DECIMALS
4
1. Round 8356.9574 to the nearest :
(i) thousands
(ii)
hundreds
(iii) tens
(iv) units
(v) tenths
(iv) hundredths (vii) thousandths
Ans. Rounded 8356.9574 to the nearest :
(i) thousands = 8000
(ii)
hundreds = 8400
(iii) tens = 8360
(iv) units = 8357
2. Round 4.96075 correct to :
(i) one significant figure
(ii) two significant figures
(iii) three significant figures
(iv) four significant figures.
Ans. Rounded 4.96075 correct to :
(i) one significant figure = 5
(ii) one significant figure = 5.0
(iii) three significant figures = 4.96
(iv) four significant figures = 4.961
3. Convert each of the following into a vulgar fraction :
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(i) 0.34
(iv) 0.1234
Ans. (i)
(ii)
(ii)
(iii)
0.367
(v)
0.6789
Let x = 0.34 = 0.343434
Multiply both sides by 100.
100x = 34.343434
Subtracting (i) from (ii)
99x = 34
34
x=
⇒
99
34
Hence, fraction =
99
Let x = 0.367 = 0.367367367
(Multiply both sides by 1000)
10000x = 367.367367367…
Subtracting (i) from (ii)
ICSE Math Class VII
6.379
1
…(i)
…(ii)
…(i)
…(ii)
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999 x = 367
Hence fraction =
⇒
x=
367
999
367
999
(iii) 6.379
Let
x = 6.379 = 6.379379379
Multiply both sides by 1000
1000x = 6379.379379
Subtracting (i) from (ii)
999 x = 6373
6373 6373
=
999
999
6379
Hence fraction =
999
⇒
…(i)
…(ii)
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(iv) 0.12.34
Let
x = 0.1234 = 0.12343434
Multiply both sides by 100 we have
100 x = 12.343434
Multiply both sides by 100 we have
and 10000 x = 1234.343434
Subtracting (i) from (ii)
(10000 – 100) x = 1234 – 12
9900 x = 1222
1222
611
=
9900 4950
611
Hence, fraction =
4950
(v) 0.6789
Let
x = 0.6789 = 0.6789789789.........
Multiply both sides by 10
10 x = 6.789789789........
∴
and 10000x = 6789.789789789.
Subtracting (i) from (ii)
(10000 – 10) x = 6789 – 6
⇒
ICSE Math Class VII
…(i)
…(ii)
x=
2
......(i)
…(ii)
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9990 x = 6783
x=
2261
3330
Hence, fraction =
2261
3330
⇒
⇒ x=
4. Simplify :
(i) 32.8 – 13 – 10.725 + 3.517
(iii) 0.1835 + 163. 2005 – 25.9 – 100
6783
9990
(ii)
4000 – 30.51 – 753.101 – 69.43
(iv)
38.00 – 30 + 200.200 – 0.230
(v) 555.555 + 55.555 – 5.55 – 0.555
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Ans. (i) 32.8 – 13 – 10.725 + 3.517 = (32.8 + 3.517) – (13 + 10.725)
= 36.317 – 32.725 = 12.592
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(ii) 4000 – 30.51 – 753.101 – 69.43 = 4000 – (30.51 + 753.101 + 69.43)
= 4000 – 853.041 = 3146.959
(iii) 0.1835 + 163.2005 – 25.9 – 100 = (0.1835 + 163.2005) – (25.9 + 100)
= 163.3840 – 125.9 = 37.484
(iv) 38.00 – 30 + 200.200 – 0.230 = (38.00 + 200.200) – (30 + 0.230)
= 238.200 – 30.230 = 207.970 = 207.97
(v) 555.555 + 55.555 – 5.55 – 0.555 = (555.555 + 55.555) – (5.55 + 0.555)
= 611.110 – 6.105 = 605.005
5. Simplify :
(i) 16.648 + 2.002 – 13.79
(iii) 83 – 64.37 – 9.047
(v) 1 – 10.3 + 10.2131
Ans. (i) Now,
16.648
+2.002
18.650
18.650
(ii)
(iv)
(ii)Now,
0.1645
+72.6400
72.8045
72.8045
–25.9993
46.8052
–13.790
4.860
ICSE Math Class VII
0.1645 + 72.64 – 25.9993
666 – 66.6 – 6.666
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(iii) 83 – 64.37 – 9.047
= 83.000 – (64.370 + 9.047)
= 83.000 – 73.417 = 9.583
(iv) 666 – 66.6 – 6.666
