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A is 60% more efficient than B. In how many days will A and B together complete a piece of work if A alone can complete the work in 15 days?1). $$9\frac{7}{13}$$2). $$8\frac{9}{13}$$3). $$8\frac{4}{13}$$4). $$10\frac{1}{13}$$ |
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Answer» Solution Time TAKEN by A alone = 15 days Efficiency of A : B = 100 : 160 = 15 : 24 => Time taken by B = 24 days $\THEREFORE$ (A + B)'s 1 day's work = $\FRAC{1}{15} + \frac{1}{24}$ = $\frac{8 + 5}{120} = \frac{13}{120}$ $\therefore$ Required time = $\frac{1}{\frac{13}{120}}$ = $\frac{120}{13} = 9\frac{3}{13}$ days |
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