# Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Exercise 7C Solutions (Including Properties and Uses)

## Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Solutions

### Exercise 7 (C)

(Q1) If a:b = c:d, prove that:

(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d

(ii) (9a + 13b) (9c – 13d) = (9c + 13d (9a – 13b)

(iii) xa + yb : xc + yd = b:d

= Solution:

Given that a/b = c/d

(i) To Prove: 5a + 7b/5a – 7b = 5c + 7d/5c – 7d

Proof: From given,

a/b = c/d

Multiply by ‘5’ and divide by ‘7’ on both sides,

5a/7b = 5c/7d

Apply componendo and dividendo,

5a + 7b/5a – 7b = 5c + 7d/5c – 7d Hence Proved.

(ii) To prove: (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b)

Proof = given a/b = c/d

Multiply on both side by 9/13.

9a/13b = 9C/13d

Apply componendo and dividendo,

9a + 13b/9a – 13b = 9c + 13d/9c- 13d

(9a+ 13b) (9c – 13d) = (9c + 13d) (9a – 13d)

Hence proved

(iii) To prove: xa + yb/xc + yd = b/d

Proof: Given,

a/b = c/d

Multiply on both side by x/y

xa/yb = xc/yd

Apply componendo

xa + yb/yb = xc + yd/yd

xa + yb/xc + yd = yb/yd

xa + yb/xc + yd = b/d (Hence proved)

(Q2) If a:b = c;, prove that

(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b)

= Solution:

Given that,

a/b = c/d

To prove: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b)

Proof: Given that,

a/b = c/d

Multiply on both side by 6/7

6a/7b = 6c/7d

Apply componenedo,

6a + 7b/7b = 6c + 7d/7d

6a + 7b/6c + 7d = 7b/7d

6a + 7b/6c + 7d = b/d —– (i)

Now, a/b = c/d

Multiply on both side by 3/4,

3a/4b = 3c/4d

Apply dividendo,

3a – 4b/4b = 3c – 4d/4d

3a – 4b/3c – 4d = 4b/4d

∴ 3a – 4b/3c – 4d = b/d —– (ii)

From equation (i) and (ii),

We get,

6a + 7b/6c + 7d = 3a – 4b/3c – 4d

Apply Alternnendo,

(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b) (Hence proved)

(Q3) Given, a/b = c/d, prove that

(3a – 5b)/(3a + 5b) = (3c – 5d)/(3c + 5d)

= Solution:

Given that,

a/b = c/d

Multiply on both side by 3/5

3a/5b = 3c/5d

Apply componendo and dividendo

3a + 5b/3a – 5b = 3c + 5d/3c – 5d

Apply alternendo,

3a – 5b/3a + 5b = 3c – 5d/3c + 5d (Hence proved)

(Q4) If 5x + 6y/5u + 6v = 5x – 6y/5u – 6v

Then prove that x:y = u:v

= Solution:

Given that,

5x + 6y/5u + 6v = 5x – 6y/5u – 6v

To prove: x:y = u:v

Prove: 5x + 6y/5u + 6v = 5x – 6y/5u – 6v

Apply alternendo on both side,

5x + 6y/5x – 6y = 5u + 6v/5u – 6v

Apply componenedo and dividendo

..> 5x + 6y + 5x – 6y/5x + 6y – (5x – 6y) = 5u + 6v + 5u – 6v/5u + 6v – (5u – 6v)

..> 5x + 6y + 5x – 6y/5x + 6y – 5x + 6y = 5u + 6v + 5u – 6v/5u + 6v – 5u + 6v

..> 10x/12y = 10u/12v

x/y = u/v

∴ x:y = u:v (Hence proved)

(Q5) If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)

= Prove that a:b = c:d

= Solution:

Given that,

(7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)

To prove: a:b = c:d

Proof: (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)

Apply invertendo on both sides,

7a + 8b/7a – 8b = 7c + 8d/7c – 8d

Apply componendo and dividendo,

7a + 8b + 7a – 8b/7a + 8b – (7a – 8b) = 7a + 8b + 7c – 8d/7c + 8d – (7c – 8d)

