6 men can complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days, whereas 18 children can complete the piece of work in 10 days.
4 men, 12 women and 20 children work together for 2 days. If only men were to complete there remaining work in 1 day, how many men would be required totally?
I am completely clueless. No idea how to think through this problem.
one man completes $\frac{1}{6\times 12}= \frac{1}{72}$ pieces of work per day
one woman completes $\frac{1}{8\times 18}= \frac{1}{144}$ pieces of work per day
one child completes $\frac{1}{18\times 10}= \frac{1}{180}$ pieces of work per day
4 men , 12 women and 20 children can do
$$W = \frac{4}{72}+ \frac{12}{144} + \frac{20}{180} = \frac{1}{18}+\frac{1}{12}+\frac 19$$
pieces in one day - after 2 days $1-2W$ pieces of work remain, so you need to find $x$ where
$$\frac{x}{72} = 1-2W $$
you should get that 36 men are required to finish the work on the third day.