
8 a
A
A–B = 10 + 15 = 25
A–C = 10 + 12 = 22
0
10
49
52
K
M
52–K = 52 + 9
= 61
49–M = 49 + 7
= 56
*Take maximum value
Final outcome:
49
61
25
41
52
J, 11
D, 8
B, 15
A, 10
0
C, 12
10
G, 6
M, 7
H, 10
E, 10
22
35
49
F, 25
b The earliest completion time for the project:
A–F–G–J–K = 10 + 25 + 6 + 11 + 9
= 61 minutes
c Working backwards to find predecessors:
H and L are predecessors of M.
25
L
D
B
A
C
H
E
M
J is a predecessor of K.
0
10
22
A–B–D = 25 + 8 = 33
A–C–E = 22 + 10 = 32
A–F = 10 + 25 = 35
*Take maximum value
J
35
F
41
J
52
K
G is a predecessor of J and L.
J
G
K, 9
L, 8
L
G
35
P df_Fol i o: 188
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
61
TOPIC 10 Directed graphs and network flow • EXERCISE 10.2
b
D, E and F are predecessors of G and H.
D
G
E
Activity letter
Immediate predecessor
A
—
B
A
C
A
D
C
E
B
H
F
B is a predecessor of D.
B
D
F
B
E
G
F
A is a predecessor of B, C and F.
H
D, E, G
I
J, H
J
D, E, G
C is a predecessor of E.
C
B
A
C
First edge A has no predecessor.
F
A has no predecessor.
A
Activity letter
Immediate predecessor
Time
A
—
10
B
A
15
C
A
12
D
B
8
E
C
10
F
A
25
G
D, E, F
6
H
D, E, F
10
J
G
11
K
J
9
L
G
8
M
H, L
7
d i Critical path = longest path
A–F–G–J–K = 10 + 25 + 6 + 11 + 9
= 61 minutes
ii Activities which have float times are not on the critical
path. These are therefore B, C, D, E, H, L and M.
9 a
Activity letter
Immediate predecessor
A
B
C
—
—
A
Activity A and B have no predecessor, so they become the
first edge.
A is a predecessor for activities B and C.
C
A
B
B is a predecessor for E and F.
C
A
E
B
F
C is a predecessor for activity D.
D
E
C
A
B
F
F is a predecessor for activity G.
D
C
A
E
B
F
G
D, E and G are predecessors for activities H and J.
D
C
A
E
H
B
G
F
F
J
H and J are predecessors for activity I.
D
C
B
A
J
E
B
A
F
F
G
H
Activity C has A as an immediate predecessor.
B
A
C
P df_Fol i o: 189
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
I
189
190
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
10 a First edge A has no predecessor.
A is a predecessor for B and D.
D is a predecessor for E.
Forward and backward scanning through the activity network
results in the following information.
The earliest starting time for activity N is 12 hours after the
start of the project.
Alternatively, look for the longest path to the start of
activity N.
The correct answer is D.
VCAA Examination Report note:
The critical path analysis in Question 4 involved standard
forward-scanning calculations. While there was some
complexity of the activity network, students should be able to
apply standard routine calculations to graphs such as this with
care.
3 I and J have immediate predecessors.
The correct answer is C.
A
B
A
D
B
A
E
D
E is a predecessor for F.
B and E are predecessors for C.
F
B
A
10.3 Critical path analysis with backward scanning
and crashing
C
E
D
C and F are predecessors for G.
10.3 Exercise
F
B
1
G
A
E
D
A, 3
C
C, 7
D, 6
E, 2
G is a predecessor for H.
B, 20
G, 10
A, 2
H, 12
E, 3
3
b Minimum time in which all tasks could be completed
follow the path:
A–B–F–G–H = 2 + 20 + 5 + 10 + 12
= 49 minutes
A, 3
C, 7
D, 6
13
E, 2
6
B, 4
4
10.2 Exam questions
3
C = 3 + 7 = 10
6
E=6+2=8
7
G = 7 + 6 = 13*
Take maximum value
G, 6
F, 3
7
(4 + 3)
1 Minimum completion time = 7 + 4 + 3 + 3 + 5 = 22
The correct answer is E
2
Activity
EST
LST
Float
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
F, 3
Forward scanning:
C, 4
D, 5
G, 6
B, 4
F, 5
0
0
0
4
4
4
4
10
7
7
10
10
13
12
15
1
2
0
6
9
4
5
10
8
8
12
11
13
13
16
1
2
0
2
5
0
1
0
1
1
2
1
0
1
1
Backward scanning:
3
6
(13 – 7)
C, 7
A, 3
0
0
6
A=6–3=3
11
D = 11 – 6 = 5
*4
B=4–4=0
Take minimum value
D, 6
13
13
E, 2
6
11
B, 4
(13 – 2)
F, 3
4
4
Critical path = B–F–G
Float time for the non-critical activities:
A : 3 hours
C : 3 hours
D : 5 hours
E : 5 hours
(7 – 3)
7
7
(13 – 6)
P df_Fol i o: 190
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
G, 6
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
2 a Enter 10 for the last node.
Enter 8 for the next node before activity J.
Enter 7 for the node before activity H.
Enter either 8 − 2 (path J, K) or 8 − 1 − 1.5 (path H, G) for
the node before activities G, K.
Reject the path J, K.
Enter 5.5 − 2.5 = 3 for the node before activity F.
Enter 5.5 − 1 = 4.5 for the node before activity C.
Enter 4.5 − 2 = 2.5 for the node before activity B.
Complete by entering 0 at the start node.
1
2.5
3
b First edge A (no predecessor):
A, 12
A is a predecessor of B and C.
B, 35
A, 12
C, 16
B is a predecessor of D.
C is a predecessor of E and F.
7
7
B, 2 4.5
B, 35
H, 1
C, 1
A, 1
F, 2.5
D, 3
4 5.5 3.5 K, 2 7.5 8
8
5.5 6
D, 20
A, 12
G, 1.5
E, 3.5
0
0
191
E, 12
C, 16
J, 2
10
10
F, 5
D and E are predecessors of G.
3
3
B, 35
A
B
C
D
E
F
G
—
A
A
B
C
C
D, E
D, 20
A, 12
The critical path is where triangle number = box number:
D–F–G–H–J.
b Float times for non-critical activities:
Float (K) = 8 − 5.5 − 2 = 0.5 hours
Float (C) = 5.5 − 3 − 1 = 1.5 hours
Float (B) = 4.5 − 1 − 2 = 1.5 hours
Float (A) = 2.5 − 0 − 1 = 1.5 hours
Float (E) = 5.5 − 0 − 3.5 = 2 hours
3 a
Activity letter
Immediate predecessor
Time
E, 12
G, 18
C, 16
F, 5
c See the figure at the foot of the page.*
The earliest completion time is 85 minutes.
d Critical path = A–B–D–G
4 Activities with float time are not on the critical path. These
activities would be C, E and F. The answer is C.
12
35
16
20
12
5
18
*3 c
12 + 35
47
47
(67 – 20)
47
55
(0 + 12)
B = 47 – 35 = 12 *Take min value
12
C = 55 – 16 = 39
12
A, 12
67
67
D, 20
B, 35
47
28
D = 47 + 20 = 67 *Take max value
E = 28 + 12 = 40
(85 – 18)
G, 18
0
0
C, 16
(12 – 12)
(12 + 16)
28
55
E, 12
F, 5
85
85
67
28
G = 67 + 18 = 85 *Take max value
F = 28 + 5 = 33
P df_Fol i o: 191
67
85
E = 67 – 12 = 55 *Take min value
F = 85 – 5 = 80
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
192
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
5
C, 4
B
A, 12
F
D and E are predecessors of J.
E, 3
D
B, 9
J
E
D, 11
M and G are predecessors of L.
