Question
A can complete a work in 18 hours while B can complete it in 12 hours. If C is 80% 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 (18 and 12) = 36 units
Work by A in one hour = 36 ÷ 18 = 2 units
Work by B in one hour = 36 ÷ 12 = 3 units
Work by A and B together = 2 + 3 = 5 units
C is 80% more efficient → C’s work = 5 + (80% of 5) = 5 + 4 = 9 units
Time taken by C = 36 ÷ 9 = 4 hours
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