New Century Maths 11
Mathematics General
Preliminary course (Pathway 2)
d
1
(a  b)h
2
1
 (7  13)6
2
1
= (20)6
2
 60 cm 2
A
Worked Solutions
Chapter 6
SkillCheck
4a
1
A  bh
2
1
  9  40
2
1
  360
2
 180 cm 2
1 a 20.831000  20 830
b 970.2 10  97.02
c 6.59 10000  65 900
d 72.5 100  0.725
e 10.4 1000  0.0104
b
f 0.0735 10  0.735
2 a 107  10 10 10 10 10 10 10
 10 000 000
4
b 10  10 10 10 10  10 000
1
1

 0.01
c 102 
10 10 100
3a
A  l l
 4 4
 16 cm
a 2  b2  c2
402  92  c 2
1600  81  c 2
1681  c 2
1681  c
41  c
41 cm  h
5a
1
of a kilogram
4
 $5.10  4
2
250g 
b
A  l l
 15  6
 90 m 2
= $1.28
b
Cost per kilogram = $4.15
c
A    r2
 $4.15  5
= $0.83
For three kilograms = $0.83  3
   52
 78.54 cm 2
Cost = $2.49
6a
Convert 210 cm into m
210  100 = 2.1 m
1
b
h
Convert 210 cm into mm
210 10 = 2100 mm
Convert 3400 kg into t
3400 1000 = 3.4 t
c
i
Convert 210 cm into km
210  100, then  1000 = 0.0021 km
Convert 4720 L into kL
4720 1000 = 4.72 kL
j
7a
Convert 6000 mg into kg
6000  1000,  1000 = 0.006 kg
Convert 8.1L into kL
8.1  1000 = 0.0081 kL
b
k
Convert 7.5 h into s
7.5  60, 60 = 27 000 s
Convert 8.1L into mL
8.1  1000 = 8100 mL
l
Convert 9.4 km into cm
9.4 1000, 100= 940 000 cm
Exercise 6-01
1a
Convert 6.3 cm into mm
6.3  10 = 63 mm
2
Convert 100 kg into g
100 1000 = 100 000g
A
b
Convert 4.36m into cm
4.36  100 = 436 mm
3a
Convert 169 cm into mm
169 10 = 1690 mm
c
Convert 7200mm into m
7200  10,  100 = 7.2 mm
b
Convert 169 cm into m
169 100 = 1.69 m
d
Convert 285g into kg
285  1000 = 0.285 kg
4
Convert 6.9mm into μm
6.9  1000 = 6900 μm
5
Convert 58 000 mL into L
58 000  1000 = 58 L
6a
Convert 57.5 kg into g
57.5 1000 = 57 500 g
e
Convert 1500 m into km
1500 1000 = 1.5 km
f
g
Convert 5.32 kg into g
5.32 1000 = 5320 g
From 7.41 a.m.  8 a.m. = 19 minutes
From 8.00 a.m.  9 a.m. = 60 minutes
From 9.00 a.m.  9.18 a.m. = 18minutes
19 + 60 + 18 = 97 minutes
2
b
Convert 97 minutes to seconds
97  60 = 5820 seconds
c
Convert 97 minutes to hours (correct to one dp)
.
11a
5400 m  5.4 km
5.4 + 3.8 + 9.5 = 18.7 km
b
12 + 38 + 58 = 108 mins
.
97  60 = 1.616  1hr and (0.616  60) minutes
=1hr and 37minutes
7 59kL  1000 = 59 000 L
8 To be measured: unit
a your mass: kg
108 mins = 1 hr and 48 mins
12
Convert 8560 L into kL
8560 1000 = 8.56 kL
13
a The most likely measurement for the height of
a 10-story building is: 40 m or B.
b your height: cm or m
b The most likely measurement for the capacity
c the length of your bedroom: m
of a backyard swimming pool is: 3000 L or C.
d the distance from Perth to Sydney: km
c The most likely measurement for the mass of
e the capacity of your kitchen sink: mL or L
an orange is: 300 g or C.
f the mass of an ant: mg
d The most likely measurement for the distance
g the width of a sheet of paper: cm or mm
from Sydney to Brisbane is: 1000 km or C.
h the amount of water in an Olympic-sized
e The most likely measurement for the diameter
swimming pool: kL
of a DVD is: 12 cm or A.
i the mass of a train: t
f The most likely measurement for the capacity
j the thickness of a strand of hair: mm or μm
of a soft drink can is: 380 mL or A.
9a
Convert 102 g into kg
102  1000 = 0.102 kg
14
2.5 km =2500m = 250 000 cm
Dad
250 000  80 = 3125 steps
Anna
250 000  55 = 4545.54 steps
b
Convert 5.2 cm into m
5.2 100 = 0.052 m
10
2 000 000  2.45tonnes = 49 000 000 tonnes
49 000 000  1 000 000  4.9 megatonnes
Difference  1420 more steps
15
Convert 24 h into s
24  60  60 = 86 400 s
Exercise 6-02
3
1 a i The smallest unit on the scale is 1m.
ii The shown measurement is 4 m.
iii The absolute error of the measurement is
4  0.5 = 3.5 to 4.5 m.
2 When using a jug marked in millimetres
 0.5 mL is the absolute measurement, so C.
3 The lengths which could be measured using a
measuring tape with a precision of 1 cm are
37 cm and 9 cm, so A and D.
b i The smallest unit on the scale is 1mm or
0.1 cm.
4 The angle sizes which are incorrectly recorded
if using a protractor marked in degrees are
ii The shown measurement is 18 mm or
1.8 cm.
iii The absolute error of the measurement
 0.5 mm.
c i The smallest unit on the scale is 5 km/hr.
ii The shown measurement is 50 km/hr.
iii The absolute error of the measurement
 2.5 km/hr.
1
103.5°, 64 ° and 88.4°, so C, D and F.
2
5 a The absolute error for a measurement of
25 mL is  0.5 mL.
b The absolute error for a measurement of
5560 mm is  0.5 mm.
c The absolute error for a measurement of 6.5 g
is  0.5 g.
d i The smallest unit on the scale is 0.5°C.
d The absolute error for a measurement of78.1 s
ii The shown measurement is 38°C.
is  0.05 s.
iii The absolute error of the measurement
 0.25°C.
e i The smallest unit on the scale is 5m/s.
e The absolute error for a measurement of
500 grams (to the nearest gram) is  0.5 g.
ii The shown measurement is 45 m/s.
f The absolute error for a measurement of 24 m
iii The absolute error of the measurement
is  0.5 m.
 2.5 m/s.
f i The smallest unit on the scale is 500 rpm.
ii The shown measurement is 7000 rpm.
iii The absolute error of the measurement
 250 rpm.
g The absolute error for a measurement of
600 mL (to the nearest milliliter) is  0.5 mL.
h The absolute error for a measurement of 500g
(to the nearest 10 grams) is  5 g.
i The absolute error for a measurement of
4
8.6 cm is  0.05cm.
absolute error
100
measurement
0.05
=
 100
15.2
= 0.33%
Percentage error =
j The absolute error for a measurement of 5000g
(to the nearest 100 grams) is  50 g
b
6
2964  3021  2938  2899
4
11 822

