RD Sharma Solutions for Class 6 Chapter 9: Ratio, Proportion and Unitary Method Exercise 9.3

RD Sharma Class 6 Solutions Chapter 9 Ex 9.3 PDF Free Download

RD Sharma Solutions Class 6 Exercise 9.3 helps students understand concepts like proportion and problems based on them. An equality of two ratios is called a proportion. The two types of proportion are continued proportion and mean proportion. The students can obtain a good score in the exam using the solutions prepared by experts in Mathematics. The students can download RD Sharma Solutions for Class 6 Maths Chapter 9 Ratio, Proportion and Unitary Method Exercise 9.3 PDF which are provided here.

RD Sharma Solutions for Class 6 Chapter 9: Ratio, Proportion and Unitary Method Exercise 9.3 Download PDF

 

rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3
rd sharma solution class 6 maths chapter 9 ex 3

 

Access RD Sharma Solutions for Class 6 Chapter 9: Ratio, Proportion and Unitary Method Exercise 9.3

1. Which of the following statements are true?

(i) 16: 24 = 20: 30

(ii) 21: 6 = 35: 10

(iii) 12: 18 = 28: 12

(iv) 51: 58 = 85: 102

(v) 40 men: 200 men = Rs 5: Rs 25

(vi) 99 kg: 45 kg = Rs 44: Rs 20

Solution:

(i) 16: 24 = 20: 30

It can be written as

16/24 = 20/30

Dividing 16/24 by 4/4 and 20/30 by 5/5

(16/24) ÷ (4/4) = (20/30) ÷ (5/5)

On further calculation

4/6 = 4/6

We get

2/3 = 2/3

Hence, 16: 24 = 20: 30 is true.

(ii) 21: 6 = 35: 10

It can be written as

21/6 = 35/10

Dividing 21/6 by 3/3 and 35/10 by 5/5

(21/6) ÷ (3/3) = (35/10) ÷ (5/5)

On further calculation

7/2 = 7/2

Hence, 21: 6 = 35: 10 is true.

(iii) 12: 18 = 28: 12

It can be written as

12/18 = 28/12

On further calculation

6/9 ≠ 14/6

Hence, 12: 18 = 28: 12 is false.

(iv) 51: 58 = 85: 102

It can be written as

51/58 = 85/102

On further calculation

51/58 ≠ 5/6

Hence, 51: 58 = 85: 102 is false.

(v) 40 men: 200 men = Rs 5: Rs 25

It can be written as

40/200 = 5/25

We get 40/200 = 1/2 and 5/25 = 1/5

Hence, 40 men: 200 men = Rs 5: Rs 25 is true.

(vi) 99 kg: 45 kg = Rs 44: Rs 20

It can be written as

99/45 = 44/20

Dividing the fraction by 9

(99/45) ÷ (9/9) = (44/20) ÷ (9/9)

On further calculation

11/5 = 11/5

Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true.

2. Find which of the following are in proportion:

(i) 8, 16, 6, 12

(ii) 6, 2, 4, 3

(iii) 150, 250, 200, 300

Solution:

(i) 8, 16, 6, 12

We know that

8: 16 = 8/16 = 1/2

6: 12 = 6/12 = 1/2

So we get 8/16 = 6/12

Therefore, 8, 16, 6, 12 are in proportion.

(ii) 6, 2, 4, 3

We know that

6: 2 = 6/2 = 3/1

4: 3 = 4/3

So we get 3/1 ≠ 4/3

Therefore, 6, 2, 4, 3 are not in proportion.

(iii) 150, 250, 200, 300

We know that

150: 250 = 150/250 = 3/5

200: 300 = 200/300 = 4/6 = 2/3

So we get 3/5 ≠ 2/3

Therefore, 150, 250, 200, 300 are not in proportion.

3. Find x in the following proportions:

(i) x: 6 = 55: 11

(ii) 18: x = 27: 3

(iii) 7: 14 = 15: x

(iv) 16: 18 = x: 96

Solution:

(i) x: 6 = 55: 11

It can be written as

x/6 = 55/11

We get

x/6 = 5/1

On further calculation

x = 5 (6) = 30

(ii) 18: x = 27: 3

It can be written as

18/x = 27/3

We get

18/x = 9/1

On further calculation

x = 18/9 = 2

(iii) 7: 14 = 15: x

It can be written as

7/14 = 15/x

We get

1/2 = 15/x

On further calculation

x = 15 (2) = 30

(iv) 16: 18 = x: 96

It can be written as

16/18 = x/96

We get

8/9 = x/96

On further calculation

x = 8/9 (96) = 256/3

4. Set up all proportions from the numbers 9, 150, 105, 1750.

Solution:

