Question
The ratio of the quantity of juice and water in vessel A and vessel B is 1:1 and 3:2 respectively. 40 liters of mixture is taken out from vessel A and added into vessel
B. After this, the ratio of juice and water in vessel B becomes 7:5. The initial quantity of juice in vessel A is equal to the difference between the final quantity of juice and water in vessel
B. Find the total quantity of mixture in vessel A initially.
Solution
ATQ, In vessel A, Juice = x Water = x In vessel B, Juice = 3y Water = 2y From vessel A, 40 liters of mixture is removed. Total parts in A = 1 + 1 = 2 Juice removed = 40 × 1/2 = 20 liters Water removed = 40 × 1/2 = 20 liters So, in vessel B after mixing: Final juice = 3y + 20 Final water = 2y + 20 Given final ratio: (3y + 20)/(2y + 20) = 7/5 5(3y + 20) = 7(2y + 20) 15y + 100 = 14y + 140 y = 40 Final juice in B = 3×40 + 20 = 140 liters Final water in B = 2×40 + 20 = 100 liters Difference = 140 – 100 = 40 liters Given: initial juice in A = difference ⇒ x = 40 Total mixture in A initially = x + x = 2x = 2 × 40 = 80 liters
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