4
 2955.5
 2956 mm
absolute error
100
measurement
5
=
 100
800
= 0.625% = 0.63%
Percentage error =
c
absolute error
100
measurement
0.005
=
 100
0.15
= 3.33%
Percentage error =
7 a The absolute error of measurement of
21 cm is  0.5 cm.
b The true measurement lies between the values
of 20.5 cm and 21.5 cm.
c
absolute error
100
measurement
0.5
=
 100
21
= 2.38%
Percentage error =
15.2 mg is the most accurate as it has the smallest
percentage error.
11 We don’t know the absolute error; the figure
could be to the nearest whole number, ten or fifty.
12
38L  0.5L correct to the nearest whole L
38.0L  0.05L correct to the nearest 0.1L
8 2.3t  0.05t
absolute error
100
measurement
0.05
=
 100
2.3
= 2.17%
Percentage error =
9  0.5 km
Exercise 6-03
1 305.725 rounded to two significant figures is
310, so B.
2 a 3812.4 correct to two significant figures:
absolute error
100
measurement
0.5
=
 100
12 683
= 0.0039%
3800.
Percentage error =
b 2068 correct to two significant figures: 2100.
c 0.006102 correct to two significant
figures: 0.0061.
10 a
d 245 698 correct to two significant
figures: 250 000.
5
e 14 569 235 correct to two significant
c 24 mm is recorded correctly, so Yes.
figures:15 000 000.
d 6.5 mm is recorded incorrectly, so No.
f 0.000 467 correct to two significant figures:
e 31 mm is recorded correctly, so Yes.
0.00047.
f 10 mm is recorded correctly, so Yes.
3 a 129.805 correct to three significantfigures:
130.
6 The following measurement has been recorded
correctly to two significant figures: 23 μm . The
b 4982 correct to three significant figures: 4980.
measurements of 104 μm , 19.6 μm , 832 μm , 28.6
c 0.010 609 correct to three significant
μm and 3 μm have been recorded incorrectly, so
the incorrect responses are: b, c, d, e and f.
figures: 0.0106.
7 Answers are to be evaluated correct to two
d 1 359 048 correct to three significant
figures: 1 360 000.
e 25 430 082 correct to three significant
figures: 25 400 000.
f 0.000 680 32 correct to three significant
figures: 0.000 680.
4 a 2914.23 correct to one significant
figure: 3000.
significant figures.
a 0.2  0.3  0.67
b 11 1990  0.0055
c 16 12  1.3
d 0.0075 = 0.087
e 9 300 000  0.085 = 790 000
b 3.2548 correct to one significant figure: 3.
f 2.7 2 = 7.29  7.3
c 11 950 correct to one significant
figure: 10 000.
g
560 = 24
h
5.6 = 2.4
d 0.005 134 correct to one significant
figure: 0.005.
i 3.4  9.9 = 34
e 20.46 correct to one significant figure: 20.
f 0.6517 correct to one significant figure: 0.7.
5 The rain gauge is marked in millimetres.
Measurements of less than 10 mm are to be
recorded to one significant figure.
Measurements between 10 mm and 99 mm are to
be recorded to two significant figures.
a 9 mm is recorded correctly, so Yes.
b 7.23 is recorded incorrectly, so No.
8 Round 7 247 777 to three significant
figures: 7 250 000.
9 a Rounded to three significant figures:
351 540 kg = 352 000 kg
b Rounded to three significant
figures: 4188 m = 4190 m
c Rounded to three significant figures:
67.14 km/h = 67.1 km/h
d Rounded to three significant figures:
14.79 mL = 14.8 mL
6
e Rounded to three significant figures:
149 573 881 km = 150 000 000 km
Exercise 6-04
1 a 42 130 000
Place a decimal point at the end of the number and
move it 7 places to the left.
= 4.213 10 7
i 230
Place a decimal point at the end of the number and
move it 2 places to the left.
= 2.3 10 2
j 0.000 072 3
Move the decimal point 5 places to the right.
= 7.23 10 5
k 610 000 000
b 0.0181
Move the decimal point 2 places to the right.
= 1.81 10 2
c 3400
Place a decimal point at the end of the number and
move it 3 places to the left.
= 3.4 103
Place a decimal point at the end of the number and
move it 8 places to the left.
= 6.1 108
l 0.000 000 08
Move the decimal point 8 places to the right.
= 8.0 10 8
2 a 53 467 892
d 20 000
Place a decimal point at the end of the number and
move it 4 places to the left.
= 2.0 10 4
e 0.0035
Move the decimal point 3 places to the right.
= 3.5 10 3
f 0.0002
Move the decimal point 4 places to the right.
= 2.0 10 4
g 0.33
Move the decimal point 1 place to the right.
= 3.3 10 1
Round to two significant figures = 53 000 000
Place a decimal point at the end of the number and
move it 7 places to the left.
= 5.3 10 7
b 146 089
Round to two significant figures = 150 000
Place a decimal point at the end of the number and
move it 5 places to the left.
= 1.5 10 5
c 2453
Round to two significant figures = 2500
Place a decimal point at the end of the number and
move it 3 places to the left.
= 2.5 103
d 0.000 457 3
h 0.004
Move the decimal point 3 places to the right.
= 4.0 10 3
Round to two significant figures = 0.000 46
Move the decimal point 4 places to the right.
= 4.6 10 4
7
e 0.002 652
Round to two significant figures = 0.0027
Move the decimal point 3 places to the right.
= 2.7 10 3
f 0.102 05
Round to two significant figures = 0.10
Move the decimal point 1 place to the right.
= 1.0 10 1
g 0.000 033 38
Round to two significant figures = 0.000 033
Move the decimal point 5 places to the right.
= 3.3 10 5
h 0.444 44
Round to two significant figures = 0.44
Move the decimal point 1 place to the right.
= 4.4 10 1
i 6.5207
Round to two significant figures = 6.5
Move the decimal point 0 places to the right.
= 6.5 10 0
3 356 587 239
Round to three significant figures = 357 000 000
Place a decimal point at the end of the number and
move it 8 places to the left.
= 3.57 108
So, B.
4 a 7.4 105
Move the decimal point 5 places to the right and
complete using place holding zeros.
= 740 000
b 3.12 101
Move the decimal point 5 places to the left.
= 0.312
c 1.85  103
Move the decimal point 3 places to the right and
complete using place holding zeros.
= 1850
d 6.6 104
Move the decimal point 4 places to the left and
complete using place holding zeros.
= 0.000 66
e 2.54 103
Move the decimal point 3 places to the left and
complete using place holding zeros.
= 0.002 54
f 4.751 108
Move the decimal point 8 places to the right and
complete using place holding zeros.
= 475 100 000
g 9.8 102
Move the decimal point 2 places to the left and
complete using place holding zeros.
= 0.098
h 3  102
Place a decimal point at the end of the number and
move it 2 places to the right, complete using place
holding zeros.
= 300
i 5.497 102
Move the decimal point 2 places to the left and
complete using place holding zeros.
= 0.054 97
j 1.216  10 4
Move the decimal point 8 places to the right and
complete using place holding zeros.
= 12 160
8
= 7.03  10 9
k 8.02 101
Move the decimal point 1 place to the left.
= 0.802
l 6.309 10
3
Move the decimal point 3 places to the right.
= 6309
5 a 13 700 000 000
Place a decimal point at the end of the number and
move it 10 places to the right.
= 1.37 1010
b
1
1

