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The diagram above shows a rectangular garden, bordered by a

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The diagram above shows a rectangular garden, bordered by a [#permalink]

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The diagram above shows a rectangular garden, bordered by a walkway consisting of white and gray rectangles. If all four of the gray rectangles have the same dimensions, and the garden measures 20 by 10 feet, what is the area of the walkway?

(1) Each gray rectangle has area 12 square feet.
(2) The outer perimeter of the walkway is 88 feet.
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Jun 2013, 12:12, edited 1 time in total.
Renamed the topic and edited the question.
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Re: 700+ A rectangular garden, bordered by a walkway consisting [#permalink]

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New post 03 Jun 2013, 11:24
The diagram above shows a rectangular garden, bordered by a walkway consisting of white and gray rectangles. If all four of the gray rectangles have the same dimensions, and the garden measures 20 by 10 feet, what is the area of the walkway?

The area of the walkway will result from \(TotArea-Gardern\), we need to define the area of the outer rectangle.
I will call \(x, y\) the sides of the small grey rectangle

(1) Each gray rectangle has area 12 square feet.
\(xy=12\) could result from sides of \(1*12=12\) or \(2*6=12\) for example.
As the overall lenght of the sides of the rectangle change accordingly, this is not sufficient.

(2) The outer perimeter of the walkway is 88 feet.
So half of the perimeter is the sum of the two sides of the outer rectangle: \(20+2x+10+2y=44\) or \(x+y=7\), this could be the case if \(x=2\) and \(y=5\) or \(x=1\) and \(y=6\),...
As we do not define a side, the area is not defined, not sufficient

1+2)
\(xy=12\)
\(x+y=7\)
This is true if y=4 and x=3 (or the opposite), but since we do not know which side of the rectangle they correspond, it's still not sufficient.
Example: side retangle \(20+2*3=26\), other side \(10+2*4=18\) \(Area_1=18*26\). But if I switch x,y the result changes
Or side retangle \(20+2*4=28\), other side \(10+2*3=16\) \(Area_2=28*16\), and \(Area_1\neq{}Area_2\)
E
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Re: The diagram above shows a rectangular garden, bordered by a [#permalink]

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New post 10 Dec 2013, 15:38
Thought it was C, but now I realize it is E. Here is my reasoning for 1+2 only.

Dark rectangle has an area of 12 sq. ft. which means it's dimensions can be: (1x12), (2x6), (3x4) Depending on what dimensions the dark rectangles are, the unshaded rectangles will be of different widths and we can get several different areas for the walkway. We are also told the perimeter of walkway is 88. 2(L+W) = 88. However, we are given partial dimensions thanks to the interior dimensions of the garden. 2*(20 + x+x) + 2*(10+y+y) = 88 where x and y represent the length and height of the dark rectangles respectively. 40+4x + 20 + 4y = 88 --> 4x+4y = 28 --> x+y = 7. Now, we can look back to where we got the possible dimensions of a 12 sq ft square and see that one of them is 3x4 which is originally why I said this problem could be solved because in statement 2 we got x+y = 7. However, depending on the length and width, the area of the rectangles changes. Take for example the top large rectangle bordering the smaller gray one. If the width = 4 and the length (or height) = 3 then the large rectangle area would = 20*3. If the dark rectangle had a width of 3 and a length (height) of 4, then the area of the large rectangle would be 20*4. Insufficient.

One question though. On the test it would be time consuming to list the two different cases of areas depending on how the dark rectangles are laid out. Is there a possibility that the areas would remain the same regardless of how the dark angles length and width were?
Re: The diagram above shows a rectangular garden, bordered by a   [#permalink] 10 Dec 2013, 15:38
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