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.
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If p is the perimeter of rectangle Q...
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Statement 1: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.
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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