 1.0 106
6
1 000 000 1.0 10
6 a 3 108
Place a decimal point at the end of the number.
Move the decimal point 8 places to the left and
complete using place holding zeros.
= 0.000 000 03
b 9.461 109
Move the decimal point 9 places to the right and
complete using place holding zeros.
= 9 461 000 000
c 2.0  10 6
Move the decimal point 6 places to the left and
complete using place holding zeros.
= 0.000 002
8 By calculator as per example.
a (6.7 104 )  (3.2 102 )  2.144 107
b (1.92 1010 )  (6 104 )  3.2  105
c (3.75 104 )  (2.5 103 )  3.5  104
d (4.8 103 )2  2.304 10 5
e (8.1104 )3  5.314 411010
9.4 102
 0.094  0.0025  3.76  101
f
3
2.5 10
g (4.2 105 )  (3 102 )  1.26  10 4
h 5.29 1010  2.3 105
9 By calculator as per example.
a (5.1103 )  (2.5 104 )  3.0 104
b (7.8 105 )  (3.7 103 )  3.6 103
c (9.6 108 )  (4.3 103 )  4.11012
d (3.2 103 )3  3.3 108
3.2 102
 5.9 104
e
3
5.4 10
f
7 7 034 307 183
Round to three significant figures = 7 000 000 000
Place a decimal point at the end of the number and
move it 6 places to the left.
3.8 105  3.4 102
g
8.5 105  9.2 102
h
6 103
 1.2 104
7
5.110
d 1.526  108
Move the decimal point 8 places to the right and
complete using place holding zeros.
= 152 600 000
3
10 By calculator as per example.
a 96  20 000 = 0.0048
b 480 000  91 000 = 43 680 000 000
c
0.235
 0.0094
25
9
d (13700)2  187 690 000
Divide through by common factor of 3.
12 36 9
: :  4 :12 : 3
3 3 3
e (4.4 104 )  (3.1102 )  13 640 000
f (1.9 107 )  (6.3 104 )  19 063 000
g 0.04 : 0.4
Divide through by common factor of 0.04.
0.04 0.4
:
 1:10
0.04 0.04
g 2.704 109  0.000 052
h
7.7 109
 350 000
2.2 104
11 a Answers will vary.
4 2
h :
5 3
Multiply each fraction by 15 to make them
whole.
12 :10
b Answers will vary.
Divide through by common factor of 2.
12 10
:  6:5
2 2
Exercise 6-05
1 a 5 : 15
Divide through by common factor of 5.
5 15
:  1: 3
5 5
b 4 : 2 : 10
Divide through by common factor of 2.
4 2 10
: :  2 :1: 5
2 2 2
c 14 : 14
Divide through by common factor of 14.
14 14
:  1:1
14 14
d 105 : 75
Divide through by common factor of 15.
105 75
:
 7:5
15 15
e 17 : 68
Divide through by common factor of 17.
17 68
:
 1: 4
17 17
f 12 : 36 : 9
i
3 1
:1
8 4
Convert the mixed number to an improper
fraction.
3 5
:
8 4
Multiply each fraction by 8 to make them
whole.
3:10
j 250 mL : 2 L
Convert to be the same unit.
250 mL : 2000 mL
Divide through by common factor of 250.
250 2000
:
 1: 8
250 250
k 3 days : 2 weeks
Convert to be the same unit.
3 days : 14 days
This cannot be simplified.
3:14
10
20 000 m2 : 100 m2
l $4.20 : 95 cents
Convert to be the same unit.
420 cents : 95 cents
Divide through by common factor of 5.
420 95
:  84 :19
5 5
m 45 cm : 3 m
Convert to be the same unit.
45 cm : 300 m
Divide through by common factor of 15.
45 300
:
 3 : 20
15 15
n $2000 : $1 000 000
Divide through by common factor of 2000.
2000 1 000 000
:
 1: 500
2000
2000
1 1
o :2
2 2
Convert the mixed number to an improper
fraction.
1 5
:
2 2
Multiply each fraction by 2 to make them
whole.
1: 5
p 1 fortnight : 3 weeks : 1 year
Convert to be the same unit.
14 days : 21 days: 365 days
Divide through by common factor of 7.
14 21 365
: :
 2 : 3 : 52
7 7 7
Divide through by common factor of 100.
20 000 100
:
 200 :1
100 100
r 350 mL : 2.4 L
Convert to be the same unit.
350 mL : 2400 mL
Divide through by common factor of 50.
350 2400
:
 7 : 48
50
50
2 2.5 : 0.5 : 0.25
Multiply by 100 to make all parts whole.
250 : 50 : 25
Divide through by common factor of 25.
250 50 25
: :
 10 : 2 :1
25 25 25
3 27 : 99
Divide through by common factor of 9.
27 99
:
 3 :11 , so C.
9 9
4 0.5 kL : 1.4 L
Convert to be the same unit.
500 L : 1.4 L
Multiply by 10 to make both parts whole.
5000 L : 14 L
Divide through by common factor of 2.
5000 14
:  2500 : 7 , so D.
2
2
5 15 : 130
Divide through by common factor of 5.
15 130
:
 3 : 26
5 5
q 2 ha: 100 m2
Convert to be the same unit.
11
6 Walkers : Bus travelers: Bike riders
1 1
1 1
: :1  
2 3
2 3
1 1 1
 : :
2 3 6
3 2 1
 : :
6 6 6
Multiply each part by 6 to make it whole.
3:2:1
To calculate the mass of the white:
12 : 11 = 600 : x
600
11
x=
12
x = 550 grams of white jelly beans
b To calculate the total mass we need to
calculate the mass of the green:
12 : 22 = 600 : x
a Bike riders : walkers = 1 : 3
b Bus travelers : whole class
2:6=1:3
c Walkers : bus travelers = 3 : 2
7 side length : perimeter (4 times the side length)
=1:4
81:3:5
x=
600
 22
12
x = 1100 grams of white jelly beans
The total mass = 600 + 550 + 1100
= 2250 g
11 teachers : students = 1 : 15
1 : 15 = x : 1200
a The longest length is 75.
Divide 75 by 5 to find the length of one
portion = 15.
Multiply this by 3 to find the length of the
third portion.
15 : 45 : 75
b Total length = 15 + 45 + 75 = 135 cm
9 Boys : girls = 23 : 57
If the number of boys = 529, calculate the
x=
1
 1200
15
x = 80 teachers
12 petrol : oil = 3 : 2
3:2= x:3
3
3
2
x = 4.5 L, so C.
x=
13 won : lost = 8 : 5
number of girls.
8 : 5 = x : 15
23 : 27 = 529 : x
529
 27
x=
23
x=
x = 621 girls
The total number of students = 529 + 621
= 1150
10 yellow : white : green
12 : 11 : 22
8
 15
5
x = 24 games
14 fiction : non-fiction = 3 : 7
3 : 7 = 13 980 : x
7
 13 980
3
x = 32 620 non-fiction books
x=
a 600 grams of yellow
12
Total number of books = 13 980 + 32 620
= 46 600
15 cement : sand : gravel = 1 : 4 : 6
a If there are three shovelfuls of cement (one
x = 3900 mL of water
3900 mL = 3.9 L
18 Thelma : Louise = 7 : 8
7 : 8 = x : $384 000
part), there will be 3 × 4 shovelfuls of sand
required; that is, 12 shovelfuls of sand.
b 2.2 tonnes = 2200 kg of gravel
x=
7
 384 000
8
x = $336 000
1 part = 2200 ÷ 6 = 366.67 kg
To calculate the kilograms of sand required:
Exercise 6-06
366.67 kg  4  1467 kg
1 Vila : Kim = 2 : 3
16 copha : coconut : rice bubbles : cocoa
=2:2:5:1
a If she used 400 g of copha, how many grams
of cocoa?
400 g represents 2 parts:
400 g ÷ 2 = 200 g (1 part)
b If she used 400 g of copha, how many grams
of rice bubbles?
200 g (1 part)
200 g × 5 = 1000 g (5 parts)
c The mass of the individual components is:
400 g : 400 g : 1000 g : 200 g
Total = 400 + 400 + 1000 + 200 = 2000 g
2000 g = 2 kg
Total parts = 5
Total cash = $1200
1200
 $240
5
Vila receives 2  $240  $480 , so B.
One part =
2 a 4 : 5, Total parts = 9
Total to be divided = $31 500
One part =
4  $3500 : 5  $3500
$14 000 : $17 500
b 13 : 2, Total parts = 15
Total to be divided = 3750 kg
17 a concentrate : water = 2 : 13
2.6 L = 2600 mL
2 : 13 = x ml : 2.6 L
2
 2600
x=
13
x = 400 mL of concentrate
b 2 : 13 = 600 ml : x L
x=
13
 600
2
31 500
 $3500
9
One part =
3750
 250 g
15
13  250 kg : 2  250 kg
3250 kg : 500 kg
c 6 : 7, Total parts = 13
Total to be divided = 34 000 m2
(to the nearest m2)
13
One part =
34 000
 2615 m 2
13
NB: Answers differ slightly to those
provided due to rounding.
6  2615 m2 : 7  2615 m2
15 690 m2 : 18 305 m2
Total value = $53 600
1
cups
2
1
1
One part = 2  5 
2
2
1
1
1 : 4 
2
2
1
cup : 2 cups
2
e 2 : 3, Total parts = 5
Total to be divided = 7500 L
7500
 1500 L
One part =
5
2 1500 L : 3 1500 L
3000 L : 4500 L
f 5 : 3, Total parts = 8
Total to be divided = 10 ha
10
 1.25 ha
One part =
8
5 1.25 ha : 3 1.25 ha
6.25 ha : 3.75 ha
