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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?

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 = \(\frac{Area of shaded region}{Area of ABCD}\) = \(\frac{3}{16}\)

Answer is D.
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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.

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 = \(\frac{Area of shaded region}{Area of ABCD}\) = \(\frac{3}{16}\)

Answer is D.

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.

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 = \(\frac{Area of shaded region}{Area of ABCD}\) = \(\frac{3}{16}\)

Answer is D.

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

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 = \(\frac{Area of shaded region}{Area of ABCD}\) = \(\frac{3}{16}\)

Answer is D.

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.