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A rectangular circuit board is designed to have a width of W inches, a

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Bunuel
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A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   05 Dec 2014, 07:34
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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

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Source: Chili Hot GMAT
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B
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   05 Dec 2014, 11:24
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Bunuel wrote:

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


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.
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   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
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   Updated on: 05 Dec 2014, 13:31
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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


Perimeter of the circuit board , \(P = 2W + 2L\)

As all the answer choices are in terms of A, P and W, let us express L in terms of P and W

\(L = \frac{P}{2} - W\)

Substituting the above value of L in equation \(2A = 2*W*L\),

\(2A = 2*(W*L) = 2* W *( \frac{P}{2}-W)\)

Or, \(2A = WP - 2W^2\)

Or, \(2W^2 -WP+2A =0\)

Answer: B

Originally posted by arunspanda on 05 Dec 2014, 11:57.
Last edited by arunspanda on 05 Dec 2014, 13:31, edited 1 time in total.
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   05 Dec 2014, 13:04
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arunspanda wrote:
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


Perimeter of the circuit board , \(P = 2W + 2L\)

As all the answer choices are in terms of A, P and W, let us express L in terms of P and W

\(L = \frac{P}{2} - W\)

Substituting the above value of L in equation \(2A = 2*W*L\),

\(2A = 2*(W*L) = 2* W *( \frac{P}{2}-W)\)

Or, \(2A = WP - 2W^2\)

Or, \(2W^2 -WP+2A =0\)

Answer: E


Highlight portion is option B not E.
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   06 Dec 2014, 08:16
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p=2l+2w
a=lw
putting the values in option B we get as follows
2w^2-w(2l+2w)+2lw=2w^2-2lw-2w^2+2lw=0

Hence proved.
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   08 Dec 2014, 06:03
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Bunuel wrote:

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


The correct answer is B.
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   08 Dec 2014, 23:22
p = 2w + 2l ............ (1)

a = wl

Looking at the OA, we see that "l" is not present. Hence replacing \(l = \frac{a}{w}\) in equation (1)

\(p = 2w + 2\frac{a}{w}\)

\(pw = 2w^2 + 2a\)

\(2w^2 + 2a - pw = 0\)

Answer = B
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   18 Apr 2016, 06:09
i just plugged it in real quick
l 2
w 4
a 8
p 12

Started with D, which immediatly made it clear that I need 2 times A within the same formula to get 0. This lead me to B.

Ans B
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Re: A rectangular circuit board is designed to have a width of W inches, a [#permalink] New post   28 Aug 2022, 19:29
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