# Rd Sharma Solutions Class 7 Chapter 9

## Rd Sharma Solutions Class 7 Chapter 9 Ratio and Proportion

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 9, Ratio and Proportion. Here students can easily find Exercise wise solution for chapter 9, Ratio and Proportion. Students will find proper solutions for Exercise 9.1, 9.2 and 9.3. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.

Ratio and Proportion Exercise 9.1 Solution

Question no – (1)

Solution :

In the given question,

x : y = 3 : 5

x/y = 3/5

= x = 3y/5

Now, 3x + 4y : 8x + 5y

= 3 × 3y/5 + 4y : 8 × 3y/5 + 5y

= 9y + 20y/5 : 24y + 25y/5

= 29 : 49

Therefore, the ratio will be 29 : 49

Question no – (2)

Solution :

In the given question,

x : y = 8 : 9

x/y = 8/9

x = 8y/9

So, 7x – 4y : 3x + 2y

= 7 × 8y/9 – 4y : 3 × 8y/9 + 2y

= 56y – 36y/9 : 24y + 18y/9

= 20 : 42

Therefore, the ratio will be 20 : 42

Question no – (3)

Solution :

Let, the number 6x and 13x

So, 78x = 312

x = 312/78

= x = 4

The numbers are = 24

And, 13 × 4 = 52

Therefore, the required numbers are 24, 52.

Question no – (4)

Solution :

Let, the number 3x and 5x

3x + 8 : 5x + 8 = 2:3

3x + 8/5x + 8 = 2/3

= 9x + 24 = 10x + 16

= 9x – 10x = 16 – 24

= – x = – 8

x = 8

The numbers are,

= 3 × 8 = 24

= 5 × 8 = 40

Therefore, 24, and 40 will be the required numbers.

Question no – (5)

Solution :

Then, 7 + x/13 + x = 2/3

= 21 + 3x = 26 + 2x

= 3x – 2x = 26 – 21

x = 5

Hence, 5 should be added to each term of the ratio to ratio become 2 : 3.

Question no – (6)

Solution :

Let, the numbers 2x, 3x and 5x

So, 2x + 3x + 5x = 800

= x = 80

The numbers are,

2 × 80 = 160,

3 × 80 = 240

And 5 ×80 = 400

Therefore, the required numbers are 160, 240, and 400.

Question no – (7)

Solution :

Let, ages of the persons = 5x and 7x

5x – 18/7x – 18 = 8/13

= 65x – 234 = 56x – 144

= 65x – 56x = – 144 + 234

= x = 90/9

x = 10

Therefore, their present ages of the persons are 50 and 70.

Question no – (8)

Solution :

Let, the numbers 7x and 11x

= 7x + 7/11x + 7 = 2/3

= 21x + 21 = 22x + 14

= 21x – 22x = 14 – 21

= – x = -7

= x = 7

The numbers are,

= 7 × 7 = 49

And, 11 × 7 = 77

Thus, the required numbers are 49 and 77.

Question no – (9)

Solution :

Let, the numbers 2x and 7x

So, 2x + 7x = 810

= x = 810/9

= x = 90

The numbers are,

= 2 × 90 = 180

And 7 × 90 = 630

Therefore, the required numbers are 180 and 630.

Question no – (10)

Solution :

Let, Ravish and Shikha get 2x and 3x

= 2x + 3x = 1350

= x = 1350/5

= x = 270

∴ Ravish get,

= 2 × 270

= 540

∴ Shikha get,

= 3 × 270

= 810

Therefore, Ravish get Rs. 540 and Shikha gets Rs 810.

Question no – (11)

Solution :

Let, P,Q,R get 2x, 3x,and 5x

2x + 3x + 5x = 2000

= x = 2000/10

= x = 200

∴ P get,

= 2 × 2000

= ₹400

∴ Q get,

= 3 × 2000

= ₹600

∴ R get,

= 5 × 2000

= ₹1000

Question no – (12)

Solution :

Let, Boys and girls in schools = 7x and 4x

7x + 4x = 550

= x = 550/11

x = 50

Boys in school,

= 7 × 50

= 350

Girls in school,

= 4 × 50

= 200

Therefore, the number boys are 350 and the girls are 200.

Question no – (13)

Solution :

Let, Income and savings = 7x and 2x

So, = x = 500/5

= x = 100

Savings,

= 2x = 500

= x = 250

Income,

= 7 × 250

= 1750

Expenditure,

= 1750 – 500

= 1250

Hence, the income is Rs 1750 and expenditure is Rs 1250.

Question no – (14)

Solution :

Let, sides of the triangle are x, 2x and 3x

x + 2x + 3x = 36

= 6x = 36

= x = 6

Sides are,

= 6, 6 × 2 = 12

And 6 × 3 = 18

Therefore, its sides are 6, 12, and 18 cm.

