6 men can complete a piece of work in 12 days, 8 women can complete the same piece of work in 18 days and 18 children can do it in 10 days. 4 men, 12 women and 20 children do the work for 2 days. If the remaining work be completed by men only in 1 day, how many men will be required ?

Answer: Option A

6 men will complete the work in 12 days
1 men will complete the work in (6 × 12) = 72 days
8 women will complete two work in 18 days
1 women will complete the work in (8 × 18) = 144 days
18 children will complete the work in 10 days
1 children will complete the work in (18 × 10) = 180 days
$$\begin{array}{rl}& \text{1 men’s 1 day’s work}=\frac{1}{72}\\ & \text{1 women’s 1 day’s work}=\frac{1}{144}\\ & \text{1 children’s 1 day’s work}=\frac{1}{180}\end{array}$$
(4 men + 12 women + 20 children)'s 2 day's work
$$\begin{array}{rl}& =2\left(\frac{4}{72}+\frac{12}{144}+\frac{20}{180}\right)\\ & =2\left(\frac{1}{18}+\frac{1}{12}+\frac{1}{9}\right)\\ & \text{L}\text{.C}\text{.M of 18, 12 and 9}=\text{36}\\ & \text{ = }\frac{2\left(2+3+4\right)}{36}\\ & =\frac{1}{2}\\ & \therefore \text{Remaining work}=\frac{1}{2}\\ & \therefore \text{Required number of men}\\ & =72\times \frac{1}{2}\\ & =36\end{array}$$