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in Commercial Mathematics by (49.2k points)
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Three men, four women and six children can complete a work in seven days. A woman 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

1 Answer

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Best answer

(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.

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