SAVE A TREE – PLEASE DO NOT PRINT ME
IB Math Studies SL2 – Semester I Final Exam Review Questions
1.
The heights and weights of 10 students selected at random are shown in the table below.
Student
1
2
3
4
5
6
7
8
9
10
Height
155
161
173
150
182
165
170
185
175
145
x cm
Weight
50
75
80
46
81
79
64
92
74
108
y kg
(a)
Plot this information on a scatter graph. Use a scale of 1 cm to represent 20 cm on the
x-axis and 1 cm to represent 10 kg on the y-axis.
(4)
(b)
Calculate the mean height.
(c)
Calculate the mean weight.
(d)
It is given that Sxy = 44.31.
(i)
By first calculating the standard deviation of the heights, correct to two decimal places, show that the
gradient of the line of regression of y on x is 0.276.
(ii)
Calculate the equation of the line of best fit.
(iii) Draw the line of best fit on your graph.
(e)
Use your line to estimate
(i)
the weight of a student of height 190 cm;
(ii)
the height of a student of weight 72 kg.
(f)
It is decided to remove the data for student number 10 from all calculations. Explain briefly what effect this will
have on the line of best fit.
(1)
(1)
(6)
(2)
(1)
(Total 15 marks)
2.
200 people of different ages were asked to choose their favourite type of music from the choices Popular, Country and
Western and Heavy Metal. The results are shown in the table below.
Age/Music
Popular
Country
Heavy
Totals
choice
and
Metal
Western
11–25
35
5
50
90
26–40
30
10
20
60
41–60
20
25
5
50
Totals
85
40
75
200
It was decided to perform a chi-squared test for independence at the 5 level on the data.
(a)
Write down the null hypothesis.
(1)
(b)
Write down the number of degrees of freedom.
(c)
Write down the chi-squared value.
(d)
State whether or not you will reject the null hypothesis, giving a clear reason for your answer.
(1)
(2)
(2)
(Total 6 marks)
3.
The eye colour and gender of 500 students are noted and the results are indicated in the table below.
Blue
Brown
Green
Male
18
152
50
Female
40
180
60
It is believed that eye colour is related to gender in a school in Banff. It is decided to test this hypothesis by using a
test at the 5% level of significance.
2
1
(a)
Write down the null hypothesis for this experiment.
(b)
Show that the number of degrees of freedom is 2.
(1)
(1)

2
(c)
Write down the
critical value for the degrees of freedom.
(d)
Calculate the