Forward scanning:
(12 + 4)
12
16
C, 4
L
G
A, 12
E, 3
0
16
E = 16 + 3 = 19
9
D = 9 + 11 = 20
*Take maximum value
20
B, 9
D, 11
9
M
H is a predecessor of M.
H
M
Backward scanning:
(17 – 4)
12
13
C, 4
16
17
(26 – 3)
C is a predecessor of G and H.
C
E, 3
A, 12
G
0
0
13 – 12 = 1
9–9=0
B, 9
D, 11
20
20
H
B is a predecessor of E and F.
Take maximum value*
9
9
(20 – 11)
E
Critical path is B–D.
Float times for the non-critical activities:
A: 1 minute
C: 1 minute
E: 1 minute
6 a
B, 11
F, 6
D, 3
A, 7
C, 4
G, 8
E, 12
Earliest completion time = 31 days
b Critical path, from the network, where the ‘triangle’
numbers are equal to the ‘box’ numbers.
A–C–E–G
c If activity E is reduced to 9 days, the A–C–E–G path is
reduced to 28 days, which is still greater than the other
possible path (A–B–D–F).
d Float time for activity D = 25 − 18 − 3
= 4 days
The correct answer is D.
7 a J, K and L are predecessors of N.
B
F
A is a predecessor of D.
D
A
A, B and C have no predecessors.
Activity letter
Immediate predecessor
Time
A
B
C
D
E
F
G
H
J
—
—
—
A
B
B
C
C
D, E
3
4
6
7
8
5
12
2
11
K
L
F
G, M
10
3
M
N
H
J, K, L
9
6
J
N
K
L
F is a predecessor of K.
P df_Fol i o: 192
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
C
b and c
D, 7
A
A, 3
E, 8
F, 5
B, 4
D
J, 11
N, 6
K, 10
E
B
L, 3
C, 6
G, 12
H, 2
C is a predecessor of F.
D and E are predecessors of G.
M, 9
C
The earliest completion time is 29 minutes.
The critical path is B–E–J–N. (Where the ‘triangle’
numbers are equal to the ‘box’ numbers.)
d Non-critical activities are:
A, C, D, F, G, H, K, L and M.
Float times are:
A∶ 5 − 0 − 3 = 2
C∶ 8 − 0 − 6 = 2
D∶ 12 − 3 − 7 = 2
F∶ 13 − 4 − 5 = 4
G∶ 20 − 6 − 12 = 2
H∶ 11 − 6 − 2 = 3
K∶ 23 − 9 − 10 = 4
L∶ 23 − 18 − 3 = 2
M∶ 20 − 8 − 9 = 3
8 a A and B are first edges and have no predecessors.
A
D
B
A
A, 5
B, 3
C, 4
9
11
D, 7
3
8
B
C
E
F
J
D
B
B is a predecessor of E.
(0 + 5)
(5 + 4)
5
5
9
11
20
20
(20 – 9)
C, 4
F, 9
D, 7
H, 3
E, 4
G, 5
(0 + 3)
P df_Fol i o: 193
J
G
E, 4
12
12
F, 9
20
20
H, 3
G, 5
K, 4
J, 6
17
17
24
24
b See the figure at the foot of the page.*
Earliest completion time = 24 hours
c Critical path A–D–G–H–K
d
Activity letter
Time EST EFT
A
5 – A = 0*
8–B=5
E
5
5
C
0
0
H
F and H are predecessors of K.
A is a predecessor of C and D.
B, 3
F
D
B
B
A, 5
G
C
0
0
11 – C = 7
12 – D = 5*
E
G is a predecessor of H and J.
A
*8b
F
3
4
4
9
6
0
5
3
9
17
Float time
8
11
12
20
24
9 + F = 18
17 + H = 20*
(24 – 4) = 20
K, 4
J, 6
(12 + 5)
12
12
3
8
17
17
20 – H = 17*
24 – J = 18
24
24
20 + K = 24*
17 + J = 23
(12 – 4) (17 – 5)
5
3
D = 12*
E=7
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
5
2
5
2
1
193
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
194
9
G
G'
F
B
A
E
12
D'
D
10
H
C
Activity letter
Immediate predecessor
—
—
A
B
B, C
3
5
7
7
1
F
D, E
2
D, 1100
E, 400
A, 1000
Time (h)
A
B
C
D
E
ii Backward scanning shows that the critical path is
A–C–D–F.
iii Float times for non-critical activities:
Float (E) = 27 − 15 − 7 = 5 minutes
Float (B) = 7 − 0 − 6 = 1 minute
B, 600
F, 100
G, 600
C, 800
13 a
H, 1600
B', 0
Activity letter
Immediate predecessor
Time (h)
A
B
C
D
E
F
G
H
J
—
—
A
A
B
C
D
E
E, F, G
11
9
2
5
12
3
3
4
7
A and B are first edges and have no predecessors.
A
B
A is a predecessor of C.
B is a predecessor of D and E.
C is also a predecessor of E.
(To skip a parallel edge use dummy edge B.)
C
A
A and B are first edges and have no predecessors.
A
B
C
A
B
B′
A is a predecessor of C and D.
B is a predecessor of E.
B
C
D
D and E are predecessors of F.
A
D
C, 7
A, 3
B
E, 1
B', 0
B, 5
D, 7
C is a predecessor of F.
D is a predecessor of G.
F, 2
11 a Create dummy activities B′, E′, since C has both A and B
as immediate predecessors and F has both D and E as
immediate predecessors.
Alternatively, A could have a dummy activity instead of B,
and D could have a dummy activity instead of E.
A, 7
B, 6
b
C, 8
B', 0
E
F, 9
D, 12
E, 7
F
C
A
G
D
B
E
E is a predecessor of H and J.
F and G are predecessors of J, as well as E, F, G–J, but E
also needs –H; use a dummy edge E′.
E', 0
i Forward scanning shows that the earliest completion
time is 36 minutes.
A, 7
0
0
7
7
C, 8
B′, 0
B, 6
6
7
15
15
D, 12
27
27
F, 9
36
36
E′, 0
E, 7
22
27
P df_Fol i o: 194
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
15 a and b
See the figure at the foot of the page.*
Earliest completion time = 35 days
Critical path = C–F–J–M–Q
See the figure at the foot of the page.*
Forward/backward scanning
b Earliest completion time = 28 hours
c Critical path = B–E–E′–J
d
Activity letter
Time
EST
A
C
D
F
G
H
14
11
2
5
3
3
4
0
11
11
13
16
21
H
K
EFT
Float time
13
18
18
21
21
28
2
5
2
5
2
3
c Float time for activity X = EFT − EST − time
= 16 − 10 − 3
= 3 days
d When J is reduced by 2 days to 5 days, the earliest
completion time is reduced to 34 days. The new critical
path becomes C–F–H–P. J is no longer a critical activity.
