CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Workbook answers
Unit 1 The number system
12 Odd one out is 368.4. All the others are
equivalent to 36.84.
Exercise 1.1
Focus
1
1.9 and 0.19
2
a
3
a
15.37
Challenge
b
105.05
c
75
×
× 100
× 10
75 000
b
25 000
× 10
250 000
10
÷ 100
2500
4
5
13 a
4
tens
5
tenths
6
hundredths
7
ones
3 hundredths
15 a
10
b
100
c
100
d
10
e
100
f
100
$150
b
$1.90
c
$75
16 3.9
17 a
18 Marcus’s number could have more tenths;
for example Arun could write 0.59 and
Marcus 0.67.
Exercise 1.2
Focus
1
34.4 rounded to the nearest whole number
is 34.
2
36.4
3
9.9
10.1
8.5
7.4
10.7
8.2
9.4
9
10
11.5
0.5
7
Practice
4
8
Ring around 0.05.
7
0.36
8
0.1 and 0.04
9
Any two regrouped versions. For example:
Practice
30 + 0.54, 3054 hundredths, 3 tens and
54 hundredths etc.
5
11 a
720
b
75
c
42.8
d
2.7
e
1.51
f
0.66
11
12
Any number with 1 decimal place from
0.5 to 1.4.
Any number with 1 decimal place from
9.5 to 10.4.
6
10 3330
1
b
3 tenths
14 23.5, 2.35, 25.3, 2.53, 32.5, 3.25, 35.2, 3.52,
5.23, 5.32
7500
1000
÷
34.34
a
4 cm
b
7 cm
c
10 cm
d
9 cm
99.5
b
100.4
6
19.5
7
a
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Challenge
8
BELT
9
20.3
9
10 2.4 and 4.4
11
Rounds to 3 Rounds to 4 Rounds to 5
3.3
3.5
3.9
3.7
4.4
4.5
More than one possible solution. Clues should
define or describe these words:
Across
Down
4 tessellates
1 isosceles
6 scalene
2 angle
3 side
4.9
5 triangle
Exercise 2.2
Unit 2 2D shape and
pattern
Focus
1
Exercise 2.1
Focus
1
Yes, two of the sides are equal length.
2
C, D and F
3
A
a
acute
b
right angle
c
acute
B
a
acute
b
obtuse
c
acute
C
a
acute
b
acute
c
acute
Practice
4
Sketch of a scalene triangle.
5
a
50° angles circled.
b
45° angles circled.
c
55° angles circled.
d
59° angles circled.
6
Equilateral circled.
3
aThe sides of an equilateral triangle are all
equal in length.
b
The angles of an equilateral triangle are
all equal.
4
Challenge
7
2
aSketch of isosceles triangle with two sides
of 6 cm and a shorter side.
b
Sketch of isosceles triangle with two sides
of 6 cm and a longer side.
8
2
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Practice
5
10 aMore than one possible answer,
for example:
a
1 line of symmetry drawn.
b
No lines of symmetry drawn.
c
No lines of symmetry drawn.
d
1 line of symmetry drawn.
e
3 lines of symmetry drawn.
6
b
c
4
7
Unit 3 Numbers and
sequences
Exercise 3.1
3
Challenge
Focus
8
Completed pattern should have two lines of
symmetry along the diagonal lines on the
diagram.
1
9
a
2
b
Circle in the centre of the chessboard.
c
Cross anywhere on the chessboard except
along the diagonals of the board.
d
Triangle in any position along the
diagonals of the board, but not in
the centre.
−2 because each term is found by
subtracting 2.
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2
3
a
b
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
1
10
2
13
3
16
4
19
c
Add 3
d
31 hexagons
105
The numbers that they both say must be
multiples of both 3 and 5. The first number
greater than 100 that is a multiple of both 3
and 5 is 105.
The multiples form diagonal lines.