(e)
Does the evidence suggest that eye colour is related to gender in this school? Give a clear reason for your answer.
(1)
2
test statistic for this data.
(2)
(2)
(Total 7 marks)
4.
(a)
For his Mathematical Studies project, Marty set out to discover if stress was related to the amount of time that
students spent travelling to or from school. The results of one of his surveys are shown in the table below.
Travel time (t mins)
Number of students
↓
high stress
moderate stress
low stress
9
5
18
t  15
17
8
28
15  t  30
18
6
7
30  t
2
He used a χ test at the 5 level of significance to find out if there was any relationship between student stress
and travel time.
(i)
Write down the null and alternative hypotheses for this test.
(2)
(ii)
Write down the table of expected values. Give values to the nearest integer.
(iii)
Show that there are 4 degrees of freedom.
(3)
(1)
2
Calculate the χ statistic for this data.
(iv)
(2)
2
The χ critical value for 4 degrees of freedom at the 5 level of significance is 9.488.
(v)
What conclusion can Marty draw from this test? Give a reason for your answer.
(2)
(b)
Marty asked some of his classmates to rate their level of stress out of 10, with 10 being very high. He also asked
them to measure the number of minutes it took them to get from home to school. A random selection of his
results is listed below.
Travel time (x)
13
24
22
18
36
16
14
20
6
12
Stress rating (y)
3
7
5
4
8
8
4
8
2
6
(i)
Write down the value of the (linear) coefficient of correlation for
this information.
(1)
(ii)
Explain what a positive value for the coefficient of correlation indicates.
(iii)
Write down the linear regression equation of y on x in the form y = ax + b
(iv)
Use your equation in part (iii) to determine the stress rating for a student who takes three quarters of an
hour to travel to school.
(v)
Can your answer in part (iv) be considered reliable? Give a reason for your answer.
(1)
(2)
(2)
(2)
(Total 18 marks)
2
5.
A bank in Canada offers the following exchange rate between Canadian dollars (CAD) and Euros (EUR). The bank sells
1 CAD for 1.5485 EUR and buys 1 CAD for 1.5162 EUR. A customer wishes to exchange 800 Canadian dollars for
Euros.
(a)
Find how many Euros the customer will receive.
(b)
The customer has to cancel his trip and changes his money back later when the rates are “sells 1 CAD = 1.5546
EUR, buys 1 CAD = 1.5284 EUR”. Use the “we sell” information to find how many Canadian dollars he
receives.
(c)
How many Canadian dollars has he lost on the transaction?
(Total 6 marks)
6.
The following is a currency conversion table:
FFR
USD
JPY
GBP
French Francs (FFR)
1
p
q
0.101
US Dollars (USD)
6.289
1
111.111
0.631
Japanese Yen (JPY)
0.057
0.009
1
0.006
British Pounds (GBP)
9.901
1.585
166.667
1
For example, from the table 1 USD = 0.631 GBP. Use the table to answer the following questions.
(a)
Find the values of p and q.
(2)
(b)
Mireille wants to change money at a bank in London.
(i)
How many French Francs (FFR) will she have to change to receive 140 British Pounds (GBP)?
(ii)
The bank charges a 2.4% commission on all transactions. If she makes this transaction, how many British
Pounds will Mireille actually receive from the bank?
(c)
Jean invested 5000 FFR in Paris at 8% simple interest per annum.
Paul invested 800 GBP in London at 6% simple interest per annum.
(i)
How much interest in FFR did Jean earn after 4 years?
(ii)
How much interest in US Dollars did Paul earn after 4 years?
(iii) Who had earned more interest after 4 years?
(iv) Explain your reasoning in part (c) (iii).
(4)
(7)
(Total 13 marks)
7.
A family in Malaysia received a gift of $4000 AUD from a cousin living in Australia.
The money was converted to Malaysian Ringgit. One Ringgit can be exchanged for 0.4504 AUD.
(a)
Calculate the amount of Ringgit received.
The money was invested for 2 years and 6 months at 5.2% p.a. compounding monthly.
(b)
Calculate the amount of interest earned from this investment. Give your answer to the nearest Ringgit.
(Total 8 marks)
8.
William invests $1200 for 5 years at a rate of 3.75% compounded annually.
(a)
Calculate the amount of money he has in total at the end of the 5 years.
(b)
The interest rate then drops to 3.25%. If he continues to leave his money in the bank find how much it will be
worth after a further 3 years.
(Total 6 marks)
9.
Miranti deposits $1000 into an investment account that pays 5% interest per annum.
(a)
What will be the value of the investment after 5 years if the interest is reinvested?
(2)
(b)
How many years would it take Miranti’s investment of $1000 to double in value?
(4)
At the beginning of each year Brenda deposits $1000 into an investment account that pays 5% interest per annum.
Interest is calculated annually and reinvested.
(c)
How much would be in Brenda’s account after 5 years?
(4)
(Total 10 marks)
3
10.
Sandra is attempting an exam question. She has to choose two correct statements from a list of five. Below is a tree
diagram showing Sandra’s possible choices. One of the probability values is missing.
correct
—
correct
incorrect
2
–
5
incorrect
3
–
5
correct
incorrect
(a)
Fill in the missing probability value on the diagram.
(b)
(i)
(ii)
3
–
4
2
–
4
2
–
4
If Sandra makes two guesses, what is the probability that she will get only one of them correct?
Sandra definitely knows the first correct statement but has to guess the second. What is the probability
that she will answer both correctly?
(Total 8 marks)
11.
The data in the table below refers to a sample of 60 randomly chosen plants.
Growth rate
Classification by environment
dark
light
shady