16 a LST for J = 29 − 8 = 21
So, LST for I = 17
So, activity time I = 24 − 17 = 7
LST for G = 24 − 4 = 20
LST for D′ = 17 − 0 = 17
So, LFT for D and E = 17
D
C
A
B
E
E'
G
G'
J
*14a
(11 + 2)
13
18
(21 – 3)
(11 + 5)
(0 + 11)
C, 2
11
13
18 – D = 13*
18 – C = 16
F, 3
16
18
21
21
13 + F = 16
16 + G = 19
21 + E′ = 21*
(21 – 3)
D, 5
G, 3
A, 11
B, 9
E, 12
0
0
13 – A = 2
9 – B = 0*
J, 7
Eʹ, 0
H, 4
(0 + 9)
(9 + 12)
9
9
21
21
28
28
(21 – 12)
28 – J = 21*
28 – H = 24
(0 + B)
4+E=9
10 + X = 13*
21 + J = 28
21 + H = 25
*15a and b
4
11
13
16
(16 – E) E, 5 (25 – G)
(0 + C)
B, 4
10
10
X, 3
(35 – P)
19 – F = 10*
16 – X = 13
0
0
H, 5
F, 9
A, 15
J, 7
(11 + J)
(0 + A) D, 3
C, 10
11 – B = 7
10 – C = 0*
16 – A = 1
15
16
(19 – 3)
10 + F = 19*
15 + D = 18
P df_Fol i o: 195
13 + G = 22
19 + H = 24*
24
25
G, 9
19
19
26
26
P, 10
N, 6
M, 4
35
35
24 + P = 34
26 + N = 32
30 + Q = 35*
Q, 5
K, 5
(19 + K) L, 3
24
27
30
30
26 + M = 30*
19 + H = 22
(35 – Q)
(30 – L)
25 – H = 20
26 – J = 19*
32 – K = 22
195
30 – M = 26*
35 – N = 29
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
196
TOPIC 10 Directed graphs and network flow • EXERCISE 10.3
LST for E = 17 − 1 = 16
So, LST for F = 15
So, activity time F = 24 − 15 = 9
LST for A and B = 0
10
10
C, 7
0
0
I, 7
E, 1
K, 5
B, 8
S
29
29
24
24
F, 9
b Critical path = A–C–I–K
c Float time for F = 24 − 8 − 9 = 7
17 a Activity time for K = 44 − 38 = 6
LST for D = 25 − 9 = 16
LST for B′ = 16
LST for F = 38 − 16 = 22
So, LST for E = 14
So, activity time E = 25 − 14 = 11
LST for H = 30 − 18 = 12
LST for G = 38 − 20 = 18
So, LFT for A = 12
G, 20
A, 12
0
0
B, 10
10
14
B', 0
C, 15
D, 4
S
C5
H3
9
9
F2
4
7
I5
14
14
F
D4
44
44
5
5
D, 9 24
b Critical path = A–H—I—K
c Float time for F = 38 − 10 − 16 = 12
18 a Immediate predecessors of C and F are A, C and D
respectively.
EST for G = 3 hours
EST for K = 3 + 2 + 9 = 14 hours
b EST for J = 17 and EST for F = 9
So, activity time X = 17 − 9 = 8 hours
25
10.3 Exam questions
1 When forward scanning is done, the latest start time of
activity J is 11 (as the minimum completion time is 18 hours).
To not affect the completion time, x must be a value less than
3 – i.e. 8 + x = 11, so x = 11 − 8 = 3.
See image bottom of the page*
The correct answer is B.
F, 3
C, 4
A, 5
D, 4
8
11
G6
B3
K, 6
J, 13
15
16
2
5
0
0
38
38
F
a 5 weeks
b Minimum time is 15 weeks.
c Critical path is A–E–F–I.
d Slack time is 12 − 9 = 3 weeks.
e Stages along the critical path can be shortened: A–E–F–I
(from the critical path).
f After stages A and F are reduced to 2 weeks, the new
critical path will be C–D–I with a minimum time for
completion of 14 weeks.
I, 8
E, 11
15
15
10
10
A2
F, 16
I, 5
5
6
30
30
H, 18
H, 3
F, 5
C, 5
8
15
12
12
5
5
B, 3
0
0
9
12
G, 6
E, 2
A, 3
J, 8
D', 0
G, 4
16
17
3
3
17
17
D, 6
A, 10
19
d Earliest completion time = 22
LST for K = 22 − 7 = 15
LST for H = 15 − 9 = 6 hours after start
G, 8
X, 8 I, 4 J, 5
B, 3
E, 2
K, 7
c Critical path = A–C–X–J
H, 9
*1
9
D, 6
3
11
H, 2
A, 3
0
Start
8
B, 5
F, 1
Finish
11
11
C, 4
J, 7
E, x
18
I, 4
11
P df_Fol i o: 196
4
G, 3
7
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.4
2
Minimum = 200
G
200
D
F
H
I
B
Finish
D1
T
Minimum = 200
100
A
Start
100
S
E
100
C
T
The correct answer is B.
VCAA Examination Report note:
Students could have answered the question without sketching
the entire network.
3 Start by completing forwards and backwards scanning.
See the Image bottom of the page*
Float time = Latest Starting Time – Earliest Starting Time –
Duration of Activity.
The activities with a float time of 10 hours are:
G∶ 19 − 6 − 3 = 10
I∶ 29 − 12 − 7 = 10
N∶ 31 − 19 − 2 = 10
Therefore, there are three activities with a float time of
10 hours.
The correct answer is D.
U
50
100
Minimum = 150
i ∴ Flow capacity = 150
ii This does meet the demand as V requires 150.
c
250
200
R
T
V
d Flow capacity
250
200
R
S
250
50
100
From
To
Flow capacity
R
S
T
T
U
S
T
U
V
V
250
200
100
100
50
200
250
T
(Minimum = 100)
Total flow capacity = 150
2 a
100
T
U
50
100
V
Flow capacity
250
R
To
F
G
H
J
H
J
K
K
Flow capacity
8
8
5
3
2
6
8
8
200
9
9
E, 3
B, 4
A, 5
0
0
From
E
E
G
G
F
F
J
H
S
*3
start
50
V
U
200
U
100
V
S
100
T
50
250
50
(Minimum = 50)
100
S
R
U
V
100
b
T
100
10.4 Exercise
R
50
100
T
250
U
100
10.4 Flow problems
1 a
100
S
D, 1
6
10
5
5 C, 7
12
12
J, 6
F, 8
G, 3
P df_Fol i o: 197
12
15
I, 7
H, 4
18
18
K, 6
29
29
39
41
N, 2
P, 8
R, 1
M, 5
O, 7
finish
31
31
Q, 11
42 S, 9
42
51
51
L, 5
19
19
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
197
TOPIC 10 Directed graphs and network flow • EXERCISE 10.4
198
3
G
8
ii
J
E
From
M
M
N
N
Q
O
R
8
6
K
5
8
8 F
H
2
b
3
G
8
J
8
6
E
K
5
To
N
Q
O
R
R
P
P
8
8 F
15
H
2
E
M
5
20 Q
6
200
Minimum = 16
C
8
Minimum = 15
K
7
H
i ∴ Flow capacity = 15
ii No, this does not meet the demand as K requires 16.
2
B
A
50
200
meets C demand
of 200
C
250
so D can only
have 250
i ∴ Flow capacity = 250
This doesn’t meet the demand as E requires 300.
ii
15
12
O
N
20
5
M
3
G
J
10
F
Flow capacity:
8
N
15
H
20
5
2
d Flow capacity
Total flow capacity = 15
15 + 10 (E−K) = 25
3 a i
From
To
A
B
A
C
B
C
C
D
D
E
M
20
Flow capacity
100
200
50
250
300
100
A
B
Q
Q
R
250
12
Minimum = 24
10
15
12
∴ Flow capacity = 24
This does meet the demand as P requires 24.
N
O
P
100
A
D
P
5
c i
50
10
Minimum = 30
200
C
12
R
K
5
P
10
20 Q
8
6
8
E
300
Flow capacity:
G
6
D
250
3
E
R
B
2
8
10
50
F
c
12
100
A
8
5
P
5
3
G
b
12
O
N
20
Flow capacity
8
Flow capacity
20
20
15
5
10
12
12
B
100
E
50
300
200
C
ii
20
300
12
O
N
P
5
5
M
20 Q
E
D
250
15
12
10
R
P df_Fol i o: 198
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.4
d i Flow capacity:
b Edge capacity flowing out of B is 3.
100
B
A
B
3
50
C
c The outflow from B is the minimum of a and b, so 3.
