## RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

**Objective questions :**

**Mark (✓) against the correct answers in each of the following :**

**Question 1.**

**Solution:**

(d) a : b = 3 : 4, b : c = 8 : 9

**Question 2**

**Solution:**

(a) A : B = 2 : 3, B : C = 4 : 5

**Question 3.**

**Solution:**

(d)

**Question 4.**

**Solution:**

(b) 15% of A = 20% of B

**Question 5.**

**Solution:**

(a)

**Question 6.**

**Solution:**

(b) A : B = 5 : 7, B : C = 6 : 11

LCM of 7, 6 = 42

**Question 7.**

**Solution:**

(c) 2A = 3B = 4C = x

**Question 8.**

**Solution:**

(a)

\(\frac { A }{ 3 }\) = \(\frac { B }{ 4 }\) = \(\frac { C }{ 5 }\) = 1(suppose)

A = 3, B = 4, C = 5

A : B : C = 3 : 4 : 5

**Question 9.**

**Solution:**

(b)

**Question 10.**

**Solution:**

(c)

**Question 11.**

**Solution:**

(c) (3a + 5b) : (3a – 5b) = 5 : 1

**Question 12.**

**Solution:**

(c) 7 : x :: 35 : 45

x = 9

**Question 13.**

**Solution:**

(b) Let x to be added to each term of 3 : 5

Then \(\frac { 3 + x }{ 5 + x }\) = \(\frac { 5 }{ 6 }\)

By cross multiplication

18 + 6x = 25 + 5x

6x – 5x = 25 – 18

x = 7

7 is to be added

**Question 14.**

**Solution:**

(d) Ratio in two numbers = 3 : 5

Let first number = 3x

Then second number = 5x

According to the condition,

\(\frac { 3x + 10 }{ 5x + 10 }\) = \(\frac { 5 }{ 7 }\)

(By cross multiplication)

25x + 50 = 21x + 70

25x – 21x = 70 – 50

4x = 20

x = 5

First number = 3 x 5 = 15

and second number = 5 x 5 = 25

Sum of numbers = 15 + 25 = 40

**Question 15.**

**Solution:**

(a)

Let x be subtracted from each of the term

\(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)

⇒ 4 (15 – x) = 3 (19 – x)

⇒ 60 – 4x = 57 – 3x

⇒ -4x + 3x = 57 – 60

⇒ -x = -3

x = 3

Required number = 3

**Question 16.**

**Solution:**

(a)

Amount = Rs. 420

and ratio = 3 : 4

Sum of ratios = 3 + 4 = 7

A’s share = \(\frac { 420 x 3 }{ 7 }\) = Rs. 60 x 3 = Rs. 180

**Question 17.**

**Solution:**

(d)

Let number of boys = x, then

x : 160 : : 8 : 5

⇒ x x 5 = 160 x 8

x = \(\frac { 160 x 8 }{ 5 }\) = 32 x 8 = 256

Number of total students of the school = 256 + 160 = 416

**Question 18.**

**Solution:**

(a)

**Question 19.**

**Solution:**

(c)

Let x be the third proportional to 9 and 12 then

9 : 12 :: x : 12

⇒ 9 x x = 12 x 12

⇒ x = \(\frac { 12 x 12 }{ 9 }\) = \(\frac { 144 }{ 9 }\) = 16

Third proportional = 16

**Question 20.**

**Solution:**

Answer = (b)

Mean proportional of 9 and 16 = √(9 x 16) = √144 = 12

**Question 21.**

**Solution:**

(a)

Let age of A = 3x

and age of B = 8x

6 years hence, their ages will be 3x + 6 and 8x + 6

\(\frac { 3x + 6 }{ 8x + 6 }\) = \(\frac { 4 }{ 9 }\)

⇒ 9 (3x + 6) = 4 (8x + 6)

⇒ 27x + 54 = 32x + 24

⇒ 32x – 27x = 54 – 24

⇒ 5x = 30

⇒ x = 6

A’s age = 3x = 3 x 6 = 18 years

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C are helpful to complete your math homework.

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