RD Sharma Solutions For Class 7 Maths Exercise 3.3 Chapter 3 Decimals

RD Sharma Class 7 Solutions Chapter 3 Ex 3.3 PDF Free Download

The pdf of RD Sharma Solutions for Exercise 3.3 of Class 7 Maths Chapter 3 Decimals are provided here. The questions present have been solved by BYJU’S experts in Maths, and this will help students solve the problems without any difficulties. This exercise contains ten questions with many sub-questions. By practising sincerely, the students can obtain worthy results in Class 7 exams by using the RD Sharma Solutions. By practising the RD Sharma Solutions for Class 7, students will be thorough about their concepts. This exercise deals with the division of decimals.

Download the PDF of RD Sharma Solutions For Class 7 Chapter 3 – Decimals Exercise 3.3

 

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rd sharma class 7 maths chapter 3
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rd sharma class 7 maths chapter 3

 

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 3 – Decimals Exercise 3.3

1. Divide:

(i) 142.45 by 10

(ii) 54.25 by 10

(iii) 3.45 by 10

(iv) 0.57 by 10

(v) 0.0043 by 10

(vi) 0.004 by 10

Solution:

(i) Given 142.45 by 10

Now shifting the decimal point by one place to the left we can get the result

142.45/10 = 14.245

(ii) Given 54.25 by 10

Now shifting the decimal point by one place to the left we can get the result

54.25/10 = 5.425

(iii) Given 3.45 by 10

Now shifting the decimal point by one place to the left we can get the result

3.45/10 = 0.345

(iv) Given 0.57 by 10

Now shifting the decimal point by one place to the left we can get the result

0.57/10 = 0.057

(v) Given 0.0043 by 10

Now shifting the decimal point by one place to the left we can get the result

0.0043/10 = 0.0043

(vi) Given 0.004 by 10

Now shifting the decimal point by one place to the left we can get the result

0.004/10 = 0.0004

2. Divide:

(i) 459.5 by 100

(ii) 74.3 by 100

(iii) 5.8 by 100

(iv) 0.7 by 100

(v) 0.48 by 100

(vi) 0.03 by 100

Solution:

(i) Given 459.5 by 100

Now shifting the decimal point by two places to the left we can get the result

459.5/100 = 4.595

(ii) Given 74.3 by 100

Now shifting the decimal point by two places to the left we can get the result

74.3/100 = 0.743

(iii) Given 5.8 by 100

Now shifting the decimal point by two places to the left we can get the result

5.8/100 = 0.058

(iv) Given 0.7 by 100

Now shifting the decimal point by two places to the left we can get the result

0.7/100 = 0.007

(v) Given 0.48 by 100

Now shifting the decimal point by two places to the left we can get the result

0.48/100 = 0.0048

(vi) Given 0.03 by 100

Now shifting the decimal point by two places to the left we can get the result

0.03 / 100 = 0.0003

3. Divide:

(i) 235. 41 by 1000

(ii) 29.5 by 1000

(iii) 3.8 by 1000

(iv) 0.7 by 1000

Solution:

(i) Given 235. 41 by 1000

Now shifting the decimal point by three places to the left we can get the result

235. 41/1000 = 0.23541

(ii) Given 29.5 by 1000

Now shifting the decimal point by three places to the left we can get the result

29.5/1000 = 0.0295

(iii) Given 3.8 by 1000

Now shifting the decimal point by three places to the left we can get the result

3.8 /1000 = 0.0038

(iv) Given 0.7 by 1000

Now shifting the decimal point by three places to the left we can get the result

0.7 /1000 = 0.0007

4. Divide:

(i) 0.45 by 9

(ii) 217.44 by 18

(iii) 319.2 by 2.28

(iv) 40.32 by 9.6

(v) 0.765 by 0.9

(vi) 0.768 by 1.6

Solution:

(i) Given 0.45 by 9

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 1

0.45 by 9 = 0.05

(ii) Given 217.44 by 18

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 2

217.44 by 18 = 12.08

3. Given 319.2 by 2.28

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 3

319.2 by 2.28 = 140

(iv) Given 40.32 by 9.6

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 4

40.32 by 9.6 = 4.2

(v) Given 0.765 by 0.9

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 5

0.765 by 0.9 = 0.85

(vi) Given 0.768 by 1.6

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 6

0.768 by 1.6 = 0.48

5. Divide:

(i) 16.64 by 20

(ii) 0.192 by 12

(iii) 163.44 by 24

(iv) 403.2 by 96

(v) 16.344 by 12

(vi) 31.92 by 228

Solution:

(i) Given 16.64 by 20

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 7

16.64 by 20 = 0.832

(ii) Given 0.192 by 12

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 8

0.192 by 12 = 0.016

(iii) Given 163.44 by 24

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 9

163.44 by 24 = 6.81

(iv) Given 403.2 by 96

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 10

403.2 by 96 = 4.2

(v) Given 16.344 by 12

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 11

16.344 by 12 = 1.362

(vi) Given 31.92 by 228

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 12

31.92 by 228 = 0.14

6. Divide:

(i) 15.68 by 20

(ii) 164.6 by 200

(iii) 403.80 by 30

Solution:

(i) Given 15.68 by 20

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 13

15.68 by 20 = 0.784

(ii) Given 164.6 by 200

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 14

164.6 by 200 = 0.823

(iii) Given 403.80 by 30

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 15

403.80 by 30 = 13.46

7. Divide:

(i) 76 by 0.019

(ii) 88 by 0.08

(iii) 148 by 0.074

(iv) 7 by 0.014

Solution:

(i) Given 76 by 0.019

Multiply both numerator and denominator by 1000 then divide

= 76000/19

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 16

76 by 0.019 = 4000

(ii) Given 88 by 0.08

Multiply both numerator and denominator by 100 then divide

= 8800/8

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 17

88 by 0.08 = 1100

(iii) Given 148 by 0.074

Multiply both numerator and denominator by 1000 then divide

= 14800/74

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 18

148 by 0.074 = 2000

(iv) Given 7 by 0.014

Multiply both numerator and denominator by 1000 then divide

= 7000/14

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 19

7 by 0.014 = 500

8. Divide:

(i) 20 by 50

(ii) 8 by 100

(iii) 72 by 576

(iv) 144 by 15

Solution:

(i) Given 20 by 50

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 20

20 by 50 = 0.4

(ii) Given 8 by 100

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 21

8 by 100 = 0.08

(iii) Given 72 by 576

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 22

72 by 576 = 0.125

(iv) Given 144 by 15

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 23

144 by 15 = 9.6

9. A vehicle covers a distance 0f 43.2 km in 2.4 litres of petrol. How much distance will travel in 1 litre of petrol?

Solution:

Given distance covered by vehicle in 2.4 litres of petrol is = 43.2 km

Distance travel by vehicle in 1 litre of petrol = 43.2/2.4

Multiply both numerator and denominator by 10 then we get 432/24

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 24

Therefore 18km can travel in 1litre of petrol.

10. The total weight of some bags of wheat is 1743 kg. If each bag weighs 49.8 kg, how many bags are there?

Solution:

Given total weight of total bags = 1743 kg

Each bag weighs = 49.8 kg

Number of bags = 1743/49.8

Multiply both numerator and denominator by 10 then we get 17430/498

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 25

Therefore number of bags are 35

11. Shikha cuts 50 m of cloth into pieces 0f 1.25 m each. How many pieces does she get?

Solution:

Given that total length of cloth = 50 m

Length of each piece = 1.25 m

Number of cloth piece = 50/1.25

Multiply both numerator and denominator by 100 then we get 5000/125

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 26

Number of cloth piece = 40

12. Each side of a rectangular polygon is 2.5cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have?

Solution:

Given that length of each side of polygon is = 2.5 cm

Perimeter of polygon = 12.5 cm

Number of sides = 12.5/2.5

Multiply both numerator and denominator by 10 then we get 125/25

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 27

Number of sides of polygon is 5

13. The product of two decimals is 42.987. If one of them is 12.46, find the other.

Solution:

Given that product of two decimals = 42.987

One of the number is = 12.46

Another number is = 42.987/12.46

Multiply both numerator and denominator by 1000 then we get 42987/12460

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 28

Another number is 3.45

14. The weight of 34 bags of sugar is 3483.3 kg. If all bags weigh equally, find the weight of each bag.

Solution:

Given that weight of 34 bags of sugar is = 3483.3 kg

Weight of each bag = 3483.3/34

Multiply both numerator and denominator by 10 then we get 34833/34

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 29

Weight of each bag = 1o2.45 kg

15. How many buckets of equal capacity can be filled from 586.5 litres of water, if each has capacity of 8.5 litres?

Solution:

Given that capacity of each bucket = 8.5 litres

Total water available = 586.5 litres

Number of buckets = 586.5/8.5

Multiply both numerator and denominator by 10 then we get 5865/85

On dividing

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals image 30

Number of buckets = 69


Access other exercises of RD Sharma Solutions For Class 7 Chapter 3 – Decimals

  1. Exercise 3.1 Solutions 10 Questions
  2. Exercise 3.2 Solutions 15 Questions

RD Sharma Solutions for Class 7 Maths Exercise 3.3 Chapter – 3 Decimals

RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals Exercise 3.3 has problems which are based on the division of decimals. Some of the topics focused prior to this exercise are listed below.

  • Dividing a decimal by 10, 100, 1000
  • Dividing a decimal by the whole number
  • Dividing a decimal by a decimal

The solutions to all questions in RD Sharma books are given here in a detailed and step by step way to help the students understand more effectively.

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