Chapter 5 – Decimals

Solutions to Exercise 5A
5A Building understanding
1 58.237: 3
c The 5 in 10.053 is 5 hundreths =
100
2
a the 2 is two tenths =
10 6
2 a The 6 in 23.612 is 6 tenths
3 10
b the 3 is three hundredths =
100 6
b The 6 in 17.46 is 6 hundredths =
7 100
c the 7 is three thousandths =
1000 6
c The 6 in 80.016 is 6 thousandths =
1000
2 36.57:
6
d The 6 in 0.693 is 6 tenths =
a the tenths digit is 5 10
e The 6 in 16.4 is 6 units
b the units digit is 6
6
f The 6 in 8.568 is 6 hundredths =
c the hundredths digit is 7 100
6
d 0.57 ≥ 0.5 so 36.57 is closer to 37 seconds g The 6 in 2.3641 is 6 hundredths =
100
6
3 a 7.6 h The 6 in 11.926 is 6 thousandths =
1000
b 12.9
3
3 a = 0.3
c 33.04 10
8
d 26.15 b = 0.8
10
e 8.42 15 1 5
c = +
100 10 100
f 99.012
= 0.15

Exercise 5A 23 2 3
d = +
100 10 100
3
1 a The 5 in 1.57 is 5 tenths = = 0.23
10
3 9
b The 5 in 3.215 is 5 thousanths = e = 0.9
1000 10

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210 CambridgeMATHS NSW Stage 4 Year 7 Third Edition

2 5 a 4.7 > 4.2

f = 0.02
100
121 1 2 1 b 3.65 < 3.68
g = + +
1000 10 100 1000
c 1.05 > 1.03
= 0.121
d 5.32 < 5.4
74 7 4
h = +
1000 100 1000 e 6.17 < 6.2
= 0.074
f 8.25 > 8.19

4 a 6
4
=6+
4 g 2.34 > 2.3
10 10
= 6.4 h 5.002 < 5.01

7 7 i 6.03 > 6.00

b 5 =5+
10 10
j 8.21 > 7.3
= 5.7

3 3 k 9.204 > 8.402

c 212 = 212 +
10 10
l 4.11 > 4.111
= 212.3

16 1 6 6 a F: 7.24 > 7.18

d 1 =1+ +
100 10 100
= 1.16 b F: 21.32 > 20.89

83 8 8 c T: 4.61 > 4.57

e 14 = 14 + +
100 10 100
= 14.83 d F: 8.09 < 8.41

51 5 1 e T: 25.8 ≤ 28.5
f 7 =7+ +
100 10 100
f T: 2.1118 ≤ 2.8001
= 7.51
g F: 7.93 < 8.42
7 7
g 5 =5+
100 100 h T: 11.11 ≥ 11.109
= 5.07
3 30
i T: =
612 6 1 2 10 100
h 18 = 18 + + +
100 10 100 1000 7 70
= 18.612 j T: =
10 100

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 Chapter 5 worked solutions 211

5 1 11.87 and 17.81 have 1 in the tens

k T: = ,5
10 2 column, but 1 < 7 in the units column:
2 20 1.718 < 1.871 < 11.87 < 17.81
l F: =
10 100
d 22.69 and 22.96 have 2 in the tens and units
columns but 22.69 has 6 in the tenths column
7 a 7.0 − 6.9 = 0.1
where 22.96 has 9.
b 7.03 − 7.00 = 0.03 26.92 has 6 in the units column, while 29.26
and 29.62 have 9.
c 19.00 − 18.98 = 0.02 Of the last pair, 29.26 has 2 in the tenths
column where 29.62 has 6:
d 17.0 − 16.5 = 16.5 − 16.0 22.69 < 22.96 < 26.92 < 29.26
= 0.5 < 29.62

e 18.000 − 17.999 = 0.001 9 a Border and Waugh both have 5 in the tens
column, but Waugh has 1 in the units as
f 5.00 − 4.99 = 0.01 against 0 for Border.
Gilchrist and Taylor have 4 in the tens column,
g 1.00 − 0.85 = 0.15
but Gilchrist has 7 in the units as against 3 for
h 99.11 − 99.00 = 0.11 Taylor.
Hughes has 3 in the tens column:
Waugh > Border > Gilchrist > Taylor >
8 a All numbers have 3 in the units column. In the Hughes
tenths column 3.05 has 0, 3.25 has 2 and the
others have 5. b Since Ponting has 5 in the tens and 6 in the
In the hundredths column 3.52 has 2 and 3.55 units, he is placed first.
has 5:
3.05 < 3.25 < 3.52 < 3.55
10 a On day 6 the tenths column is 6, as against 4
b 3.06 and 3.6 have 3 in the units column but and 5 the other days.
3.06 has 0 in the tenths column.
b On days 3 and 4 the tenths column
30.3 and 30.6 have 3 in the tens column and 0
is 4.
in the units, but 3 < 6 in the tenths column:
The hundredths column has 9 on day 3 and 7
3.06 < 3.6 < 30.3 < 30.6
on day 4, so the level is lowest on day 4.
c 1.718 and 1.871 have 1 in the units column but
c On days 2, 5 and 6 the river level is higher
1.718 has 7 in the tenths column where 1.871
than the previous day.
has 8.

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212 CambridgeMATHS NSW Stage 4 Year 7 Third Edition

11 When you insert a ‘0’ at the start or the end A

13 a = 0.A
you do not change the place value of any other 10
digits, but when you insert a ‘0’ in the middle A
b = 0.0A
somewhere, you do change the place value. For 100
example:
6 A A
In 15.26, the value of the 6 is but in c + = 0.AA
100 10 100
6 A A
15.206, the value of the 6 is . d A+ + = A.A0A
1000 10 1000
In 2.32, the value of the 2 is 2 units but if you
insert a 0 in 20.32, the value of the 2 is 2 tens.
14 a i 2 possible arrangements: 0.1, 1.0

12 a C is the smallest digit so C.C and C.A are the ii 10 possible arrangements:
smallest numbers. 0.12, 0.21, 1.02, 1.20, 2.01,
Since C < A, C.C is the smallest number. B.C 2.10, 10.2, 12.0, 20.1, 21.0
and B.A are next, with B.C < B.A.
iii 60 possible arrangements:
A is the largest digit, so A.C and A.B are last,
0.123, 0.132, 0.213, 0.231, 0.312,
with A.C < A.B:
C.C < C.A < B.C < B.A < A.C < A.B 0.321, 1.023, 1.032, 1.203, 1.230,
1.302, 1.320, 2.013, 2.031, 2.103,
b As before, where there is C in the units
column these come first. 2.130, 2.301, 2.310, 3.012, 3.021,
Since B < A,C.BC < C.AB. B.CA and B.BB 3.102, 3.120, 3.201, 3.210, 10.23,
have B in the units column but C < B, so B.CA
10.32, 12.13, 12.31, 13.02, 13.20,
comes first.
These are followed by those with A. Since C 20.13, 20.31, 21.03, 21.30, 23.01,
is the smallest digit, A.CA is next, followed by 23.10, 30.12, 30.21, 31.02, 31.20,
A.BC and then A.AA. BA.CA and AB.AB are 32.01, 32.10, 102.3, 103.2, 120.3,
last since they have digits in the tens column.
B < A so AB.AB is the biggest number: 123.0, 130.2, 132.0, 201.3, 203.1,
C.BC < C.AB < B.CA < B.BB < A.CA < A. 210.3, 213.0, 230.1, 231.0, 301.2,
BC < A.AA < BA.CA < AB.AB 302.1, 310.2, 312.0, 320.1, 321.0,

b 408 ways

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