Question
A can finish a job in 30 hours while B can finish it in
20 hours. If C is (300/7)% more efficient than A and B together, then find the time taken by C alone to complete the work.Solution
ATQ, Let the total work be LCM of (30 and 20) = 60 units
Work done by A in one hour = 60 ÷ 30 = 2 units
Work done by B in one hour = 60 ÷ 20 = 3 units
Work done by A and B together in one hour = 2 + 3 = 5 units
Work done by C in one hour = 5 + (300/7)% of 5 = 5 + 3 = 8 units
Time taken by C alone = 60 ÷ 8 = 7.5 hours
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