10 6, 10, 14
c
No
11 4 and 9
a
7, 9, 11, 13
b
7, 9, 11, 13, 15, 17
c
Add 2
Exercise 3.2
Focus
1
4
−8
5
79, 70, (61, 52,) 43
6
a
b
Pattern number
Number of sticks
1
6
2
11
3
16
4
21
c
Add 5
d
51 sticks
square numbers
2
1030
Challenge
8
Number of
hexagons
b
Practice
7
9
Pattern number
a
triangular numbers
3
a
36 and 49
b
15 and 21
Practice
4
a
square
b
10 + 15 = 25
15 + 21 = 36
4
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5
64
6
6
a
9
7
a
64, 4, 100, 81, 9
Challenge
b
Yes. 49 is a square number so she can
make a 7 by 7 square.
7
b
c
4
49
Challenge
8
9
16
42 = 16
4 × 6 = 24
25
52 = 25
5 × 7 = 35
36
62 = 36
a
1 + 9 = 10
b
c
4 + 36 = 40
d
25 + 25 = 50 or 1 + 49 = 50
e
16 + 64 = 80
g
36 + 64 = 100
4 + 16 = 20
Prime number 1 is 19.
Prime number 2 is 13.
8
3 × 5 = 15
Factors of 15 are 1, 3, 5 and 15. A prime
number has only two factors.
9
A counter example, for example:
•
1 more than 24 is 25 which is a
square number
•
1 more than 54 is 55 which is a multiple
of 5.
67, 71, 73
10
f
Triangular
numbers
9 + 81 = 90
1
10 55
6
3
13 17 19
2
10
Even
numbers
4 12 16 20
9
8
14
18
Exercise 3.3
Focus
1
15
Prime
numbers
5 7 11
P
2
Prime
numbers
Composite
numbers
2
4
5
Exercise 4.1
6
3
Unit 4 Averages
Focus
3
A number with only two factors is called a
prime number.
1
a
8
2
a
109, 117, 118, 120, 121
b
118
3
a
24
b
23
4
a
11
b
12
Practice
4
2
4
6
8
13
3
23
29
71
65
1
51
45
7
5
15
92
25
1
2
31
37
16
14
11
5
c
7
3
Practice
5
6
7
5
b
a
2+7
b
19 + 31 or 7 + 43 or 13 + 37 or 47 + 3
First put the numbers in order from smallest
to greatest.
Then find the middle number. The middle
number is the median of the data.
A and D ticked.
Any set of numbers where the most frequent
is 8 and the middle number is 9 when the
numbers are put in order.
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8
a
1
b
2
c
The median would describe the scores
better because although three people gave
the book a score of 1, everyone else gave
the book a higher score. The median of 3
represents this better.
3
10
c
30, 33 or 29 (if learner gives another
number that is not already in the set then
it is a correct answer, but it will mean
the data has no mode. This situation is
considered fully in later stages).
b
5.4
6.3
106
e
Any three different numbers except 4.
10 aA set of five numbers which contain
3, 3, 4 and two other larger numbers.
Two more sets of five numbers
which contain 3, 3, 4 and two other
larger numbers.
They all include 3, 3 and 4. The other two
numbers are greater than 4.
11 Learners should explain or demonstrate that
for the mode to be 3, two of the numbers must
be 3. This means that the middle of the set
of numbers when they are put in order will
be 3 and so in this situation the median will
always be 3.
Unit 5 Addition and
subtraction
Exercise 5.1
7.3
4
d
c
4.2
7.6
a
b
3.4
5.8
Challenge
9
2.4
3.7
3
a
4
0.5 + 0.5 = 1
8.4
b
5.9
b
3.33
Practice
5
a
6
6.55
7
Accept any reasoned answer.
Learners are critiquing (TWM.07) when they
evaluate the two methods and say which is the
most efficient.
8
A small triangle has a value of 2.
11.49
Challenge
9
0.42 + 0.58 or 0.52 + 0.48
10 3.43
11
+
2
2
8
2
7
2
5
0
0
12 More than one possibility, for example:
0.1
Focus
1
0.5
0.6
0.9
0.9
0.2
0.3
0.5
0.4
0.8
0.1
0.8
0.7
0.2
6
0.7
0.3
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13 15 cents
Unit 6 3D Shapes
Exercise 5.2
Exercise 6.1
Focus
1
a
−8
b
−1
2
a
−2
b
2
3
a
−4
b
−6
820
b
3955
−7
b
−3
c
−6
c
3
Focus
1
A and C circled.
2
a
More than one solution possible, e.g.
