If 9 men working 71/2 hours a day can finish a piece of work in 20 days, then how many days will be taken by 12 men, working 6 hours a day to finish the work ? (It is being given that 2 men of latter type work as much as 3 men of the former type.)

Answer: Option C

Let the required number of days be x
2 men of latter type = 3 men of former type
12 men of latter type
= $$\left(\frac{3}{2}\times 12\right)$$
= 18 men of former type
More men, Less days (Indirect proportion)
Less working hours, More days (Indirect proportion)
$$\begin{array}{c}\text{Men 18}:9\text{hfill}\\ \text{Working hrs 6}:\frac{15}{2}\text{hfill}\end{array}\}::20:x$$
$$\begin{array}{rl}& \therefore 18\times 6\times x=9\times \frac{15}{2}\times 20\\ & \iff 108x=1350\\ & \iff x=\frac{25}{2}\\ & \iff x=12\frac{1}{2}\end{array}$$