A and B together can complete a work in 7.2 days, while B and C together can complete the same work in 12 days. If A is twice efficient than C, then find the time taken by A alone to complete the work.

Let the total work be LCM of (72 and 12) = 72 units Number of units of work done by A and B together in one day = 72/7.2 = 10 units Number of units of work done by B and C together in one day = 72/12 = 6 units Number of units of work done by (A + 2B + C) together in one day = 10 + 6 = 16 units Since, A is twice efficient than C. So, number of units of work done by (2C + 2B + C) together in one day = 10 + 6 = 16 units 2C + 2B + C = 16 2(B + C) + C = 16 2 × 6 + C = 16 C = 4 units So, number of units of work done by C in one day = 4 units Number of units of work done by B in one day = 6 – 4 = 2 units Number of units of work done by A in one day = 10 – 2 = 8 units So, time taken by A alone to complete the work = 72/8 = 9 days