C
6 i a
3
2
A
B
5
A
250
4
C
E
2
100
A
3
B
50
E
(Minimum = 50)
C
D
To
Flow capacity
A
A
A
B
C
C
D
D
B
C
D
E
B
E
C
E
4
5
3
3
2
4
2
6
(Minimum = 250)
Total flow capacity = 300
ii Flow capacity
250
15
12
O
N
P
(Minimum = 12, excess 3)
6
D
From
E
250
20
B
4
50
200
M
M
N
b
P
5
A
12
Q
R
(Minimum = 12, excess 3)
3
Flow capacity
A
A
A
B
B
C
C
D
B
C
D
E
C
E
D
E
4
5
3
3
2
4
2
6
B
16
A
D
27
c
34
A
3
B
16
c The outflow from B is the minimum of a and b, so 16.
5
B
4
3
2
5
A
C
4
2
E
a The inflow of B is 4 + 2 = 6.
3
D
6
E
6
2
D
7
b Edge capacity flowing out of B is 16.
3
5
B
23
B
4
C
a The inflow of B is 23.
6
D
To
5
23
E
From
P
(Minimum = 5)
Total flow capacity = 29
4
4
C
2
10
N
3
2
5
20
B
4
8
C
From
To
Flow capacity
A
A
A
A
B
C
D
D
B
C
D
E
E
E
B
E
4
7
3
5
3
8
2
6
B
4
2
P df_Fol i o: 199
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
199
200
TOPIC 10 Directed graphs and network flow • EXERCISE 10.4
5
d
B
4
5
A
E
E
6
2
12
7
(Minimum = 5)
A
3
B
4
D
4
2
(Minimum = 3)
Not required
A
7
C
3
8
F
From
To
Flow capacity
A
A
A
A
C
D
D
D
F
B
C
D
E
F
B
E
F
E
4
7
12
5
7
2
6
4
8
E
A
E
2
3
4
(Minimum = 3)
Total flow capacity = 18
D
d
5
(Minimum = 5)
A
E
B
4
3
2
(Minimum = 3)
A
ii a
2
5
4
A
B
C
(5, but 2 → B ⇒ 4)
6
D
4
A
(Minimum = 6 to E and D)
F
2
6
(3, but 2 → C ⇒ 1)
D
A
F
7
7
C
B
3
4
(3)
but A → B doesn’t use all 6, so extra 2 from C not
required ∴ extra 2 from D is not required.
B→3
C→4
D→3
Flow capacity is 10.
5
E
(Minimum = 8)
Total flow capacity = 22
7 Considering outflow at A and inflow at H, the largest possible
flow is 130.
Cut 4
C
3
(Minimum flow = 3)
E
A
E
6
3
(Minimum flow = 3)
Total flow capacity = 4 + 3 + 3
= 10
D
A
E
8
7
(Minimum = 7)
C
50
E
A
Cut 3
60
30
130
130
B
G
H
70
50
D
B
A
F
70
(Minimum flow = 4) Capacity is met on this flow, so
flow from B and C not required.
C
4
P df_Fol i o: 200
7
4
A
c
8
4
2
A
b
E
2
12
2
3
E
F
40
Cut 2
Cut 1
Note that cut 4 is in fact not a cut as it fails to stop all flow.
Some possible cuts are:
Cut 1 = 70 + 40 = 110
Cut 2 = 50 + 30 + 40 = 120
Cut 3 = 60 + 70 = 130
Any other cut is more than 110.
Minimum cut = maximum flow = 110
8 a Cut 2 is invalid as it does not stop all flow from A to E.
b Cut 1 = 9 + 7 + 12 = 28
Cut 3 = 4 + 4 + 7 + 5 + 8 = 28
Cut 4 = 4 + 4 + 7 + 12 = 27
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
c
b
B
12
7
A
8
5
12
D
4
9
E
4
Minimum cut is 4 + 12 + 8 = 24.
Therefore the maximum flow is 24.
C
9 a
i There would be a traffic jam because inflow > outflow.
Node E can only handle (80 + 50) = 130.
ii At node H, the traffic should flow smoothly as the
inflow (100) is less than the capacity of flows leading
from H.
(50 + 30 + 60 = 140)
iii
40
B
I
A
10
N
75
11
P
25
55
b Maximum flow = 71
(35 + 11 + 25)
10 a
Q
O
120
C
E
G
H
30
100
150
80
45 Freeway 60
75 Freeway 50
F
J
50
K
60
As node H has the potential to carry another 40 cars,
join a road between E and H. Need to ensure capacity
through the whole network is 140 (input at node A has
a capacity of 140). E−G−H has a maximum of 100, so
add an edge from E to H to add capacity of 40 to meet
demand.
R
35
25
50
D
20
40
20
M
45
45
Q
35
201
60
60
95
M
110
N
P
145
10.4 Exam questions
R
25
25
45
100
b Maximum flow = 240
(60 + 110 + 25 + 45)
11 a Cut 4 is invalid since it does not stop all flow from A to G.
b Cut 1 = 6 + 9 + 10 = 25
Cut 2 = 5 + 8 + 13 = 26
Cut 3 = 11 + 13 = 24
Cut 5 = 10 + 7 + 5 + 8 + 6 + 6 = 42
c Minimum cut = maximum flow = 11 + 5 + 4 = 20
12 a i
Q
O
12
1
13
M
14
11
N
P
10
R
2
1
12
ii Maximum flow = 31
(12 + 1 + 10 + 8)
b i
Q
8
12
8
1 Maximum flow is 33 L/min.
Cuts B, C and D have the same capacity as the maximum flow.
The correct answer is C.
2 The flow must be heading towards the sink.
3 + 6 + 5 = 14
The correct answer is C.
3 Exploring all the options for the three given values for x
shows the following.
When x = 1: The capacity of the minimum cut (cut B) is 24.
This is not stated.
When x = 2: The capacity of the minimum cut (cut B) is 25.
When x = 3: The capacity of the minimum cut (cut B) is 26.
This is not stated.
The maximum flow is cut B if x = 2.
The correct answer is B.
O
10.5 Bipartite graphs and allocation problems
10.5 Exercise
7
9
M
9
N
P
4
R
4
5
10
ii Maximum flow = 18
(8 + 4 + 6)
6
O
13 a
B
A
D
20
75
I
45
45
C
50
E
G
H
30
100
150
80
45 Freeway 60
75 Freeway 50
F
J
50
K
60
Yes, a freeway is one road from one point to another. So,
we have only two freeways (DE and GH).
1 Electricity produced = supply 4000 kWh, 5000 kWh, and
6000 kWh
Total = 15 000 kWh
Towns supplied = demand
Town A = 20% of 15 000
= 3000 kWh
Town B = 25% of 15 000
= 3750 kWh
Town C = 15% of 15 000
= 2250 kWh
Town D = 15 000 − (3000 + 3750 + 2250)
= 6000 kWh
P df_Fol i o: 201
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
202
Sam
Knitting
Frances
Jogging
Mal
Cooking
A
p1
3000
B
4000
3750
p2
S
D
5000
6000
C
6
6000
p3
D
2 a
s1
30 000
A
s2
40 000
S
B
10 000
25 000
60 000
B
D
15 000
30 000
C
A
B
C
S
D
E
D
7
B
D
b Based on information in question 3, ‘Brian and Chris
between them have more different dishes than David and
Earl’.
The correct answer is D.
Brian and Chris have fish, soup, beef and dessert while
David and Earl have dessert, beef and fish.
