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

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mandy
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A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   Updated on: 06 Nov 2018, 03:39
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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) \(2w^2 – pw + 2k = 0\)
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E

Originally posted by mandy on 14 Jun 2005, 08:55.
Last edited by Bunuel on 06 Nov 2018, 03:39, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Bunuel
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   25 Jul 2013, 03:29
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mandy 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


Notice that we can discard options A, and C right away. The sum of 3 positive values Cannot be 0.

Now, assume:
Width = w = 1 inch and length = 1 inch;
Perimeter = p = 4 inches;
Area = k = 1 square inches.

Plug the values of w, p, and k into the answer choices: only for E 2w^2 - pw + 2k = 2 - 4 + 2 = 0.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   28 Nov 2017, 05:56
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mandy 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's see how the 3 are related.

Perimeter = 2*(Length + Width)
(p - 2w)/2 = Length

Area = Length * Width
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] New post   14 Jun 2005, 09:04
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mandy wrote:
. :? Hello
need help thanks
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) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0



C :

K = [(P-2W)/2] *W
thus 2k = PW - 2W^2
= 2w^2 - PW + 2k = 0
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   14 Jun 2005, 11:40
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mandy wrote:
. :? Hello
need help thanks
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) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0


Another way to solve the problem is to choose a rectangle of your own.

I chose one , in which each side was 1 unit.
hence w =1 , p =4 and k = 1.

Only E satisfied these values.

HMTG.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   25 Jul 2013, 03:18
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mandy wrote:
. :? Hello
need help thanks
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) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0


Let l be the length of the board, then p = 2l + 2w, k = w*l
By plugging, we can find that the answer is E:
\(2w^2 - w(2l + 2w) + 2w*l = 0\), thus
2w - 2l - 2w + 2l = 0
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   25 Jul 2013, 03:24
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mandy 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


Similar question to practice: an-equilateral-triangle-is-inscribed-in-a-circle-if-the-130556.html
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   09 Apr 2016, 10:24
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mandy 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


lets say length is X.

area K = WX, hence X= K/W
perimeter P = 2W+2X

P = 2W + 2K/W
PW = 2W^2 + 2K

2w^2 - pw + 2k = 0 -- E
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   11 Oct 2017, 01:24
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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] New post   29 Nov 2017, 18:13
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mandy 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] New post   12 Oct 2019, 07:35
If you're just terrible at geometry, just remember that:

P=2(l+w)
A=wl=k inches^2

Just with this, you know there must be 2w in the equation, which eliminates A & B, which will then lead you to logic: C is too big, D is too small, E is just balanced: Bang, it's found in 45 seconds :')
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   28 Feb 2021, 00:01
Weird question. I didn't even really understand what it was asking initially...took me 4 minutes.

I did it algebraically.

A = lw
P = 2l + 2w

We can quickly eliminate A and C because those will always be positive.

Plug in the equations above for the remainder.

E is the answer.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   15 Mar 2021, 04:34
Another way to solve it, allthough already explaind Plugin values,
say w= 2
so P= 8
and K= 4
Lets put thesde values in options, only E satisfies.
As pointed out by expert choice A&C can be directly discarded as sum of positive values never be zero.

Amar P
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   23 May 2021, 12:55
Start with defining a rectangle that has width of W (mentioned in the question) and length of L (NOT mentioned in the question).

The question stem provides that the AREA of the rectangle is equal to K, so with our variables above we can say that LW = K. Since L is NOT mentioned in the question we can set L to be in terms of variables that ARE mentioned in the question stem which is K/W, so now we have a rectangle with length of K/W and width of W.

Finally the question stem states that the perimeter is equal to P. So 2K/W + 2W = P, which means that E is the correct answer.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink] New post   29 Mar 2024, 04:36
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