Aptitude: Time and Work

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.

• Wow, thanks for explaining so clearly. – user1502 Jun 20 '15 at 19:55

Hint: compute how much of the work one man, one woman, and one child can each complete in one day. Then add up how much is completed by the group in two days. The rest has to be finished by the men.

• In one day, a man can do 6/12, women 8/18, and children 18/10 amount of work. So one day's total work is 247/90, that's already over 1. – user1502 Jun 20 '15 at 19:41
• No, if it takes six men 12 days, each man-day is 1/72. – Ross Millikan Jun 20 '15 at 19:43
• @user1502: No, it takes 6×12 man-days to complete one piece of work, so a man does only 1/72 of a piece in one day. – Henning Makholm Jun 20 '15 at 19:43
• Ok, but now 2 days work by them would be 2 * 19/720, what now? – user1502 Jun 20 '15 at 19:51