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Source: — Data Sufficiency |
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by GMATGuruNY » Tue Jul 22, 2014 3:17 am
If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A
Statement 1: P = (4/3)A
Case 1: A=9
If the rectangle is a square with an area of 9, then s=3 and P = 4s = 12.
This works, since statement 1 requires that P = (4/3)(A) = (4/3)(9) = 12.
Thus:
In Case 1, the rectangle is a square with a side of 3.

Case 2: A=81
If the rectangle is a square with an area of 81, then s=9 and P = 4s = 36.
This doesn't work, since statement 1 requires that P = (4/3)(A) = (4/3)(81) = 108.
Thus:
In Case 2, the rectangle is NOT a square.
INSUFFICIENT.

Statement 2: P = 4√A
Case 1: A=9
If the rectangle is a square with an area of 9, then s=3 and P = 4s = 12.
This works, since statement 2 requires that P = 4√A = 4√9 = 12.
Thus:
In Case 1, the rectangle is a square with a side of 3.

Case 2: A=81
If the rectangle is a square with an area of 81, then s=9 and P = 4s = 36.
This works, since statement 2 requires that P = 4√A = 4√81 = 36.
Thus:
In Case 2, the rectangle is a square with a side of 9.

The cases above illustrate that, in statement 2, the rectangle must be a square.
SUFFICIENT.

The correct answer is B.
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by GMATinsight » Tue Jul 22, 2014 10:16 am
abhasjha wrote:If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A
If 'L' is the length and 'W' is width then
Perimeter, P = 2x(L+W)
Area, A = LxW

Question : Is the Region Square?

Question Rephrased : Is L = W?

Statement 1) P = 4/3*A

i.e. 3P = 4A
i.e. Perimeter must be chosen as multiple of 4 and Area must be chosen as multiple of 3

Case 1:P = 12, A = 9 i.e. all sides can be equal i.e. L = W = 3 i.e. It can be a Square

Case 2:P = 24, A = 18
L+W = 12 for this to be a square L = W = 6 but LW = 18 is not satisfied therefore
For above equation to be true L and W are Certainly NOT equal i.e. It's not a case of Square

INSUFFICIENT

Statement 2) P = 4√A
i.e. 2(L+W) = 4√(LW)
i.e. (L+W) = 2√(LW) [Squaring both sides]
i.e. (L+W)^2 = 4(LW)
i.e. L^2+W^2 + 2LW = 4(LW)
i.e. L^2+W^2 = 2(LW)
i.e. L^2+W^2 - 2(LW) = 0
i.e. (L-W)^2 = 0
i.e. L = W [It's a square]

SUFFICIENT

Answer: Option B
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