Four men and 6 women can complete a certain piece of work in 5 days whereas three men and 4 women can complete it in 7 days. How many men should assist 25 women to complete \(2\frac{1}{2}\) times the same work in 5 days?
5
Step 1: Set up work-rate equations. Let M = work done by 1 man in 1 day, W = work done by 1 woman in 1 day. Assume total work = 1 unit.
From scenario 1: \((4M + 6W) \times 5 = 1\), so \(20M + 30W = 1\).
From scenario 2: \((3M + 4W) \times 7 = 1\), so \(21M + 28W = 1\).
Step 2: Solve the system. Equating: \(20M + 30W = 21M + 28W\) gives \(M = 2W\).
Substituting in \(20M + 30W = 1\): \(40W + 30W = 1\), so \(W = \frac{1}{70}\) and \(M = \frac{1}{35}\).
Step 3: New scenario. Let n men assist 25 women to complete \(2\frac{1}{2} = \frac{5}{2}\) units of work in 5 days.
\((nM + 25W) \times 5 = \frac{5}{2}\)
\(nM + 25W = \frac{1}{2}\)
Substitute \(M = \frac{1}{35}, W = \frac{1}{70}\):
\(\frac{n}{35} + \frac{25}{70} = \frac{1}{2}\)
\(\frac{2n}{70} + \frac{25}{70} = \frac{35}{70}\)
\(2n + 25 = 35\)
\(2n = 10 \Rightarrow n = 5\)
Answer: 5 men.
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