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05 Dec 2014, 07:34
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B
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Difficulty:
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23%
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Tough and Tricky questions: Geometry.
A rectangular circuit board is designed to have a width of W inches, a length of L inches, a perimeter of P inches, and an area of A square inches. Which of the following equations must be true?
A) 2W^2 + PW + 2A = 0
B) 2W^2 − PW + 2A = 0
C) 2W^2 − PW − 2A = 0
D) W^2 + PW + A = 0
E) W^2 − PW + 2A = 0
Kudos for a correct solution.
Source: Chili Hot GMAT
05 Dec 2014, 11:24
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Bunuel
P = 2(L+W)-----------------1)
A= LW------------------------2)
option A is not possible why ?? because all the terms are positive.
lets try option B , put value of P and A from 1 and 2 we have,
2W^2-2(L+W)W + 2(LW)
2W^2 - 2LW - 2W^2 + 2LW=0.
hence answer is B.
intheend14
GMAT 1: 740 Q49 V41
Posts: 125
05 Dec 2014, 11:32
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Choice B is the answer.
We know area = w*l and perimeter = 2w + 2l
Let's see what some of these terms equal to: 2w^2 = 2w^2, pw = 2w^2 + 2wl and a = w*l
We need all the terms to cancel out and we can see that each term in pw goes with either area or 2w^2, so we probably need either pw to be negative or 2A and 2W^2 to be negative--Choice B does this
2w^2 -pw + 2A = 2w^2 - 2w^2 - 2wl + 2wl, which all cancels out.
B
We know area = w*l and perimeter = 2w + 2l
Let's see what some of these terms equal to: 2w^2 = 2w^2, pw = 2w^2 + 2wl and a = w*l
We need all the terms to cancel out and we can see that each term in pw goes with either area or 2w^2, so we probably need either pw to be negative or 2A and 2W^2 to be negative--Choice B does this
2w^2 -pw + 2A = 2w^2 - 2w^2 - 2wl + 2wl, which all cancels out.
B
