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If p is the perimeter of rectangle Q...

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by nchaswal » Tue May 10, 2016 12:54 pm

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If P is the perimeter of rectangle Q, what is the value of p?

1) Each diagonal of rectangle Q has length 10
2) The area of rectangle Q is 48.
It is GMAT. So what?
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Source: — Data Sufficiency |

by GMATGuruNY » Tue May 10, 2016 1:26 pm
nchaswal wrote:If P is the perimeter of rectangle Q, what is the value of p?

1) Each diagonal of rectangle Q has length 10
2) The area of rectangle Q is 48.
Statement 1:
Here, L² + W² = 10² = 100.

Test one case that also satisfies Statement 2.
Case 1: L=6 and W=8, with the result that LW=48 and L² + W² = 6² + 8² = 36+64 = 100.
In this case, p = 6+6+8+8 = 28.
Test one case that DOESN'T also satisfy Statement 2.
Case 2: L=1 and W=√99, with the result that LW≠48 but L² + W² = 1² + √99² = 1+99 = 100.
In this case, p = 1+1+√99+√99 = 2 + 2√99.

Since p can be different values, INSUFFICIENT.

Statement 2:
Here, LW = 48.
Case 1 also satisfies Statement 2.
In Case 1, p = 28.

Case 3: L=1 and W=48, with the result that LW = 1*48 = 48
In this case, p = 1+1+48+48 = 98.

Since p can be different values, INSUFFICIENT.

Statement combined:
Case 1 satisfies both statements (L² + W² = 100 and LW = 48).
No other length and width will yield both a diagonal of 10 and an area of 48.
Thus, p = 28.
SUFFICIENT.

The correct answer is C.
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Hi All,

The question asks us to figure out the PERIMETER of rectangle Q. For that, we'll need the length (L) and width (W) of the rectangle.

1) Each diagonal of rectangle Q has length 10.

From this Fact, we can create one equation:

L^2 + W^2 = 10^2

Unfortunately, there are lots of different values for L and W here (and most of them are non-integers), so there are lots of possible perimeters. Here are two possibilities:
L = 6, W = 8... Perimeter = 28
L = 1, W = (Root99)... Perimeter = 2 + 2(Root99)
Fact 1 is INSUFFICIENT

2) The area of rectangle Q is 48.

From this Fact, we can create one equation:

(L)(W) = 48

Again though, there are lots of different values for L and W here, so there are lots of possible perimeters. Here are two possibilities:
L = 6, W = 8... Perimeter = 28
L = 1, W = 48... Perimeter = 98
Fact 2 is INSUFFICIENT

Combined, we know...
L^2 + W^2 = 10^2
(L)(W) = 48

We have a 'system' of equations here - two variables and two unique equations. Since rectangles cannot have "negative sides", there's just one solution to this system (and it happens to be 6 and 8, although you don't have to do that work).
Combined, SUFFICIENT

Final Answer: C

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