Practice
4
a
5
−35
6
a
7
5 °C
b
Challenge
8
a
−1 °C
9
b
−2 °C
addition
calculation
subtraction
calculation
positive
answer
D
A
negative
answer
E
B C
3
A – iv
B–i
C – iii
D – ii
4
a
Shape A is a triangular prism.
It has 5 faces, 9 edges and 6 vertices.
10 a
−4 °C
b
5
b
Shape B is a tetrahedron or a trianglebased pyramid.
11 a−401 −302 −203 −104 −5 94 193
The ones digits are 0, 1, 2, 3, 4, 5, 4, 3.
They increase by 1 each time until the
numbers become positive and then they
decrease by 1 each time.
b
Learner’s own answers. Sample questions:
•What would happen if I started
at −400? (The ones digits would be
0, 1, 2, 3, 4, 5, 4.)
•What would happen if I started
at −100? (The ones digits would be
0, 1, 8, 7, 6, 5, 4.)
When the numbers are negative the ones
digits increase by 1 each time. When
the numbers are positive the ones digits
decrease by 1 each time.
7
It has 4 faces, 6 edges and 4 vertices.
c
Shape C is a cuboid.
It has 6 faces, 12 edges and 8 vertices.
Practice
5
A, C and D crossed out.
6
Any three colours may be used, but faces with
the same letter as shown should be coloured
the same.
x
y
z
y
z
x
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7
9
a
A and H
B and G
C and D
E and F
J and K
L and M
N and I
10 a
b
More than one solution possible, e.g.
Sketch of a triangle or square.
b
Sketch of two triangles joined along one
side or one square and one triangle joined
along one side.
c
Sketch of four triangles (square-based
pyramid from ‘above’).
Unit 7 Fractions,
decimals and
percentages
Exercise 7.1
Focus
1
Challenge
8
The possible nets are:
1
4
2
2
Show divisions to give the answer .
3
a
5
27
b
16
c
20
Practice
4
Learner should draw two different shapes,
each made up of three of the original shape.
One possible answer:
15 rectangles are in the whole shape.
8
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5
8, 16, 24, 8, 16
Possible answers could include:
• All calculations include a fraction
in sevenths.
• All answers are even numbers.
• Some calculations have the same answer,
for example:
•
•
6
7
a
✗
b
✓
c
✗
2
1
of 56 = 7 of 28 because 28 is half of
7
2
1
56 and is double .
7
7
b
0
10
20
2
a
9% and 9 squares shaded
b
50% and 50 squares shaded
c
25% and 25 squares shaded
3
3
5
8
of 24 litres = 15 litres
1
2 of 40
2
5 of 60
Answer equal to 20
50
60
70
80
90 100
P
of 30 litres = 9 litres
Answer less than 20
40
15, 35, 50, 95, 75, 17, 8, 25, 5, 55
2
4
of 56 = of 28 because 28 is half of
7
7
4
2
56 and is double .
7
7
10
30
1%
99%
35%
75%
8%
40%
70%
25%
100%
17%
12%
44%
15%
95%
5%
20%
38%
50%
34%
30%
90%
60%
55%
65%
4%
Practice
4
Incorrect:
9
10
, 0.9, 9 tenths and
3
4 of 16
Answer more than 20
2
3 of 30
Marcus has confused 9% and 90%.
Per cent means out of a hundred so
Challenge
9% =
8
180
9
Zara has confused multiplication and division.
She should say ‘To find
10 and multiply by 7.’
The answer is 70.
7
10
of 100, I divide by
7
10
0.3
9
0.5
3
10
Focus
10
1
2
0.6
Exercise 7.2
0
or 0.09 or 9 hundredths.
0.4
11 Use Zara’s, Arun’s or Marcus’s method as they
give the correct answer (32).
Do not use Sofia’s method which gives the
wrong answer (2).
a
100
0.2
5
10 23 stickers
1
9
0.7
20
30
40
50
60
70
80
90 100
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6
5
a
10
b
50
or
100
or
1
2
and 50% in the circles.