high
low
total
(a)
3
8
11
8
9
17
14
18
32
total
25
35
60
(i)
Find the probability of a plant being in a shady environment.
(ii)
Find the probability of a plant having a low growth rate and being in a dark environment.
(iii)
Find the probability of a plant not being in a dark environment.
(5)
(b)
A plant is chosen at random from the above group.
Find the probability that the chosen plant has
(i)
a high growth rate or is in a dark environment, but not both
(ii)
a light environment, given that it has a high growth rate.
(c)
The 60 plants in the above group were then classified according to leaf type. It was found that 15 of the plants
had type A leaves, 37 had type B leaves and 8 had type C leaves.
Two plants were randomly selected from this group. Find the probability that
(i)
both plants had type C leaves
(ii)
neither of the plants had type B leaves.
(4)
(5)
(Total 14 marks)
4
12.
The Venn diagram below shows the number of students studying Science (S), Mathematics (M) and History (H) out of a
group of 20 college students. Some of the students do not study any of these subjects, 8 study Science, 10 study
Mathematics and 9 study History.
U
1
S
M
4
A
2
1
3
3
H
(a)
(i)
(ii)
(iii)
(b)
Draw a sketch of the Venn diagram above and shade the region which represents S   H.
(c)
How many students belong to the region labelled A?
Describe in words the region labelled A.
How many students do not study any of the three subjects?
(5)
(1)
Calculate n(S  H).
(2)
This group of students is to compete in an annual quiz evening which tests knowledge of Mathematics, Science and
History. The names of the twenty students are written on pieces of paper and then put into a bag.
(d)
One name is randomly selected from the bag. Calculate the probability that the student selected studies
(i)
all three subjects;
(ii)
History or Science.
(2)
(e)
A team of two students is to be randomly selected to compete in the quiz evening. The first student selected will
be the captain of the team. Calculate the probability that
(i)
the captain studies all three subjects and the other team member does not study any of the three subjects;
(ii)
one student studies Science only and the other student studies History only;
(iii) the second student selected studies History, given that the captain studies History and Mathematics.
(5)
(Total 15 marks)
13.
There are 49 mice in a pet shop.
30 mice are white.
27 mice are male.
18 mice have short tails.
8 mice are white and have short tails.
11 mice are male and have short tails.
7 mice are male but neither white nor short-tailed.
5 mice have all three characteristics and
2 have none.
Copy the diagram below to your examination script.
S
W
3
U
W represents white mice.
M represents male mice.
S represents short-tailed mice.
5
M
(a)
Complete the diagram, using the information given in the question.
(4)
5
(b)
Find
(i)
(ii)
n(M W)
n(M′  S)
(3)
Two mice are chosen without replacement.
(c)
Find P (both mice are short-tailed).
(2)
(Total 9 marks)
14.
The sets U, P, R and S are defined as follows:
U = {all quadrilaterals}
P = {all parallelograms}
R = {all rectangles}
S = {all squares}
(a)
Draw a Venn Diagram illustrating the relationships of the above sets.
(4)
(b)
Draw a separate Venn Diagram for each of the examples below.
Indicate by shading each of the following:
(i)
(P  S)
(ii)
(R  S)  P
(4)
(Total 8 marks)
15.
The sets A, B and C are subsets of U. They are defined as follows:
U = {positive integers less than 16}
A = {prime numbers}
B = {factors of 36}
C = {multiples of 4}
(a)
List the elements (if any) of the following:
(i)
A;
(ii)
B;
(iii) C;
(iv) A  B  C.
(4)
(b)
(i)
(ii)
Draw a Venn diagram showing the relationship between the sets U, A, B and C.
Write the elements of sets U, A, B and C in the appropriate places on the Venn diagram.
(4)
(c)
From the Venn diagram, list the elements of each of the following
(i)
A  (B  C);
(ii)
(A  B);
(iii) (A  B)  C.
(3)
(d)
Find the probability that a number chosen at random from the universal set U will be
(i)
a prime number;
(ii)
a prime number, but not a factor of 36;
(iii) a factor of 36 or a multiple of 4, but not a prime number;
(iv) a prime number, given that it is a factor of 36.
(6)
(Total 17 marks)
6
16.
The following table shows the age distribution of teachers who smoke at Laughlin High School.
Ages
Number of smokers
20 ≤ x < 30
5
30 ≤ x < 40
4
40 ≤ x < 50
3
50 ≤ x < 60
2
60 ≤ x < 70
3
(a)
Calculate an estimate of the mean smoking age.
(b)
On the following grid, construct a histogram to represent this data.
(Total 4 marks)
17.
For the set of {8, 4, 2, 10, 2, 5, 9, 12, 2, 6}
(a)
calculate the mean;
(b)
find the mode;
(c)
find the median.
(Total 4 marks)
A random sample of 200 females measured the length of their hair in cm. The results are displayed in the cumulative
frequency curve below.
200
175
Cumulative frequency
18.
150
125
100
75
50
25
0
0
5
10 15 20 25 30 35 40 45 50
length (cm)
(a)
Write down the median length of hair in the sample.
(b)
