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# The three squares above share vertex A with AF = FE and AE =

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The three squares above share vertex A with AF = FE and AE = [#permalink]

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09 Jan 2013, 20:34
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The three squares above share vertex A with AF = FE and AE = ED. If a point X is randomly selceted from square region ABCD, what is the probability that X will be contained in the shaded region?

A. 1/16
B. 1/12
C. 1/4
D. 3/16
E. 1/3
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Jan 2014, 05:20, edited 1 time in total.
Edited the question.
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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09 Jan 2013, 21:37
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fozzzy wrote:
The three squares above share vertex A with AF = FE and AE = ED. If a point X is randomly selceted from square region ABCD, what is the probability that X will be contained in the shaded region?

A) 1/16
B) 1/12
c) 1/4
D) 3/16
4) 1/3

Lets say AD = 4, So, AE = 2 & AF = 1.

Area of ABCD = 16
Area of Square with side AE = 4
Area of Square with side AF = 1

Area of shaded region = 4-1 = 3

Probability = $$\frac{Area of shaded region}{Area of ABCD}$$ = $$\frac{3}{16}$$

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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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10 Jan 2013, 18:56
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the probability of the point to lie in shaded region is giveny by

area of the shaded region / area of the original square

To find the area of shaded region = area of the square formed by side AE - area of the square formed by side AF

If we assume that the side AD = 1 unit; then from given info we can take AF = 1/4 and AE = 1/2

Then the area of the square formed by side AE would be 1/4 and the area of the square formed by side AF would be 1/16
Therefore, area of the shaded region = 1/4 minus 1/16 = 3/16

We can take this 3/16 directly as probability because the area of the original square is 1 unit from out assumption of orginal side to be 1 unit.

thanks
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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09 Jan 2013, 21:42
MacFauz wrote:
fozzzy wrote:
The three squares above share vertex A with AF = FE and AE = ED. If a point X is randomly selceted from square region ABCD, what is the probability that X will be contained in the shaded region?

A) 1/16
B) 1/12
c) 1/4
D) 3/16
4) 1/3

Lets say AD = 4, So, AE = 2 & AF = 1.

Area of ABCD = 16
Area of Square with side AE = 4
Area of Square with side AF = 1

Area of shaded region = 4-1 = 3

Probability = $$\frac{Area of shaded region}{Area of ABCD}$$ = $$\frac{3}{16}$$

The figure is so deceiving easy way to trick a person! Thanks for the solution!
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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10 Jan 2013, 08:19
Say each side of the largest square is 4. This gives a totall area of 16.
Since AF=FE and AE=ED, AF and FE must both equal 1 and ED must equal two. Based on this, the area of the 2nd smallest square is 4 and the smallest square is one...giving us an area of 3 for the shaded region.

Therefore the probability of x being in the shaded region is 3/16. D.
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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23 Jan 2014, 05:05
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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04 Jul 2015, 06:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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27 Nov 2015, 11:25
MacFauz wrote:
fozzzy wrote:
The three squares above share vertex A with AF = FE and AE = ED. If a point X is randomly selceted from square region ABCD, what is the probability that X will be contained in the shaded region?

A) 1/16
B) 1/12
c) 1/4
D) 3/16
4) 1/3

Lets say AD = 4, So, AE = 2 & AF = 1.

Area of ABCD = 16
Area of Square with side AE = 4
Area of Square with side AF = 1

Area of shaded region = 4-1 = 3

Probability = $$\frac{Area of shaded region}{Area of ABCD}$$ = $$\frac{3}{16}$$

how do we know AE = 2 and AF = 1 instead of the reverse order?
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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]

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27 Nov 2015, 11:39
sagnik242 wrote:
MacFauz wrote:
fozzzy wrote:
The three squares above share vertex A with AF = FE and AE = ED. If a point X is randomly selceted from square region ABCD, what is the probability that X will be contained in the shaded region?

A) 1/16
B) 1/12
c) 1/4
D) 3/16
4) 1/3

Lets say AD = 4, So, AE = 2 & AF = 1.

Area of ABCD = 16
Area of Square with side AE = 4
Area of Square with side AF = 1

Area of shaded region = 4-1 = 3

Probability = $$\frac{Area of shaded region}{Area of ABCD}$$ = $$\frac{3}{16}$$

how do we know AE = 2 and AF = 1 instead of the reverse order?

Hi,
A per Q, AF = FE. So, AE = 2*AF.
If we assume AF = 1, it implies AE = 2.
Hope that helps.
Re: The three squares above share vertex A with AF = FE and AE =   [#permalink] 27 Nov 2015, 11:39
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