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Three-fourths of the area of a rectangular lawn 30 feet wide

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Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the enclosure has full width and reduced length rather than full length and reduced width, how much less fence will be needed?

A) 2 1/2
B) 5
C) 10
D) 15
E) 20

B

Source: Official Guide 2020

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by Brent@GMATPrepNow » Sat May 04, 2019 6:02 am
AbeNeedsAnswers wrote:Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the enclosure has full width and reduced length rather than full length and reduced width, how much less fence will be needed?

A) 2 1/2
B) 5
C) 10
D) 15
E) 20

B

Source: Official Guide 2020
Here's a diagram of the 30 x 40 lawn
Image

If we keep the full width (of 30 feet), then the length of the enclosure = 3/4 of 40 = 30 feet
Image
So, the enclosure is a 30 by 30 square.
The PERIMETER = 30 + 30 + 30 + 30 = 120 feet


If we keep the full length (of 40 feet), then the width of the enclosure = 3/4 of 30 = 22.5 feet
Image
So, the enclosure is a 40 by 22.5 rectangle.
The PERIMETER = 40 + 40 + 22.5 + 22.5 = 125 feet

If the enclosure has full width and reduced length rather than full length and reduced width, how much less fence will be needed?
125 feet - 120 feet = 5

Answer: B

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by swerve » Sat May 04, 2019 10:24 am
For each case, we determine the unknown dimension of the fence and calculate the perimeter of the fence.

Case 1:
\(30\cdot x=\frac{3}{4}\cdot 30 \cdot 40\)
\(x=30\,\Rightarrow\,P_1=4\cdot 120\)

Case 2:
\(y\cdot 40 = \frac{3}{4}\cdot 30 \cdot 40\)
\(y=22.5\,\Rightarrow\, P_2=2(40+22.5)=125\)

Finally, we determine the difference between perimeters

\(P_2-P_1=125-120=5\)

Therefore, the correct answer is __B__

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by Scott@TargetTestPrep » Mon May 20, 2019 6:03 pm
AbeNeedsAnswers wrote:Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the enclosure has full width and reduced length rather than full length and reduced width, how much less fence will be needed?

A) 2 1/2
B) 5
C) 10
D) 15
E) 20

B

Source: Official Guide 2020
If the enclosure has full width and reduced length, then the width = 30 ft and the reduced length = 40 x 3/4 = 30 ft. So we need 2(30) + 2(30) = 60 + 60 = 120 feet of the fence for the enclosure.

On the other hand, if the enclosure has full length and reduced width, then the length = 40 ft and the reduced width = 30 x 3/4 = 22.5 ft. So we need 2(40) + 2(22.5) = 80 + 45 = 125 feet of the fence for the enclosure.

Therefore, we see that we save 5 feet of fence in the first option.

Answer: B

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Answer

by Rich.C@EMPOWERgmat.com » Wed May 22, 2019 12:08 pm
Hi All,

We're told that 3/4 of the AREA of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. We're asked if the enclosure has full width and REDUCED length rather than full length and REDUCED width, how much LESS fence will be needed. This question is built around some standard Geometry and Arithmetic rules - and you might find that a couple of drawings can help you to stay organized.

To start, the area of the FULL lawn is (30)(40) = 1200 square-feet. Three-quarters of that needs to be enclosed by a fence....
(3/4)(1200) = 3600/4 = 900 square-feet

With full width, we would need the length to be 900/30 = 30 feet, meaning that we would have a 30 foot by 30 foot space. The length of THAT fence would be:
30 + 30 + 30 + 30 = 120 feet

With full length, we would need the width to be 900/40 = 90/4 = 22.5 feet, meaning that we would have a 22.5 foot by 40 foot space. The length of THAT fence would be:
22.5 + 22.5 + 40 + 40 = 125 feet

Thus, the difference in fence length would be 125 - 120 = 5 feet.

Final Answer: B

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Rich
Contact Rich at Rich.C@empowergmat.com
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