62 500 m 2 : 37 500 m 2
3 boys : girls = 4 : 5
Total parts = 9
Total students = 846
5
Number of girls =  846  470
9
4 copper : nickel = 7 : 4
Total parts = 11
5 Tan : Brad = 5 : 3
Total parts = 8
d 1 : 4, Total parts = 5
Total to be divided = 2
Total mass of alloy = 500 kg
7
Amount of copper =  500  318 kg
11
4
Amount of nickel =  500  182 kg
11
5
 53 600  $33 500
8
5
Brad’s share =  53 600  $20 100
8
Tan’s share =
6 3:2:2:1
Total parts = 8
3
 $200 000  $75 000
8
2
Share 2 =  $200 000  $50 000
8
2
Share 3 =  $200 000  $50 000
8
1
Share 4 =  $200 000  $25 000
8
Share 1 =
7 oil : petrol = 1 : 49
Total parts = 50
Total volume of fuel = 20 L= 20 000 mL
Volume of oil =
1
 20 000  4000 mL or 0.4 L
50
Volume of petrol =
49
 20 000  19 600 mL or 19.6 L
50
8 cement : sand : aggregate = 1 : 4 : 6
Total parts = 11
Total mass of concrete = 550 kg
Mass of cement =
1
 550  50 kg
11
14
4
 550  200 kg
11
6
Mass of aggregate =  550  300 kg
11
Mass of sand =
9 fruit juice : lemonade : ginger ale = 2 : 2 : 1
Volume of herbicide =
100
 85 000  84 158 mL
101
13 70% of $1080 = $756 to be divided
between savings and living expenses.
Total parts = 5
savings : living = 2 : 5
Volume of punch = 6 L = 6000 mL
Total parts = 7
2
Volume of fruit juice =  6000  2400 mL
5
2
a Share (savings) =  756  $216
7
5
b Share (living) =  756  $540
7
2
Volume of lemonade =  6000  2400 mL
5
1
Volume of ginger ale =  6000  1200 mL
5
10 Rosie : Lisa : Ally = 3 : 2 : 1
Total parts = 6
Total cost of rent = $450
3
Rosie’s share =  $450  $225
6
2
Lisa’s share =  $450  $150
6
1
Ally’s share =  $450  $75
6
11 formula : water = 1 : 14
14 protein : fat : carbohydrate = 3 : 0.5 : 9.5
a Multiply each by 10 to make the parts whole
30 : 5 : 95
This ratio can be simplified: divide through
by 5.
6 : 1 : 19
b Total parts = 26
Total mass = 3.5 kg = 3500 g
6
 3500 g  808 g
Mass of protein =
26
c protein : fat : carbohydrate = 6 : 1 : 19
If fat = 20 g, she consumes the following:
Total parts = 15
6 × 20 g = 180 g
Total volume of formula = 300 mL
1
 300  20 mL
Volume of formula =
15
14
 300  280 mL
Volume of water =
15
19 × 20 g = 380 g
Total consumed = 20 + 180 + 360 = 520g
d Mass of carbohydrate =
507
 19  370.5 g
26
12 wetting agent : herbicide = 1 : 100
Exercise 6-07
Total parts = 101
1 Units used to measure each of the following:
Total volume = 85 L = 85 000 mL
a cost of a mobile phone call - $/m
Volume of wetting agent =
1
 85 000  842 mL
101
b wage rate - $/hour
c typing speed – words/minute
15
d postage rate for a parcel - $/g
e cost of potatoes - $/kg
f speed of an athlete – m/s
2 Answers may vary. Suggested answers include:
a population density
b cost of a newspaper classified ad
c fuel consumption
d internet download speed
e birth rate
f cost of carpeting
1
3 31.2 cm in (  52 weeks)
4
31.2
 2.4 cm/week
Rate =
13
4 42 runs in 8 overs
42
 5.25 runs/over
Rate =
8
5 population grows by 16 500 over 6 years
16 000
 2750 people/year
Rate =
6
6 $4.55 for 160 g
Rate =
4.55
 0.028 cents/g = $28.44/kg
160
7 6.5 megabytes in 20 seconds
Rate =
6.5
 0.325 MB/s
20
1
a In 3 hours, the bus will have travelled:
2
1
3  96  336 km
2
b If the bus travels 480 km the time taken is:
480
 5 hours
96
10 Rate is 64 w/m.
Time taken to type 1500 words is
1500
 23.44  24 minutes
64
11 3 mg of fat per 60 gram serve
a fat content =
b in a 100 gram serve there would be
0.05  100 mg = 5mg
12 Rate = 50 L/min
Pool capacity = 38 KL (38 000 L)
Time to fill =
38 000
 760 minutes
50
= 12 hours and 40 minutes
13 Rate = 4.9 runs/ over in 30 overs they score:
30  4.9  147 runs
14 Exchanged A$750 for US$474.50
a $A1 =
474.50
 0.63 cents
750
8 Population = 23 194 000
2
Area (km ) = 7 682 300
Rate =
23 194 000
 3.02 people/km 2
7 682 300
9 Bus travels at 96 km/h.
3
 0.05mg/g
60
b US$200 =
750
 200  $316.12
474.50
15 Recommended dose = 80 mg/kg of body
weight
If the child weighs 30kg:
16
30  80  2400 mg
2400 mg = 2.4 g
So, A.
16 Rate = 40 drops/minute
Each drop = 1.3 mL
Per minute = 40 1.3  52 mL per minute
In one day: 24  60  1440 minutes
Volume in one day:
52 1440 = 74 880 mL
= 74.88 L
17 Death rate = 5.2 per 1000
125
= 25 mg/ml
5
45
 1.8 mL
25
21 25 mg per mL
120
 4.8 mL
25
22 75 mg per 10 mL
per 1 mL = 7.5 mg
975
 130 mL
7.5
23 a
5 L = 5000 ml
125
100  2.5%
5000
How many people died during the year?
21 435 680
 5.2
1000
 111 465.54
 111 466 (to the nearest whole person)
18 a
11 7.5  82.5 m 2
5 kg are required per 100 m 2
5
= 0.05 kg
100
0.05  82.5  4.125 kg
Per 1 m 2 =
b A 25 kg bag of fertilizer is purchased.
25
 6 times
4.1
b
2.5% of 2.4 L
= 2.5% of 2400 mL
= 60 ml
24 Capsule contains 20 mg/mL
1
 0.05 mL for 1 mg
20
3 mg = 3  0.05  0.15 mL
25 0.6 g in 300 mL
a 600 mg in 300 mL
600
 2 mg/mL
300
This will feed the lawn 6 times.
19 a 1L = 1000 mL
1000
 2.5
400
2.5  240  600 mL/L
b
600
 100  60%v/v
1000
b
2
100  0.2%
1000
26 10g/L
a 10g in 1000 mL
5g in 500 mL
b 240g/300 mL
20 125 mg per 5 ml
17
240
 0.8 g/mL
300
in 500 mL there are 5 g required
c 15 mL/s
5
 6.25 mL of solution
0.8
d $10.80/h
Exercise 6-08
1 a $46 212/year
46 212
 $3851/ month
12
b 75 km/h
75
 1.25 km / min
60
c 5 L/h
5 × 24  120 L/day
d 3m/s
3 × 60  180 m/min
15 × 60 × 60  54 000 mL/hr = 54 L/hr
$10.80 = 1080 cents
1080
 18 c / min
60
e 60 km/h
60
 1 km / min
60
1000 m/60 seconds
1000
2
 16 m / s
60
3
f 3t/day = 3000 kg/day
3000
 125 kg / hr
24
125
 2.08 kg / min
60
3 20.12 m in 45 seconds
e 0.8 cm/min
0.8 × 60  48 cm/h
f 78 words per minute
78
 1.3 words / sec
60
20.12
 44.71 m / s
0.45
44.71× 60 × 60  160 960 m/hr = 161 km/hr
4 1224 km/h = 1224 km/3600 s
(1h = 60 × 60 = 3600 s)
g 8.2 L/100 km
8.2
 0.082 L / km
100
h 1750 mL/s
1750 × 60  105 000 mL/min
2 a 24c/min
24 × 60  1440 cents = $14.40/hr
b 8 m/s
8 × 60 × 60  28 800 m/hr = 28.8 km/hr
1224
 0.34km / s = 340 m/s , so C.
3600
5 20 seconds to fill a 10L bucket
10
 0.5 L / s
20
(1h = 60 × 60 = 3600 s)
0.5× 60 × 60  1800L /hr
6 100 m in 9.725 s
100
 10.3 m/s
9.725
18
10.3 m = 0.0103 km
(1h = 60 × 60 = 3600 s)
0.0103× 60 × 60  37.08 km/hr
7 298 m/42 s
298
 7.1 m/s
42
(1h = 60 × 60 = 3600 s)
7.1× 60 × 60  25 542.9 m/hr
= 25.54 km/h
8a
250.7  1.852  464 300 m
= 464.3 km/h
b
i
464.3 km  464 300 m
so, 464 300 m/h
464 300
 7738 m/min
60
7738
 129 m/s
ii 7738 m/min =
60
9 75 km/hr = 75 000 m/hr
75 000
 1250 m/min
60
1250
 20.8 m/s
60
In 5 seconds the distance travelled will be
5 × 20.8  104.2 m
10 Reaction time is 0.9 s
a 60 km/h = 60 000 m/h
60 000
 1000 m/min
60
1000
 16.67 m/s
60
16.67  0.9 = 15 m
b 84 km/h = 84 000 m/h
84 000
 1400 m/min
60
1400
 23.33 m/s
60
23.33  0.9 = 21 m
c 100 km/h = 100 000 m/h
100 000
 1666.67 m/min
60
1666.67
 27.78 m/s
60
27.78  0.9 = 25 m
Exercise 6-09
1 The total paid for the calculators after a discount
of 12% was $2508.
150  $19  $2850
100  12  88%
$2850  88%
 $2850  0.88
 $2508
2 The population after an increase of 1.2%
was 21 692 908 people.
100  1.2  101.2%
21 435 680  101.2%
 $2850 1.012
 21 692 908.16
Round to the nearest whole person
21 692 908
3 Phuong’s income after an increase of 8%
was $734.94.
100  8  108%
$680.50  108%
 $680.50  1.08
 $734.94
19
4 The total paid for the television after a discount
Value of change
Original price
7.41

