Question

    'A', 'B' and 'C', alone can complete a piece of work

    alone in 12 days, 15 days and 20 days, respectively. 'A' started working alone. In how many days will the work be completed if 'A' is assisted by both 'B' and 'C' on every 4th day?
    A 12 days Correct Answer Incorrect Answer
    B 10 days Correct Answer Incorrect Answer
    C 7(4/5) days Correct Answer Incorrect Answer
    D 9(1/5) days Correct Answer Incorrect Answer

    Solution

    Let the total work be '60x' units. So, efficiency of 'A' = 60x ÷ 12 = '5x' units/day efficiency of 'B' = 60x ÷ 15 = '4x' units/day efficiency of 'C' = 60x ÷ 20 = '3x' units/day Work done by 'A', 'B' and 'C' together in 4 days = Work done by 'A' alone in 3 days + Work done by 'A', 'B' and 'C' together on 4th day Work done by 'A', 'B' and 'C' together in 4 days = 5x X 3 + (5x + 4x + 3x) X 1 = 15x + 12x = '27x' units So, Work done by 'A', 'B' and 'C' together in next 4 days = '27x' units Amount of work done in first 8 days = 27x X 2 = '54x' units Remaining work = 60x - 54x = '6x' units So, time taken by 'A' to complete the remaining work = 6x ÷ 5x = 1(1/5) days So, total time taken = 8 + 1(1/5)  = 9(1/5) days

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