A, B and C can do a piece of work in 10, 15 and 30 days respectively. In how many days can A do the work if he is assisted by B and C on every 5th day?

A’s 4 days’ work = (1/10×4) = 4/10 = 2/5 work (A + B + C) 1 day work = (1/10+ 1/15+1/30) =(3 +2 +1 )/30 = 1/5work Work done in 5 days = 2/5 + 1/5 = 3/5 Remaining work = 1-3/5 =2/5 work A can do 2/5 work in 4 days. ∴ Whole work will be done in 5 + 4 = 9 days Alternate Method: A B C Time taken 10 15 30 Total work = LCM 30 units 3 : 2 : 1 ATQ, A 's work in first 4 days = 3`xx` 4 = 12 units & work in 5th day by all = 3+2+1 = 6 units So total work in 5 days = 12+6=18 units Hence remainning work = 30 - 18 = 12 units , after that again A worked alone for next 4 days @ 3 units per day So he will finish this 12 units in 4 days

Hence total days = 5+4 = 9 days

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