Question
A alone can complete 50% of a work in 20 days while B
takes 16 days more than A to complete it. If B and C together can complete the work in 24 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 = 20/0.5 = 40 days Time taken by B alone to complete the entire work = 40 + 16 = 56 days Let the total work = L.C.M of 56, 40 and 24 = 840 units Then, efficiency of B = (840/56) = 15 units/day Combined efficiency of B and C = (840/24) = 35 units/day So, efficiency of C alone = 35 – 15 = 20 units/day So, time taken by C alone to complete the entire work = (840/20) = 42 days
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99² +? ² = 18826
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