4 Produces ∴ supply
• produces 1000 copies per month
400 → Factory 1
600 → Factory 2
Distributed ∴ demand
• Queensland = 350
Victoria = 1000 − 350
= 650
f1
X
Y
Z
3
0
4
0
2
1
3
3
0
A
X
B
Y
C
Z
W X
A ⎡4 3
⎢
B ⎢9 4
C ⎢⎢5 6
D ⎣4 8
Y
7
6
7
3
Z
3⎤
⎥
5⎥
8⎥⎥
5⎦
W X
Y
Z
0
0
1
5
3
2
2
0
0
1
3
2
Subtract smallest in row A = 3
in row B = 4
in row C = 5
in row D = 3
A
B
C
D
1
5
0
1
Number of lines = number of columns
A
W
B
X
C
Y
Q
400
350
S
D
600
650
f2
Photography
Z
7⎤
⎥
5⎥
2⎦
Y
3
4
5
There is only 1 possible allocation:
A → Y, B → X, C → Z
Total time = 3 + 2 + 2
= 7 hours
F
C
Will
Number of lines = number of columns
b Send 30 000 from S1 to A, 30 000 from S2 to A, 10 000
from S2 to B, 5000 from B to B, 5000 from B to C and
25 000 from C to C. (This may not be the cheapest
method.)
3 a
A
S
X
A ⎡6
⎢
B ⎢2
C ⎣3
subtracts in row A = 3
in row B = 2
in row C = 2
A
C
B
Chess
Katherine
2250
V
5 From the bipartite graph it can be said that Mal and Frances,
in total, have more hobbies than Sam and Will.
The correct answer is B.
Allocation: D → Y, C → W, B → X, so A → Z
Total time = 3 + 4 + 5 + 3
= 15 hours
8 ⎡6
9 9 4⎤
⎢
⎥
⎢10 9 9 7⎥
⎢4
9 6 3⎥⎥
⎢
⎣5
8 8 6⎦
D
Z
Row reduction
P df_Fol i o: 202
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Subtract smallest in row 1 = 4
2=7
3=3
4=5
⎡2 5 5 0 ⎤
⎢
⎥
⎢3 2 2 0 ⎥
⎢1 6 3 0 ⎥
⎢
⎥
⎣0 3 3 1 ⎦
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
Minimum total allocation = 5 + 17 + 6 + 8
= 36
10 a
3
0
3
0
0
1
Smallest uncovered number = 1
⎡2 3 3 1 ⎤
⎥
⎢
⎢4 1 1 2 ⎥
⎢1 4 1 1 ⎥
⎥
⎢
⎣1 2 2 3 ⎦
2
2
0
3
0
0
1
0
3
0
0
0
1 1
2
23
29
17
27
9 ⎤
⎥
14⎥
13⎥⎥
8 ⎦
14
6
14
22
Row reduction
Subtract smallest number in row 1 = 5
2=6
3 = 13
4=8
⎡0
18 9
4⎤
⎢
⎥
23 0
8⎥
⎢5
⎢8
4
1
0⎥⎥
⎢
⎣12 19 14 0⎦
Column reduction
Subtract smallest number in column 1 = 0
2=4
3=0
4=0
P df_Fol i o: 203
0 14
9
4
5 19
0
8
0
0
1
0
12 15
14
0
𝑍
13⎤
⎥
16⎥
18⎥⎥
24⎦
3 1
7
0
B
0
1
2
1
C
6
0
0
5
D
2
6
0
4
W
X
Y
Z
A → Z, B → W
C → X (as no other to X)
∴
D→Y
Total time 15 + 13 + 13 + 20 = 61 hours
b A
W
B
X
C
Y
D
Z
4 lines are required.
Minimum total allocation:
4 + 6 + 9 + 5 = 24
Solved by the Hungarian algorithm.
9 ⎡5
⎢
⎢11
⎢21
⎢
⎣20
𝑌
20
17
13
20
A
A
B
C
D
Subtract overall smallest number = 1
1
𝑋
14
16
13
26
W X Y Z
0
0
1 4 1
0 1 1
𝑊
A ⎡16
⎢
B ⎢15
C ⎢⎢19
D ⎣22
Subtract smallest in row A = 13
in row B = 15
in row C = 13
in row D = 20
Column reduction
Subtract smallest in column 1 = 0
2=2
3=2
4=0
2
3
203
11
Game Doll Truck
1 ⎡30
50
35⎤
⎢
⎥
2 ⎢45
50
30⎥
3 ⎣40
60
30⎦
Subtract smallest in row 1 = 30
row 2 = 30
row 3 = 30
Game Doll Truck
1
2
3
0
15
10
20 5
20 0
30 0
Minimum cost:
1 → G = $30
It is cheaper to buy the doll from store 2 than from store 3.
∴ 3−T
Store
1
2
3
Item
Game
Doll
Truck
∴ 30 + 50 + 30 = $110
Solved by column reduction.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
204
12 A ⎡7
⎢
B ⎢3
C ⎣6
7⎤
⎥
5⎥
5⎦
3
3
5
Subtract smallest in row A = 3
in row B = 3
in row C = 5
A
B
C
4
0
0
0
4
2
Allocation. A → 3, B → 3, C → 5
∴ 3 + 3 + 5 = 11
The correct answer is B.
13 a
1
0
0
⎡0
⎢
⎢3
⎢0
⎢
⎣0
𝐷1 𝐷2 𝐷3 𝐷4 𝐷5
Car 1 ⎡20 15 17 16 18⎤
⎢
⎥
2 ⎢17 15 19 17 16⎥
3 ⎢18 19 16 19 16⎥
⎢
⎥
4 ⎢19 19 17 21 17⎥
5 ⎣24 19 17 17 17⎦
Row reduction
Subtract smallest in row 1 = 15
2 = 15
3 = 16
4 = 17
5 = 17
D1 D2 D3 D4 D5
Car 1
2
3
4
5
5
2
2
2
7
0
0
3
2
2
2
4
0
0
0
1 3
2 1
3 0
4 0
0 0
Column reduction
Subtract smallest in column 1 = 2
2=0
3=0
4=0
5=0
𝐷1
Car 1 ⎡ 3
⎢
2 ⎢ 0
3 ⎢ 0
⎢
4 ⎢ 0
5 ⎣ 5
𝐷2
0
0
3
2
2
b Car 1
D1
Car 2
D2
Car 3
D3
Car 4
D4
𝐷3
2
4
0
0
0
𝐷4
1
2
3
4
0
D5
5
7
6
3
2
5
1
6
7⎤
⎥
0⎥
3⎥⎥
4⎦
Column reduction
Subtract smallest in column 1 = 0
2=3
3=1
4=0
0
3
0
0
2
4
3
0
1
4
0
5
7
0
3
4
Minimum number of lines required = 4
Solved by column reduction.
Minimum total allocation = 10 + 26 + 17 + 14
= 67
b ⎡12
⎢
⎢11
⎢12
⎢
⎢9
⎣14
10
11
16
10
11
11
13
13
9
11
13
12
16
11
11
11⎤
⎥
12⎥
12⎥
⎥
9 ⎥
11⎦
Row reduction
Subtract smallest number in row 1 = 10
2 = 11
3 = 12
4=9
5 = 11
𝐷5
3 ⎤
⎥
1 ⎥
0 ⎥
⎥
0 ⎥
0 ⎦
c i C1 → D2, C2 → D1, C5 → D4, C3 → D3,
C4 → D5 or C1 → D2, C2 → D1, C5 → D4,
C4 → D3
ii Total = $82 000
Car 5
P df_Fol i o: 204
14 a ⎡10 15 12 17⎤
⎢
⎥
⎢17 21 19 14⎥
⎢16 22 17 19⎥
⎢
⎥
⎣23 26 29 27⎦
Row reduction
Subtract smallest in row 1 = 10
2 = 14
3 = 16
4 = 23
⎡2
⎢
⎢0
⎢0
⎢
⎢0
⎣3
0
0
4
1
0
1
2
1
0
0
3
1
4
2
0
1⎤
⎥
1⎥
0⎥
⎥
0⎥
0⎦
Solved by row reduction.
