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RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7 PDF Download

Q: 5. How can I add and subtract decimal numbers?

Ans. To add or subtract decimal numbers, you can follow these steps:1. Align the decimal points of the numbers.2. Add or subtract the digits in each place value column, starting from the right.3. Carry over any values greater than 9 to the next column.4. If one number has more digits after the decimal point, continue the pattern of zeros and add or subtract as usual.5. Simplify the result, if necessary.For example, to add 3.12 and 1.5, we align the decimal points and add the digits in each place value column: 3 + 1 = 4, 1 + 5 = 6, and 2 + 0 (padding zero) = 2. The sum is 4.62.

Q.1. Divide:
(i) 131.6 by 10
(ii) 32.56 by 10
(iii) 4.38 by 10
(iv) 0.34 by 10
(v) 0.08 by 10
(vi) 0.062 by 10
Ans. We have the following:
(i) 131.6 ÷ 10 = 131.610=13.16131.610=13.16 [Shift the decimal point to the left by 1 place]
(ii) 32.56 ÷ 10 = 32.5610=3.25632.5610=3.256 [Shift the decimal point to the left by 1 place]
(iii) 4.38 ÷ 10 = 4.3810=0.4384.3810=0.438 [Shift the decimal point to the left by 1 place]
(iv) 0.34 ÷ 10 = 0.3410=0.0340.3410=0.034 [Shift the decimal point to the left by 1 place]
(v) 0.08 ÷ 10 = 0.0810=0.0080.0810=0.008 [Shift the decimal point to the left by 1 place]
(vi) 0.062 ÷ 10 = 0.06210=0.00620.06210=0.0062 [Shift the decimal point to the left by 1 place]

Q.2. Divide:
(i) 137.2 by 100
(ii) 23.4 by 100
(iii) 4.7 by 100
(iv) 0.3 by 100
(v) 0.58 by 100
(vi) 0.02 by 100
Ans. We have the following:
(i) 137.2 ÷ 100 = 137.2100=1.372137.2100=1.372 [Shifting the decimal point to the left by 2 places]
(ii) 23.4 ÷ 100 =23.4100=0.23423.4100=0.234 [Shifting the decimal point to the left by 2 places]
(iii) 4.7 ÷ 100 = 4.7100=0.0474.7100=0.047 [Shifting the decimal point to the left by 2 places]
(iv) 0.3 ÷ 100 = 0.3100=0.0030.3100=0.003 [Shifting the decimal point to the left by 2 places]
(v) 0.58 ÷ 100 = 0.58100=0.00580.58100=0.0058 [Shifting the decimal point to the left by 2 places]
(vi) 0.02 ÷ 100 = 0.02100=0.00020.02100=0.0002 [Shifting the decimal point to the left by 2 places]

Q.3. Divide:
(i) 1286.5 by 1000
(ii) 354.16 by 1000
(iii) 38.9 by 1000
(iv) 4.6 by 1000
(v) 0.8 by 1000
(vi) 2 by 1000
Ans. We have the following:
(i) 1286.5 ÷ 1000 = 1286.51000=1.28651286.51000=1.2865 [Shift the decimal point to the left by 3 places]
(ii) 354.16 ÷ 1000 = 354.161000=0.35416354.161000=0.35416 [Shift the decimal point to the left by 3 places]
(iii) 38.9 ÷ 1000 = 38.91000=0.038938.91000=0.0389 [Shift the decimal point to the left by 3 places]
(iv) 4.6 ÷ 1000 = 4.61000=0.00464.61000=0.0046 [Shift the decimal point to the left by 3 places]
(v) 0.8 ÷ 1000 = 0.81000=0.00080.81000=0.0008 [Shift the decimal point to the left by 3 places]
(vi) 2 ÷ 1000 = 21000=0.00221000=0.002 [Shift the decimal point to the left by 3 places]

Q.4. Divide:
(i) 12 by 8
(ii) 63 by 15
(iii) 47 by 20
(iv) 101 by 25
(v) 31 by 40
(vi) 11 by 16
Ans. (i) 12 ÷ 8 = 12/8=3/2
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 12 ÷ 8 = 1.5
(ii) 63 ÷ 15 = 63/15=21/5

