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?

Answer: Option E

Given,
A is 60% more efficient of B Means,
$$\begin{array}{rl}& \text{A}=\text{B}+60\%\text{of B}\\ & \text{A}=\text{B}+\frac{60\text{B}}{100}\\ & \text{A}=\frac{100\text{B}+60\text{B}}{100}\\ & \text{A}=\frac{160\text{B}}{100}\\ & \text{A}=\frac{8\text{B}}{5}\end{array}$$
A can complete whole work in 15 days. So,
One day work of A = $$\frac{1}{15}$$
One day work of A = $$\frac{8\text{B}}{5}$$ = $$\frac{1}{15}$$
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 = $$\frac{8\text{B}}{5}$$
A can complete whole work in 15 days
Work rate of A = $$\frac{100}{15}$$ = 6.66% per day
Work rate of $$\frac{8\text{B}}{5}$$ = 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