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(A) Let the work done by a man in one day be $x$.Since a woman does double the work of a man,the work done by a woman in one day is $2x$.Since a child

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$7$

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(A) Let the work done by a man in one day be $x$.Since a woman does double the work of a man,the work done by a woman in one day is $2x$.Since a child does half the work of a man,the work done by a child in one day is $\frac{x}{2}$.The total work done by $3$ men,$4$ women,and $6$ children in one day is $3x + 4(2x) + 6(\frac{x}{2}) = 3x + 8x + 3x = 14x$.Given that they complete the work in $7$ days,the total work is $14x 	imes 7 = 98x$.We need to find the number of women $(n)$ required to complete the same work in $7$ days.The work done by $n$ women in $7$ days is $n 	imes (2x) 	imes 7 = 14nx$.Equating the total work: $14nx = 98x$.Dividing both sides by $14x$,we get $n = 7$.Therefore,$7$ women are required to complete the work in $7$ days.

Three men,four women and six children can complete a work in $7$ 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 this work in $7$ days?

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