 100
285
 2.6%
Percentage change 
of 15% was $391.
100  15  85%
$460  85%
 $460  0.85
 $391
8 The population after an increase of 10% and a
decrease of 4% was 2112 people.
5 The value of Lisa’s computer after a devaluation
of 9% was $2193.10.
100  9  91%
$2410  91%
 $2410  0.91
 $2193.10
6 The cost for the meal after a surcharge of
12% was added was $83.44.
100  10  110%
2000  110%
 $2000 1.1
 2200
100  4%  96%
2200  96%
 2200  0.96
 2112, so B.
100  12  112%
9 105% × 103%
$74.50  112%
= 1.05 × 1.03
 $74.50 1.12
 $83.44
7 a The price after the 8% rise was $307.80.
100  8  108%
= 1.0815
1.0815 × 100 = 108.15%
So an increase of 8.15%; D
$285  108%
 $285 1.08
 $307.80
b The price after the 5% discount was $292.41.
100  5  95%
10 a $1000 × 108% = $1080
$1080 × 108% = $1166.40
$1166.40 × 107% = $1248.05
$307.80  95%
After the three years the investment was worth
 $307.80  0.95
 $292.41
$1248.05.
c The overall change in price was $7.41.
$292.41  $285  $7.41
The percentage change on the original price
was 2.6%.
b
Value of change
Original price
248.05