Find the interquartile range for the length of hair in the sample.
(c)
Given that the shortest length was 6 cm and the longest 47 cm, draw and label a box and whisker plot for the data
on the grid provided below.
(1)
(2)
0
5
10 15 20 25 30 35 40 45 50
length (cm)
(3)
(Total 6 marks)
7
19.
The heights (cm) of seedlings in a sample are shown below.
6
7
8
9
10
11
(a)
(b)
(c)
(d)
3,
2,
3,
2,
3,
2,
7
5,
6,
5,
6,
2
key 6 3
8
6,
7,
6
8,
8
represents
63 cm
8
State how many seedlings are in the sample.
Write down the values of
(i)
the median;
(ii)
the first and third quartile.
Calculate the range.
Using the scale below, draw a box and whisker plot for this data.
60
70
80
90
100
110
120
(Total 6 marks)
The number of hours that a professional footballer trains each day in the month of June is represented in the following
histogram.
10
9
8
7
number of days
20.
6
5
4
3
2
1
0
1
2
3
4
5
7
6
number of hours
(a)
Write down the modal number of hours trained each day.
(b)
Calculate the mean number of hours he trains each day.
8
9
10
(Total 8 marks)
8
21.
The table below shows the number and weight (w) of fish delivered to a local fish market one morning.
weight (kg)
frequency
cumulative frequency
0.50 ≤ w < 0.70
16
16
0.70 ≤ w < 0.90
37
53
0.90 ≤ w < 1.10
44
c
1.10 ≤ w < 1.30
23
120
1.30 ≤ w < 1.50
10
130
(a)
(i)
Write down the value of c.
(ii)
On graph paper, draw the cumulative frequency curve for this data. Use a scale of 1 cm to represent 0.1 kg
on the horizontal axis and 1 cm to represent 10 units on the vertical axis. Label the axes clearly.
(iii)
Use the graph to show that the median weight of the fish is 0.95 kg.
(1)
(4)
(1)
(b)
th
(i)
The zoo buys all fish whose weights are above the 90
How many fish does the zoo buy?
percentile.
(ii)
A pet food company buys all the fish in the lowest quartile. What is the maximum weight of a fish bought
by the company?
(2)
(3)
(c)
A restaurant buys all fish whose weights are within 10% of the median weight.
(i)
Calculate the minimum and maximum weights for the fish bought by the restaurant.
(2)
(ii)
22.
Use your graph to determine how many fish will be bought by the restaurant.
(3)
(Total 16 marks)
In an experiment a vertical spring was fixed at its upper end. It was stretched by hanging different weights on its lower
end. The length of the spring was then measured. The following readings were obtained.
Load (kg) x
0
1
2
3
4
5
6
7
8
Length (cm) y
23.5
25
26.5
27
28.5
31.5
34.5
36
37.5
(a)
Plot these pairs of values on a scatter diagram taking 1 cm to represent 1 kg on the horizontal axis and 1 cm to
represent 2 cm on the vertical axis.
(4)
(b)
(i)
Write down the mean value of the load ( x ).
(1)
(ii)
Write down the standard deviation of the load.
(1)
(iii) Write down the mean value of the length ( y ).
(1)
(iv) Write down the standard deviation of the length.
(1)
(c)
Plot the mean point ( x , y ) on the scatter diagram. Name it L.
(1)
It is given that the covariance Sxy is 12.17.
(d)
(i)
Write down the correlation coefficient, r, for these readings.
(1)
(ii)
Comment on this result.
(2)
(e)
Find the equation of the regression line of y on x.
(2)
(f)
Draw the line of regression on the scatter diagram.
(2)
(g)
(i)
Using your diagram or otherwise, estimate the length of the spring when a load of 5.4 kg is applied.
(1)
(ii)
Malcolm uses the equation to claim that a weight of 30 kg would result in a length of 62.8 cm. Comment
on his claim.
(Total 18 marks)
9
23.
The table below shows the percentage, to the nearest whole number, scored by candidates in an examination.
Marks (%)
0–9
10–19
20–29
30–39
40–49
50–59
60–69
70–79
80–89 90–100
Frequency
2
7
8
13
24
30
6
5
3
2
The following is the cumulative frequency table for the marks.
Marks (%)
Cumulative frequency
< 9.5
2
< 19.5
9
< 29.5
s
< 39.5
30
< 49.5
54
< 59.5
84
< 69.5
t
< 79.5
95
< 89.5
98
< 100
100
(a)
Calculate the values of s and of t.
(2)
(b)
Using a scale of 1 cm to represent 10 marks on the horizontal axis, and 1 cm to represent 10 candidates on the
vertical axis, draw a cumulative frequency graph.
(c)
Use your graph to estimate
(i)
the median mark;
(ii)
the lower quartile;
(iii) the pass mark, if 40% of the candidates passed.
(3)
(4)
(Total 9 marks)
24.
A bag contains two red sweets and three green sweets. Jacques takes one sweet from the bag, notes its colour, then eats
it. He then takes another sweet from the bag.
Complete the tree diagram below to show all probabilities.
Red
2
5
Red
Green
Red
2
4
Green
Green
(Total 4 marks)
10
IB Math Studies SL2 – Semester I Final Exam Review Questions Mark Scheme
1.
(a)
x
y
155
50
161
75
173
80
150
46
80
100
182
81
165
79
170
64
185
92
175
74
145
108
y
100
90
80
70
weight
(kg) 60
50
40
30
20
10
0
20
40
60
120 140 160
Height (cm)
180
200
x
(A2)(A2)
4
Notes: Award (A1) for axes correctly labelled, and (A1) for correct scales.
Award (A1) for 4, 5 6, or 7 correctly plotted points, (A2) for 8 or more.
(b)
Mean height = 166.1 = 166 (3 s.f.)
(A1)
1
(c)
Mean weight = 74.9 (3 s.f.)
(A1)
1
(d)
(i)
Sx = 12.68
(A1)
Sxy
44.31

