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    Question

    Quantity-I: A mixture contains 50% milk and the rest

    360 litres water. When (x −15) litres of water and (x+10) litres of milk are added to this mixture, the ratio of the quantity of milk to that of water in the resultant mixture becomes 5:6. Find the value of ‘x’. Quantity-II: The ratio of the quantity of sugar to that of water in a mixture (sugar + water) is 8:6, respectively. 35% of this mixture is replaced with 6 litres of sugar and 8 litres of water such that the ratio of the quantity of sugar to that of water in the resultant mixture becomes 7:5. If the measure of 25% of the initial quantity of water in the mixture is ‘y’ litres, then find the value of ‘y’. In the question, two Quantity 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. 
    A Quantity-I > Quantity-II Correct Answer Incorrect Answer
    B Quantity-I < Quantity-II Correct Answer Incorrect Answer
    C Quantity-I ≤ Quantity-II Correct Answer Incorrect Answer
    D Quantity-I = Quantity-II or No relation Correct Answer Incorrect Answer
    E Quantity-I ≥ Quantity-II Correct Answer Incorrect Answer

    Solution

    ATQ, Quantity I: Initial quantity of milk = 360×(0.50/0.50 )=360 litres Final quantity of milk = 360+x+10=(370+x) litres Final quantity of water = 360+x−15=(345+x) litres According to the question, (370+x)/(345+x) = 5/6 6(370+x)=5(345+x) 2220+6x=1725+5x x=160 So, Quantity I = 160 Quantity II: Let the initial quantity of sugar and water be 8x litres and 6x litres, respectively Final quantity of sugar = (0.65×8x)+6=(5.2x+6) litres Final quantity of water = (0.65×6x)+8=(3.9x+8) litres According to the question, (5.2x+6)/(3.9x+8) = 7/5 5(5.2x+6)=7(3.9x+8) 26x+30=27.3x+56 1.3x=26 x = 26/1.3  x = 20 Therefore, ‘y’ = 0.25 × 6x = x So, Quantity II = 20 So, Quantity I > Quantity II

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