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) = 33/ 43
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) = 32/ 22
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