(a) 7
Let 1 woman’s 1 days’ work = x. Then
1 man’s 1 days’ work = \(\frac{X}{2}\) and 1 child’s 1 days’ work = \(\frac{X}{4}\)
Given, (3 men’s + 4 women’s + 6 children’s) 1 days’ work = \(\frac{1}{7}\)
\(\Rightarrow\) \(\frac{3X}{2}\)+ 4x + \(\frac{6X}{4}\) = \(\frac{1}{7}\) \(\Rightarrow\) \(\frac{6X+16X+6X}{4}\) = \(\frac{1}{7}\)
\(\Rightarrow\) \(\frac{28X}{4}\) = \(\frac{1}{7}\) \(\Rightarrow\) x = \(\frac{1}{49}\)
\(\therefore\) 1 woman can alone complete the work in 49 days To complete the work in 7 days, number of women required = \(\frac{49}{7}\) = 7.