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Perimeter

This topic has 4 expert replies and 3 member replies
sparkles3144 Master | Next Rank: 500 Posts Default Avatar
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Perimeter

Post Fri May 10, 2013 2:07 pm
A rectangle has a perimeter of 28 inches. What are the dimensions of the rectangle?

(1) The area of the rectangle is 40 inches squared.

(2) The length is 6 inches more than the width

I know how to solve this.
However, I am wondering if I can assume that l and w are integers.
It is not stated.

Thanks!

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Matt@VeritasPrep GMAT Instructor
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Post Fri Jun 10, 2016 9:56 am
Gurpreet singh wrote:
Hi Brent,

In case the question stem had asked for the length instead of dimension will the statement 1 still be sufficient? since we are getting 2 values.
Thanks
Gurpreet
If I remember correctly, the GMAT actually considers the length to be the longer side of the rectangle. (I learned this the hard way, doing an OG problem back in 2009 when I was studying for the test; I was quite surprised.) I had never heard this anywhere before, so I'm still skeptical, but I'm quite sure I remember an OG DS problem that turned on this.

By their logic, then, if I tell you I have a rectangle with dimensions 10 and 4, the length MUST be 10 and the width MUST be 4.

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Post Fri Jun 10, 2016 10:15 am
Matt@VeritasPrep wrote:
Gurpreet singh wrote:
Hi Brent,

In case the question stem had asked for the length instead of dimension will the statement 1 still be sufficient? since we are getting 2 values.
Thanks
Gurpreet
If I remember correctly, the GMAT actually considers the length to be the longer side of the rectangle. (I learned this the hard way, doing an OG problem back in 2009 when I was studying for the test; I was quite surprised.) I had never heard this anywhere before, so I'm still skeptical, but I'm quite sure I remember an OG DS problem that turned on this.

By their logic, then, if I tell you I have a rectangle with dimensions 10 and 4, the length MUST be 10 and the width MUST be 4.
Now I'm intrigued - Matt, do you know which edition of the OG that question shows up in? (I'm assuming it's 10th or earlier.)

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Gurpreet singh Senior | Next Rank: 100 Posts Default Avatar
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Post Thu Jun 09, 2016 6:10 pm
Hi Brent,

In case the question stem had asked for the length instead of dimension will the statement 1 still be sufficient? since we are getting 2 values.
Thanks
Gurpreet


Brent@GMATPrepNow wrote:
sparkles3144 wrote:
A rectangle has a perimeter of 28 inches. What are the dimensions of the rectangle?

(1) The area of the rectangle is 40 inches squared.

(2) The length is 6 inches more than the width

I know how to solve this.
However, I am wondering if I can assume that l and w are integers.
It is not stated.

Thanks!
No, we cannot assume that the lengths are integers.
Here's one solution:

Target question: What are the dimensions of the rectangle?

Given: A rectangle has a perimeter of 28 inches.
Let L = length of rectangle
Let W = width of rectangle

If the perimeter is 28, then 2L + 2W = 28
We can simplify this by dividing both sides by 2 to get: L + W = 14

Statement 1: The area of the rectangle is 40 inches squared.
In other words, LW = 40

Since we already know that L + W = 14, we can solve for W to get W = 14 - L
Now take the equation LW = 40, and replace W with (14 - L) to get . . .
L(14 - L) = 40
Expand: 14L - L^2 = 40
Rearrange to get: L^2 - 14L + 40 = 0
Factor: (L - 4)(L - 10) = 0
So, L = 4 or 10

If L = 4, then W = 10, which means the dimensions are 4 by 10
If L = 10, then W = 4, which means the dimensions are still 4 by 10
In other words, the the dimensions must be 4 by 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The length is 6 inches more than the width
In other words, L = W + 6

Since we already know that L + W = 14, we can solve for W to get W = 14 - L
Now take the equation L = W + 6, and replace W with (14 - L) to get . . .
L = (14 - L) + 6
Rearrange: 2L = 20
Solve: L = 10
If L = 10, then W = 4, which means the dimensions are 4 by 10
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

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Post Fri Apr 01, 2016 10:24 am
Thanks a lot Smile . A key point I missed out and was scratching my head .. Thanks a lot for your guidance again

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Post Fri Apr 01, 2016 10:21 am
Rohit_Prakash88@yahoo.in wrote:
Hello Together,

Please clarify as I believe answer should be option B not D because When solve statement 1 we get 2 values of W and L i.e when l = 10 then W = 4 and when W = 4 then l =10 since this statement does lock L and W to have one particular value I believe it is insufficient and Option B should be the right choice.

Please reply as I am in too much of a fix
Yes, we do get two different values for W (W = 4 or W = 10).
However, the target question does not ask us to determine the value of W. The target question asks us to determine the dimensions of the rectangle.
A rectangle with dimensions 4 x 10 is identical to a rectangle with dimensions 10 x 4

Cheers,
Brent

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Post Fri Apr 01, 2016 10:13 am
Hello Together,

Please clarify as I believe answer should be option B not D because When solve statement 1 we get 2 values of W and L i.e when l = 10 then W = 4 and when W = 4 then l =10 since this statement does lock L and W to have one particular value I believe it is insufficient and Option B should be the right choice.

Please reply as I am in too much of a fix

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Post Fri May 10, 2013 2:20 pm
sparkles3144 wrote:
A rectangle has a perimeter of 28 inches. What are the dimensions of the rectangle?

(1) The area of the rectangle is 40 inches squared.

(2) The length is 6 inches more than the width

I know how to solve this.
However, I am wondering if I can assume that l and w are integers.
It is not stated.

Thanks!
No, we cannot assume that the lengths are integers.
Here's one solution:

Target question: What are the dimensions of the rectangle?

Given: A rectangle has a perimeter of 28 inches.
Let L = length of rectangle
Let W = width of rectangle

If the perimeter is 28, then 2L + 2W = 28
We can simplify this by dividing both sides by 2 to get: L + W = 14

Statement 1: The area of the rectangle is 40 inches squared.
In other words, LW = 40

Since we already know that L + W = 14, we can solve for W to get W = 14 - L
Now take the equation LW = 40, and replace W with (14 - L) to get . . .
L(14 - L) = 40
Expand: 14L - L^2 = 40
Rearrange to get: L^2 - 14L + 40 = 0
Factor: (L - 4)(L - 10) = 0
So, L = 4 or 10

If L = 4, then W = 10, which means the dimensions are 4 by 10
If L = 10, then W = 4, which means the dimensions are still 4 by 10
In other words, the the dimensions must be 4 by 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The length is 6 inches more than the width
In other words, L = W + 6

Since we already know that L + W = 14, we can solve for W to get W = 14 - L
Now take the equation L = W + 6, and replace W with (14 - L) to get . . .
L = (14 - L) + 6
Rearrange: 2L = 20
Solve: L = 10
If L = 10, then W = 4, which means the dimensions are 4 by 10
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

_________________
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