Two men can complete a piece of work in 6 days. Two women can complete the same work in 9 days, whereas 3 children can complete the same piece of work in 8 days. 3 women and 4 children worked together for 1 day. If only men were to finish the remaining work in 1 day, how many men would be required?

Given:

2 men can finish the work = 6 days

2 woman can finish the work = 4 days

3 children can finish the work = 8 days

Formula used:

Total work = efficiency × time

Formula used:

Work done per day by a group = 1 / (Total days taken by the group)

Calculation:

Work done by 2 men in 1 day = \(\frac{1}{6}\)

Work done by 1 man in 1 day = \(\frac{1}{6} \div 2 = \frac{1}{12}\)

Work done by 2 women in 1 day = \(\frac{1}{9}\)

Work done by 1 woman in 1 day = \(\frac{1}{9} \div 2 = \frac{1}{18}\)

Work done by 3 children in 1 day = \(\frac{1}{8}\)

Work done by 1 child in 1 day = \(\frac{1}{8} \div 3 = \frac{1}{24}\)

Work done by 3 women in 1 day = \(3 \times \frac{1}{18} = \frac{1}{6}\)

Work done by 4 children in 1 day = \(4 \times \frac{1}{24} = \frac{1}{6}\)

Total work done by 3 women and 4 children in 1 day = \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\)

Remaining work = \(1 - \frac{1}{3} = \frac{2}{3}\)

Let x men complete the remaining work in 1 day:

\(x \times \frac{1}{12} = \frac{2}{3}\)

⇒ \(x = \frac{2}{3} \times 12 = 8\)

∴ The required number of men = 8