14a/7a + 8b – 7a + 8b = 14c/7c + 8d – 7c + 8d

14a/16b = 14c/16d

a/b = c/d

∴ a:b = c:d (Hence proved)

(Q6) (i) If x = 6ab/(a + b), Find the value of:

x + 3a/x – 3a + x + 3b/x – 3b

= Solution:

Given that,

x = 6ab/(a + b)

Find: x+3a/x–3a + x+3b/x–3b

x = 6ab/(a+b)

= (3×2)ab/a+b

x = 3a×2b/a+b

x/3a = 2b/a+b

Apply componenedo and dividendo

x+3a/x-3a = 2b+a+b/2b – (a+b)

x + 3a/x – 3a = 2b + a + b/2b – a – b

x + 3a/x – 3a = 3b + a/b + a

x + 3a/x – 3a = a + 3b/- (a – b) —— (i)

x = 6ab/a+b

x = 3b×2a/a+b

x/3b = 2a/a + b

Apply componendo and dividendo,

x + 3b/x – 3b = 2a + a + b/2a – (a + b)

x + 3b/x – 3b = 2a + a + b/2a – a – b

x + 3b/x – 3b = 3a + b/a – b —– (ii)

Now, x + 3a/x – 3a + x + 3b/x – 3b

From equation (i) and (ii)

a + 3b/- (a – b) + 3a + b/a – b

Take L.C.M

– a – 3b + 3a + b/(a – b) = 2a – 2b/(a – b)

= 2 (a – b)/(a – b)

= 2

(ii) Solution:

Given:

a = 4√6/√2 + √3

a = 2×2×√2×√3/√2 +√3

a/2√2 = 2√3/√2 + √3

Apply componendo and dividendo,

a + 2√2/a – 2√2 = 2√3+√2+√3/2√3-√2-√3

a+2√2/a-2√2 = 3√3+√2/√3 – √2 —— (i)

a = 4√6/√2 + √3

a = 2×2×√2×√3/√2 + √3

a/2√3 = 2√2/√2 + √3

Apply componendo and divideno,

a+2√3/a-2√3= 2√2+√2+√3/2√2-√2-√3

a+2√3/a-2√3 = 3√2+√3/√2 – √2 —— (ii)

Now, a+2√2/a-2√2 + a+2√3/a-2√3

3√3+√2/√3-√2 + 3√2+√3/-(√3-√2)

3√3+√2/√3-√2 – (3√2+√3)/(√3-√2)

3√3+√2-3√2-√3/(√3-√2)

..> +2√3 – 2√2/√3 – √2 = 2(√3 – √2)/(√3 – √2)

= 2

(Q7) If (a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d), prove that, a:b = c:d

Solution:

Given that,

(a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d)

Prove that: a:b = c:d

Proof: (a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d)

(a+b+c+d)/(a+b-c-d) = (a-b+c-d)/(a-b-c+d)

Apply componendo and = dividendo,

a+b+c+d+a+b-c-d/a+b+c+d-(a+b-c-d) = a-b+c-d+a-b-c+d/a-b+c-d-(a-b-c+d)

= a+b+c+d+a+b-c-d/a+b+c+d-a-b+c+d = a-b+c-d+a-b-c+d/a-b+c-d-a+b+c-d

2a+2b/2c+2d = 2a-2b/2c-2d

2(a+b)/2(c+d) = 2(a-b)/2(c-d)

(a+b)/(c+d) = (a-b)/(c-d)

Apply invertendo,

a+b/a-b = c+d/c-b

Apply componendo and dividendo,

a+b+a-b/a+b-a+b = c+d+c-d/c+d-c+d

2a/2b = 2c/2d

a/b = c/d

a:b = c:d (Hence proved)

Here is your solution of Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Exercise 7C Solutions

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

Updated: September 3, 2021 — 3:15 pm