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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

by guerrero » Wed Sep 11, 2013 12:37 pm
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

OAE
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by [email protected] » Wed Sep 11, 2013 1:43 pm
Hi guerrero,

I'm going to give you a nudge and have you try this question again:

1) Draw a picture of a rectangle.
2) TEST VALUES. Pick a value for the width and the length
3) Calculate anything else that you need to calculate

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by vinay1983 » Wed Sep 11, 2013 7:01 pm
Suppose,

L= 2 in, W= 4 in

Then Area=8 in and Perimeter = 12 inches, let us test the equations

(A) w^2 + pw + k = 0 16+48+8 not equal to zero
(B) w^2 - pw + 2k = 0 16-48-16 not equal to zero
(C) 2w^2 + pw + 2k = 0 32+48+16 not equal to zero
(D) 2w^2 - pw - 2k = 0 32-48-16 not equal to zero
(E) 2w^2 - pw + 2k = 0 32-48+16 equals zero

Option E

Also we could have eliminated options A and C since if the equations have to be equal to zero, then we need at least one expression as negative.


Hope it helps!
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by guerrero » Wed Sep 11, 2013 10:40 pm
[email protected] wrote:Hi guerrero,

I'm going to give you a nudge and have you try this question again:

1) Draw a picture of a rectangle.
2) TEST VALUES. Pick a value for the width and the length
3) Calculate anything else that you need to calculate

GMAT assassins aren't born, they're made,
Rich
thanks Rich , I used height as "H" & Width as "W" . Equated the assumed value( H )in Perimeter and Area formula and got the required equation in ~1 Min .

Did not realize it was that simple :(
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by sanjoy18 » Wed Sep 11, 2013 10:49 pm
Let Length =a width = w (given)

hence perimeter = 2*(a+w)=p ----(A)
areq= a*w=k
>> a= k/w -----------(B)

2* ((k/w)+w)=p using (B)
> 2W^2 -pw+ 2k=0

hence E
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by ganeshrkamath » Wed Sep 11, 2013 11:41 pm
guerrero 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

OAE
Since w,p, and k are all positive, eliminate A and C.
p = 2w + 2l (l is the length)
k = lw
2k = (2l) w
2k = (p - 2w)w
2k = pw - 2w^2
2w^2 - pw + 2k = 0

Choose E

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by [email protected] » Thu Sep 12, 2013 11:07 am
Hi vinay1983, guerrero et al,

TESTing values makes these types of questions really easy to solve.

Here's another tactic that can save you some time:

Notice how the answer choices have similar "pieces"? For example, "pw" shows up in each answer. Rather than do THAT calculation 5 separate times, you can do it ONCE and plug the result into all 5 answers. This approach, when available, can create a significant shortcut and save you time.

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by Scott@TargetTestPrep » Mon Dec 18, 2017 7:52 am
guerrero 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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