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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
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Last edited by Bunuel on 23 Jan 2014, 05:20, edited 1 time in total.
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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 = 41 = 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.
thanks Mohan



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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 = 41 = 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.



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Re: The three squares above share vertex A with AF = FE and AE = [#permalink]
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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
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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 = 41 = 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?



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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 = 41 = 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.




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