4
Fraction
6
7
10
8
30%
10
9
or equivalent
10%
50%
1
10
15
10
100
150
75
30
4
100
40
1
25
50
20%
0.4
40%
4
100
200
10
1
60
50
10
1
100%
2
120
100
1
0.25
25%
0.8
80%
0.5
50%
0.9
90%
0.7
70%
10
8
A different
correct
answer
10
30%
1
2
9
10
Focus
7
10
1.9, 2.7, 3.4, 3.5, 5.3
Practice
1 2 3 4 6
, , , ,
7 7 7 7 7
2
10
4
4
3
18
5
5
a
2 =
b
3 =
0.5 =
5
10
< 0.5
=
(accept 2
1
2
2
7
4
5
for 2 )
5
2
6
3
4
1
0
1
2
3
2
0.6 >
1
6
8
1
a
1
d
1 or 1
2
7
10
0.2
4
Exercise 7.3
10
50%
2
3
Any
correct
answer
3
4
0.5
10
Equivalent fractions Percentage
10
3
75%
4
Fraction
2
0.75
10
b
b
30%
5
0.6 and 60%
25%
a
0.3
3
75%
1
60%
4
10
a
0.6
10
6
b
10%
3
7
a
0.1
10
Challenge
b
Percentage
1
0.8 and 80% in the circles.
a
Decimal
25
100
a
4
2
1
4
2
b
1
e
2
4
5
1
3
1
c
3
f
1 or 1
3
3
1
6
2
and 0.25 circled.
<
b
=
c
>
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Challenge
9
2
5
4
=
or
10
2
=
5
=
4
10
5
40
and 25% =
100
=
40
100
a
False
b
True
25
c
False
d
True
100
e
True
= 40%
6
25% < 40%
10
1
34
16
10
4
15
3.25
2.75
2
25
2.4
7
2 10
Challenge
7
3
24
2
8
1.8
11 a
2.7
1.6
b
5
1
Learner’s own answers. Answers will depend
on learner’s prediction and experiment.
2
B and D circled.
5
Mia could not be using bag E because bag E
does not have any letter ‘A’s and she has taken
‘A’s out of the bag.
even chance
Practice
certain
likely
3
No, Marcus is not correct. There is an even
chance of flipping either a head or a tail on
the next flip.
4
There are several possible solutions.
All solutions have half of the cards with
squares on and an equal number of triangles
and circles.
The likelihood of Arun taking a green apple
is impossible.
The outcomes recorded and the sentences
written will depend on the learner’s choice of
shapes and the results of their experiment.
A, C, E and F circled.
Challenge
Practice
F
impossible unlikely
11
There is one outcome for ‘1’ and two other
equally likely outcomes, so the chance of
the spinner landing on ‘1’ is unlikely, not an
even chance.
Focus
The likelihood of Arun taking a red apple
is certain.
D
8
3
unlikely
4
1
8
4
Focus
3
1
orange, blue, pink.
2
Exercise 8.1
2
4
Exercise 8.2
0.3 45% 0.5 impossible
1
3
Unit 8 Probability
1
red,
1
0.2 30% 0.7 1
More than one possible solution. For example,
the spinner could be coloured or labelled
1
3.5
7
2
Answers will depend on learner’s results.
(Winning and losing are not equally likely.
For example, there are more black counters
than white counters so it is more likely that
the learner will take a black counter. This
means that there is a greater chance of moving
right rather than left and a greater chance of
winning rather than losing.)
B
even
chance
E
likely
AC
certain
5
aLearner should draw 5 balls in the bag
with at least one each of red, green
and yellow.
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
b
6
There must be at least one each of red,
green and yellow because red, green and
yellow balls were taken out of the bag.
The learner should justify their choice of
colour for the other two balls, for example,
‘I have drawn an orange ball in the bag
as just because no orange balls have been
pulled out of the bag yet it does not mean
there are not orange balls’ or ‘I have
drawn two red balls and two yellow balls
because more of them have been taken out
of the bag so it is likely that there are more
red and yellow balls than green balls.’
Spinner drawn should show bronze, silver
and gold in equal proportions. Answers will
depend on learner’s experiments.
Unit 9 Addition and
subtraction of fractions
Exercise 9.1
a
2
a
3
a
4
a
10
11
1
8
3
8
12 Accept any reasoned answer, for example:
A – the denominators are even. In B and C,
the denominators are odd.
B – the only answer that is a proper fraction.
C – the denominators are the same. In A and
B one denominator is a multiple of the other
denominator.
