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Source: — Data Sufficiency |
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by GMATGuruNY » Mon Jul 17, 2017 1:15 pm
amontobin wrote:What is the area of rectangular garden R?

(1) The length of the garden is twice the width.
(2) The perimeter is 84 yards.
Statement 1: L = 2W
Case 1: L=2 and W=1, with the result that A = LW = 2*1 = 2.
Case 2: L=4 and W=2, with the result that A = LW = 4*2 = 8.
Since the area can be different values, INSUFFICIENT.

Statement 2: 2L + 2W = 84, implying that L+W = 42.
Case 1: L=41 and W=1, with the result that A = LW = 41*1 = 41.
Case 2: L=40 and W=2, with the result that A = LW = 40*2 = 80.
Since the area can be different values, INSUFFICIENT.

Statements combined:
Since we have two variables (L and W) and two distinct linear equations with these variables (L=2W and L+W = 42), we can solve for the two variables.
Since we can solve for L and W, the area of the rectangle can be determined.
SUFFICIENT.

The correct answer is C.
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by Jay@ManhattanReview » Mon Jul 17, 2017 11:55 pm
amontobin wrote:What is the area of rectangular garden R?

(1) The length of the garden is twice the width.
(2) The perimeter is 84 yards.
Say the length and the width of the rectangular garden be x and y, respectively.

Area of the rectangular garden = xy

Thus, we have to find out the xy.

Statement 1: The length of the garden is twice the width.

We have x = 2y

Thus, Area = xy = 2y*y = 2y^2. Cannot get the value. Not sufficient.

Statement 2: The perimeter is 84 yards.

Perimeter of the garden = 2*(x + y)

Thus, 2*(x + y) = 84

=> x+y = 42

We cannot get the unique value of xy. Not sufficient.

Statement 1 & 2 combined:

From Satement 1, we have x = 2y and from Statement 2, we have x+y = 84

Thus, 2y + y = 42 => 3y = 42 => y = 14 => x = 2*14 = 28 => Area = xy = 28*14 = A unique value. There is no need to compute it as we are satisfied that we get the unique value. Suffcient.

The correct answer: C

Hope this helps!

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