Question
In the question, two Quantities I and II are given. You
have to solve both the Quantity to establish the correct relation between Quantity-I and Quantity-II and choose the correct option. Quantity I: The ratio of the present ages of 'R' and 'S' is 3:1. The ratio of the age of 'S' after (y + 1) years to the age of 'R' 10 years ago is 9:7. If 'R' is (y − 3) years older than 'M', and 7 years from now the sum of the ages of 'S' and 'M' will be 40 years, then find the value of 'y'. ' Quantity II: The age of 'X' 10 years from now is equal to the present age of 'V'. The ratio of the present ages of 'V' and 'T' is (n + 3):19. If the age of 'T' is 34 years more than the age of 'X', and the ratio of the age of 'X' 7 years from now to the age of 'V' 4 years from now is 7:8, then find the value of 'n'.Solution
ATQ, Quantity I: Let the present age of R and S be '3x' years and 'x' years respectively. ATQ; (x + y + 1) /(3x - 10) = 9/7 7(x + y + 1) = 9(3x - 10) 7x + 7y + 7 = 27x - 90 20x - 7y = 97 ---------------(I) Present age of M = (3x - y + 3) years Again; x + 7 + 3x - y + 3 + 7 = 40 4x - y + 17 = 40 4x - y = 23 -----------------(II) Subtracting equation (I) from 5 X equation (II) , we get; (20x - 5y) - (20x - 7y) = 115 - 97 20x - 5y - 20x + 7y = 18 2y = 18 'y' = 9 So, Quantity I = 9 Quantity II: Â Let present age of V and T be '(n + 3) y' years and '19y' years respectively Present age of Raj = {(n + 3) y - 10} years ATQ; 19y - {(n + 3) y - 10} = 34 19y - (n + 3) y + 10 = 34 19y - ny - 3y = 24 16y - ny = 24 y(16 - n) = 24 ------------------ (I) Again; 
8y(n + 3) - 24 = 7y(n + 3) + 28 y(n + 3) = 52 ------------------- (II) Dividing equation (I) by equation (II) , we get; 
208 - 13n = 6n + 18 19n = 190 'n' = 10 So, Quantity II = 10 Quantity I < Quantity II
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