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1. A is 60% more efficient than B. In how many days will A and B working together complete a piece of work which A alone takes 15 days to finish?

A. $\frac{124}{13}$ days

B. $\frac{113}{13}$ days

C. $\frac{108}{13}$ days

D. $\frac{131}{13}$ days

Solution:

Given,
A is 60% more efficient of B Means,

\begin{aligned} A = B + 60 % of B \\ A = B + \frac{60 B}{100} \\ A = \frac{100 B + 60 B}{100} \\ A = \frac{160 B}{100} \\ A = \frac{8 B}{5} \end{aligned}

A can complete whole work in 15 days. So,
One day work of A =

\frac{1}{15}

One day work of A =

\frac{8 B}{5}

\frac{1}{15}

One day work of B =

\frac{5}{120}

\frac{1}{24}

One day work, (A + B) =

\frac{1}{15} + \frac{1}{24}

One day work, (A + B) =

\frac{24 + 15}{360}

\frac{39}{360}

So, Time taken to finish the work by A and B together =

\frac{360}{39}

\frac{120}{13}

days

Alternatively
We can solve it through percentage method
Given,
A =

\frac{8 B}{5}

A can complete whole work in 15 days
Work rate of A =

\frac{100}{15}

= 6.66% per day
Work rate of

\frac{8 B}{5}

= 6.66%
Work rate of B = 4.16% per day
Work rate of (A + B) = 6.66 + 4.16 = 10.82% per day
So, A and B can complete 100% work in

\frac{120}{13}

days

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