Minimum total allocation = 10 + 11 + 12 + 9 + 11
= 53
15 a
J1 J2 J3
A ⎡30 40 50
⎢
B ⎢70 30 40
C ⎢⎢60 50 60
D ⎣20 80 50
Row reduction
J4
60⎤
⎥
70⎥
30⎥⎥
70⎦
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
Subtract smallest number in row A = 30
B = 30
C = 30
D = 20
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
𝐽1 𝐽2 𝐽3 𝐽4
𝐴 ⎡0
10 20 30⎤
⎢
⎥
𝐵 ⎢40 0
10 40⎥
𝐶 ⎢⎢30 20 30 0 ⎥⎥
𝐷 ⎣0
60 30 50⎦
Column reduction
Subtract smallest number in column J1 = 0
J2 = 0
J3 = 10
J4 = 0
A
B
C
D
J1
J2
J3
J4
0
40
30
0
10
0
20
60
10
0
20
20
30
40
0
50
b Hungarian algorithm
Smallest uncovered number = 10
𝐽1 𝐽2 𝐽3 𝐽4
A ⎡10 10 10 30⎤
⎢
⎥
B ⎢60 10 10 50⎥
⎢
C ⎢50 30 30 10⎥⎥
D ⎣10 60 20 50⎦
Overall smallest number = 10
𝐽1 𝐽2 𝐽3 𝐽4
A ⎡0
0
0 20⎤
⎢
⎥
B ⎢50 0
0 40⎥
⎢
C ⎢40 20 20 0 ⎥⎥
D ⎣ 0 50 10 40⎦
(One possible result)
A
J1
B
J2
C
J3
J4
c D
d i A → J2, B → J3, C → J4, D → J1 or A → J3, B → J2,
C → J4, D → J1
ii Total = 40 + 40 + 30 + 20
= 130 minutes
16
𝐽1 𝐽2
T ⎡100 50
⎢
U ⎢ 60 45
V ⎢⎢ 40 70
W ⎣ 70 50
Row reduction
𝐽3
35
70
50
70
𝐽4
55⎤
⎥
55⎥
30⎥⎥
70⎦
205
Subtract smallest in row T = 35
U = 45
V = 30
W = 50
𝐽1
T ⎡65
⎢
U ⎢15
V ⎢⎢10
W ⎣20
𝐽2 𝐽3
15 0
0 25
40 20
0 20
𝐽4
20⎤
⎥
10⎥
0 ⎥⎥
20⎦
Column reduction
Subtract smallest number from column 1 = 10
2=0
3=0
4=0
T
U
V
W
J1
J2
J3
J4
55
5
0
10
15
0
40
0
0
25
20
20
20
10
0
20
Hungarian algorithm
Smallest uncovered number = 5
𝐽1 𝐽2 𝐽3 𝐽4
T ⎡60 25 5 25⎤
⎢
⎥
U ⎢5
5 25 10⎥
V ⎢⎢ 5 50 25 5 ⎥⎥
W ⎣10 5 20 20⎦
Smallest number = 5
J1
J2
J3 J4
T
55
20
0
20
U
0
0
20
5
V
0
45
20
0
a T → J3, U → J1, V → J4, W → J2
b Time = 35 + 60 + 30 + 50
= 175 minutes
W
5
0
15
17 a
A
K ⎡60
⎢
L ⎢45
M ⎢⎢60
N ⎣42
C
78
80
35
66
F
67
70
70
54
5
G
37⎤
⎥
90⎥
86⎥⎥
72⎦
First modify the minimisation problem, by subtracting each
number by the overall largest value, 90.
A
⎡
K 30
⎢
L ⎢45
M ⎢⎢30
N ⎣48
C
12
10
55
24
F
23
20
20
36
G
53⎤
⎥
0⎥
4 ⎥⎥
18⎦
Row reduction
Subtract smallest from row K = 12
L=0
M=4
N = 18
P df_Fol i o: 205
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
206
TOPIC 10 Directed graphs and network flow • EXERCISE 10.5
A
K ⎡18
⎢
L ⎢45
M ⎢⎢26
N ⎣30
C
0
10
51
6
G
41⎤
⎥
0⎥
0 ⎥⎥
0⎦
F
11
20
16
18
b Column reduction
Subtract smallest from column A = 18
C=0
F = 11
G=0
K
L
M
N
A
C
F
G
0
27
8
12
0
10
51
6
0
9
5
7
41
0
0
0
c To solve, continue with Hungarian algorithm.
Smallest uncovered number = 5
A C F G
K ⎡5
5 5 51⎤
⎢
⎥
L ⎢27 10 9 5 ⎥
M ⎢⎢ 8 51 5 5 ⎥⎥
N ⎣12 6 7 5 ⎦
Subtract smallest number from all = 5
A
0
22
3
7
K
L
M
N
C
0
5
46
1
F
0
4
0
2
G
46
0
0
0
Repeat.
Smallest uncovered number = 1
A C F G
K ⎡1
1 1 48⎤
⎢
⎥
L ⎢22 5 4 1 ⎥
M ⎢⎢ 4 47 1 2 ⎥⎥
N ⎣7
1 2 1⎦
Subtract smallest number from all = 1
K
L
M
N
A
C
F
G
0
0
0
21
3
6
4
46
0
3
47
0
1
0
0
1
Ken
Algebra
Louise
Calculus
Mark
Functions
18
𝑃1 𝑃2
𝑉 1 ⎡13 17
⎢
𝑉 2 ⎢ 8 12
𝑉 3 ⎢⎢ 9 17
𝑉 4 ⎣21 16
𝑃3
14
17
14
13
𝑃4
23⎤
⎥
9⎥
11⎥⎥
14⎦
𝑃1
𝑉 1 ⎡0
⎢
𝑉 2 ⎢0
𝑉 3 ⎢⎢0
𝑉 4 ⎣8
𝑃3
1
9
5
0
𝑃4
10⎤
⎥
1⎥
2 ⎥⎥
1⎦
Row reduction
Subtract smallest number from row 1 = 13
2=8
3=9
4 = 13
𝑃2
4
4
8
3
Column reduction
Subtract smallest number from column 1 = 0
2=3
3=0
4=1
P1 P2 P3 P4
V1
V2
V3
V4
0
0
0
8
1
1
5
0
1
9
5
0
9
0
1
0
Smallest uncovered number = 1
𝑃1 𝑃2 𝑃3 𝑃4
𝑉1 ⎡1
1
1 10⎤
⎥
⎢
𝑉2 ⎢1
1
9
1⎥
5
5
2 ⎥⎥
𝑉 3 ⎢⎢ 1
1
1
2⎦
𝑉 4 ⎣10
Subtract smallest number = 1
P1 P2 P3 P4
V1
V2
V3
V4
0
0
0
9
0
0
4
0
0
8
4
0
9
0
1
1
a V1 → P2, V2 → P4, V3 → P1, V4 → P3 or
V1 → P3, V2 → P4, V3 → P1, V4 → P2
b Total = 17 + 9 + 9 + 13
= 48 km
10.5 Exam questions
Ken → Algebra
Louise → Geometry
Mark → Functions
Nancy → Calculus
d Average score
60 + 90 + 70 + 66
=
4
= 71.5%
Nancy
Geometry
1 The graph would be bipartite (two parts) and planar (no edges
which cross).
(The graph would not be ‘connected’ as not all vertices
connect to each other. It is not a ‘tree’ as the vertices are not
connected to each other.)
A
Choc
D
Straw
G
Vanilla
I
Lemon
P df_Fol i o: 206
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
(Note that the lines from A to Strawberry and from D to
Chocolate do NOT have to be drawn as crossed. Hence, the
graph is planar.)
