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ATQ,
Let the total work be 120 units (LCM of 15, 20 and 24)
Efficiency of 'X' = (120 ÷ 15) = 8 units/day
Efficiency of 'Y' = (120 ÷ 20) = 6 units/day
Efficiency of 'Z' = (120 ÷ 24) = 5 units/day
Required time = 120 ÷ (8 + 6 + 5) = (120 ÷ 19) = 6(6/19) days
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