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!
Perimeter
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No, we cannot assume that the lengths are integers.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!
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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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
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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Yes, we do get two different values for W (W = 4 or W = 10).[email protected] 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
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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Thanks a lot . A key point I missed out and was scratching my head .. Thanks a lot for your guidance again
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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
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:No, we cannot assume that the lengths are integers.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!
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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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.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
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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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.)Matt@VeritasPrep wrote: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.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
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.