x men y work

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x men y work

by maihuna » Mon May 04, 2009 10:17 am
If x men working x hrs/day can do x units of work in x days, then y men working y hrs/day would be able to complete how many units of works in y days:

x^2/y^3
x^3/y^2
y^2/x^3
x^2/y^3
y^3/x^2
x/y^3
Source: — Problem Solving |

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by PussInBoots » Mon May 04, 2009 11:51 am
P = how much of 1 unit of work can 1 worker finish in 1 hour, unit / (worker * hour)

We have: P * x hrs/day * x days * x workers = x units, hence P = 1 / x^2
We need to find: P * y hrs/day * y days * y workers = y^3 / x^2

"y^3/x^2" is the answer

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by maihuna » Tue May 05, 2009 5:32 am
PussInBoots wrote:P = how much of 1 unit of work can 1 worker finish in 1 hour, unit / (worker * hour)

We have: P * x hrs/day * x days * x workers = x units, hence P = 1 / x^2
We need to find: P * y hrs/day * y days * y workers = y^3 / x^2

"y^3/x^2" is the answer
Hi puss , nope thats not correct...

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by 2010gmat » Tue May 05, 2009 6:12 am
i believe puss is right...y^3/x^2 shud be the ans

is there any catch??

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by lilu » Tue May 05, 2009 6:19 am
2010gmat wrote:i believe puss is right...y^3/x^2 shud be the ans

is there any catch??
Same here, don't see why y^3/x^2 is not the answer....
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by maihuna » Tue May 05, 2009 7:43 am
Ohh sorry met it is fine, puss putted corret, sorry for confusion