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

Question

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

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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)

michalky · Answer

C : 

K =   *W
thus 2k = PW - 2W^2
= 2w^2 - PW + 2k = 0

HowManyToGo · Answer

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.

thewind · Answer

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

Bunuel · Answer

Similar question to practice: an-equilateral-triangle-is-inscribed-in-a-circle-if-the-130556.html

HKD1710 · Answer

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

Alexey1989x · Answer

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.

ScottTargetTestPrep · Answer

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 =  w

2k = pw - 2w^2

2w^2 - pw + 2k = 0

Answer: E

KingAdam44 · Answer

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 :')

CEdward · Answer

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.

amarprajapati92 · Answer

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

dylanl1218 · Answer

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

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