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Poots
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Geometry Problem 22 Oct 2008, 17:42
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Please explain this.

Thanks!



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LiveStronger
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Geometry Problem 22 Oct 2008, 17:48
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B - 24/25

4s = 2(l+w)
l/w = 2/3
so,
s = 5/6w

Question asks for (l*w)/(s^2) = (2/3)/(25/36) = 24/25
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LiveStronger
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Geometry Problem 22 Oct 2008, 17:49
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Poots: I think you are posting your questions in the wrong section :P
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Geometry Problem 22 Oct 2008, 18:06
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Oh, you're so right, LS. I am posting them wrong. See what GMAT does to me?!
Do you know how I can move them or is all hope lost?
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bigtreezl
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Geometry Problem 22 Oct 2008, 23:40
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just make the sides of the rectangle 2 and 3, the area is 6
that means that the perimeter is 10 (2+3+2+3)
the perimeter of the square is 10 also so each side is 10/4
the area of the square is 10/4*10/4=100/16=25/4
so the ratio of area of R to S is 6 / 25/4 or 24/25
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amitdgr
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Geometry Problem 23 Oct 2008, 04:35
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LiveStronger
B - 24/25

4s = 2(l+w)
l/w = 2/3
so,
s = 5/6w

Question asks for (l*w)/(s^2) = (2/3)/(25/36) = 24/25
Show more

l/w = 2/3 ---> shouldn't it be w/l = 2/3 ?? length > width ,
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linau1982
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Geometry Problem 23 Oct 2008, 08:47
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Rectangular's sides are 2, 3, 2, 3, therefore perimeter is 10
If the square has the same perimeter - 10 - that means that the lengh of one side is 2,5
Area of rectangular is 2*3=6
Area of square 2,5*2,5=6,25
Although we do not have answer choice for that, and we need integers in ratio, therefore we multiply both areas by 4 and get 24:25
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Geometry Problem 23 Oct 2008, 10:26
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amitdgr
LiveStronger
B - 24/25

4s = 2(l+w)
l/w = 2/3
so,
s = 5/6w

Question asks for (l*w)/(s^2) = (2/3)/(25/36) = 24/25

l/w = 2/3 ---> shouldn't it be w/l = 2/3 ?? length > width ,
Show more

either way will work amit. Make sure you calculate the right value for 's' though :)



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!