 100
1000
 24.8%
 25% increase
Percentage change 
11 a Increase by 20% = 120%
20
Decrease by 20% = 80%
Exercise 6-10
1.2 × 0.8 = 0.96 = 96% = 4% decrease
1a Perimeter = 8 + 2 + 10 + 5 + 2 + 3
= 30 m
b Decrease by 20% = 80%
Increase by 20% = 120%
0.8 × 1.2 = 0.96 = 96% = 4% decrease
c Increase by 8% = 108%
Increase by 8% = 108%
b Perimeter = (15 × 3) + 3 + 4 + 9 + 4 + 3
= 68 cm
c Perimeter = (15 × 3) + 12 + 5 + 4
= 66 cm
d Perimeter = 12 + 12 + 10 + 10 + 10
= 54 cm
1.08 × 1.08 = 1.1664
= 16.64% = 16.64% increase
d Increase by 10% = 110%
Decrease by 15% = 85%
1.1 × 0.85 = 0.935
= 93.5% = 6.5% decrease
e Perimeter = (100 × 4) + (300 × 2) + (80 × 4) +
(200 × 2)
= 400 + 600 + 320 + 400
= 1720 cm
f x2 = 72 + 362
x2 = 1345
x = 1345
e Decrease by 4% = 96%
x = 36.67 m
Decrease by 6% = 94%
Perimeter = 36.67 + 7 + 36 = 79.67 m
0.96 × 0.94 = 0.9024
= 90.24% = 9.76% decrease
g x2 = 72 + 52
x2 = 74
f Decrease by 5% = 95%
Increase by 2% = 102%
0.95 × 1.02 = 0.969
= 96.9% = 3.1% decrease
12 a Decrease by 5% = 95%
x=
74
x = 8.6 m
Perimeter = 5 + 5 + 12 + 8.6
= 30.6 cm
h Perimeter = (8 × 6) + (13 × 4) + (36 × 2)
Decrease by 4% = 96%
= 48 + 52 + 72
Decrease by 2% = 98%
0.95 × 0.96 × 0.98 = 0.894
89.4% × 3.5 t = 3.13 t
b 89.4% = 10.6% decrease
= 172 mm
i Perimeter = 30 + 30 + 20 + 20
= 100 cm
j x2 = 752 − 602
x2 = 2025
21
x=
f By Pythagoras the hypotenuse is 10 cm.
2025
x = 45 mm
2 r
2
2  4
=
2
Circumference of the curve =
Perimeter = 75 + 85 + 45 + 25
= 230 mm
k Perimeter = 10 × 4
= 40 cm
= 12.6 cm
Perimeter = 10 + 15 + 12.6 + 9
l Perimeter = 7 + 3 + 7 + 13 + 13 + 15
= 58 cm
= 46.6 cm
2 a Circumference = 2 r
= 2    8.5
g Perimeter =
2 r
 10  10
2
=
2   10
 20
2
= 53.4 m
2 r
4
b Perimeter =
2
=  2 4
=  10  20
= 51.4 cm
= 10.3 m
2 r
66
c Perimeter =
4
=
=
2  6
 12
4
12  
 12
4
= 3  12
= 21.4 cm
d Perimeter =
=
2 r
646
2
2  2
646
2
= 2  16
= 22.3 cm
e Perimeter = 0.75  (2 15)  30
= 100.7 mm
h Perimeter = 2 r  26  26
= 2   11  52
= 121.1 mm
i Perimeter =
=
2 r 2 r