The correct answer is C.
2 Looking at batting position number 1, Bo’s average score is
the highest, so he would be placed in the first position.
In batting position number 2, we can see that both Bo and
Cameron have an equal highest average score, but Bo has
already been used, so we will assign Cameron to this position.
This leaves Alex in batting position number 3 (he also has the
highest position 3 batting average).
Player
Batting position
Alex
Bo
Cameron
3
1
2
[1 mark]
3 Using the Hungarian algorithm, the minimum total time is
24 minutes.
This is produced when:
Job 1 = Chamath
Job 2 = Alan
Job 3 = Deidre
Job 4 = Ewen
Job 5 = Brianna
OR
Job 1 = Chamath
Job 2 = Alan
Job 3 = Brianna
Job 4 = Ewen
Job 5 = Deidre.
The correct answer is B.
c EST for activity E is 37.
The correct answer is C.
d Earliest completion time for the network is 69.
The correct answer is D.
2 Activity time for B:
Float = EFT–EST–time
2=7−0−T
T=5
EFT for activity C:
0 = EFT − 3 − 6
EFT = 9
The correct answer is A.
3 Flow starts at V and ends at S. V is the source and S is the
sink.
The correct answer is D.
4 The outflow from node T is:
15
35
12
T
4
There is more than one possible outflow.
The correct answer is E.
5 Minimum cut is shown in E.
J
14
12
7
K
13
16
23
16
Q
L
N
4
10
3
O
17
2
A − 38
B − 34 – but not minimum cut, because flow not cut off
C − 43
D − 60
E − 34
The correct answer is E.
6 Cuts stop all flow from the source to the sink.
P
10.6 Review
10.6 Exercise
Multiple choice
1 a The activities that come before activity E are A, B, C and
D.
The correct answer is C.
b A and C are the first edges (with no predecessors).
R
A
2
3
1
C
5
2
T
V
A–B
A–D
C–D
U
4
4
3
A
S
B
R
A′, 0
C
2
D
3
1
5
B, D–E
A
2
T
V
B
U
E
4
A′, 0
C
4
3
D
S
P df_Fol i o: 207
207
E–F
The correct answer is A.
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
208
Short answer
R
2
3
10
1
13
5
6
T
10
B
2
V
4
A
F
D
10
3
U
15
C
E
14
4
3
11 a
4
A
S
10
4
13
B
6
F
D
10
R
3
15
C
2
3
5
2
T
b The maximum flow is 19. (13 + 6)
12 a
10
14
1
V
E
U
G
16
4
4
3
A
H
13
10
4
B
S
13
3
2
6 10
3
R
15
C
2
3
5
2
T
E
b i The maximum flow is 33. (6 + 10 + 4 + 3 + 10)
ii
10
14
1
V
F
D
U
G
16
4
2
4
3
A
13
B
6 10
S
a and c do not show cuts.
The correct answer is A.
7 ⎡15 11 16 6 ⎤
⎢
⎥
2 12⎥
⎢13 5
⎢4
2 11 14⎥⎥
⎢
⎣12 7 12 10⎦
Row reduction: subtract smallest number from row.
1=6
2=2
3=2
4=7
⎡ 9 5 10 0 ⎤
⎢
⎥
⎢11 3 0 10⎥
⎢ 2 0 9 12⎥
⎢
⎥
⎣5 0 5
3⎦
The correct answer is C.
8 Column reductions: subtract smallest number from column.
1=2
2=0
3=0
4=0
⎡7 5 10
0⎤
⎢
⎥
9
3
0
10
⎢
⎥
⎥
⎢0 0
9
12
⎢
⎥
⎣3 0
5
3⎦
The correct answer is A.
9 It is not true that Melbourne has twice as many flights leaving
it as does Brisbane, because there are 3 flights leaving
Melbourne and only 1 flight leaving Brisbane.
The correct answer is D.
H
13
3
10
4
F
D
3
C
15
E
14
V U B
F
A ⎡5 7
5 9⎤
⎢
⎥
B ⎢6 10 10 7⎥
⎢
C ⎢7 5
3 8⎥⎥
D ⎣7 8
8 9⎦
Convert the table for a minimisation problem by subtracting
each number from the largest overall number (10).
V U B F
A ⎡5 3 5 1 ⎤
⎢
⎥
B ⎢4 0 0 3 ⎥
C ⎢⎢3 5 7 2⎥⎥
D ⎣3 2 2 1 ⎦
b Row reduction
Subtract smallest number from row A = 1
B=0
C=2
D=1
13 a
V
A ⎡4
⎢
B ⎢4
C ⎢⎢1
D ⎣2
U
2
0
3
1
B
4
0
5
1
F
0⎤
⎥
3⎥
0⎥⎥
0⎦
c Column reduction
Subtract smallest number from column V = 1
U=0
B=0
F=0
P df_Fol i o: 208
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
V
A ⎡3
⎢
B ⎢3
C ⎢⎢0
D ⎣1
F
0⎤
⎥
3⎥
0⎥⎥
0⎦
U B
2 4
0 0
3 5
1 1
A, 3
F
3
3
0
1
2
0
3
1
0
3
0
0
5
A
B
B−C
A, B−D
D
B'
C
C−G
D−H
D, G−E
B', 0 G, 2
A, 3
5
C, 4
5
5
D, 8
d
G
16
14
13
13
H, 3
9
11
J, 4
EFT
Float time
A
B
C
D
E
F
G
H
J
K
0
0
5
5
13
14
9
13
16
20
5
5
11
13
19
20
13
16
20
23
2
0
2
0
5
5
2
0
0
0
e To shorten the minimum completion time of the project,
activities that are on the critical path should be crushed.
Therefore, activities B, D, H, J, K should be crushed.
f If activity D is reduced to 5 hours, total duration for
activities C and G will become greater than the duration of
the activity D (as 4 + 2 = 6 and 6 > 5). Therefore, the new
critical path will go through C and G instead of D. That is,
the new critical path is B–C–G–H–J–K.
Originally, activity D took 8 hours to complete. The
new critical path will contain activities C and G instead of
D, with the combined duration of 4 + 2 = 6 hours. This is
reduction of 8 − 6 = 2 hours. Therefore, the total
completion time will be reduced by 2 hours. That is, the
new minimum completion time is 23 − 2 = 21 hours.
15 a
4
12
T, 4
14
U, 3
S', 0
S, 7
P, 4
Y, 7
12
12
5
5
V, 3
17
17
M, 2
F
5
5
J
23
23
16
16
O, 4
H
20
E, 1 19 F, 1 20 K, 3
EST
Q', 0
B'
J, 4
Activity letter
Q, 5
E
9
C, 4
5
5
10
E−F
H−J
B
23
K, 3
B, 5
0
0
D
H, 3
B', 0 G, 2
0
0
H
C
A
F, 1
c Critical path = B–B′–D–H–J–K
E
G
B'
20
E, 1
B, 5
14 a A, B first edge (no predecessors)
B
D, 8
0
Extended response
D
14
13
A, 3
ii A → F, B → U, C → V, D → B or A → F, B → B,
C → V, D → U
A
J, 4
b Earliest completion time = 23 hours
Smallest uncovered number is 1.
Add 1 to each covered number (twice at the
intersections).
V U B F
A ⎡3 2 4 1⎤
⎢
⎥
B ⎢4 1 1 5⎥
⎢
C ⎢1 4 6 2⎥⎥
D ⎣1 1 1 1 ⎦
Overall smallest number is 1.
Subtract 1 from each number in the matrix.