∴ 63 ÷ 15 = 4.2
(iii) 47 ÷ 20 = 47/20

∴ 47 ÷ 20 = 2.35
(iv) 101 ÷ 25 = 101/25

∴ 101 ÷ 25 = 4.04
(v ) 31 ÷ 40

∴ 31 ÷ 40 = 0.775
(vi) 11 ÷ 16 = 11/16

∴ 11 ÷ 16 = 0.6875

Q.5. Divide:
(i) 43.2 by 6
(ii) 60.48 by 12
(iii) 117.6 by 21
(iv) 217.44 by 18
(v) 2.575 by 25
(vi) 6.08 by 8
(vii) 0.765 by 9
(viii) 0.768 by 16
(ix) 0.175 by 25
(x) 0.3322 by 11
(xi) 2.13 by 15
(xii) 6.54 by 12
(xiii) 5.52 by 16
(xiv) 1.001 by 14
(xv) 0.477 by 18
Ans. (i) We have:
43.2 ÷ 6
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 43.2 ÷ 6 = 7.2
(ii) We have:
60.48 ÷ 12

∴ 60.48 ÷ 12 = 5.04
(iii) We have:
117.6 ÷ 21

∴ 117.6 ÷ 21 = 5.6
(iv) We have:
217.44 ÷ 18

∴ 217.44 ÷ 18 = 12.08
(v) We have:
2.575 ÷ 25

∴ 2.575 ÷ 25 = 0.103
(vi) We have:
6.08 ÷ 8
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 6.08 ÷ 8 = 0.76
(vii) We have:
0.765 ÷ 9

∴ 0.765 ÷ 9 = 0.085
(viii) We have:
0.768 ÷ 16

∴ 0.768 ÷ 16 = 0.048
(ix) We have:
0.175 ÷ 25
= 0.175/25
= (0.175×1000) / (25×1000)
= 175 / (25×1000)
= 7/1000
=0.007
(x) We have: 0.3322 ÷ 11
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 0.3322 ÷ 11 = 0.0302
(xi) We have:
2.13 ÷ 15

∴ 2.13 ÷ 15 = 0.142
(xii) We have:
6.54 ÷ 12

∴ 6.54 ÷ 12 = 0.545
(xiii) We have:
5.52 ÷ 16

∴ 5.52 ÷ 16 = 0.345
(xiv) We have:
1.001 ÷ 14

∴ 1.001 ÷ 14 = 0.0715
(xv) We have:
0.477 ÷ 18
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 0.477 ÷ 18 = 0.0265

Q.6. Divide:
(i) 16.46 ÷ 20
(ii) 403.8 ÷ 30
(iii) 19.2 ÷ 80
(iv) 156.8 ÷ 200
(v) 12.8 ÷ 500
(vi) 18.08 ÷ 400
Ans. (i) 16.46 ÷ 20
= 16.46/20
= (16.46 × 100) / (20×100)
= 1646 / (2×1000)
= 823 / 1000
= 0.823
(ii) 403.8 ÷ 30 = 403.8 / 30
= (403.8×10) / (30×10)
= 4038 / (3×100)
=1346 / 100
=13.46
(iii) 19.2 ÷ 80
= 19.2/80
= 19.2×10/80×10
= 192/800
= 192/8×100
= 24/100
= 0.24
(iv) 156.8 ÷ 200
= 156.8/200
=156.8×10/200×10
=1568/2000
=784/1000
=0.784
(v) 12.8 ÷ 500
= 12.8/500
= (12.8×10) / (500×10)
= 128/5000
= 25.6/1000
= 0.0256
(vi) 18.08 ÷ 400
= 18.08/400
= 18.08×100 / 400×100
= 1808/40000
= 452/10000
=0.0452

Q.7. Divide:
(i) 3.28 by 0.8
(ii) 0.288 by 0.9
(iii) 25.395 by 1.5
(iv) 2.0484 by 0.18
(v) 0.228 by 0.38
(vi) 0.8085 by 0.35
(vii) 21.976 by 1.64
(viii) 11.04 by 1.6
(ix) 6.612 by 11.6
(x) 0.076 by 0.19
(xi) 148 by 0.074
(xii) 16.578 by 5.4
(xiii) 28 by 0.56
(xiv) 204 by 0.17
(xv) 3 by 80
Ans. (i) 3.28 ÷ 0.8
= 3.28/0.8
= (3.28×10) / (0.8×10)
= 32.8/8
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 3.280.8=32.88=4.13.280.8=32.88=4.1
(ii) 0.288 ÷ 0.9
= 0.288/0.9
= (0.288×10) / (0.9×10)
= 2.88/9
Now, we have:

