**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 :**

Let, x be added

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 : **

👉 Chapter 10 👈