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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 Updated on: 06 Nov 2018, 02:39
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A
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E

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Question Stats:

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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) \(2w^2 – pw + 2k = 0\)

Originally posted by mandy on 14 Jun 2005, 07:55.
Last edited by Bunuel on 06 Nov 2018, 02:39, edited 2 times in total.
Renamed the topic, edited the question and added the 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 25 Jul 2013, 02:29
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2
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]

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New post 14 Jun 2005, 10:40
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2
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]

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New post 14 Jun 2005, 08:04
1
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]

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New post 25 Jul 2013, 02:18
1
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]

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New post 25 Jul 2013, 02:24
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 09 Apr 2016, 09:24
1
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]

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New post 11 Oct 2017, 00:24
1
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 28 Nov 2017, 04:56
2
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]

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New post 29 Nov 2017, 17:13
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 &nbs [#permalink] 29 Nov 2017, 17:13
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