∴ 0.2880.9=2.889=0.320.2880.9=2.889=0.32
(iii) 25.395 ÷ 1.5
= 25.395 / 1.5
= (25.395×10) / (1.5×10)
= 253.95/15
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 25.395/1.5
= 253.95/15
= 16.93
(iv) 2.0484 ÷ 0.18
= 2.0484/0.18
= (2.0484×100) / (0.18×100)
= 204.84/18
Now, we have:

∴ 2.0484/0.18
= 204.84/18
= 11.382.04840.18=204.8418=11.38
(v) 0.228 ÷ 0.38
= 0.228/0.38
= (0.228×100) / (0.38×100)
= 22.8/38
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 0.2280.38=22.838=0.60.2280.38=22.838=0.6
(vi) 0.8085 ÷ 0.35
= 0.8085/0.35
= (0.8085×100) / (0.35×100)
= 80.85/35
Now, we have:

∴0.80850.35=80.8535=2.310.80850.35=80.8535=2.31
(vii) 21.976 ÷ 1.64 = 21.976/1.64
= (21.976×100) / (1.64×100)
= 2197.6/164
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 21.9761.64=2197.6164=13.421.9761.64=2197.6164=13.4
(viii) 11.04 ÷ 1.6
= 11.04/1.6
= (11.04×10) / (1.6×10)
= 110.4/16
Now, we have:

∴11.04/1.6
= 110.4/16
= 6.9
(ix) 6.612 ÷ 11.6
= 6.612/11.6
= (6.612×10) / (11.6×10)
= 66.12/1166
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 6.612/11.6
= 66.12/116
= 0.57
(x) 0.076 ÷ 0.19
= 0.076/0.19
= (0.076×100) / (0.19×100)
= 7.6/19
Now, we have:

∴ 0.076/0.19
= 7.6/19
= 0.4
(xi) 48 ÷ 0.074
= 148/0.074
= 148×1000/0.074×1000
= 148000/74
= 2×1000
=2000
(xii) 16.578 ÷ 5.4
= 16.578/5.4
= (16.578×10)/(5.4×10)
= 165.78/54
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 16.578/5.4
= 165.78/54
= 3.07
(xiii) 28 ÷ 0.56
= 28/0.56
= (28×100) / (0.56×100)
= 2800/56
= 1×100/2
= 50
(xiv) 204 ÷ 0.17
= 2040.17
= (204×100) / (0.17×100)
= 20400/17
= 12×100
= 1200
(xv) 3 ÷ 80 = 3/80
Now, we have:
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
∴ 380380 = 0.0375

Q.8. The total cost of 24 chairs is Rs 9255.60. Find the cost of each chair.
Ans. Cost of 24 chairs = Rs 9255.60
∴ Cost of one chair = Rs (9255.60/24)
= Rs (9255.60×10) / (24×10)
= Rs (92556/240)
= Rs 385.65
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
Hence, the cost of one chair is Rs 385.65.

Q.9. 1.8 m of cloth is required for a shirt. How many such shirts can be made from a piece of cloth 45 m long?
Ans. Cloth required for 1 shirt = 1.8 m
∴ Number of shirts that can be made from 45 m of cloth
= 45/1.8
=15/0.6
=5/0.2
= 50/2
= 25
Hence, 25 shirts can be made from a piece of cloth of length 45 m.

Q.10. A car covers a distance of 22.8 km in 2.4 litres of petrol. How much distance will it cover in 1 litre of petrol?
Ans. Distance covered by the car with 2.4 litres of petrol = 22.8 km
∴ Distance covered with 1 litre of petrol = (22.8/2.4) km
= (228/24) km
= (228÷12) / (24÷12) km
= (19/2) km
= RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7 km
Hence, the distance covered by the car with 1 litre of petrol is 912912 km.

Q.11. A tin holds 16.5 litres of oil. How many such tins will be required to hold 478.5 litres of oil?
Ans. Capacity of 1 tin of oil = 16.5 litres
∴ Number of tins required to hold 478.5 litres of oil = (478.516.5)=(4785165)=(4785÷15165÷15)=31911=29478.516.5=4785165=4785÷15165÷15=31911=29
Hence, 29 oil tins will be required to hold 478.5 litres of oil.

Q.12. The weight of 37 bags of sugar is 3644.5 kg. If all the bags weigh equally, what is the weight of each bag?
Ans. Weight of 37 bags of sugar = 3644.5 kg
∴ Weight of 1 bag of sugar = (3644.5/37)
= 98.5 kg
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
Hence, each bag of sugar weighs 98.5 kg.

