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ericuva
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GMATPREP EXAM 1 (1) 18 Sep 2007, 22:21
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Does anyone have an efficient (test scenario) way to solve the problem below?

Thanks



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GMATBLACKBELT
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GMATPREP EXAM 1 (1) 19 Sep 2007, 00:06
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Wow aside from elim A and B. U got me on this one....
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GK_Gmat
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GMATPREP EXAM 1 (1) 19 Sep 2007, 00:11
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ericuva
Does anyone have an efficient (test scenario) way to solve the problem below?

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This one is straightfwd.


Ar. of circle = pi*r^2
Since the radius of circle is r, circumference = 2pi*r
Therefore, amt. of wire used to make the square = perimeter of the circle= 4s = 40 - 2pi*r
Therefore, s = 10- 1/2*pi*r and ar. of sq. = (10 - 1/2*pi*r)^2

Ar. of circle + ar. of sq. = pi*r^2 + (10 - 1/2*pi*r)^2
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mastergmat1
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GMATPREP EXAM 1 (1) 19 Sep 2007, 11:06
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This is combination of Algebric equation and geometry

if one half is = L

Second half = ( 40 -L )

2 * pi * r = L

4 A = 40 - L

Substituting for L , you can find out the value of Side of the square in terms of R

A = 10 - Pi * r/ 2

Adding areas of Square = Pi r ^2 + ( 10 - Pi * R/ 2)
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ashkrs
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GMATPREP EXAM 1 (1) 19 Sep 2007, 11:22
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ericuva
Does anyone have an efficient (test scenario) way to solve the problem below?

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Plane simple geometry E

if its cut in x (circle) and 40-x for square.

2 *pi * r = x

total area = pi*r^2 + [(40-x)/2]^2

substitute x and get the answer.



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