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fencing material is used

by sanju09 » Sun Feb 20, 2011 11:15 pm
80 linear feet of fencing material is used to construct a rectangular enclosure. What is the area of the enclosure?

(1) The length of one particular side of the enclosure is 10 feet.

(2) It is possible to divide the enclosure into three square enclosures of equal size.

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by Anurag@Gurome » Sun Feb 20, 2011 11:53 pm
(1) Let the width of the enclosure = W feet
Then 2*(10 + W) = 80 or W = 30 feet
Since we know the length and width, area can be calculated.
So, (1) is SUFFICIENT.

(2) Let each side of the square is S.
Then 8S = 80 or S = 10 feet, which implies 3S or length of enclosure = 30 feet
Since we know the length and width, area can be calculated.
So, (2) is SUFFICIENT.

The correct answer is D.
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by ankur.agrawal » Mon Feb 21, 2011 6:22 am
Anurag@Gurome wrote:(1) Let the width of the enclosure = W feet
Then 2*(10 + W) = 80 or W = 30 feet
Since we know the length and width, area can be calculated.
So, (1) is SUFFICIENT.

(2) Let each side of the square is S.
Then 8S = 80 or S = 10 feet, which implies 3S or length of enclosure = 30 feet
Since we know the length and width, area can be calculated.
So, (2) is SUFFICIENT.

The correct answer is D.
Please Elaborate on 8S=80. Pls show by making figure.
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by mk101 » Mon Feb 21, 2011 8:08 am
ankur.agrawal wrote:
Anurag@Gurome wrote:(1) Let the width of the enclosure = W feet
Then 2*(10 + W) = 80 or W = 30 feet
Since we know the length and width, area can be calculated.
So, (1) is SUFFICIENT.

(2) Let each side of the square is S.
Then 8S = 80 or S = 10 feet, which implies 3S or length of enclosure = 30 feet
Since we know the length and width, area can be calculated.
So, (2) is SUFFICIENT.

The correct answer is D.
Please Elaborate on 8S=80. Pls show by making figure.
Image
Image

pardon me but could not draw that cleanly... just mark sides of each of the square as s and you will see why 8S =80
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by gmatmachoman » Mon Feb 21, 2011 10:49 am
Anurag@Gurome wrote:(1) Let the width of the enclosure = W feet
Then 2*(10 + W) = 80 or W = 30 feet
Since we know the length and width, area can be calculated.
So, (1) is SUFFICIENT.

(2) Let each side of the square is S.
Then 8S = 80 or S = 10 feet, which implies 3S or length of enclosure = 30 feet
Since we know the length and width, area can be calculated.
So, (2) is SUFFICIENT.

The correct answer is D.
@Anurag Ji,

Shouldn't we take 10S = 80 as we are using the extra 2 segments to create the 2 partitions??
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by Anurag@Gurome » Mon Feb 21, 2011 8:07 pm
gmatmachoman wrote: @Anurag Ji,

Shouldn't we take 10S = 80 as we are using the extra 2 segments to create the 2 partitions??
Image

It is given that the fencing material used is 80 feet, which is the perimeter of the square.
Perimeter of the square = 8S = 80 feet implies S = 10 feet

Does that help?
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by sanju09 » Mon Feb 21, 2011 11:49 pm
Anurag@Gurome wrote:
gmatmachoman wrote: @Anurag Ji,

Shouldn't we take 10S = 80 as we are using the extra 2 segments to create the 2 partitions??
Image

It is given that the fencing material used is 80 feet, which is the perimeter of the square.
Perimeter of the square = 8S = 80 feet implies S = 10 feet

Does that help?
I don't understand which square has got perimeter as 80. When the rectangular enclosure is considered as three squares put adjacently as shown by Anurag, then by perimeter we never literally mean the perimeter of the three squares so formed, as only 8 sides of three identical squares are to be taken as equal to 80 linear feet of fencing material. Well done Anurag!
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by ankur.agrawal » Mon Feb 21, 2011 11:54 pm
Thanks Anurag.

This means the two partitions which has been used to create partition is not counted in the calculation of perimeter.
Anurag@Gurome wrote:
gmatmachoman wrote: @Anurag Ji,

Shouldn't we take 10S = 80 as we are using the extra 2 segments to create the 2 partitions??
Image

It is given that the fencing material used is 80 feet, which is the perimeter of the square.
Perimeter of the square = 8S = 80 feet implies S = 10 feet

Does that help?
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by ankur.agrawal » Mon Feb 21, 2011 11:58 pm
Thanks Sanju . Good Question.
sanju09 wrote:
Anurag@Gurome wrote:
gmatmachoman wrote: @Anurag Ji,

Shouldn't we take 10S = 80 as we are using the extra 2 segments to create the 2 partitions??
Image

It is given that the fencing material used is 80 feet, which is the perimeter of the square.
Perimeter of the square = 8S = 80 feet implies S = 10 feet

Does that help?
I don't understand which square has got perimeter as 80. When the rectangular enclosure is considered as three squares put adjacently as shown by Anurag, then by perimeter we never literally mean the perimeter of the three squares so formed, as only 8 sides of three identical squares are to be taken as equal to 80 linear feet of fencing material. Well done Anurag!
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