Question
A can do a work in 30 hours while B can do it in 20
hours. If C is 60% 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 (30 and 20) = 60 units
Work by A in one hour = 60 ÷ 30 = 2 units
Work by B in one hour = 60 ÷ 20 = 3 units
Work by A and B together = 2 + 3 = 5 units
C is 60% more efficient → C’s work = 5 + (60% of 5) = 5 + 3 = 8 units
Time taken by C = 60 ÷ 8 = 7.5 hours
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