Attached is a question from GMAT Prep Test 2.
Please advise how to achieve the result.
Answer: B
Thanks,
K
GMAT Test 2_PS Square #15
This topic has expert replies
- Attachments
-
- GMAT Test 2_PS Square #15.docx
- (54.72 KiB) Downloaded 76 times
GMAT/MBA Expert
- Anurag@Gurome
- GMAT Instructor
- Posts: 3835
- Joined: Fri Apr 02, 2010 10:00 pm
- Location: Milpitas, CA
- Thanked: 1854 times
- Followed by:523 members
- GMAT Score:770
Area of square garden = A sq ftkwah wrote:Attached is a question from GMAT Prep Test 2.
Please advise how to achieve the result.
Answer: B
Thanks,
K
Perimeter of square garden = P ft
A = 2P + 9
Let us assume that each side of the square garden = s² sq ft
Then A = s² and P = 4s
So, A = 2P + 9 implies s² = 2 *(4s) + 9
s² = 8s + 9
s² - 8s - 9 = 0
s² - 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
The correct answer is B.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
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.
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.