What is the area of rectangular garden R?
(1) The length of the garden is twice the width.
(2) The perimeter is 84 yards.
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Statement 1: L = 2Wamontobin 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.
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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Jay@ManhattanReview
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Say the length and the width of the rectangular garden be x and y, respectively.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.
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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