This is a basic question from ML aggarwal book of class 10th, chapter 7, ratio and proportion., ICSE board

(i) we have given (x – 9): (3x + 6) is the duplicate ratio of 4: 9 and we have to, find the value of x

. (ii)and If (3x + 1): (5x + 3) is the triplicate ratio of 3: 4, find the value of x.

(iii) If (x + 2y): (2x – y) is equal to the duplicate ratio of 3: 2, find x: y.

Chapter 13

Solution:(i) (x – 9)/ (3x + 6) = (4/9)

^{2}So we get

(x – 9)/ (3x + 6) = 16/81

By cross multiplication

81x – 729 = 48x + 96

81x – 48x = 96 + 729

So we get

33x = 825

x = 825/33 = 25

(ii) (3x + 1)/ (5x + 3) = 3

^{3}/ 4^{3}So we get

(3x + 1)/ (5x + 3) = 27/64

By cross multiplication

64 (3x + 1) = 27 (5x + 3)

192x + 64 = 135x + 81

192x – 135x = 81 – 64

57x = 17

So we get

x = 17/57

(iii) (x + 2y)/ (2x – y) = 3

^{2}/ 2^{2}So we get

(x + 2y)/ (2x – y) = 9/4

By cross multiplication

9 (2x – y) = 4 (x + 2y)

18x – 9y = 4x + 8y

18x = 4x = 8y + 9y

So we get

14x = 17y

x/y = 17/14

x: y = 17: 14