Gradient =
= 0.276
2
Sx
(12.68) 2
(M1)(AG)
y – 74.9 = 0.276(x – 166)
y = 0.276x + 29.1
OR
y = 0.276x + 29.1
(M1)
(A1)
(iii)
Line on graph.
Note: Award (A1) for the y-intercept at 29.1, and (A1) for a straight line through
(166, 74.9).
(A2)
(i)
y = 0.276 × 190 + 29.1
= 81.5 kg
72 = 0.276x + 29.1
(A1)
(ii)
(e)
(ii)
x=
(G2)
6
72  29.1
0.276
= 155 cm.
(A1)
11
OR
From the graph
(i)
(ii)
(f)
(A1)
y = 81 (1)
x = 155 (1)
Note: Follow through with candidate’s line.
(A1)
(A1)
The “line of best fit” becomes closer to the remaining points.
OR
Gradient becomes steeper and the line is more accurate ‘best fit’.
OR
Any reasonable explanation. (Line becomes y = 1.10x – 113)
(R1)
2
(R1)
(R1)
1
[15]
2.
(a)
Choice of music is independent of age.
(A1)
(C1)
(b)
(3 – 1)(3 – 1)
=4
(A1)
(C1)
(c)
2
χ = 51.6
(A2)
Note: 52 is an accuracy penalty (A1)(A0)(AP).
(d)
p-value < 0.05 for 5 level of significance
2
or 51.6  χ crit
Reject the null hypothesis (do not accept the null hypothesis).
Note: Do not award (R0)(A1).
(C2)
(R1)(ft)
(R1)(ft)
(A1)(ft)
(C2)
[6]
3.
(a)
Eye colour and gender are independent.
OR
There is no relationship (association) between eye colour and gender.
(A1)
1
(2 – 1)(3 – 1)
=2
(M1)
(AG)
1
(c)
5.991 (5.99)
(A1)
1
(d)
4.48
(G2)
2
(b)
(e)
2
2
For comparing  test statistic with  critical value
No, eye colour is not related to gender
2 test statistic < 2 critical value
OR
For comparing their p-value with 0.05
No, eye colour is not related to gender
p-value of 0.106 > 0.05
(A1)
(R1)
(A1)
(R1)
2
[7]
4.
(a)
(i)
(ii)
H0 : level of stress is independent of travel time
H1 : level of stress is not independent of travel time
(or reasonable equivalents)
12.1 5.24 14.6
20.1 8.68 24.2
11.8 5.08 14.2
(A1)
(A1)(ft) 2
(M1)(A1)(G2)
12
Note: (M1) for attempting to calculate expected values by hand eg
44  32
12.1etc.
116
(iii)
(iv)
(v)
12 5 15
20 9 24
12 5 14
Nearest integers
(A1)(G3)
df = (r – 1) (c – 1) = (3 – 1)(3 – 1) = 4
2
 = 9.83(1)
2
OR  = 9.277 ..... if calculated from integer values
2
For  = 9.83 Do not accept H0 :
(Level of stress is not independent of travel time or reasonable equivalent)
because
2
2
 calc
  crit
or p-value < 0.05
3
(M1)(AG)
(G2)
1
2
OR (M1)(A1)
(A1)(ft)
(R1)(ft)
OR
2
For  = 9.278 Accept H0 :
2
2
χ calc
 χ crit
because
or p-value > 0.05
Note: a correct reason must be given for the (A1) to be awarded.
(b)
(i)
(ii)
(iii)
(iv)
(v)
r = 0.667
Stress rating increases as travel time increases
(or reasonable equivalent eg y increases as x increases).
Note: Do not accept “positive correlation”
y = 0.181x + 2.22
for 0.181x and
for 2.22
Note: For y = 2.22x + 0.181, award (A0)(A1)(ft)
Putting x = 45
0.181× 45 + 2.22
= 10.365 (10.4)
Notes: Allow 10 or 11 only if the method is shown and is correct.
Allow follow through only if method shown.
(A1)(ft)
(R1)(ft) 2
(A1)
1
(R1)
1
(A1)
(A1)
2
(M1)
(A1)(ft)(G2)
not reliable …
(A1)
Because result is outside the data range or because the
correlation coefficient not high or the sample is small or
responses are subjective.
(R1)
Note: Award (R1) for any of the above. A correct reason must be given to award the (A1).
2
2
[18]
5.
(a)
800 × 1.5162 for multiplying by 1.5162
= 1212.96 EUR (accept 1213)
(M1)
(A1)
(C2)
(b)
1212.96 / 1.5546
= 780.24 (accept 780)
(M1)
(A2)
(C3)
(c)
19.76 CAD
(A1)
(C1)
[6]
13
6.
(a)
(b)
p = 0.159
q = 17.5
(A1)
(A1)
(i)
(M1)
140(9.901)
= 1386.14
= 1390 FFR (3 s.f.)
OR
140
= 1386.14
0.101
(ii)
(c)
(i)
(ii)
(iii)
(iv)
2
(A1)
(M1)
= 1390 FFR (3 s.f.)
Also accept 1387 FFR (you must go up to get 140 GBP)
(A1)
2
140(0.024) = 3.36 GBP Commission
140 – 3.36 = 136.64 or 137 (3 s.f.)
Note: Commission can be calculated in FFR also for (Ml).
(M1)
(A1)
2
I = 5000(0.08)(4)
= 1600 FFR
I = 800 (0.06)(4)
= 192 GBP
192(1.585) = 304.32 USD or 304 USD (3 s.f.)
(M1)
(A1)
2
(M1)
(A1)
2
Paul
Jean: 1600 FFR
1600 ÷ 6.289 = 254.40 or 254 (3 s.f.)
Paul: 304.32 or 304 (3 s.f.)
Note: Accept any same currency comparison.
(A1)
1
(R1)
(R1)
2
[13]
7.
(a)
$4000 / 0.4504 = 8880.99 Ringgit (8880 to 3 s.f.)
Note: Allow 8881 (nearest Ringgit).
(b)
8880.99 × (1.0043) 30 – 8880.99
(M1)(A1)
(C2)
(M1)(A1)(
A1)(M1)
= 1230.09
(A1)
Note: Award (M1) for compound calculation, (A1) for multiplier, (A1) for 30 seen
and (M1) for deducting principle.
Use of 1.0043 (no recurrence) is premature rounding: award at
most (M1)(A0)(A1)(M1)(A0), but final (A1) ft can be awarded below for 1220.
= 1230 Ringgit
(A1)
(C6)
Note: Final (A1) is for rounding candidate’s answer to nearest Ringgit.
This is awarded only if the interest is calculated.
[8]
8.
(a)
For attempting to find 5 years by compound interest formula or any
alternative method.
For using (1.0375)
$1442.52 accept 3 s.f.
(M1)
(M1)
(A1)
(G3)
Note: Accept $1440 or $1443.
(b)
For using answer in part (a) in an expression.
3
For multiplying by (1.0325)
$1587.79 accept $1588 or $1590
(M1)
(M1)
(A1)
(G3)
[6]
14
9.
(a)
r 


A = C 1 
 100 
n
5 


= 1000 1 
100


5
(M1)
= $1276.28
(b)
(A1)
5 


2000 = 1000 1 
 100 
2 = 1.05
n
2
n
(M1)
n
(M1)
n
1.05
10
1.05 = 1.6
20
20
1.05 = 2.7
15
15
1.05 = 2.07
14
14
1.05 = 1.98
n = 15 years
10
(M1)
(A1)
4
(A1)
4
(c)
Year
Deposit
Start year balance
1
$1000
$1000
2
$1000
$2050
3
$1000
$3152.50
4
$1000
$4310.125
5
$1000
$5525.63125
After 5 years $5801.91
End of year balance
$1050
$2152.50
$3310.125
$4525.63125
$5801.91
(M3)
[10]
10.
(a)
correct
incorrect
correct
1
4
2
5 incorrect
3
4
3
5
2
4
correct
incorrect
(b)
(i)
2
4
2 3 3 2
  
5 4 5 4
(A2)
(C2)
(A1)(A1)
Note: Award (A1) for each correct product.
12
=
(= 0.6)
20
(A1)
(C3)
15
(ii)
2 1

5 4  1 = (0.25)
3
1 4

10 10
(A1)(A1)(
(C3)
A1)
Note: Award (A1) for
2 1
3
1
seen.
 seen and (A1) for

5 4
10 10
[8]
11.
Accept all answers given as percentages. If a correct fraction is seen as answer, ignore subsequent cancellations or
decimal reductions (including AP’s) if incorrect.
Note In each of (a)(i), (a)(iii), (b)(i) and (b)(ii) award (A1) for numerator and (A1)
for denominator. Ft if the denominator is incorrect.
(a)
(i)
P(shady) =
32  8

  or 0.533 
60  15

(A1)(A1)
Note: Award (A1) for numerator, (A1) for denominator.
(b)
8  2

  or 0.133 
60  15

(ii)
P(dark and low growth rate) =
(iii)
P(not dark) =
(i)
P (high growth rate or dark environment)
60  11 49
(= 0.817)

60
60
11  25 – 6  8  14  8  30  1

 or
    or 0.5 
60
60
60
2




8
P (in light, given high growth rate) =
(= 0.32)
25
=
(ii)
(c)
(i)
P(CC) =
8
7
56  14



 0.0158 

60 59 3540  885

(A1)
(A1)(A1)
5
(A1)(A1)
(A1)(A1)
4
(M1)(A1)
or (G2)
Note: Award (M1) for multiplication of two fractions, (A1) for the answer.
(ii)
Three alternative valid approaches might be seen:
23 22

60 59
15 14 15 8
8 15 8
7
OR







60 59 60 59 60 59 60 59
8 37 
 37 36 37 15 15 37 37 8
OR 1 – 








 
60
59
60
59
60
59
60
59
60
59 

506  253

=
or 0.143 
 or
3540  1770

(M1)(A1)
(A1)
or (G3) 5
23 22
Note: If =
is doubled, award (M1)(A0)(A0).

