A rectangular plot of land is designed to have width w inches, perimet

Question

A rectangular plot of land 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

Kudos for a correct  solution .

camlan1990 · Answer

length l = k/w

p = 2l + 2w = 2k/w+2w => 2w^2 -pw+2k = 0

Answer: E

KS15 · Answer

Let length=
width=w, perimeter=2(l+w), area=lw
Using the given options, only E satisfies both sides
Answer E

kunal555 · Answer

let lenght = b

now,
p=2b+2w or 2(b+w)
pw= 2w(b+w)
k= bw

putting these values in option 1 we get
 w^2 +2w(w+b) +wb=0
or w(w+b) + 2w(w+b) = 0

But we need an option in which we get

2w(w+b) -2w(w+b)=0
or 2w^2 -pw +2k =0

Answer:- E

GMATinsight · Answer

Let, Length, L = 4
Width, W =2
i.e. Perimeter, P = 2(L+W) = 2(4+2) = 12
and Area, k = L*W = 4*2 = 8

Option A:  w^2 + pw + k = 0  Since all the terms are positive so their sum can't be zero i.e. INCORRECT OPTION
Option B:  w^2 - pw+2k=0  =  2^2 - 12*2+2*8 is NOT equal to 0 i.e. INCORRECT OPTION
Option C:  2w^2 + pw+2k=0  Since all the terms are positive so their sum can't be zero i.e. INCORRECT OPTION
Option D:  2w^2 - pw-2k=0  =  2^2 - 12*2-2*8 is NOT equal to 0 i.e. INCORRECT OPTION
  Option E:  2w^2 - pw+2k=0  =  2*2^2 - 12*2+2*8 = 8-24+16 = 0  i.e. CORRECT OPTION

Answer: Option E

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION: When you look at what the authors are hoping to do to you, you can see that, by providing you with no numbers, multiple variables, and messy equations as answer choices, they’re hopeful to distract and intimidate you with uncertainty. As with most questions, however, you know more than you think you do, so getting the question in your own terms will be helpful as you inventory your assets to get to work. You know the shape of the rectangle, and with any rectangle you know how to calculate its area and perimeter: rectangle.jpg Area: k = Lw Perimeter: p = 2L + 2w At this point, by listing the things that you know to be true, you have plenty of assets with which to work to get started on the problem. The problem is no longer in the exam’s abstract terms, but rather in your concrete, actionable terms, and you’re prepared to use those to find an answer. Simply starting by listing what you know can be a hugely helpful tool in building your own confidence and setting up next steps. Here, because you “manufactured” the term L for length, but it does not appear in the answer choices, you should see that a next step should be to eliminate that term. With two equations (perimeter and area) you can do so, solving for L in one equation and substituting that term in the other: k = Lw \frac{k}{w} = L p = 2L + 2w p = 2(\frac{k}{w}) + 2w Now you have all of the same terms as the answer choices, and the answer choices should serve as a guide for you to work toward. You’ll need to get rid of the denominator, so you should multiply both sides by w to get: pw = 2k + 2w^2 Now, you know that you have to set your equation equal to 0, so you can subtract pw from both sides to accomplish that: 2k + 2w^2-pw = 0 Then, rearrange your equation to look like those in the answer choices (w2 term first, pw term second, k term last): 2w^2-pw+2k = 0 And you’ve matched answer choice E, the correct answer .

Loser94 · Answer

Length be L
width =w
perimeter p=2(L+W)
L=P/2-w
area =L*W=(p/2-W)*w=K
2w^2+2k-pw

Anwser- E

Fdambro294 · Answer

Right off the bat you can eliminate (A) and (C) because there is no way Adding and Multiplying (+)Positive Values can result in = 0

Then Plug-in Smart Numbers and Calculate the other 3 Options.

Let ---- w = 2 Length = 1 then Perimeter = p = 6 and Area = k = 2

Then, calculate the Values for each of the Terms in the Answer Choices: w'2 = 4 p*w = 12 2*k = 4

Only Answer Choice E works - 2 * (4) - 12 + 4 = 8 - 12 + 4 = 0

Answer E

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