13 a
14
a
6
a
7
8
9
11
8
7
5
b
8
9
5
b
12
17
7
b
8
c
6
b
12
c
8
8
Unit 10 Angles
Exercise 10.1
1
B, C and E circled.
8
2
a
40°
b
160°
c
290°
3
3
a
10°
b
80°
c
130°
b
B
c
B
8
12
1
Practice
12
4
1
5
8
8
21
b
15
1
5
b
6
c
12
c
20
hour
9
a
b
A
right angle
16
12
8
12
acute angle
obtuse angle
reflex angle
acute angle
obtuse angle
6
Reflex, less, 90°, Obtuse, less, 180°
7
a
50°
b
40°
c
90°
d
100°
Challenge
10
a
5
1
Practice
5
b
5
Focus
Focus
1
Challenge
3
12
2
3
+
−
2
3
3
12
=
=
3
12
8
12
+
−
8
12
3
12
=
=
11
8
Angles X and Y are both 32°.
180° − 116° = 64° and 64° ÷ 2 = 32°
9
60°
12
5
12
10 Learner’s own answer (poster).
12
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Unit 11 Multiplication
and division
b
28 672
448
Exercise 11.1
Focus
1
28
1792 km
2
128
16
2
16
4
12 75 × 20 or 20 × 75
13 4.32 km
8
8
64
Exercise 11.2
4
Focus
3
152 × 7 = 1064
1
103, 112 and 121 circled.
4
79 × 60 = 4740 beats
2
40, 80, 120 and 160
Practice
3
27
5
396 seats
4
6
a
Less than
10
Between
10 and 20
More than
20
7
13 × 13 = 169 and 31 × 31 = 961
81 ÷ 9
84 ÷ 6
105 ÷ 5
b
4698
3528
The digits in the two answers are the same.
961 is the reverse of 169.
8
5
Sofia has made mistakes.
•
•
•
She has not estimated her answer before
calculating it.
She has not carried figures on the middle
two lines of working.
To improve her work, Sofia should check
her answer against an estimate.
The correct answer is 5168.
Challenge
9
120 ÷ 8
10 a
Practice
6
Zara is right because a remainder of 1 can be
shown as a fraction with 1 as the numerator
and the number you are dividing by (the
divisor), which is 4, as the denominator.
7
a
8
42 × 30 or 30 × 42
b
15 276
11 a
30 494
a
18
63
2
5
2
5
b
b
22
70
3
4
7
8
6
c
13
c
157
7
1
6
Challenge
9
8092
Yes. Remainder 1 means there is 1 left over.
136 is 1 more than 135 which is a multiple
of 5.
4
10 5
11 24 packs (do not accept 185 ÷ 8 = 23.125)
119
7
13
68
17
4
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Exercise 11.3
Unit 12 Data
Focus
1
96, 1044, 132
Exercise 12.1
2
3
Focus
3
divisible by 4
1
a
5
b
2
a
Strawberry 8
c
8
False
Chocolate 11
Vanilla 3
604
divisible by 8
Mint 7
A bar chart showing favourite
flavours of ice cream
b
400
Number of people
116
28
101
not odd
48
96
37
101
54
78
3
Practice
5
No.
Learner gives a counter example, for example
14 ends in 4 but 14 is not divisible by 4, or
explains that the tens digit must be even for
the number to be divisible by 4. If the tens
digit is odd, the number is not divisible by 4.
c
Each square on the key is coloured a
different colour.
d
Grid coloured according to the colours
in the key: sparrow 10, robin 5, pigeon 2,
crow 3.
a
67 432, 444, 7696, 1748, 624
4
b
67 432, 7696, 624
582 176
10 23 + 57 = 80 or 53 + 27 = 80
Ice cream flavours
The grid has 20 squares and there are
20 pieces of data (birds).
7
9
2
b
Practice
24 and 48
4
First grid circled.
