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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
BUMP! Any quicker ways to solve this?
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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
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manimgoindowndown wrote:
BUMP! Any quicker ways to solve this?


Statement 1 :
AB = BC as the figure is square
Hence 3l+2b = 4l ==> l=2b and 4l = S (S = side of the square)

% = (14*l*b) *100/S^2 ==> as you know all the relationships, you can make this one variable equation and get the answer.

Statement 2 :
EF = FG as the figure is square
Hence 3l = 2l + 2b ==> l=2b and 4l = S (S = side of the square)

% = (14*l*b) *100/S^2 ==> as you know all the relationships, you can make this one variable equation and get the answer.

Answer hence is D
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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
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(1) SUFFICIENT: Let the length (longer dimension) of each rectangular tile be called L, and the width (shorter dimension) of each tile W. Then each horizontal side of square ABCD has total length 2W + 3L, and each vertical side has total length 4L.

Because ABCD is a square, these total lengths must be equal: 2W + 3L = 4L, which reduces to L = 2W. Therefore, each side of square ABCD is equal to 4L = 8W, and the total area of square ABCD is (8W)(8W) = 64W2.

The total area of the tiles is 14(L × W) = 14(2W × W) = 28W2. The desired fraction is thus (28W2)/(64W2) = 28/64. There is no need to reduce this fraction; the statement is sufficient.

(2) SUFFICIENT: Let the length (longer dimension) of each rectangular tile be called L, and the width (shorter dimension) of each tile W. Then each horizontal side of square ABCD has total length 3L, and each vertical side has total length 4L – 2W.

Because EFGH is a square, these total lengths must be equal: 3L = 4L – 2W, which reduces to L = 2W. Therefore, each side of square ABCD is equal to 3L = 6W.

In turn, ABCD must also be a square, since each of its sides is 2W longer than the corresponding side of EFGH (i.e., longer by W on each side). Therefore, each side of ABCD is equal to 6W + 2W = 8W, and the total area of square ABCD is (8W)(8W) = 64W2.

The total area of the tiles is 14(L × W) = 14(2W × W) = 28W2. The desired fraction is thus (28W2)/(64W2) = 28/64. There is no need to reduce this fraction; the statement is sufficient.

The correct answer is D.
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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
Hey Bunuel,
could u please help me understand how to reach at an answer..am really unable to understand it..
thanks..
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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
Okay, think of it like a real life problem. I have actually sort of seen this in floor tile designs:

n the diagram above, the fourteen rectangular tiles are all identical. What percent of the area of rectangle ABCD is covered by the tiles?

(1) ABCD is a square. - obvious, inner area is also a square due to identical difference between outer and inner square. As each side has 4 division, we get x/4; the smaller part of tile being x/8 (half of x/4). Suff

(2) EFGH is a square. - so outer part has to be a square. Again we reach the x/4 and x/8 combination

Sufficient.

Ans D

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Re: In the diagram above, the fourteen rectangular tiles are all [#permalink]
In the diagram above, the fourteen rectangular tiles are all identical. What percent of the area of rectangle ABCD is covered by the tiles?

(1) ABCD is a square.

(2) EFGH is a square.

Hi Bunuel, I did this using a different approach. Please correct me if I'm wrong.

St 1. The figure ABCD is a square. Hence, with the same tile measurement, the square ABCD can be divided into 32 figures.
So the percentage should be (14/32)*100

St 2. As EFGH is a square, we can derive that ABCD is also a square. Using the same approach as st 1, we can find the percentage.

Hence, D it is.
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In the diagram above, the fourteen rectangular tiles are all [#permalink]
I tried to solve this problem intuitively. Since the question asked the percentage of area occupied by the tiles, the actual dimensions of the ABCD does not matter. What matters is the relative dimension of the tile area EFGH with respect to the ABCD, i.e., Is it possible to get different dimensions of EFGH, within the constraints of ABCD being square and 14 tiles of same measurement can be fit into it. If you carefully see, had ABCD been a rectangle, it would have been possible to get different measurements for EFGH. Since ABCD is square, there is no way to increase or reduce the dimensions of EFGH without changing the number of tiles that can be fit into it.

Hence ABCD being square is sufficient.
Hence EFGH is also sufficient.
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In the diagram above, the fourteen rectangular tiles are all [#permalink]
vards wrote:
Hey Bunuel,
could u please help me understand how to reach at an answer..am really unable to understand it..
thanks..


You can consider it this way - all 14 tiles are identical, and assuming they have width d, both length and width of the smaller rectangle are 2d each lesser than that of the larger rectangle. Key takeaway here is that both length and width of the smaller rectangle are smaller than the corresponding ones for the larger rectangle by the "same amount".

Hence it follows that if any one of the rectangles is a square, the other one will be a square as well. Hence, D is the answer.
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In the diagram above, the fourteen rectangular tiles are all [#permalink]
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