60 59
If non-replacement is treated incorrectly in (i), ft to (ii) if consistent
[14]
16
12.
(a)
(i)
(ii)
(iii)
10 – 6 = 4
OR
4
(M1)(A1)
(A2)
Students who study Mathematics only.
20 – (4 + l + 2 + l + 3 + 3 + 4) = 2
OR
2
(A1)
(M1)(A1)
(A2)
5
(A1)
1
(b)
M
S
H
(c)
(d)
n(S  H) = 8 + 9 – 3 = 14
OR
n(S  H) = 14
(i)
(ii)
(e)
(i)
2  1

   0.1
20  10

14  7

P(History or Science) =
   0.7 
20  10

P(captain studies all, other student studies none) =
2
(A1)
(A1)
2
2

20 19
4  1

 0.0105 

380  95

2
(M1)
(A1)
P(one studies Science only and the other studies History only)
=
4
3
 ×2
20 19
Note: Award (M1) for
=
(iii)
(A2)
P(studies all 3 subjects) =
=
(ii)
(M1)(A1)
(M1)
4
3

20 19
24  6

 0.0632 

380  95

P(HistoryHistory and Maths) =
(A1)
8
19
(A1)
5
[15]
17
13.
(a)
(A1)(A1)
(A1)(ft)
(A1)(ft)
Note: Award (A1) for 2 (must be in a box), (A1) for 7, (A1)(ft) for 6 and 4, (A1)(ft)
for 9 and 13. Observe the assignment of (ft) marks strictly here. Example A common
error is likely to be 11 instead of 6 (A0). In this case follow through to 4 and 18
(A1)(ft) for the final pair. Here the 4 follows from the total of 27 for n(M).
4
n(M  W) = 14
(A1)(ft)
n(M  S) = 22+11
OR 15+18
(A1)(ft)
= 33
(A1)(ft)
Note: Award (A2) if answer 33 is seen. Award (A1) for any of 22, 11, 15 or 18 seen
but 33 absent.
3
(b)
(i)
(ii)
(c)
P (both mice short-tailed) =
18 17 306
 
( 0.130 ).
49 48 2352
(M1)(A1)(ft)
(G1)
Note: Allow alternatives such as 153/1176 or 51/392.
Award (M1) for any of
18
17
18 17
18 17
and
or

or

seen.
49
48
49 49
49 48
2
[9]
14.
(a)
U
P
R
S
(A4)
4
Note: Award (A1) for rectangle labelled with U, (A1) for R placed correctly with
respect to S, (A1) for S placed correctly with respect to P, (A1) for R placed
correctly with respect to P.
18
(b)
(i)
U
U
P
P
R
S
S
OR
(A2)
(ii)
U
U
P
R
R
S
S
OR
(A2)
4
Note: Award (A2) for correct shading in each ft from circle placement in (a).
[8]
15.
(a)
(i)
(ii)
(iii)
(iv)
(b)
A = {2, 3, 5, 7, 11, 13}
B = {1,2, 3, 4, 6, 9, 12}
C = {4, 8, 12}
ABC=
(A1)
(A1)
(A1)
(A1)
4
(i)(ii)
U
1 6
7
A 5
13
11
15
10
2
3
9
C
12
4
B
U
B
A
8
14
C
(A2)(A2)
4
Also accept with elements
correctly placed
Note: Allow ft from part (a).
Award (A1) for rectangle labelled, (A1) for circles in correct places. Award (A1) for
correct elements in intersections, award (A1) for the rest of the elements correctly
placed. IF the wrong Universal set is used, penalise only one mark.
(c)
(i)
(ii)
(iii)
(d)
Note: Award follow through from part (b) only.
A  (B  C) = {2, 3}
(A  B) = {1, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Note: Award follow through from part (c) (ii).
(A  B)  C = {4, 8, 12}
(A1)
(A1)
(A1)
3
Note: Award follow through from either part (a) or part (b).
(i)
(ii)
6 2
 or 0.4 or 40%
15 5
4
or 0.267 or 26.7% (3 s.f)
15
(A1)
(A1)
19
Note: In parts (iii) and (iv), award (A1) for the correct numerator, (A1) for the
correct denominator.
(iii)
(iv)
6
2
or
or 0.4 or 40%
15 5
2
or 0.286 or 28.6% (3 s.f.)
7
(A2)
(A2)
6
[17]
16.
(a)
mean =
(25)5  (35)4  (45)3  (55)2  (65)3
17
(M1)
Note: Award (M1) for using mid-interval values.
mean = 41.5
(A1)
(b)
5
frequency
4
(# of
smokers) 3
2
1
20
30
40
50
60
70
ages
(A2)
Note: Award (A1) for correct intervals, (A1) for correct bar lengths
[4]
17.
(a)
Mean =
60
10
=6
(A1)
(C1)
(b)
Mode = 2
(A1)
(C1)
(c)
2, 2, 2, 4, 5, 6, 8, 9, 10, 12

