Different Problem on Time and Work

“This problem is just a little bit different from previous time and work problems.In previous problems ,a work done by a person .Here a work done by more than one person. "

Question: 10 men and 15 women together can complete a work in 6 days.It takes 100 days for one man alone to complete the same work.How many days will be required for one woman alone to complete the same work?

Solution:

Consider, Workdone by 1man in 1 day=W[M]
Workdone by 1Woman in 1 day=W[W]

Given that,

10 men and 15 women =10M+15W
Therefore,Workdone by 10 men and 15 women in 1 day =W [10M+15W]
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1.Workdone by 10 men and 15 women in 1 day,
W[ 10M+15W]=1/6[Since 10 men and 15 women together can complete a work in 6 days ]
2.Workdone by 1man in 1 day,
W[M] =1/100 [one man alone complete the same work in 100 days]
3.Workdone by 10 man in 1 day,
W[10M]=10/100=1/10
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We can find ,how many days for one woman alone complete the same work using above.

W[10M+15W] =1/6
W[10M] + W[15W] =1/6
1/10 + W[15W] =1/6
W[15 W] =(1/6) –(1/10)[Take LCM and solve]
W[15 W]=(5-3)/30
W[15 W]=2/30
W[15 W]=1/15
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Workdone by 15 woman in 1 day =1/15
Workdone by 1 woman in 1 day,
(i.e)W[1W]= 1/(15*15)
W[1W]=1/225.
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Then as per concept,
If (a person 1day work )=[1/n],
then(a person) finish the work in “n days” .
So,one woman alone finish the work in 225 days