Question no – (15)

Solution :

Let, Raman and Aman get 2x and 3x

2x + 3x = 5500

= 5x = 5500

= x = 1100

Raman will get,

= 2 × 1100

= 2200

Aman will get,

= 3 × 1100

= 3300

Hence, Raman will get Rs 2200 and Aman will get Rs 3300.

Question no – (16)

Solution :

Zinc and copper in alloy = 7x and 9x

So, 9x = 11.7

= x = 11.7/9 = 1.3

Weight of zinc in the alloy,

= 1.3 × 7

= 9.10 kg

Weight of Zinc,

= 9.10 kg

Therefore, the weight of zinc in the alloy is 9.10 kg.

Question no – (17)

Solution :

Let, the number are 7x and 8x

So, 8x = 40

= x = 5

The antecedent,

= 7 × 5

= 35

Hence, the antecedent will be 35.

Question no – (18)

Solution :

The two parts = 2x and 7x

So, 2x + 7x = 357

= x = 357/9

= x = 39

One part will be,

= 2 × 39

= Rs. 78

Second part will be,

= 7 × 39

= Rs. 273

Question no – (20)

Solution :

If one out of six student fail term ratio = 1 : 6

Let, fails students in the class be x

Hence, 1 : 6 = x : 42

= 1/6 = x/42

= 1/x = 6/42

= x = 7

7 student fails out of 42 students.

Passed Students,

= (42 – 7)

= 35

Therefore, 35 students will pass.

Ratio and Proportion Exercise 9.2 Solution

Question no – (1)

Solution :

(i) 3 : 4 or 9 : 6

= 3 : 4 = 3/4 = 12/10

= 9 : 6 = 9/16

Therefore, 3 : 4 > 9 : 6

(ii) 15 : 16 or 24 : 25

15 : 16 = 15/16 = 15 × 25/16 × 25 = 375/400

24 : 25 = 24/25 = 24 × 16/25 + 16 = 384/400

Hence, 24 : 25 > 15 : 16

(iii) 4 : 7 or 5 : 8

= 4/7 = 4 × 8/7 × 8 = 32/56

= 5/8 = 5 × 7/8 × 7 = 35/56

So, 5/8 > 4/7

Thus,  5 : 8 > 4 : 7

(iv) 9 : 20 or 8 : 13

= 9/20 = 9 × 13/20 × 13 = 117/260

= 8/13 = 8 × 20/13 × 20 = 160/260

= 8/13 > 9/20 8

Hence, 8 : 13 > 9 : 20

(v) 1 : 2 or 13 : 12

= 1/2 = 1 × 27/2 × 27 = 27/54

= 13/27 = 13 × 2/27 × 2 = 26/54

Therefore, 1 : 2 > 13 : 27

Question no – (2)

Solution :

Two equivalent ratios of 6 : 8 are,  3 : 4 and 18 : 24

Explanation :

First, 6 : 8

= 6/2 and 8/2

= 3 : 4

Second, 6 : 8

= 6 × 3 = 18

= 8 × 3 = 24

= 18 : 24

Question no – (3)

Solution :

= 12/20 = 3/5 = 9/15

Ratio and Proportion Exercise 9.3 Solution

Question no – (1)

Solution :

(i) 33, 44, 66, 88,

= 33 : 44 = 66 : 88

= 3 : 4 = 3 : 4

Hence they are proportion

(ii) 46, 69, 69, 46

= 46 : 69 ≠ 69 : 46

So, they are not proportion

(iii) 72, 84, 186, 217

= 72 : 84 and 186 : 217

= 72/84

= 6/7

= 186/217 = 6/7

= 6/7

= 6/7 = 6/7

Hence they are in proportion.

Question no – (2)

Solution :

(i) 16 : 18 = x : 96

= 16 : 18 =x : 96

= 16/18 = x/96

= x = 16 × 96/18

x = 256/3

Therefore, the x will be 256/3

(ii) x : 92 = 87 : 116

= x/92 = 87/116

= x = 87 × 92/116

x = 69

Hence, the x will be 69

Question no – (3)

Solution :

Let,  Income = 7x

Expenditure = 6x

Then, 7x = 1400

= x = 200

Then saving = 7x – 6x = x

=  Rs 200

Therefore, the savings of the family is Rs 200.

Question no – (4)

Solution :

As per the given question,

Distance ratio = 1 : 4000000

Distance between two towns,

= 5 × 4000000

= 20000000 cm

= 200 km

Therefore, the actual distance between the two towns is 200 km.

Question no – (5)

Solution :

According to the question,

Income = 10x

Savings = x

x = 6000

Income per year,

= 10 × 6000

= ₹60,000

Income per month,

= 60000/12

= 5000

Therefore, his monthly income is Rs. 5000.

Question no – (6)

Solution :

Let, the height of the pole = x

Given, The pole costs a shadow of length 20 metre

and 6m cost a shadow of length 8 metre

x : 20 = 6 : 8

= x/20 = 6/8

= x = 20 × 6/8

= 15 m

Therefore. the height of the electric pole will be 15 meter.

Next chapter solution :

Updated: June 8, 2023 — 3:19 pm