Median = 5  6
(M1)
2
= 5.5
(A1)
(C2)
[4]
18.
Unit penalty (UP) is applicable where indicated.
(a)
26cm
(b)
33 −19 for identifying correct quartiles.
=14cm.
(ft) on their quartiles.
(A1)
(UP)
(C1)
(A1)
(A1)(ft)
(UP)
(C2)
20
(c)
0
5
10
15
20
25
30
35
40
45
50
length (cm)
correct median
correct quartiles and box
endpoints at 6 and 47, joined to box by straight lines.
(A1)(ft)
(A1)(ft)
(A1)
(C3)
[6]
19.
(a)
19 seedlings
(A1)
(b)
(i)
(A1)
median 88 cm
st
rd
1 quartile 78 cm, 3 quartile 103 cm (both correct)
(ii)
(c)
(C1)
(A1)
(C2)
112  63 = 49 cm
(A1)
Note: Accept 63 and 112 both seen, if they appear in the answer space for (c) or under
working for (c) (but not just implied or written on the box plot).
(C1)
(d)
(C2)
Notes:
Box with correct median and quartiles marked.
Both correct whiskers joined to box with straight lines
Allow maximum errors of  2.
Perfectly ruled lines are not essential.
(A1)(ft)
(A1)(ft) (C2)
[6]
20.
(a)
6 hours (accept (5.5–6.5))
(b)
(4  4  5  5  6  9  7  8  8  4)
30
(A2)
(C2)
(M1)(A2)(
A1)
=
183
30
= 6.1
(A2)
(C6)
Note: Award (M1) for method, (A2) for all 5 terms in numerator correct.
((A1) for 3 or 4 terms in the numerator correct), (A1) for denominator.
[8]
21
21.
(a)
(i)
(ii)
c = 97
(A1)
(not drawn to scale)
Weights of fish
140
cumulative frequency
1
120
100
80
60
40
20
0
0.5
0.7
0.9
1.1
weight (kg)
1.3
1.5
(A4)
4
Note: Award (A1) for correct scales and axes labelled, (A2) for 5 correct points,
(A1) for 3 to 4 correct, (A1) for the curve.
(iii)
(b)
(i)
(ii)
(c)
(i)
(ii)
median is 0.95 kg
line drawn correctly on diagram
(M1)(AG)
th
90 percentile ...
130 × 0.9 = 117
The zoo buys 13 fish (±2).
(G2)
2
First quartile = 32/33 fish.
Maximum weight = 0.79 kg (0.03).
1
(M1)
(A1) or
(M1)
(M1)(A1)
maximum : 0.95 × 1.10 = 1.045 kg (1.05 to 3 s.f.)
minimum : 0.95 × 0.90 = 0.855 kg
number of fish bought by restaurant = 88 – 46 = 42(4)
(A1)
(A1)
3
2
(M1)(M1)
3
(A1)
[16]
22.
(a)
(A4)
Note: Award (A1) for correct scales and labels, (A3) for correct points, (A2) for 7
or 8 correct, (A1) for 5 or 6 correct.
(b)
(i)
4
4
(G1)
22
(ii)
(iii)
(iv)
2.58
30
4.78
(G1)
(G1)
(G1)
4
Note: If wrong version of s.d. used in (ii), can (ft) in (iv) (5.07).
(c)
L correctly plotted on graph and named
(A1)(ft) 1
(d)
(i)
(ii)
(G1)
(R1)(ft)(R1)(ft)
3
(G1)(G1)
2
r = 0.986 (0.987)
(very) strong positive correlation
(e)
y = 1.83x + 22.7 (y = 1.825x + 22.7)
Award (G1) for y= 1.83x(1.825x), (G1) for 22.7
(f)
Line drawn on graph.
(A1)(A1)(ft)
Note: Award (A1) for passing through the mean point, (A1) for y intercept between
22 and 23.
(g)
(i)
32.6 cm
Note: Allow margin of error of 0.2 from value on candidate’s diagram.
Not possible to find an answer as the value lies too far outside the
given set of data.
(ii)
2
(A1)(ft)
(R1)
2
[18]
(a)
(b)
s = 17, t = 90
(A1)(A1)
(M1)(M2)
2
3
100
90
80
Cumulative frequency
23.
70
60
50
40
30
20
10
0
70
90 100
50
60
80
Marks
Note: Award (M1) for both axes with correct scales and correctly labelled.
Award (M2) for 8, 9, 10 points plotted correctly, (M1) for 5, 6, 7 points plotted
correctly, (M0) for 4 or less.
Accept a polygon or a curve.
10
20
30
40
23
(c)
(i)
(ii)
(iii)
Median mark = 48 (1)
Lower quartile = 36 (1)
Pass mark if 40% pass = 51 (1)
Note: Follow through with candidate’s own graph.
Award (M0)(A1) ft if candidate correctly finds the grade (44) where 40% fail.
(A1)
(A1)
(M1)(A1)
4
[9]
24.
1
4
2
5
3
5
Red
Red
3
4
2
4
Green
Red
Green
2
4
Green
(A4)
(C4)
Note: Award (A1) for each correct bold entry
24