= 666.000 – (66.600 + 6.666) = 666.000 – 73.266 = 592.734
(v) 1 – 10.3 + 10.2131
= 1.0000 + 10.2131 – 10.3000 = 11.2131 – 10.3000 = 0.9131
6. Find the products :
(i) 2.3 × 0.23 × 0.01
(ii)
0.2 × 0.02 × 0.002
(iii) 1.1 × 0.11 × 11.11
(iv) 2.3 × 0.23 × 0.023 (v) .0.04 × 0.89 × 0.037
Ans. (i) 2.3 × 0.23 × 0.01= 0.00529
(ii) 0.2 × 0.02 × 0.002 = 2 × 2 × 2 = 8 = 0 .000008
(iii) 1.1 × 0.11 × 11. 11 = 1.34431
(iv) 2.3 × 0.23 × 0.023 = 0.012167
(v) 0.04 × 0.89 × 0.037 ⇒ 0.04 × 0.89 × 0.037 = 0.0013172
7. Evalute :
(i) 0.01 × 0.001 (ii)
4.75 × 0.08 × 32
(iii) (2.1)2 × (1.5)2 (iv) 9.025 × (0.2)2 × 42
(v) 12.003 × (0.2)5
Ans. (i) 0.01 × 0.001 = 0.00001
(ii) 4.75 × 0.08 × 32 = 4.75 × 0.08 × 9 = 4.75 × 0.72
= 3.4200 = 3.42
2
2
(iii) (2.1) × (1.5) = 2.1 × 2.1 × 1.5 × 1.5
= 4.41 × 2.25 = 9.9225
(iv) 9.025 × (0.2)2 × 42 = 9.025 × 0.04 × 16
= 9.025 × 64 = 5.77600 = 5.776
(v) 12.003 × (0.2)5
= 12.003 × 0.2 × 0.2 × 0.2 × 0.2 × 0.2
= 12.003 × 0.00032 = 0.00384096
8. Divide :
(i) 0.1575 ÷ 0.21
(ii)
0.0144 ÷ 0.12
(iii) 6.612 ÷ 11.6
(iv) 1.2798 ÷ 5.4
(v)
0.1164 ÷ 0.012
(vi) 0.068 ÷ 0.17
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Ans. (i)
0.1575 0.1575 × 100 15.75
=
=
= 0.75
0.21
0.21× 100
21
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(ii)
0.0144 0.0144 × 100 1.44
=
=
= 0.12
0.12
0.12 ×100
12
Hence, 1.44 ÷ 2 = 0.12
6.612 6.612 ×10 66.12
=
=
= 0.57
11.6
11.6 ×10
116
Hence, 6.612 ÷ 11.6 = 0.57
(iv) 1.2798 ÷ 5.4
Converting the divisor 5.4 into whole number by multiplying it with 10, then
(iii)
1.2798 1.2798 × 10 12.798
=
=
= 0.237
5.4
5.4 × 10
54
Hence, 1.2798 ÷ 5.4 = 0.237
=
(v)
(vi)
0.1164 0.1164 × 1000 116.4
=
=
= 9.7
0.012
0.012 × 1000
12
Hence, 0.1164 ÷ 0.012 = 9.7
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0.068 0.068 ×100 6.8
=
=
= 0.4
0.17
0.17 × 100 17
Hence, 0.068 ÷ 0.17 = 0.4
9. Find the value of
(i) 0.5 of Rs 7.60 + 1.62 of Rs 30
(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
(iii) 6.25 of 8.4 – 4.7 of 3.24
(iv) 0.98 of 235 – 0.09 of 3.2
Ans. (i) 0.5 of Rs 7.60 + 1.6 + 1.62 of Rs 30
= Rs 3.80 + Rs 848 .60
= Rs 52.40
(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
= 16.79 kg + 0.432 kg = 17.222 kg
(iii) 6.25 of 8.4 – 4.7 of 3.24
= 52.500 – 15.228 = 37.272
(iv) 0.98 of 235 – 0.09 of 3.2 = 230.30 – 0.288 = 230.012
10. Simplify :
6.3
(i)
0.3 × 0.1
ICSE Math Class VII
(ii)
0.04 × 0.09
0.03 × 0.03
5
(iii)
1.3 × 2.4
0.39
(iv)
(0.08) 2
(0.4)3
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(7.5) 2 × 0.02
(v)
1.5 × 0.375
Ans. (i)
(vi)
2.5 × 40.4
50
(vii)
 4.5 × 0.25 

 – (0.03 × 10.4)
 0.225 
6.3
63 × 10 × 10
= 21 × 10 = 210
=
0.3 × 0.1
3 × 1 × 10
(ii) 0.04 × 0.09 = 4 × 9 × 100 × 100 = 4
0.03 × 0.03 3 × 3 × 100 × 100
(iii) 1.3 × 2.4 = 13 × 24 × 100 = 24 = 8
0.39
39 × 10 × 10
3
(iv)
(0.08)2
3
(0.4)
=
8 × 8 × 10 × 10 × 10
1
0.08 × 0.08
=
= 0.1
=
0.4 × 0.4 × 0.4 4 × 4 × 4 × 100 × 100 10
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2
75 × 75 × 2 × 10 × 1000
(v) (7.5) × 0.02 = 7.5 × 7.5 × 0.02 =
=2
15 × 375 × 10 × 10 × 100
1.5 × 0.375
1.5 × 0.375
(vi)
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2.5 × 40.4
25 × 404
=
= 2.02
50
50 × 10 × 10
 4.5 × 0.25 
(vii) 
 – (0.03 ×10.4)
 0.225 
 45 × 25 × 1000   3 104 
×
=
–

 10 × 100 × 225   100 10 
312
= 5.000 – 0.321 = 4.688
1000
Find the difference between 6.85 and 0.685.
