# Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(C)

Some important properties of proportion like invertendo, alternendo, componendo, dividendo and other direct applications are the concepts discussed in this exercise. The exercise problems pertain to these topics and are very important for exams. To get complete knowledge about these concepts and attain a strong grip over other chapters, the Selina Solutions for Class 10 Maths is the right tool for it. Further, the solutions to this exercise are available in the Concise Selina Solutions for Class 10 Maths Chapter 7 Ratio and Proportion Exercise 7(C) PDF in the links attached below.

## Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(C) Download PDF

### Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion

Exercise 7(A) Solutions

Exercise 7(B) Solutions

Exercise 7(D) Solutions

### Access Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7(C)

1. 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:

(i) Given, a/b = c/d

(ii) Given, a/b = c/d

On cross-multiplication we have,

(9a + 13b)(9c â€“ 13d) = (9c + 13d)(9a â€“ 13b)

(iii) Given, a/b = c/d

– Hence Proved

2. If a : b = c : d, prove that:

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

Solution:

Given, a/b = c/d

(6a + 7b)(3c â€“ 4d) = (3a â€“ 4b)(6c + 7d)

– Hence Proved

3. Given, a/b = c/d, prove that:

(3a â€“ 5b)/ (3a + 5b) = (3c â€“ 5d)(3c + 5d)

Solution:

4. If ;

Then prove that x: y = u: v

Solution:

10x/ 12y = 10u/ 12v

Thus,

x/y = u/v â‡’ x: y = u: v

5. If (7a + 8b) (7c â€“ 8d) = (7a â€“ 8b) (7c + 8d);

Prove that a: b = c: d

Solution:

The given can the rewritten as,

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

Solution:

Given, x = 6ab/ (a + b)

â‡’ x/3a = 2b/ a + b

Now, applying componendo and dividendo we have

Again, x = 6ab/ (a + b)

â‡’ x/3b = 2a/ a + b

Now, applying componendo and dividendo we have

From (1) and (2), we get

(ii) If a = 4âˆš6/ (âˆš2 + âˆš3), find the value of:

Solution:

Given, a = 4âˆš6/ (âˆš2 + âˆš3)

a/2âˆš2 = 2âˆš3/ (âˆš2 + âˆš3)

Now, applying componendo and dividendo we have

Again, a = 4âˆš6/ (âˆš2 + âˆš3)

a/2âˆš3 = 2âˆš2/ (âˆš2 + âˆš3)

Now, applying componendo and dividendo we have

From (1) and (2), we have

7. If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a: b = c: d.

Solution:

Rewriting the given, we have

Now, applying componendo and dividendo

Applying componendo and dividendo again, we get

– Hence Proved