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# A rectangular circuit board is designed to have width w inch

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A rectangular circuit board is designed to have width w inch [#permalink]

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14 Jun 2005, 08:55
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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

Last edited by Bunuel on 25 Jul 2013, 03:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: ps CIRCUIT BOARD [#permalink]

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14 Jun 2005, 09:04
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: ps CIRCUIT BOARD [#permalink]

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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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14 Jun 2005, 16:26
i think choosing own rectangle = easiest and safest

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Re: ps CIRCUIT BOARD [#permalink]

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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 inch [#permalink]

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25 Jul 2013, 03:24
Expert's post
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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 inch [#permalink]

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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.

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 inch [#permalink]

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25 Jul 2013, 04:47
In order to solve this problem algebraicaly we need the following formulaes:

$$Perimeter = 2*w + 2*h$$
$$Area = w * h$$

We begin by observing what information we have. The left-hand side in both of the above formulae is known, and defined as p and k. We also have the width, w. We can then use what we know to find the height, by rearranging terms:

$$p= 2*w + 2*h$$ ===> $$h = \frac{p-2*w}{2}$$.

Slot this into the formulae for the area to obtain

$$k= w * \frac{p-2*w}{2}$$ =====> $$2w^2 - pw + 2k = 0$$

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

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25 Jul 2013, 21:10
For a rectangle, if one side = w & perimeter = p, then the other side is (p-2w)/2
So, area k = w x (p-2w)/2
Solving the above, we get answer = E
(2w^2 - pw + 2k = 0)
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Re: A rectangular circuit board is designed to have width w inch [#permalink]

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19 Nov 2013, 07:24
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

Smart numbers here guys

w = 2
l = 3

So P = 10
A = 6

Start by E always on these types of questions.
E is the correct answer choice.

And also try to use different numbers just in case.
Here we used different numbers for length and width
Needless to say try to avoid 0,1, or any negative numbers

Cheers
J

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

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16 May 2014, 03:35
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

I supposed the rectangle to be a square and w = 1 , then P = 4 and k = 1
(A) & (C) always positive so we can safely eliminate those choices
and a quick calculation allows to eliminate (B) and (D)
hence the correct answer is (E)

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

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16 Feb 2016, 13:50
let k/w=rectangle length
let p/2-w=rectangle length
k/w=p/2-w
2w^2-pw+2k=0

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

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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 inch [#permalink]

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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

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

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29 Nov 2017, 18:13
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

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Re: A rectangular circuit board is designed to have width w inch   [#permalink] 29 Nov 2017, 18:13
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