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Retired Moderator Joined: 07 Jun 2014
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The length of a rectangle is two more than twice its width, [#permalink]
Expert's post 00:00

Question Stats: 85% (01:10) correct 14% (01:12) wrong based on 21 sessions
The length of a rectangle is two more than twice its width, and the area of the rectangle is 40. What is the rectangle’s perimeter?

[Reveal] Spoiler: OA
28

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Sandy
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Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Kudos [?]: 2971 , given: 394

Re: The length of a rectangle is two more than twice its width, [#permalink]
Expert's post
Explanation

Convert this word problem into two equations with two variables. “The length is two more than twice the width” can be written as:

$$L = 2W + 2$$

Since the area is 40 and area is equal to length × width:

$$LW = 40$$

Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation:

$$(2W + 2)W = 40$$
$$2W^2 + 2W = 40$$

Since this is now a quadratic (there are both a W2 and a W term), get all terms on one side to set the expression equal to zero:

$$2W^2 + 2W - 40 = 0$$

Simplify as much as possible—in this case, divide the entire equation by 2—before trying to factor:

$$W^2 + W - 20 = 0$$

$$(W - 4)(W + 5) = 0$$

W = 4 or –5

Since a width cannot be negative, the width is equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve:

Perimeter = 2L + 2W

Perimeter = 2(10) + 2(4)

Perimeter = 28

Note that it might have been possible for you to puzzle out that the sides were 4 and 10 just by trying values. However, if you did this, you got lucky—no one said that the values even had to be integers!
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Re: The length of a rectangle is two more than twice its width, [#permalink]
We have system of equations: l=2w+2 and lw=40 -> w(2w+2)=40 -> w(w+1)=20 -> w=4 -> l=10 -> perimeter = 2(4+10) = 28.
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Re: The length of a rectangle is two more than twice its width, [#permalink]
sandy wrote:
Explanation

Convert this word problem into two equations with two variables. “The length is two more than twice the width” can be written as:

$$L = 2W + 2$$

Since the area is 40 and area is equal to length × width:

$$LW = 40$$

Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation:

$$(2W + 2)W = 40$$
$$2W^2 + 2W = 40$$

Since this is now a quadratic (there are both a W2 and a W term), get all terms on one side to set the expression equal to zero:

$$2W^2 + 2W - 40 = 0$$

Simplify as much as possible—in this case, divide the entire equation by 2—before trying to factor:

$$W^2 + W - 20 = 0$$

$$(W - 4)(W + 5) = 0$$

W = 4 or –5

Since a width cannot be negative, the width is equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve:

Perimeter = 2L + 2W

Perimeter = 2(10) + 2(4)

Perimeter = 28

Note that it might have been possible for you to puzzle out that the sides were 4 and 10 just by trying values. However, if you did this, you got lucky—no one said that the values even had to be integers!

I pretty much solved using this same method EXCEPT, I used the quadratic equation formula to solve for W. Gave me the same answer, took a little longer but thats what came to my mind immediately. Re: The length of a rectangle is two more than twice its width,   [#permalink] 05 Aug 2018, 17:22
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