V U B F
A ⎡2 1 3 0⎤
⎢
⎥
B ⎢3 0 0 4⎥
C ⎢⎢0 3 5 1⎥⎥
D ⎣0 0 0 0 ⎦
B
K, 3
C, 4
U B
A
F, 1
H, 3
B', 0 G, 2
V
4
0
5
1
E, 1
B, 5
d i Hungarian algorithm
A
B
C
D
D, 8
209
R, 3
14
14
24
24
Z, 8
X, 3
W, 2 16
Critical path = Q–Q′–S–M–V–Y–Z
21
C
F, J−K
P df_Fol i o: 209
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
32
32
210
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
b Activity 0:
Float time = EFT − EST − time
= 10 − 0 − 4
=6
The remaining missing float times belong to activities on
the critical path. Critical path activities have a float
time of 0.
c The EST values are the numbers in the triangles at the start
of each activity. Refer to the figure in 1a.
d The EFT values are the numbers in the boxes at the end of
each activity. Refer to the figure in 1a.
e Float time = EFT − EST − time
Activity
Immediate
predecessor
T
EST
LST
Float
time
O
P
Q
—
—
—
4
4
5
0
0
0
10
5
5
6
1
0
R
Q
3
5
14
6
S
P, Q
7
5
12
0
T
U
O
S, T
4
3
4
12
14
17
6
2
M
V
S
R, M
2
3
12
14
14
17
0
0
W
R, M
2
14
21
5
X
Y
W
U, V
3
7
16
17
24
24
5
0
Z
X, Y
8
24
32
0
16 a Source = estuary, E
Sink = lake, L
b
Node
Inflow
A
51
B
87
C
49
D
31
F
108
G
23
Moths
Tour
information
Entry
Butterflies
(rainforest)
Arachnids
(scorpions)
Glowworms
(caves)
Arachnids
(spiders)
Exit
b
Butterflies
(temperate)
D, 2
A, 12 B, 13 C, 12
G, 6
H, 6
E, 4
I, 5
J, 12
c Arrival rate for entry → tour information = 12
From here the capacity of tour information → rainforest
butterflies = 13
If we look at just ‘13’, then we could say the room is the
most crowded of all.
If we look at ‘13’ as a capacity, then with ‘12’ entering, its
maximum flow is 12, so it can cope with the people.
But the crowd will then cut considerably after the
temperate butterflies.
F, 2
2
12
4
2
d i
D, 2
A, 12 B, 13 C, 12
G, 6
E, 4
H, 6
I, 5 J, 12
F, 2
Outflow
47
74
48
29
88
20
c The outflow for every node is less than the inflow. This may
be due to the death of some fish on their way upstream or it
may be due to the fish slowing down the further upstream
they go.
d Flow capacity = 51 + 62 + 49
= 162 fish/hour
e Maximum flow = minimum cut
= 19 + 88 + 12
= 119 fish/hour
17 a Doors leading from one section to the next, indicating the
direction in which they open:
ii The minimum cut indicates that 5 people per minute
should be admitted for a smooth flow through the
building.
18 a Add a ‘dummy’ position, giving each candidate 0 votes
for it.
A
B
C
D
E
PM
Treasurer
FM
Speaker
Dummy
48
1
8
2
41
35
10
35
9
11
31
5
31
15
17
22
4
22
39
8
0
0
0
0
0
PM
Treasurer
FM
Speaker
Dummy
0
47
40
46
7
13
38
13
39
37
17
43
17
33
31
26
44
26
9
40
48
48
48
48
48
b Subtract each element from the largest in the matrix (48).
The resultant matrix is:
A
B
C
D
E
P df_Fol i o: 210
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
c Row reduction
Subtract smallest number from row:
A=0
B = 38
C = 13
D=9
E=7
A
B
C
D
E
PM
Treasurer
FM
Speaker
Dummy
0
9
27
37
0
13
0
0
30
30
17
5
4
24
24
26
6
13
0
33
48
10
35
39
41
d Column reduction
Subtract smallest number from column:
PM = 0
Treasurer = 0
FM = 4
Speaker = 0
Dummy = 10
PM
A
B
C
D
E
e
Treasurer
0
9
27
37
0
13
0
0
30
30
FM
13
1
0
20
20
0
9
27
37
0
13
0
0
30
30
13
1
0
20
20
26
6
13
0
33
g PM is not given to the top vote getter.
Because of the tie between A and C, either the treasurer or
the foreign minister is not given to the top vote getter.
D definitely deserved speaker and B definitely deserved
dummy.
h Although apparently democratic, it ‘rewards’ candidates
who are relatively unpopular over the range of positions
(particularly candidate E) who only get a lot of votes for
one position, and ‘punishes’ those (particularly candidate
A) who do get lots of votes across the positions. One could
argue, in this case, that E got PM as she could do the ‘least
damage’ in that job!
10.6 Exam questions
1 Form a minimum spanning tree, ensuring that x and y are
included:
30
29
52
42
y
x
22
Speaker
Dummy
26
6
13
0
33
38
0
25
29
31
PM Treasurer FM Speaker Dummy
A
B
C
D
E
211
38
0
25
29
31
Smallest uncovered number is 13.
Add 13 to all covered numbers (twice at the intersections).
PM Treasurer FM Speaker Dummy
⎤
A ⎡13
13
13
26
38
⎢
⎥
B ⎢35
13
14
19
13
⎥
⎥
C ⎢53
13
13
26
38
⎢
⎥
D ⎢63
43
33
13
42
⎥
⎦
E ⎣13
30
20
33
31
Overall the smallest number is 13.
Subtract 13 from each element in the matrix.
PM Treasurer FM Speaker Dummy
⎡
⎤
A 0
0
0
13
25
⎢
⎥
B ⎢22
0
1
6
0
⎥
⎥
C ⎢40
0
0
13
25
⎢
⎥
D ⎢50
30
20
0
29
⎥
⎦
E ⎣0
17
7
20
18
36
40
Adding up the lengths:
30 + 29 + 42 + 40 + 22 + 36 + 52 + x + y = 251 + x + y
The answer that matches the options is when x = 50 and
y = 60 for a minimum length of 361 km.
The correct answer is B.
2. a
F
A
E
B
C
D
[1 mark]
b 2
[1 mark]
c. i C − D − E − B − A or the other way A − B − E − D − C
[1 mark]
ii Hamiltonian Cycle
[1 mark]
3 a By counting the number of different routes, you will find
that there are 10 different routes possible.
[1 mark]
b Capacity of cut 1 = 20 + 12 + 20 = 52
[1 mark]
4 a This dummy activity could be drawn as a directed edge
from the end of activity B to the start of activity C. [1 mark]
f B → dummy (It’s the only 0 in the Dummy column).
D → speaker (It’s the only 0 in the Speaker column).
E → prime minister (It’s the only 0 in the row).
A → treasurer, C → foreign minister or A → foreign
minister, C → treasurer.
P df_Fol i o: 211
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual
TOPIC 10 Directed graphs and network flow • EXERCISE 10.6
212
See network drawing below.
See the Image bottom of the page*
b As activity C has an EST of 7 and activity D has an EST of
7, and activity C is a predecessor of activity D, then the
duration of activity C is 2 months.
[1 mark]
c Float = LST – EST
A, F, H and E.
[1 mark]
d Reduce the completion of time of activity B from 5 to 2
months, so that the project will be completed in a minimum
time of 17 months.
[1 mark]
5 This directed network has two critical paths, ADHK and
BFJK. This means the activities that are not on the critical
path can be delayed.
These activities are C, E, G and I. Therefore, 4 activities can
be delayed without affecting the minimum completion time of
the project.
The correct answer is B.
*4 a
C, ...
A, 2
E, 9
D, 7
Bʹ
G, 4
B, 5
P df_Fol i o: 212
I, 2
H, 9
F, 3
Jacaranda Maths Quest 12 General Mathematics VCE Units 3 & 4 Seventh Edition Solutions Manual