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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
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B
8
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C
14
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D
12
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Solution

The correct option is A 7
Let 1 woman's 1 days' work = x Then 1 man's 1 days' work = x2 and 1 child's 1 days' work = x4
Given (3 men's+ 4 women's + 6 children's) 1 day's work = 17
3x2+4x+6x4=176x+16x+6x4=17
28x4=17x=149 1 women can alone complete the work in 49 days To complete the work in 7 days
number of women required = 497=7

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