6 men and 8 boys can finish a work in 10 days while 26 men and 48 boys can do the same in 2 days. How many days will 15 men and 20 boys take in completing the same work?

Explanation: Let 1 man's 1 day work = \(m\) and 1 boy's 1 day work = \(b\). \((6m + 8b) \times 10 = 1\) (Total work). \((26m + 48b) \times 2 = 1\). From \((6m + 8b) \times 10 = 1\), we get \(60m + 80b = 1\). From \((26m + 48b) \times 2 = 1\), we get \(52m + 96b = 1\). Subtracting the second from the first: \((60m+80b) - (52m+96b) = 0 \Rightarrow 8m - 16b = 0 \Rightarrow m = 2b\). Substitute \(m=2b\) into \(60m + 80b = 1\): \(60(2b) + 80b = 1 \Rightarrow 120b + 80b = 1 \Rightarrow 200b = 1\). So \(b = \frac{1}{200}\) and \(m = \frac{2}{200} = \frac{1}{100}\). Work of \(15\) men and \(20\) boys in one day: \(15m + 20b = 15(\frac{1}{100}) + 20(\frac{1}{200}) = \frac{15}{100} + \frac{10}{100} = \frac{25}{100} = \frac{1}{4}\). Time taken is \(1 \div \frac{1}{4} = 4\) days.