Q.13. If 69 buckets of equal capacity can be filled with 586.5 litres of water, what is the capacity of eacch bucket?
Ans. Capacity of 69 buckets of water = 586.5 litres
∴ Capacity of one such bucket = (586.5/69) litres = 8.5 litres.
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
Hence, the capacity of each water bucket is 8.5 litres.

Q.14. Monica cuts 46 m of cloth into peices of 1.15 m each. How many pieces does she get?
Ans. Length of one piece of cloth = 1.15 m
∴ Number of pieces she gets from 46 m of cloth = (46/1.15)
= (46×100) / (1.15×100) = (4600/115) = 40
Hence, Monica has 40 pieces of cloth each of length 1.15 m.

Q.15. Mr Soni bought some bags of cement, each weighing 49.8 kg. If the total weight of all the bags is 1792.8 kg, how many bags did he buy?
Ans. Total weight of all the bags of cement = 1792.8 kg
Weight of each bag = 49.8 kg
Number of bags = (Total weight/Weight of each bag)
= (1792.8/49.8)
= (17928/498)
=36
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
Hence, Mr. Soni bought 36 bags of cement.

Q.16. How many pieces of plywood, each 0.35 cm thick, are required to make a pile 1.89 m high?
Ans. Thickness of the pile of plywood pieces = 1.89 m = 189 cm
Thickness of one piece of plywood = 0.35 cm
∴ Required number of plywood pieces = (189/0.35)
=(189×100) / (0.35×100)
= (18900/35)
= 540
RS Aggarwal Solutions: Decimals (Exercise 3D) | Mathematics (Maths) Class 7
Hence, 540 pieces of plywood are required to make a pile of height 1.89 m.

Q.17. The product of two decimals is 261.36. If one of them is 17.6, find the other.
Ans. Product of the given decimals = 261.36
One decimal = 17.6
The other decimal = 261.36 ÷ 17.6
= (261.36/17.6)
= (261.36×10) / (17.6×10)
=(2613.6/176)
= 14.85
Hence, the other decimal is 14.85.

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FAQs on RS Aggarwal Solutions: Decimals (Exercise 3D) - Mathematics (Maths) Class 7

1. What are the different types of decimal numbers?

Ans. Decimal numbers can be classified into two types: terminating decimals and non-terminating decimals. Terminating decimals have a finite number of digits after the decimal point, while non-terminating decimals have an infinite number of digits after the decimal point.

2. How can I convert a decimal number into a fraction?

Ans. To convert a decimal number into a fraction, you can follow these steps: 1. Write down the decimal number as the numerator. 2. Determine the denominator by writing 1 followed by the same number of zeros as the number of decimal places in the original number. 3. Simplify the fraction if possible. For example, to convert 0.75 into a fraction, we write 0.75 as the numerator and 100 as the denominator. Simplifying the fraction gives us 3/4.

3. What is the significance of place value in decimal numbers?

Ans. Place value plays a crucial role in decimal numbers. Each digit in a decimal number has a specific place value, which determines its worth. The place values are based on powers of 10. For example, in the number 354.78, the digit 3 is in the hundreds place, the digit 5 is in the tens place, the digit 4 is in the ones place, the digit 7 is in the tenths place, and the digit 8 is in the hundredths place.

4. How can I compare decimal numbers?

Ans. To compare decimal numbers, you can follow these steps: 1. Start by comparing the whole number parts of the decimals. The larger whole number indicates the larger decimal. 2. If the whole number parts are the same, compare the decimal parts from left to right. The first digit that is larger in one decimal determines the larger number. 3. If the decimal parts are the same, but one decimal has additional digits, the decimal with more digits is larger. For example, to compare 2.34 and 2.5, we compare the whole number parts first (both are 2), then compare the tenths place (3 < 5), so 2.5 is larger.

5. How can I add and subtract decimal numbers?

Ans. To add or subtract decimal numbers, you can follow these steps: 1. Align the decimal points of the numbers. 2. Add or subtract the digits in each place value column, starting from the right. 3. Carry over any values greater than 9 to the next column. 4. If one number has more digits after the decimal point, continue the pattern of zeros and add or subtract as usual. 5. Simplify the result, if necessary. For example, to add 3.12 and 1.5, we align the decimal points and add the digits in each place value column: 3 + 1 = 4, 1 + 5 = 6, and 2 + 0 (padding zero) = 2. The sum is 4.62.

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