593 132 or 593 136
8
6
a
6
Challenge
8
0
not divisible by 8
odd
10
Mint
divisible by 8
12
Strawberry
4
14
Vanilla
64
Chocolate
36
Favourite
sport is
football
Favourite
colour is red
Marcus
Sofia
Lou
Zara
Rajiv
Pablo
Arun
Sarah
14
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Frequency
1
IIII I
6
2
IIII II
7
3
IIII
5
4
III
3
5
II
6
III
d
Learner’s own answers
e
Learner’s own answer
Challenge
a
–10 °C
b
2 °C (1 °C also acceptable)
2
c
–34 °C (–33 °C also acceptable)
3
d
The temperatures represented in the
graph should match the temperatures in
the table.
e
1
2
3
4
5
6
Number of pencils
9
The bars go up and then back down.
a
Colour
6
a
The rainfall goes up then back down.
b
The rainfall goes down then back up.
(The learner might also describe the dip
on the graph for February.)
c
Possible answers include:
The highest amount of average monthly
rainfall in Perth is about 165 mm, in
Tehran it is only about 37 mm.
The month with the most rainfall in
Perth is June. The months with the most
rainfall in Tehran are January, March
and December.
aWaffle diagram of 24 squares. The squares
coloured according to the learner’s key:
12 beach, 1 city, 1 mountain, 4 forest,
6 ocean.
b
c
15
50%
1
Frequency Fraction
Red
100
Yellow
50
Green
30
Orange
20
June, July and August are the months with
the most rainfall in Perth and some of the
months with the least rainfall in Tehran.
7
October
Month
1
0
December
2
November
3
September
4
July
5
January
Frequency
6
August
7
10
5
0
–5
–10
–15
–20
–25
–30
–35
May
8
A bar chart showing the average
temperatures in Cambridge Bay, Canada
June
Dot plot showing the number of
pencils in children’s pencil cases
b
April
8
March
Number of Tally
pencils
February
a
Average
temperature (°C)
5
b
100
200
50
200
30
200
20
200
=
=
=
=
50
100
25
100
15
100
10
100
Percentage
50%
25%
15%
10%
Key completed with four different colours
for the four colours of sweets.
Grid coloured according to key:
red 50 squares, yellow 25 squares,
green 15 squares, orange 10 squares.
Exercise 12.2
Focus
1
2
a
8
b
d
The graph goes up, then back down.
a
7 cm
b
14
5 weeks
c
c
36
6 cm
4
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Practice
a
5
d
The tallest category is 140 cm to less
than 150 cm so none of the children
were 150 cm.
e
4
b
c
70
A frequency diagram showing the age
of people visiting a swimming pool
False
10
9
There are 5 children who are 110 cm or
more, but less than 120 cm. It is possible
that one or more of the 5 children could
be 110 cm, but none of them might be
exactly 110 cm.
Day 1: 22 mm, Day 2: 67 mm, Day 3: 83 mm,
Day 5: 115 mm, Day 6: 130 mm, Day 7: 141 mm
8
7
6
Frequency
3
b
2
3
4
Days
Height (mm)
3
2
0
6
1
4
1
Height of plant
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
5
5
6
7
10
20
30
40
Age (years)
50
60
The information in the graph should match
this table:
Time
Shadow length (cm)
9 a.m.
21
10 a.m.
14.5
11 a.m.
9
12 p.m.
5.5
1 p.m.
8.5
2 p.m.
13.5
Estimate between 90 mm and 100 mm.
Challenge
5
16
a
Age group (years)
Tally
Frequency
0 to less than 10
IIII
4
10 to less than 20
IIII IIII
9
20 to less than 30
III
3
30 to less than 40
IIII
5
40 to less than 50
II
2
50 to less than 60
III
3
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
A line graph showing the height
of a stick’s shadow
3
a
6:4
b
Learner’s own answer (drawing) which
shows a quarter of the beads as red.
c
Learner’s own answer (drawing)
which shows 2 pink beads for every 3
green beads.
22
21
20
19
18
Practice
17
4
16
Shadow length (cm)
15
5
14
13
12
6
11
10
9
7
8
7
6
3
10
Correct
Not correct
ADE
BCF
Bruno.
Picture of a necklace with 3 black beads for
every 1 white bead.
Pattern A
Pattern B
E, G, H, I, J
A, C, D, F
Statement B is not used.
