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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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