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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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