Time taken by A alone to complete the entire work = 15/0.5 = 30 days Time taken by B alone to complete the entire work = 30 + 10 = 40 days Let the total work = L.C.M of 30, 40 and 20 = 120 units Then, efficiency of B = (120/40) = 3 units/day Combined efficiency of B and C = (120/20) = 6 units/day So, efficiency of C alone = 6 – 3 = 3 units/day So, time taken by C alone to complete the entire work = (120/3) = 40 days
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