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# A rectangular circuit board is designed to have width w inches, perime

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Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [0], given: 12716

A rectangular circuit board is designed to have width w inches, perime [#permalink]

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11 Oct 2017, 00:02
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Difficulty:

35% (medium)

Question Stats:

76% (01:17) correct 24% (01:24) wrong based on 50 sessions

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A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) $$w^2 + pw + k = 0$$

(B) $$w^2 – pw + 2k = 0$$

(C) $$2w^2 + pw + 2k = 0$$

(D) $$2w^2 – pw – 2k = 0$$

(E) $$2w^2 – pw + 2k = 0$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135946 [0], given: 12716

Senior Manager
Joined: 05 Dec 2016
Posts: 253

Kudos [?]: 27 [0], given: 49

Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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11 Oct 2017, 00:24
Let x = length
p=2w+2x ===> x=1/2p-w

k=wx ===> x=k/w

Substituting:

k/w=1/2p-w
k=1/2pw-w^2
2w^2-pw+2k=0

Kudos [?]: 27 [0], given: 49

Manager
Joined: 12 Feb 2017
Posts: 71

Kudos [?]: 30 [0], given: 47

Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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11 Oct 2017, 06:17
width of rectangle= w
perimeter of rectangle= p
area of rectangle= k

perimeter of rectangle= 2(length+width)
p= 2(length+w)
p-2w= 2*length
Therefore,
length= (p-2w)/2

Area of rectangle= length*width
k= (p-2w)/2 * w
k=( p*w - 2*w^2 )/2
2*k= p*w - 2*w^2

Rearranging the terms we get,
2w^2-pw+2k=0

Kudos if it helps.

Kudos [?]: 30 [0], given: 47

Intern
Joined: 06 Oct 2017
Posts: 3

Kudos [?]: 0 [0], given: 7

Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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11 Oct 2017, 09:44
As there the length of the rectangle is not included in any of the given answers, this suggests that it should be expressed and formulated in its relation to the width, perimeter or area of the rectangle.

L - length of the rectangle
W - width of the rectangle
P - perimeter
k - area

P = 2(L+W)
P = 2L + 2W
P - 2W = 2L
Therefore the length can also be expressed as
L = (P-2W)/2

Applying this in the formula for area yields the following:
k = L*W
k = (P-2W)/2 * W
k = (PW - 2W^2)/2
2k = PW - 2W^2
Rearranging all on the same side and reversing the signs results in

2W^2-PW+2k = 0

Kudos [?]: 0 [0], given: 7

Manager
Joined: 22 May 2017
Posts: 90

Kudos [?]: 7 [0], given: 223

GMAT 1: 580 Q41 V29
GMAT 2: 580 Q43 V27
Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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12 Oct 2017, 06:41

In these type of question, try to put values in first equation and then in second equation.

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Kudos [?]: 7 [0], given: 223

Manager
Joined: 22 May 2017
Posts: 90

Kudos [?]: 7 [0], given: 223

GMAT 1: 580 Q41 V29
GMAT 2: 580 Q43 V27
Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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12 Oct 2017, 06:42

In these type of question, try to put values in first equation and then in second equation.

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In last equation. Not second.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app

Kudos [?]: 7 [0], given: 223

VP
Joined: 26 Mar 2013
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Kudos [?]: 302 [0], given: 166

Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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12 Oct 2017, 15:37
Bunuel wrote:
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) $$w^2 + pw + k = 0$$

(B) $$w^2 – pw + 2k = 0$$

(C) $$2w^2 + pw + 2k = 0$$

(D) $$2w^2 – pw – 2k = 0$$

(E) $$2w^2 – pw + 2k = 0$$

Let w =2 & Lenght =3, P =10 & k =6

Most of choices have 2k & pw...then calculate them

2k = 12 & pw = 20 ...........To get zero we need more 8 positive and deduct from 20

8 - 20 +12 = 0 ...........this resemble only one form

Kudos [?]: 302 [0], given: 166

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
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Kudos [?]: 990 [0], given: 5

Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]

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13 Oct 2017, 08:27
Bunuel wrote:
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) $$w^2 + pw + k = 0$$

(B) $$w^2 – pw + 2k = 0$$

(C) $$2w^2 + pw + 2k = 0$$

(D) $$2w^2 – pw – 2k = 0$$

(E) $$2w^2 – pw + 2k = 0$$

We can let n = the length of the rectangle and create the following equation:

2w + 2n = p

2n = p - 2w

n = (p - 2w)/2

Since area = n x w:

k = (n)(w)

k = [(p - 2w)/2]w

2k = pw - 2w^2

2w^2 - pw + 2k = 0

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Kudos [?]: 990 [0], given: 5

Re: A rectangular circuit board is designed to have width w inches, perime   [#permalink] 13 Oct 2017, 08:27
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