Question
A alone can complete 40% of a work in 16 days while B
takes 12 days more than A to complete it. If B and C together can complete the work in 20 days, then find the time taken by C alone to complete the same work.Solution
Time taken by A alone to complete the entire work = 16/0.4 = 40 days Time taken by B alone to complete the entire work = 40 + 12 = 52 days Let the total work = L.C.M of 40, 52 and 20 = 520 units Then, efficiency of B = (520/52) = 10 units/day Combined efficiency of B and C = (520/20) = 26 units/day So, efficiency of C alone = 26 – 10 = 16 units/day So, time taken by C alone to complete the entire work = (520/16) = 32.5 days
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