S Chand ICSE Maths Solutions Class 10 Chapter 6 Ratio Proportion Revision Exercise

Hi students, Welcome to Amans Maths Blogs (AMB). In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 6 Ratio Proportion Revision Exercise. This is the chapter of introduction of ‘Ratio and Proportion‘ included in ICSE Class 10 Maths syllabus.

S Chand ICSE Maths Solutions for Class 10 Chapter 6 Revision Exercise

S Chand ICSE Maths Solutions Class 10 Chapter 6 Ratio Proportion Revision Exercise: Ques No 1

Given a/b = c/d, prove that (3a – 5b)/(3a + 5b) = (3c – 5d)/(3c + 5d).

S Chand ICSE Maths Solutions:

Ques No 2

Two numbers are in the ratio of 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.

Solutions:

Ques No 3

If a : b = 5 : 3, then find the value of (5a + 8b) : (6a – 7b).

Solutions:

Ques No 4

If (3a + 4b) / (3c + 4d) = (3a – 4b) / (3c – 4d), then prove that a / b = c / d.

Solutions:

Ques No 5

The work done by (x – 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio 3 : 10. Find the value of x.

Solutions:

Ques No 6

What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?

Solutions:

Ques No 7

What number must be added to each of the numbers 6, 15, 20 and 43 to make them in proportion?

Solutions:

Ques No 8

If (3x + 5y) / (3x – 5y) = 7 / 3, then find the value of x / y.

Solutions:

Ques No 9

If x = [√(a + 3b) + √(a – 3b)] / [√(a + 3b) – √(a – 3b)], then prove that 3bx2 – 2ax + 3b = 0.

Solutions:

Ques No 10

If (8a – 5b) / (8c – 5d) = (8a + 5b) / (8c + 5d), then prove that a / b = c / d.

Solutions:

Ques No 11

What least number must be added to each of the numbers 5, 11, 19 and 37 to make them in proportion?

Solutions:

Ques No 12

Given that (a³ + 3ab²) / (b³ + 3a²b) = 63 / 62, using componendo and dividendo, find the value of a : b.

Solutions:

Ques No 13

If x, y, z are in continued proportion, then prove that (x + y)² / (y + z)² = x / z.

Solutions:

Ques No 14

Given x = [√(a² + b²) + √(a² – b²)] / [√(a² + b²) + √(a² – b²)], use componendo and dividendo to prove that b² = 2a²x / (x² + 1).

Solutions:

Ques No 15

Using componendo and dividendo, find the value of x. [√(3x + 4) + √(3x – 5)] / [√(3x + 4) + √(3x – 5)] = 9

Solutions:

Ques No 16

6 is the mean proportional between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.

Solutions:

Ques No 17

If x = [√(a + 1) + √(a – 1)] / [√(a + 1) + √(a – 1)], using properties of proportion show that x² – 2ax + 1 = 0.

Solutions:

Ques No 18

Using the properties of proportion, solve for x, given that (x⁴ + 1) / 2x² = 17 / 8.

Solutions:

Ques No 19

If (x² + y²) / (x² – y²) = 17 / 8, find the value of (i) x : y and (ii) (x³ + y³) / (x³ – y³).

Solutions:

Ques No 20

If a, b, c, d are in continued proportion, prove that (a + b + c)(a – b + c) = a² + b² + c². 

Solutions:

Ques No 21

Given that (x³ + 12x) / (6x² + 8) = (y³ + 27y) / (9y² + 27), find the value of x : y.

Solutions:

Ques No 22

If x / a = y / b = z / c, then show that x³/a³ + y³/b³ + z³/c³ = 3xyz/abc.

Solutions: