A can do a certain job in $120\text{ days}$. B is $60\%$ more efficient than A. To do the same job, B alone would take how many days?

Let the efficiency of A be $100$ units/day.
Since B is $60\%$ more efficient than A, efficiency of B = $160$ units/day.

The ratio of efficiency of A to B is:

$$E_A:E_B=100:160=5:8$$

Since efficiency and time taken are inversely proportional:

$$\frac{T_B}{T_A}=\frac{E_A}{E_B}$$

$$\frac{T_B}{120}=\frac{5}{8}$$

$$T_B=\frac{5}{8}\times 120=5\times 15=75\text{ days}$$