Quantity-I: A mixture contains 40% milk and the rest 240 litres water. When (x−10) litres of water and (x+8) litres of milk are added to this mixture, the ratio of the quantity of milk to that of water in the resultant mixture becomes 7:8. Find the value of ‘x’. Quantity-II: The ratio of the quantity of alcohol to that of water in a mixture (alcohol + water) is 6:4, respectively. 30% of this mixture is replaced with 5 litres of alcohol and 7 litres of water such that the ratio of the quantity of alcohol to that of water in the resultant mixture becomes 8:6. If the measure of 25% of the initial quantity of water in the mixture is ‘y’ litres, then find the value of ‘y’.

ATQ, Quantity I: Initial quantity of milk = 240×(0.40/0.60)=160 litres Final quantity of milk = 160+x+8=(168+x) litres Final quantity of water = 240+x−10=(230+x) litres According to the question, (168+x)/(230+x) = 7/8 8(168+x)=7(230+x) 1344+8x=1610+7x x=266 So, Quantity I = 266 Quantity II: Let the initial quantity of alcohol and water be 6x litres and 4x litres, respectively Final quantity of alcohol = (0.70×6x)+5=(4.2x+5) litres Final quantity of water = (0.70×4x)+7=(2.8x+7) litres According to the question, (4.2x+5)/(2.8x+7) = 8/6 6(4.2x+5)=8(2.8x+7) 25.2x+30=22.4x+56 2.8x=26 26/2.8 =9.29 Therefore, ‘y’ = 0.25 × 4x = x So, Quantity II = 9.29 So, Quantity I > Quantity II