Difference between 6.85 and 0.685 = 6.85 – 0.685 = 6.165
Take out the sum of 19.38 and 56.025 from 200.111.
Sum = 19.38 + 56.025 = 75.405
Difference of 200.111 and 75.405 = 200.111 – 75.405 = 124.706
Add 13.95 and 1.003 ; and from the result, subtract the sum of 2.794 and 6.2.
= 13.95 + 1.003 = 14.953
Sum of 13.95 + 1.003 = 14.953
Sum of 2.794 + 6.2 = 8.999
Difference = 14.953 – 8994 = 5.959.
What should be added to 39.587 to give 80.375 ?
Sum = 80.375
=5–
11.
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13.
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14.
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15.
Ans.
16.
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17.
Ans.
18.
Ans.
Given number = 39.587
Hence the number which is to be added = 80.375 – 39.587 = 40.788
What should be subtracted from 100 to give 19.29 ?
Sum = 100. The number = 19.29
Thus, the number which is to be subtracted
= 100 – 19.29 = 80.71
What is excess of 584. 29 over 213.95 ?
Given number = 231. 95
Hence, required difference = 584.29 – 213.95 = 370.34
What should be added to 63.42 to get 71 ?
Let x should be added
63.42 + x = 71
x = 71 – 63.42 = 7.58
⇒
Thus, required addition = 7.58
What number subtracted from 18.5 gives 6.2376 ?
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18.5000
– 6.2376
12.2624
Thus, 12.2624 must be subtracted from 18.5 to get 6.2376
19. If 2.25 m2 of a cloth cost Rs. 326.25, find the cost of 1m2 of this cloth.
Ans. Cost of 2.25 m2 cloth = Rs 326.25
∴ Cost of 1m2 cloth = Rs 326.25 ÷ 2.25
326.25 32625 × 100
=
2.25
100 × 225
32625
= Rs
= Rs 145
225
20. A piece of cloth is 24.5 m long. How many pieces each of length 1.75 m can be cut
from it ?
Ans. Full length of cloth = 24.5 m
Length of one piece = 1.75 m
=
24.5 2450
=
= 14
1.75 175
21. If the speed of a car is 66.53 km /hr, find the distance covered by it in 3.7 hours.
Ans. Speed of a car = 66.53 km /hr
Number of pieces which can be cut from it = 24.5 ÷ 1.75 =
ICSE Math Class VII
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Distance covered = ? Time = 3.7 hours
D = S × T = 66.53 × 3.7 = 246.161 km
∴
22. If a car covers a distance of 239.166 km in 4.3 hours, find the average speed of the
car.
Ans. Distance covered = 239.166 km. Time taken = 4.3 hours
Distance 239.166
= 55.62 km/hr
=
Time
4.3
A room is 3.25 m in length and 2.72 m in breadth. Calculate the area of the floor of
the room.
Length of room = 2.72m
Hence,
Area = l × b = 3.25 × 2.72 m2
= 8.8400 m2 = 8.84 m2
A car moves at a constant speed of 42.5 km/hr. Find the distance it would cover in
3.5 hours.
Speed of car = 42.5 km /hr. Time taken = 3.5 hour
∴ Hence, distance covered = Speed × Time = 42.5 × 3.5 km = 148.75 km
The cost of one pen is Rs. 42.25. Find the cost of one dozen such pens.
Cost of 1 pen = Rs 42.25
∴Cost of one dozen 12 pens = Rs 42.25 × 12 = Rs 507.00 = Rs 507
Find the cost of 17.5 m cloth at the rate of Rs 112.50 per metre.
Cost of one metre cloth = Rs 112.50
∴Cost of 17.5 m cloth = Rs 112.50 × 17.5 = Rs 1968.750 = 1968.750
One kilogram of oil costs Rs 73.40. Find the cost of 9.75 kilogram of the oil.
Cost of 1 kg oil = Rs 73.40
∴Cost of 9.75 kg oil = Rs 73.40 × 9.75 = Rs 715.6500 = Rs715.65
Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.
Weight of 8 objects =51.2 kg
∴Weight of 1 object = 51.2 ÷ 8 kg = 6.4 kg
18.5 m of cloth costs Rs. 666. Find the cost of 3.8 m cloth.
Cost of 18.5 m cloth = Rs 666
∴ Cost of 1 m cloth = Rs 666 ÷ 18.5 and cost of 3.8 m cloth
= Rs (666 ÷ 18.5) × 3.8
= Rs (666 ÷ 185) × 3.8
= Rs 36 × 3.8 = Rs 136.80
Average speed =
23.
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25.
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29.
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30. If a man travels a distance of 276.75 km by his car in 4.5 hours, find his average
speed.
Ans. Distance covered in 4.5 hours = 276.75 km
Thus, distance covered in 1 hour =
276.75
km
4.5
27675 × 10 27675
=
= 61.5 km
100 × 45
450
Hence, spead of car = 61.5 km/ hour
=
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