Question
A alone can complete 40% of a work in 18 days while B
takes 15 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 = 18/0.4 = 45 days Time taken by B alone to complete the entire work = 45 + 15 = 60 days Let the total work = L.C.M of 45, 60 and 20 = 180 units Then, efficiency of B = (180/60) = 3 units/day Combined efficiency of B and C = (180/20) = 9 units/day So, efficiency of C alone = 9 – 3 = 6 units/day So, time taken by C alone to complete the entire work = (180/6) = 30 days
Statements: K ≤ L, L = M, M < N
Conclusion: I. N > L II. M ≥ K
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