## GMAT Test 2_PS Square #15

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### GMAT Test 2_PS Square #15

by kwah » Tue Apr 03, 2012 7:09 pm
Attached is a question from GMAT Prep Test 2.

Thanks,
K
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GMAT Test 2_PS Square #15.docx

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by [email protected] » Tue Apr 03, 2012 7:56 pm
kwah wrote:Attached is a question from GMAT Prep Test 2.

Thanks,
K
Area of square garden = A sq ft
Perimeter of square garden = P ft
A = 2P + 9

Let us assume that each side of the square garden = sÂ² sq ft
Then A = sÂ² and P = 4s
So, A = 2P + 9 implies sÂ² = 2 *(4s) + 9
sÂ² = 8s + 9
sÂ² - 8s - 9 = 0
sÂ² - 9s + s - 9 = 0
s(s - 9) + 1(s - 9) = 0
(s + 1)(s - 9) = 0
s = 9, (s = -1 is not possible, as side cannot be negative)

Therefore, perimeter of square garden = 4 * 9 = 36 ft

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by gmatmath » Fri Apr 06, 2012 1:47 am
Let us assume the side of the square to be 'a' cm.
So, its area A = a^2 sq.cm
Perimeter P = 4a.

Given A = 2P + 9
We will now plug in the values of A and P in the above equation.
We get,

a^2 = 2(4a) + 9
a^2 = 8a + 9
==> a^2 - 8a - 9 = 0
==> a^2 - 9a + a - 9 = 0 [factors 0f 9 such that difference is 8 and product is 9 is -9 and 1]
==> a(a - 9) + 1(a - 9) = 0
==> (a - 9)(a + 1) = 0
==> a = 9 or -1.

We will consider 'a' = 9 as the measurement of length cannot be < 0.

Hence, the perimeter P = 4a = 4(9) = 36cm.

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