2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work? ans is 8 or 12.5?

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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work? ans is 8 or 12.5?

Mathematics
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I think the answer would be 12.5
because for it to be done in 8 day you need 3 men and 2 boys.. so if you have 2 men 1 boy it's going to take a few extra days. Do you get it? @yashiii
@yashiii you there?

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yup....i think i solved it incorrectly .my ans was 8 ..hehhe
Oh ok, i have to go. Good luck solving the problem. sorry
If I call with x the working rate of a man, and with y the working rate of a boy, then I can write this algebraic system: \[\Large \left\{ \begin{gathered} 2x + 3y = \frac{W}{{10}} \hfill \\ 3x + 2y = \frac{W}{8} \hfill \\ \end{gathered} \right.\] where W is the work to be done
  • phi
2 men and 3 boys can do a piece of work in 10 days 2 men and 1 boy will take longer (less help) so if it's a choice between 8.5 and 12.5, pick 12.5
a similar approach is to solve \[10(2x+3y)=1\\ 8(3x+2y)=1\] which is not that elegant but works
but as @phi said, common sense tells you it is longer than 8 it is in fact 12.5
actually i want correct ans...those ans options were imaginary .:(
how did you guess at the correct answer of 12.5?
now i got ans 13.5 , so any one can pls tell me the accurate ans
12.5 is correct
@satellite73 , on any website someone solved it ,but other people were saying that ans is wrong. so m confused
we can do it step by step, although i guess there probably a better method than i used
put men's rate as x, boys at y the first two statements translates as \[10(2x+3y)=1\\ 8(3x+2y)=1\]
solving this for x and y gives \[x=\frac{7}{200},y=\frac{1}{100}\]
@Michele_Laino , you have ne idea
acha..okay..got it now :)
oops i means solve \[T\left(\frac{14}{200}+\frac{1}{100}\right)=1\]
and that gives \(T=12.5\)
okay
I agree with @satellite73 we have to solve that system above, so we can find the requested time
like i said, there may be a snappier way to do it, i don't know it
@satellite73 , @Michele_Laino , thankyou so much both of you :)
:)

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