A and B together can complete a work in 8 days. B alone can complete the work in 24 days. In how many days A alone can complete the same work?
Let total work is L.C.M. (8,24) = 24 units
B alone can complete the work in 24 days, => B's efficiency = $$\frac{24}{24}=1$$ unit/day
Let A's efficiency = $$x$$ units/day
Similarly (A+B)'s efficiency = $$\frac{24}{8}=3$$ units/day
Now, (A+B)'s 1 day's work = $$x+1=3$$
=> $$x=3-1=2$$ units/day
$$\therefore$$ Days required by A alone to complete the work = $$\frac{24}{2}=12$$ days
=> Ans - (C)
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