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

 

selina solutions concise maths class 10 chapter 7c
selina solutions concise maths class 10 chapter 7c
selina solutions concise maths class 10 chapter 7c
selina solutions concise maths class 10 chapter 7c
selina solutions concise maths class 10 chapter 7c

 

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

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 1

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

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 2

On cross-multiplication we have,

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

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

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 3

– Hence Proved

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

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

Solution:

Given, a/b = c/d

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 4

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

– Hence Proved

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

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

Solution:

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 5

4. If Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 6;

Then prove that x: y = u: v

Solution:

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 7

10x/ 12y = 10u/ 12v

Thus,

x/y = u/v ⇒ x: y = u: v

5. If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d);

Prove that a: b = c: d

Solution:

The given can the rewritten as,

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 8

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

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 9

Solution:

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

⇒ x/3a = 2b/ a + b

Now, applying componendo and dividendo we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 10

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

⇒ x/3b = 2a/ a + b

Now, applying componendo and dividendo we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 11

From (1) and (2), we get

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 12

(ii) If a = 4√6/ (√2 + √3), find the value of:

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 13

Solution:

Given, a = 4√6/ (√2 + √3)

a/2√2 = 2√3/ (√2 + √3)

Now, applying componendo and dividendo we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 13

Again, a = 4√6/ (√2 + √3)

a/2√3 = 2√2/ (√2 + √3)

Now, applying componendo and dividendo we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 14

From (1) and (2), we have

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 15

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

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 16

Now, applying componendo and dividendo

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 17

Applying componendo and dividendo again, we get

Concise Selina Solutions Class 10 Maths Chapter 7 ex. 7(C) - 18

– Hence Proved


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