5
4
Challenge
3
8
a
False
9
a
A – black
2
1
Unit 13 Ratio and
proportion
Focus
2
or
3
8
b
white to brown = 6 : 10
c
brown to white = 10 : 6
a
1 : 2
c
17
6
3
7
c
b
4:4:2
True
D – black
b
3 in every 5 or
c
2 in every 5 or
10 a
c
2:4:4
4
10
or 40%
3
5
2
5
d
2
10
or 20%
11 Zara has confused ratio and proportion. She
saw one triangle and three circles which is the
ratio of triangles to circles equal to 1 : 3. She
should have written 1 in every 4 shapes which
is a proportion.
Exercise 13.1
16
False
C – grey
2 p.m.
1 p.m.
12 p.m.
11 a.m.
9 a.m.
10 a.m.
Time
The shadow was approximately 11 cm at half
past 1.
a
b
B – black
0
1
or 30%
b
d
4 : 3
2
3
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
b
Area 14 cm2
Unit 14 Area and
perimeter
Challenge
Exercise 14.1
8
9 m and 19 m
Focus
9
a
Rectangle 6 cm by 1 cm
b
Rectangle 4 cm by 3 cm
c
Rectangle 8 cm by 3 cm
1
a
Perimeter 18 cm
Two possible answers:
10 Missing sides are 30 m and 12 m.
Area 1206 m2.
b
Unit 15 Multiplying and
dividing fractions and
decimals
Two possible answers:
Exercise 15.1
c
Focus
Two possible answers:
1
5
4
+ 14
+ 14
0
2
3
a
9 km
c
4 cm, 10 cm
b
5m
2
1
4
2
4
3
4
+ 14
4
4
5
4
8
The area of rectangle A is 24 m .
1
8
The area of rectangle B is 20 m .
2
3
+ 14
7
2
The total area is 44 m2.
+ 14
1
8
1
8
1
8
1
8
1
8
1
8
1
27
Practice
4
5
6
More than one possible answer, for example,
a rectangle that is 6 cm by 4 cm, which has a
perimeter of 20 cm and an area of 24 cm2.
a
Perimeter = 12 m, Area = 9 m2
b
Perimeter = 16 m, Area = 16 m2
c
Perimeter = 20 m, Area = 25 m
18
1
6
÷4=
Perimeter = 84 cm
5
6
aGood estimates would be between 8 cm2
and 20 cm2 for the area and 14 cm to 22 cm
for the perimeter.
1
24
Practice
2
Area = 320 cm2
7
4
7
a
5
b
3
1
12
c
7
5
1
6
6×
1
3
ticked.
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
START
8
3
1
8
1
÷2
2
7
35
1
×7
5
1
4
1
×4
7
5
7
3
4
1
×3
2
3
4
1
6
1
÷2
3
1
÷4
3
a
5
b
7
c
7
b
31.5
c
60.8
Practice
4
a
5
58.1
6
1.8 × 6
4.8
Answer less than 10
2.4 × 4
3
2
7
6
1
12
Answer equal to 10
1.8 × 5
1
10
1
÷2
5
6
7
1
×7
6
END
Answer more than 10
2.5 × 4
Challenge
Challenge
9
1
a
6
7
1
b
9
10 Arun has multiplied the numerator and the
11
8
6.5 × 4 = 26
6
9
157.5 cm
4
12
10 105
12
1
4
5
÷3=
1
12
3
12
12
=
2
11 Sally runs the greater distance.
12
Zina runs 1.6 × 7 = 11.2 km
Unit 16 Time
Exercise 16.1
1.4
a
1
Sally runs 1.9 × 6 = 11.4 km
Focus
2
−
metre
Exercise 15.2
1
Estimate 2 × 4 = 8 and 3 × 4 = 12 so the
answer must be between 8 and 12.
7
denominator by 7. The correct answer is .
2
0.7 × 6
7 ÷ 10
Focus
×
6
1
a
30 seconds
b
30 minutes
c
1 day and 12 hours
d
2 hours
e
1.5 minutes
a
10:15
b
11:10
c
20 minutes
= 29 × 7 ÷ 10
d
50 minutes
= 203 ÷ 10
e
1 hour and 10 minutes
= 7 × 6 ÷ 10
= 42 ÷ 10
= 4.2
b
2.6 × 4 = 12.4 crossed out.
2
2.9 × 7
29 ÷ 10
× 7
= 20.3
19
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
3
Clock should show 1.45. Written answer 1.45, 13:45 or a quarter to 2.
