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# (i) Find two numbers in the ratio of 8: 7 such that when each is decreased by 12 ½, they are in the ratio 11: 9. (ii) The income of a man is increased in the ratio of 10: 11. If the increase in his income is Rs 600 per month, find his new income.

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

In this ques a number in ratio a:b given and both numerator amd denominator are decreased ny a fix number then ratio becomes c:d. We have to find the number.

Question 14

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1. Solution:

(i) Ratio = 8: 7

Consider the numbers as 8x and 7x

Using the condition

[8x – 25/2]/ [7x – 25/2] = 11/9

Taking LCM

[(16x – 25)/ 2]/ [(14x – 25)/ 2] = 11/9

By further calculation

[(16x – 25) × 2]/ [2 (14x – 25)] = 11/9

(16x – 25)/ (14x – 25) = 11/9

By cross multiplication

154x – 275 = 144x – 225

154x – 144x = 275 – 225

10x = 50

x = 50/10 = 5

So the numbers are

8x = 8 × 5 = 40

7x = 7 × 5 = 35

(ii) Consider the present income = 10x

Increased income = 11x

So the increase per month = 11x – 10x = x

Here x = Rs 600

New income = 11x = 11 × 600 = Rs 6600

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