6
2
2
2  6 2  3

6
2
2
=   6   3  6
= 18.85 + 6 + 9.42
= 34.3 m
3

j Perimeter =   (2 r )   4
4

3

=   (2    5)   4
4

3

=   (10   )   4
4

22
=  23.56  4
2 Convert 5 km2 to ha.
5 km2 = 5 000 000 m2
= 94.2 cm
5 000 000
 500 ha, so A.
10 000
 40

k Perimeter = 
 (2 r )   16  16
 360

3 a Convert 5 m2 to cm2
 40

= 
 (2   16)   32
 360

5 100 100  50 000 cm 2
= 11.17  32
b Convert 2500 cm2 to mm2.
2500  10  10  25 mm 2
= 43.2 m
c Convert 72 000 m2 to hectares.
72 000 10 000  7.2 ha
2 r 2 r

55
2
2
l Perimeter =
=
d Convert 6800 cm2 to m2.
2   10 2    5

 10
2
2
6800  100  100  0.68 m 2
e Convert 3.09 km2 to m2.
=  10    5 10
3.09 1000 1000  3 090 000m 2
=  15  10
f Convert 3.6 km2 to hectares.
= 57.1 cm
3 600 000 m 2  10 000  360 ha
3 Ali = 2 r  30  30
g Convert 4.73 m2 to mm2.
= 2   14  60
4.73  100  100  10  10
= 247.96 m
 4 730 000 mm2
Billy = 2 r  80  80
h Convert 540 ha to km2.
= 2   10  160
3 500 000 m 2  1000  1000  5.4 km 2
= 222.83 m
247.96 m − 222.83 m = 25.13 m difference
4 801 600 km 2  801 600 000 000 m2
Convert to ha, divide by 10 000.
Correct to the nearest 0.1 m = 25.1 m
8.016 × 107 ha
5 a 2300 × 1880 = 4 209 000 mm2
Exercise 6-11
2
Convert to m2.
2
1 Convert 300 m to cm .
2
300 × 100 × 100 = 3 000 000 cm , so C.
4 209 000  10  10  100  100  4.209 m2
23
b 2300 mm × 1830 mm
1
 80
2
=
= 2.3 m × 1.83 m
= 4.209 m
= 40 m2
2
The answers are the same.
g Area = b  h
= 15  25
6 a Area = l  l
= 375 m2
= 88
h Area =  r 2
= 64 m2
=  12002
b Area = l  b
= 4 523 893.4 cm2
= 8.3 4.2
= 34.9 m2
1
a  b h
2
i Area =
c Area =  r 2
=   10
2
= 314.2 cm2
d Area =
=
=
1
a  b h
2
1
  43  90  65
2
1
 18  32 12
2
=
=
1
 50 12
2
= 300 m2
j By Pythagoras’ theorem, the base is 4 m: triad
1
Area = bh
2
1
133  65
2
=
1
 3 4
2
=
1
12
2
= 4322.5 m2
e Area =
1
bh
2
= 6 m2
=
1
 68
2
1
 48
=
2
= 24 m2
1
f Area = xy
2
=
1
10  8
2
k Area =
=
 r2
2
  7.52
2
= 88.4 m2
l Area =  r 2   r 2
=  122    92
24
= 197.9 cm2
=
1
 8  7.2
2
=
1
 57.6
2
m The base can be found using Pythagoras.
Base =
7.2  3.5
2
2
= 6.3 m
Area =
=
=
1
bh
2
1
 6.3  3.5
2
1
 22.05
2
= 11.0 m2
n Area =
=
=
1
a  b h
2
1
  9.2  6.8  5.3
2
1
16  5.3
2
= 28.8 m2
Total area = 12 + 28.8 = 40.8 m2
p Area 1 =
1
bh
2
=
1
 8 12
2
=
1
 96
2
= 48 m2
Area 2 =
1
bh
2
=
1
 12 10
2
=
1
120
2
= 42.4 m2
o Diagonal can be found using Pythagoras.
Base =
42  62
= 60 m2
1
bh
2
= 7.2 m
Area 3 =
1
bh
2
=
1
18  9
2
1
 46
2
=
1
162
2
1
 24
2
= 81 m2
Area 1 =
=
=
= 12 m2
1
Area 2 = bh
2
Total area = 48 + 60 + 81
= 189 m2
7 a There are two significant figures in 12.
b Area =  r 2
25
=   62
= 200  60
= 113.1 cm2
= 12 000 m2
Round to 2 significant figures: 110 cm2
Area 2 =  r 2
=   302
8 2 km = 2000 m
Area = l  b
= 2827.43 m2
= 2000  450
Total area = 12 000 + 2827.43
= 900 000 m2
Divide by 10 000 to convert to ha.
900 000
 90 ha
10 000
9 Area 1 =
r
2
= 14 827.43 m2
Round to 2 significant figures: 15 000 m2
12 a Area of frame = Area 1 – area 2
Area 1 = l  b
2
= 90  70
  22
=
2
= 6300 cm2
Area 2 = l  b
= 6.3 m2
Area 2 = l  b
= 49
= 58  38
= 2204 cm2
Area = 6300 cm2 – 2204 cm2
= 36 m2
Total area = 6.3 + 36 = 42.3 m2
10 Area of square = l  l
= 20  20
= 400 cm2
a Largest disc = radius of 10 cm
Area =  r 2
= 4096 cm2
Convert to m2: ÷ 100, ÷ 100 = 0.4096 m2
b 0.4096 m2 at a cost of $135 per metre
0.4096 × 135 = $55.30
Exercise 6-12
1 a Area 1 =
1
 20 15  150
2
=   102
= 314.16 cm2
b Area remaining = 400 cm2 − 314.16 cm2
= 85.84 cm2
11 Area 1 = l  b
Area 2 =
1
 (15  11)  18  234
2
Area 3 =
1
 11 8  44
2
Total area = 150 + 234 + 44 = 428 m2
26
Round to 2 significant figures: 430 m2
b Area 1 =
1
 12  14  84
2
Area 2 =
1
 (12  5)  (10  16  8)  289
2
Area 3 =
1
 5 10  25
2
Area 4 =
1
 24  18  216
2
Area 4 =
1
 (20  25)  42  945
2
Area 5 =
1
 40  20  400
2
Area 6 =
1
 25  24  300
2
Total area = 200 + 300 + 1610 + 945 +
400 + 300
= 3755 m2
Round to 2 significant figures: 3800 m2
Area 5 =
Area 6 =
1
 (18  12)  16  240
2
e Area 1 =
1
 12  18  108
2
Total area = 184 + 289 + 25 + 216 + 240 +
108
1
 35 14  245
2
Area 2 =
1
 32  30  480
2
Area 3 =
1
 (35  14)  30  735
2
Area 4 =
1
 30  37  555
2
Area 5 =
1
 (14  18) 16  256
2
1
 9  18  81
2
= 962 m2
Round to 2 significant figures: 960 m2
c Area 1 =
1
 60 105  3150
2
Area 2 =
1
 45  58  1305
2
Area 6 =
Area 3 =
1
 60  58  1740
2
Total area = 245 + 480 + 735 + 555 +
256 + 81
Total area = 3150 + 1305 + 1740
= 2352 m2
= 6195 m2
Round to 2 significant figures: 6200 m2
d Area 1 =
Round to 2 significant figures: 2400 m2
2a
1
 20  20  200
2
Area 2 =
1
 20  30  300
2
Area 3 =
1
 (30  40)  46  1610
2
27
Area 1 =
1
 8 16  64
2
Area 2 =
1
 16  15  120
2
Area 1 =
1
 7 18  63
2
1
Area 2 =  7  5  17.5
2
Area 1 =
1
12  28  168
2
Total area = 64 + 120 + 63 + 17.5
Area 2 =
1
 (28  36)  6  192
2
Area 3 =
1
 36  20  360
2
Area 4 =
1
 38  26  494
2
= 264.5 m2
Round to the nearest metre: 265 m
2
b
Total area = 168 + 192 + 360 + 494
= 1214 m2
Exercise 6-13
1 a 7 1003  7 000 000 cm 2
1
Area 1 = 18  36  324
2
Area 2 =
1
 (36  15) 14  357
2
1
Area 3 = 15  20  150
2
Area 4 =
Area 5 =
1
 40  12  240
2
1
 40  40  800
2
Total area = 324 + 357 + 150 + 240 + 800
= 1871 m2
c
b 50 103  50 000 mm 2
c 89 000  1003 = 0.089 m 2
d 0.468 1003  468 000 cm 2
e 2400  103  2.4 cm 2
f 5 600 000  1003  5.6 m2
g 9 100 000  103  9100 cm 2
h 12 1003  12 000 000 cm 2
2 V = Ah
=
1
1.2 1.5  2
2
= 1.8 m3
3 V = Ah
28
= 2.5  2.5  2.5
= 2.67 m3
= 2.53
Difference = 7.60 m3 – 2.67 m3
= 4.93 m3
= 15.625 m3, so B.
4V=
=
8 V  11.5 1.4
1
Ah
3
= 16.1 m3
1
 35  35  67
3
= 27 358
1
cm3, so B.
3
5 a V  5  5 10
= 250 cm3
1
b V   0.9  1.8  2.1
2
= 1.701 m3
1
9 V   230  230  137
3
= 2 415 766.67 m3
= 2 420 000 m3
10 a V 
1
 (1.4  1.8)  1 2.5
2
= 4 m3
b $16.50 × 4 = $66 × 4 = $264
c V   12.32  64.9
= 30 846.4 cm3
Exercise 6-14
1 L = 1000 mL
1
d V   (15  18)  8  2.3
2
= 3036 cm3
1
e V   620 150
3
= 31 000 mm3
f V  0.8  6  4.2
= 20.16 m3
6 a V    3.12 11.6
= 350.21 cm3
b 1 cm3 = 1 mL
7 Volume 1    0.542  8.3
= 7.60 m3
Volume 2    0.322  8.3
1000 cm3 = 1000 mL = 1L
1 cm3 = 1mL
1 m 3 = 1kL = 1000L
1 ML = 1 million litres
1 a 680 cm3 to mL
1 cm3 = 1mL, so 680 mL.
b 8500 cm3 to L
1 cm3 = 1mL, so 8500 mL
1 L = 1000 mL, so 8.5 L
c 22 m3 to L
1 m 3 = 1kL = 1000L, so 22 000 L
d 8000 L to m3
1 m 3 = 1000L, so 8 m3
29
e 3.5 m3 to mL
= 0.864 m3
1 m 3 = 1000L
b 1 m 3 = 1000L, so 864 L
1 L = 1000 mL, so 3500 L = 3500 000 mL
= 3.5 × 106 mL
f 690 L to cm3
4 V    42 1.5
= 75.4 m3
a 1 m 3 = 1000L, so 75 400 L
1 L = 1000 mL, so 690 000 mL
b 1kL = 1000L, so 75.4 kL
3
1 cm = 1mL, so 690 000 cm
3
5 a V    42 10
3
g 55 m to L
= 502.65 × 8
1 m 3 = 1000L, so 55 000 L
= 4021.24 cm3
3
h 4300 m to kL
1000 cm3 = 1000 mL = 1L
3
1 m = 1kL, so 4300 kL
i 9500 L to m
So, 4021.24 cm3 = 4.02 L
3
3
1 m = 1kL = 1000L, so 9.5 m
3
= 515.3 × 8
j 8.5 × 104 cm3 to L
= 4122. 4 cm3
3
1 cm = 1mL
1 L = 1000 mL, so 85 000 mL = 85 L
-3
k 4.3 × 10 kL to cm
b V    4.052 10
Difference = 4122. 4 cm3 − 4021.24 cm3
= 101.16 cm3 = 101.16 mL
3
1kL = 1000L, so 4.3 L
1 L = 1000 mL, so 4300 mL
1 cm3 = 1mL, so 4300 cm3
l 106 m3 to ML
1 m 3 = 1kL = 1000L, so 109 L
1 ML = 1 million litres, so 109 ÷ 106
3
= 10 (1000) ML
2 2 m3 = 2000 L, so B.
3 a Change all measurements to m
80 cm = 0.8 m
90 cm = 0.9 m
V  1.2  0.8  0.9
Correct to three significant figures: 101 mL
6 V  57  81 54
= 249 318 cm3
= 249.3 L
Correct to the nearest litre = 249 L
7 5 kL = 5 m3, so A
8 a 2.031 × 106 ML to litres
1 ML = 1 million litres
So, 2.031 × 1012 litres
b 1 m 3 = 1kL = 1000L
So, 2.031 × 109 m 3
9 V   12  3.4
30
= 10.681 m3
6.4 × 1000 = 6400 kg
1 m 3 = 1kL = 1000L
c Convert 340 mL to L
So, 10 681 L
340 ÷ 1000 = 0.34 L
10 5 × 1011 L
2 1 ML = 1 million litres, so
1kL = 1000L
43 ML = 43 000 kL
So divide 5 × 108 L by 10003
So, 0.5 km3
3 a The absolute error is  0.005.
b Percentage error =
2
11 V   1.3  3.3
2
0.05
 100  0.14%
3.66
4 To two significant figures:
= 8.495 m3
a 38.915 becomes 39
1 m 3 = 1000L, so 8495 L
b 1036 becomes 1000
c 0.00872 becomes 0.0087
Sample HSC problem
a 20.7 contains 3 significant figures
b Percentage error =
0.05
100  0.24%
20.7
c i 20.7 m/1.2 seconds
d 6 587 200 becomes 6 600 000
5 $8 350 000 000 = 8.35 × 109
6 4.6 × 10-6 = 0.000 004 6
7 a 15 : 60 = 1 : 4
17.25 m/s
1 1
3 13 6 13
b 1 : 3 = : = : = 6 : 13
2 4
2 4 4 4
Correct to 3 significant figures:
c 5 mins : 100 sec = 300 : 100 = 3 : 1
17.3 m/s
d 50 kg : 1 t = 50 : 1000 = 1 : 20
ii 17.3 m/s
8 a milk : flavouring : ice-cream
1038 m/min
(17.3 × 60)
62 280 m/hr
(1038 × 60)
10 : 2 : 4
62.3 km/hr
(62 280 ÷ 1000)
5:1:2
8 parts
1
 500 mL = 62.5 mL of flavouring
8
Revision
1 a Convert 28.5 km to m
28.5 × 1000 = 28 500 m
b
1
 x mL = 110 mL
8
b Convert 6.4 t to kg
31
x  8 110
x = 880 mL
c Perimeter =
 72
4
 7  22  15
= 55.00 cm
9 Phoebe : Patrick
14 a 8400 mm2 = 84 cm2 (÷102)
3:2
5 parts
3
2
 $44 380 :  $44 380
5
5
= $26 628 : $17 752
b 5.6 ha = 10 000 m2 × 5.6 = 56 000 m2
15 a Area = (4 × 2) + (2 × 6) + (3 × 8)
= 8 + 12 + 24
= 44 m2
10 10 L in 25 sec
0.4 L / sec
b Area =
0.4 × 60 = 24 L / min
c Area =   9.52    4.22
40 kL = 40 000 L
= 228.11 cm2
40 000 ÷ 24 = 1666.67 min = 1667 min
Rounds to 228 cm2
11 50 m/21.28 sec
2.3496 m/s
1
(32  48)  26 = 1040 cm2
2
16
a i 2.350 m/s
ii 8458.65 m/hr
8.459 km/hr
b Will take 425.53 seconds (7.093 mins)
12 a $2400 × 82% = $1968
$1968 × 105% = $2066.40
b 0.82 × 1.05 = 0.861
0.861 × 100 = 86.1%
Area 1 =
1
 (16  10)  20  260
2
Area 2 =
1
10  32  160
2
Area 3 =
1
 28  15  210
2
Area 4 =
1
15  24  180
2
A decrease of 13.9%
13 a Perimeter = 9 + 8 + 9 + 2 + 3 + 4 + 3 + 2
= 40 m
b Slope is 65 m by Pythagoras.
Perimeter = 80 + 52 + 119 + 65
= 316 cm
Total area = 260 + 160 + 210 + 180
= 810 m2
b 810 m2 = 0.810 ha (÷ 1000)
32
17 a Convert 20.7 cm3 to mm3
20.7 × 103 = 20 700 mm3
b Convert 1 650 000 cm3 to m3
1 650 000 ÷ 1003 = 1.65 m3
18 a V  1  72  8.4
3
= 137.2 cm2
b V  1  (32  54)  28  95
2
= 114 380 cm3
c V  1  0.45  0.48  2
2
= 0.216 m3
19 a1 cm3 = 1 mL
A container of volume 894 cm3 will hold 894 mL.
b1 m 3 = 1000 L
A container of volume 6.5 m3 will hold 6500 L.
20 a V    1.22  1.8
= 8.143 m3
Correct to two significant figures: 8.1 m3
b 1 m 3 = 1000 L
8.1 m3 = 8100 L
33