The ratio of the quantity of juice and water in vessel

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