Question

Three men , four women and six children can compete a work in seven days. A women does double the work a man does and a child does half the work a man does . How many women alone can complete the work in 7 days ?

• A. 7
• B. 8
• C. 14
• D. 12

Answer 1

The correct option is 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 day's 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 women can alone complete the work in 49 days To complete the work in 7 days
number of women required = $\frac{49}{7}=7$