Question
Vessel A contains only milk and water in the ratio 7:3.
Vessel B contains only milk and water in the ratio 4:1. 40 litres of mixture is poured from A to B. Then half of the mixture in B is poured back into A. If the ratio of milk to water in A now is 11:4, find the initial total quantity of mixture in vessel B, given that vessel A initially had 200 litres.Solution
ATQ, Let the quantity of mixture in vessel B be 5x litres. Milk in B = 5x × (4/5) = 4x litres Water in B = 5x – 4x = x litres Milk in A initially = 200 × (7/10) = 140 litres Water in A initially = 200 – 140 = 60 litres From A → B (40 L): Milk = 40 × (7/10) = 28 L Water = 40 – 28 = 12 L So milk in B = (4x + 28) litres Water in B = (x + 12) litres Half of B back to A: Milk = (4x + 28)/2 = (2x + 14) litres Water = (x + 12)/2 = (0.5x + 6) litres ATQ, {(140 – 28) + (2x + 14)} / {(60 – 12) + (0.5x + 6)} = 11/4 (126 + 2x) / (54 + 0.5x) = 11/4 Cross multiply: 504 + 8x = 594 + 5.5x 2.5x = 90 x = 36 Initial mixture in B = 5x = 5 × 36 = 180 litres
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