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ATQ,
Let the total capacity of the container = L.C.M of 20 and 25 = 100 units
So, efficiency of pipe 'X' alone = 100 ÷ 20 = 5 units/minute
Combined efficiency of pipes 'X' and 'Y' = 100 ÷ 25 = 4 units/minute
So, efficiency of pipe 'Y' alone = 4 - 5 = -1 unit/minute (outlet)
So, time taken by pipe 'Y' alone to empty 60% of the container = (100 × 0.6) ÷ 1 = 60 minutes
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