Question 61

A, B and C. each working alone, can finish a piece of work in 27, 33 and 45 days respectively. A starts by working alone for 12 days, then B takes over from A and works for 11 days. At this stage C takes over from B and completes the remaining work. In how many days the whole work was completed ?

Solution

A's 1 day's work = $$\frac{1}{27}$$

=> A's 12 days work = $$\frac{1}{27} \times 12 = \frac{4}{9}$$

Remaining work = $$1 - \frac{4}{9} = \frac{5}{9}$$

Similarly, B's 11 days work = $$\frac{1}{33} \times 11 = \frac{1}{3}$$

Remaining work = $$\frac{5}{9} - \frac{1}{3} = \frac{2}{9}$$

Now, days taken by C to complete the work = 45 days

=> Days taken to complete $$\frac{2}{9}$$ th of the work = $$\frac{2}{9} \times 45 = 10$$ days

$$\therefore$$ Days in which the whole work is completed = 12 + 11 + 10 = 33 days


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