Practice
4
a, b Activities and times matched and ordered:
She picked up a pencil (0.5 seconds).
She wrote her name (2 seconds).
She swam one width of a swimming pool (0.5 minutes).
She boiled a kettle (3 minutes).
She wrote a chapter of a story (0.5 hours).
5
Room 1: 17 minutes
Room 2: 21 minutes
Room 3: 17 minutes
Room 4: 14 minutes
Room 5: 13 minutes
Tom left Room 6 at 23:11.
6
a
There are 24 time zones.
b
More than one possible answer including:
Russia and the USA have more than one time zone.
7
10.37 Josh
11:37 Adam
12.37 Meena
13:37 Jess
Challenge
8
a
Bus A
Bus B
Bus C
Bus D
Village
11:51
12:48
13:55
15:42
Town
12:08
13:05
14:12
15:59
City
12:32
13:29
14:36
16:23
Harbour 12:47
13:44
14:51
16:38
b
9
20
Learner’s own answers.
Lima
2 hours
São Paulo
7 hours
5 hours
Cape Town
9 ½ hours
7 ½ hours
2 ½ hours
Chennai
14 hours
12 hours
7 hours
4 ½ hours
Tokyo
14 ½ hours
12 ½ hours
7 ½ hours
5 hours
½ hours
Adelaide
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
10 a
12 a
15:13 Wednesday
b
01:07 Sunday
b
50 × 4 × 12 = 200 × 12 = 2400
c
05:28 Tuesday
c
25 × 4 × 19 = 100 × 19 = 1900
d
18:39 Thursday
13 a
5
b
3
c
5
d
4
e
8
f
3
Unit 17 Number and
the laws of arithmetic
Exercise 17.1
Unit 18 Position and
direction
Focus
Exercise 18.1
a
2
4
1
(0, 2) (0, 3) (1, 2), (1, 3) (2, 2) (2, 3)
3
5 × 5 × 5 × 3 (numbers can be in any order)
2
a
4
a
True
46
b
b
c
Focus
1
False
c
0
True
16
y
5
4
Practice
3
5
6
7
8
9
a
Missing numbers are 2, 10, 70.
2
b
Missing numbers are 20, 1, 8, 168.
1
48 × 19 = 48 × 20 − 48 × 1
= 960 − 48
= 912
Possible answers include:
2 + 3 + 5 = 10
2 × 3 × 5 = 30
2 + 3 × 5 = 17
3 + 2 × 5 = 13
5 + 2 × 3 = 11
a
8 + 12 ÷ 3 = 12
b
5 × 9 − 3 = 42
c
7+6÷3=9
d
10 ÷ 2 − 2 = 3
Any reasoned answer, for example:
Tara’s method because it is easy provided you
know that 25 × 4 = 100.
Challenge
10 a
Yes, because 3 × 4 = 12.
b
Yes, because 4 × 2 = 8.
c
No, this does not equal 12 × 8.
11 (Learners may work differently.)
21
5 × 4 × 17 = 20 × 17 = 340
a
39 × 7 = 40 × 7 − 1 × 7 = 280 − 7 = 273
b
38 × 8 = 40 × 8 − 2 × 8 = 320 − 16 = 304
c
29 × 7 = 30 × 7 − 1 × 7 = 210 −7 = 203
0
3
0
1
2
3
4
5 x
b
(4, 1) and (4, 4)
a
3 squares right
b
2 squares down
c
3 squares left and 1 square up
Practice
4
(4, 1)
5
a
6
True, False, False, True
7
right 2, down 2 left 5, down 2 right 1, down 2
(30, 0)
b
(10, 50)
right 12, up 1 right 3, up 3
c
(20, 20)
left 2, up 0
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CAMBRIDGE PRIMARY MATHEMATICS 5: TEACHER’S RESOURCE
Challenge
8
(7, 5) (7, 6) (6, 5) (6, 6) and (7, 7)
9
(3, 0) (3, 1) (3, 3) (3, 4) (3, 5) (2, 4) (2, 0)
(4, 4) (4, 0)
10 More than one solution, for example:
9 right, 1 down
2 left, 4 down
3 right, 4 down
5 right, 4 up
Start
Finish
22
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