3 men or 7 women can do a piece of work in 32 days. Find the number of days required by 5 men and 7 women to do a piece of work twice as large
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Let work done by 1 man in 1 day =m=m
work done by 1 woman in 1 day =w=w
Work done by 3 men in 1 day =132=132
3m=132m=1963m=132m=196
Work done by 7 men in 1 day =132=132
7w=132w=12247w=132w=1224
Work done by 7 men and 5 women in 1 day
=7m+5w=796+5224=49+15672=64672=221=7m+5w=796+5224=49+15672=64672=221
Therefore, 7 men and 5 women can do the same work in 212212 days
To do work twice as large, time required
=212×2=21=212×2=21 days
work done by 1 woman in 1 day =w=w
Work done by 3 men in 1 day =132=132
3m=132m=1963m=132m=196
Work done by 7 men in 1 day =132=132
7w=132w=12247w=132w=1224
Work done by 7 men and 5 women in 1 day
=7m+5w=796+5224=49+15672=64672=221=7m+5w=796+5224=49+15672=64672=221
Therefore, 7 men and 5 women can do the same work in 212212 days
To do work twice as large, time required
=212×2=21=212×2=21 days
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