Question
A can finish a job in 24 hours while B can finish it in 16 hours. If C is 50% more efficient than A and B together, then find the time taken by C alone to complete the work.
Solution
ATQ, Total work = LCM of (24 and 16) = 48 units
Work by A in one hour = 48 ÷ 24 = 2 units
Work by B in one hour = 48 ÷ 16 = 3 units
Work by A and B together = 2 + 3 = 5 units
C is 50% more efficient → C’s work = 5 + (50% of 5) = 5 + 2.5 = 7.5 units
Time taken by C = 48 ÷ 7.5 = 6.4 hours
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