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by resilient » Sun Apr 06, 2008 11:52 am
A certain square has sides of length X. If the length of each side is doubled, then which of the following represents teh areea of the resulting square, in terms of x?

a.x^2
b.2x^2
c.4x
d.4x^2
e.x^2+4



qa is d


I choose b, heres why:

Plugging in numbers for the original square say x =2 the new square is x=4, so 4x4=16. Now, going into the answer choices plug in x as 2 and look for 16. choice d gives 64 when choice b gives 16. Why is d correct then?










posted this question before but coulnt find it.
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by mmukher » Sun Apr 06, 2008 4:35 pm
The answer choices are given in terms of x. I would solve it like this.

Square 1 : side = x, area = x^2

Square 2: side = 2x, area = (2x)^2 = 4x^2


It works even when you plug in your numbers (2) in 4x^2 = 4(2)^2 = 16
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yes

by resilient » Sun Apr 06, 2008 5:50 pm
I need a solution wher ai plug in numbers from the beginning and avoid the algebra!
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by rros0770 » Sun Apr 06, 2008 6:28 pm
Your initial setup is fine Engin if you're looking to plug in numbers and skip the algebra. It's just your order of operations (PEMDAS)...


Using x=2, Choice B yields 8, not 16

Using x=2, Choice D yields 16, not 64

you have to SQUARE the X value, before multiplying it by the other number.

So for Choice D, using X=2:

4(2^2)=

4(4)=

16

Hope that helps
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by Stuart@KaplanGMAT » Sun Apr 06, 2008 8:51 pm
The problem is that you're making mistakes when you plug numbers into the choices.

with x=2:

(b) 2x^2 = 2 * x^2 = 2 * 4 = 8

(d) 4x^2 = 4 * x^2 = 4 * 4 = 16

You interepreted them as:

(b) 2x^2 = (2x)^2 = (2*2)^2 = 4^2 = 16

(d) 4x^2 = (4x)^2 = (4*2)^2 = 8^2 = 64
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by cjiang16 » Mon Apr 07, 2008 11:22 pm
thanks, stuart
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