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Math Expert V
Joined: 02 Sep 2009
Posts: 61304
Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 53% (02:30) correct 47% (02:20) wrong based on 713 sessions

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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. 5/2
B. 5
C. 10
D. 15
E. 20

PS98502.01
OG2020 NEW QUESTION

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Posts: 4329
Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Bunuel 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. 5/2
B. 5
C. 10
D. 15
E. 20
PS98502.01
OG2020 NEW QUESTION

Here's a diagram of the 30 x 40 lawn If we keep the full width (of 30 feet), then the length of the enclosure = 3/4 of 40 = 30 feet 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 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

Cheers,
Brent
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Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Total area of lawn=30*40= 1200
area of fenced= 0.75*1200=900

dimensions of the fenced area when there is full width and reduced length-
width=30
length=900/30=30
length of fence= 2*(30+30)=120

dimensions of the fenced area when there is full length and reduced width-
length= 40
width=900/40=22.5
length of fence= 2*(40+22.5)=125

Difference in the length of fence in 2 Cases= 125-120=5
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Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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2
1
Bunuel 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. 5/2
B. 5
C. 10
D. 15
E. 20

PS98502.01
OG2020 NEW QUESTION

For each case, we determine the unknown dimension of the fence and calculate the perimeter of the fence.

Case 1:
$$30\cdot x=\frac{3}{4}\cdot 30\cdot 40$$

$$x=30 \implies P_1=4\cdot 30=120$$

Case2:
$$y\cdot 40=\frac{3}{4}\cdot 30\cdot 40$$

$$y=22.5 \implies P_2=2(40+22.5)=125$$

Finally, we determine the difference of the perimeters.

$$P_2-P_1=125-120=5$$

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e-GMAT Representative V
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Posts: 3239
Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Solution

Given:
• Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence.

To find:
• How much less fence will be needed if the enclosure has full width and reduced length rather than full length and reduced width.

Approach and Working:

• Area of lawn = 30 × 40
o 3/4th of the area of lawn = ¾(30 × 40) = 30 * 30

Case-1) When full width will be fenced, and reduced length will be fenced.

• Width = 30 feet
o 30 * L = 30 * 30
• Hence, length = 30 feet
• Length of fence needed = 2(30 +30) = 120 feet

Case-2) When full length will be fenced, and reduced width will be fenced

• Length = 40 feet
o 40 * W = 30 * 30
 W = 22.5 feet
o Length of fence needed = 2(40 +22.5) = 125 feet

Difference in length of fence needed = 125- 120 = 5 feet.

Hence, option B is the correct answer.

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Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Bunuel 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. 5/2
B. 5
C. 10
D. 15
E. 20

PS98502.01
OG2020 NEW QUESTION

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.

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Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee  [#permalink]

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Bunuel 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. 5/2
B. 5
C. 10
D. 15
E. 20

PS98502.01
OG2020 NEW QUESTION

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?

Case 1: the enclosure has full width and reduced length
(40-2w)*30 = 3/4 * 30 * 40
40 -2w = 30
w = 5
Dimensions = (30,30)
Perimeter = 120 feet

Case 2:the enclosure has full length and reduced width
40 * (30-2w) = 3/4 * 30 * 40
30 - 2w = 45/2 = 22.5
w = 3.75
Dimensions = (40,22.5)
Perimeter = 125 feet

Difference in perimeter = difference in fence needed = 125 - 120 = 5 feet

IMO B Re: Three-fourths of the area of a rectangular lawn 30 feet wide by 40 fee   [#permalink] 18 Sep 2019, 23:33
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