The proportions from the numbers are

9: 150 = 3: 50

9: 105 = 3: 35

9: 1750

150: 9 = 50: 3

150: 105 = 10: 7

150: 1750 = 3: 35

105: 9 = 35: 3

105: 150 = 7: 10

105: 1750 = 3: 50

1750: 9

1750: 150 = 35: 3

1750: 105 = 50: 3

Hence, the proportions that are formed are

9: 150 :: 105: 1750

150: 9 :: 1750: 105

1750: 150 :: 105: 9

9: 105 :: 150: 1750

5. Find the other three proportions involving terms of each of the following:

(i) 45: 30 = 24: 16

(ii) 12: 18 = 14: 21

Solution:

(i) 45: 30 = 24: 16 can be written as 3: 2 in simplest form

So the other three proportions involving terms are

45: 24 = 3: 16 can be written as 15: 8 in simplest form

30: 45 = 16: 24 can be written as 2: 3 in simplest form

16: 3 = 24: 45 can be written as 8: 15 in simplest form

(ii) 12: 18 = 14: 21 can be written as 2: 3 in simplest form

So the other three proportions involving terms are

12: 14 = 18: 21 can be written as 6: 7 in simplest form

21: 18 = 14: 12 can be written as 7: 6 in simplest form

18: 12 = 21: 14 can be written as 3: 2 in simplest form

6. If 4, x, 9 are in continued proportion, find the value of x.

Solution:

We know that 4, x, 9 are in continued proportion

It can be written as

4: x :: x: 9

We get

4/x = x/9

On further calculation

x2 = 9 (4) = 36

So we get

x = 6

7. If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third term.

Solution:

It is given that in a proportion the first, second and fourth terms are 32, 112 and 217

Consider x as the third term

We can write it as

32: 112 :: x: 217

On further calculation

32/112 = x/217

So we get

x = 32/112 (217) = 62

8. Show that the following numbers are in continued proportion:

(i) 36, 90, 225

(ii) 48, 60, 75

(iii) 16, 84, 441

Solution:

(i) 36, 90, 225

Consider the fraction 36/90

By dividing the fraction by 18

We get

36/90 = 2/5

Consider the fraction 90/225

By dividing the fraction by 45

We get

90/225 = 2/5

Hence, 36: 90 :: 90: 225.

(ii) 48, 60, 75

Consider the fraction 48/60

By dividing the fraction by 12

We get

48/60 = 4/5

Consider the fraction 60/75

By dividing the fraction by 15

We get

60/75 = 4/5

Hence, 48: 60 :: 60: 75.

(iii) 16, 84, 441

Consider the fraction 16/84

By dividing the fraction by 4

We get

16/84 = 4/21

Consider the fraction 84/441

By dividing the fraction by 21

We get

84/441 = 4/21

Hence, 16: 84 :: 84: 441.

9. The ratio of the length of a school ground to its width is 5: 2. Find its length if the width is 40 metres.

Solution:

It is given that

Ratio of length of a school ground to its width = 5: 2

Width of the school ground = 40 m

So the length of the school ground = 5/2 (40) = 100 m

Hence, the length of the school ground is 100 m.

10. The ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop was 2: 9. If the total sale of eggs in the same week was Rs 360, find the sale of eggs on Sunday.

Solution:

It is given that

Ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop = 2: 9

We know that the sale of eggs in a week is Rs 9 and on Sunday is Rs 2

If eggs of Rs 1 is sold in a week, the cost of egg on Sunday = Rs 2/9

If the total sale of eggs in the same week was Rs 360, the sale of eggs on Sunday = 2/9 (360) = Rs 80

Hence, the sale of eggs on Sunday is Rs 80.

11. The ratio of copper and zinc in an alloy is 9: 7. If the weight of zinc in the alloy is 9.8 kg, find the weight of copper in the alloy.

Solution:

It is given that

Ratio of copper and zinc in an alloy = 9: 7

We know that

If the weight of zinc is 7 kg then the weight of copper is 9 kg

If the weight of zinc is 1 kg then the weight of copper = 9/7 kg

So if the weight of zinc is 9.8 kg then the weight of copper = 9/7 (9.8) = 12.6 kg

Hence, the weight of copper in the alloy is 12.6 kg.

12. The ratio of the income to the expenditure of a family is 7: 6. Find the savings if the income is Rs 1400.

Solution:

It is given that

Ratio of the income to the expenditure of a family = 7: 6

We know that saving = total income – expenditure

So we get

Ratio of saving to the income = [7 – 6]: 7 = 1: 7

It is given that income = Rs 1400

So the saving of the family = 1/7 (1400) = Rs 200

Hence, the saving of the family is Rs 200.

13. The ratio of story books in a library to other books is 1: 7. The total number of story books is 800. Find the total number of books in the library.

Solution:

It is given that

Ratio of story books in a library to other books = 1: 7

We know that 1 is a story book our of 1 + 7 = 8 books

So if there is 1 story book then the total number of books = 8

If there is 800 story books then the total number of books = 8 (800) = 6400

Hence, the total number of books in the library is 6400.


Access other exercise solutions of Class 6 Maths Chapter 9: Ratio, Proportion and Unitary Method

Exercise 9.1 Solutions

Exercise 9.2 Solutions

Exercise 9.4 Solutions

Objective Type Questions

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