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

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

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New post 11 Oct 2017, 00:02
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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

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

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New post 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

Answer E.

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

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New post 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

Hence Answer is Option E.


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

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New post 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

Answer: E

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

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New post 12 Oct 2017, 06:41
Image

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


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

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New post 12 Oct 2017, 06:42
Adityagmatclub wrote:
Image

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.

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

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New post 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

Answer: E

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

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New post 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

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