In a certain housing development, each garden has...

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In a certain housing development, each garden has the shape of a rectangle whose width is 2/3 its length and the length of the largest garden in the development is 2 times the length of the smallest garden in the development. If the perimeter of the smallest garden is 30 feet, what is the area of the largest garden?

A. 45
B. 54
C. 81
D. 108
E. 216

The OA is E.

I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.

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by GMATGuruNY » Sat Feb 17, 2018 4:50 am
LUANDATO wrote:In a certain housing development, each garden has the shape of a rectangle whose width is 2/3 its length and the length of the largest garden in the development is 2 times the length of the smallest garden in the development. If the perimeter of the smallest garden is 30 feet, what is the area of the largest garden?

A. 45
B. 54
C. 81
D. 108
E. 216
Each garden has the shape of a rectangle whose width is 2/3 its length.
The perimeter of the smallest garden is 30 feet
.
Thus:
W = (2/3)L
2(L + W) = 30, implying that L+W = 15.
Substituting W = (2/3)L into L+W = 15, we get:
L + (2/3)L = 15
3L + 2L = 45
5L = 45
L = 9.

The length of the largest garden is 2 times the length of the smallest garden.
Each garden has the shape of a rectangle whose width is 2/3 its length.

Since the smallest garden has a length of 9, the length of the largest garden = 2*9 = 18.
Thus, the width of the largest garden = (2/3)L = (2/3)(18) = 12.

What is the area of the largest garden?
LW = 18*12 = 216.

The correct answer is E.
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by [email protected] » Sat Feb 17, 2018 10:38 am
Hi LUANDATO,

We're told that in a certain housing development, each garden has the shape of a rectangle whose width is 2/3 its length, the length of the largest garden in the development is 2 times the length of the smallest garden in the development and the perimeter of the smallest garden is 30 feet. We're asked for the area of the largest garden. This question can be solved with some basic Arithmetic and a bit of logic.

Since the ratio of the width to length is 2:3, we can figure out the area of the smallest garden rather quickly (using 'brute force')

IF... W=2, L=3, then perimeter = 2+2+3+3 = 10
IF... W=4, L=6, then perimeter = 4+4+6+6 = 20
IF... W=6, L=9, then perimeter = 6+6+9+9 = 30

Since the smallest garden has a perimeter of 30, it's dimensions MUST be 6x9. We're told that the largest garden has a length that is TWICE that of the smallest garden - and since the ratio of the width to length is always 2:3, that means the width ALSO must be twice the width of the smallest garden. Since the area of the smallest garden is (6)(9) = 54, we just have to double those two dimensions to get the area of the largest garden:

(54)(2)(2) = 216

Final Answer: E

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by Scott@TargetTestPrep » Sun Jun 23, 2019 10:23 am
BTGmoderatorLU wrote:In a certain housing development, each garden has the shape of a rectangle whose width is 2/3 its length and the length of the largest garden in the development is 2 times the length of the smallest garden in the development. If the perimeter of the smallest garden is 30 feet, what is the area of the largest garden?

A. 45
B. 54
C. 81
D. 108
E. 216

The OA is E.
We are given that the length of the largest garden in the development is 2 times the length of the smallest garden in the development. If we let the length of the smallest garden = S, then we can let the length of the largest garden = 2S.

We are also given that the perimeter of the smallest garden is 30 feet. Since the width is 2/3 the length, the width = (2/3)S. Thus, we can create the following equation:

2S + 2(2/3)S = 30

2S +(4/3)S = 30

Multiplying the entire equation by 3 gives us:

6S + 4S = 90

10S = 90

S = 9

Thus, the length of the largest garden is 2 x 9 = 18, and the width of the largest garden is (2/3) x 18 = 12.

Therefore the area of the largest garden is 18 x 12 = 216.

Answer: E

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