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In the diagram above, WZ = XZ, and circular arc

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In the diagram above, WZ = XZ, and circular arc [#permalink] New post 20 Feb 2013, 16:48
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  35% (medium)

Question Stats:

80% (03:24) correct 20% (02:48) wrong based on 83 sessions
Attachment:
right isosceles triangle with sector.JPG
right isosceles triangle with sector.JPG [ 14.66 KiB | Viewed 1412 times ]

In the diagram above, WZ = XZ, and circular arc XY has a center at W. If a point is selected from anywhere within this figure, what is the probability that it is selected from the shaded region?

(A) \frac{{\pi}-2}{{\pi}}

(B) \frac{{\pi}-1}{2{\pi}}

(C) \frac{2{\pi}-3}{4{\pi}}

(D) \frac{4{\pi}-1}{8{\pi}}

(E) \frac{6{\pi}-3}{8{\pi}}


For the complete solution to this problem, as well as a detailed discussion of geometric probability, please see this post:
http://magoosh.com/gmat/2013/geometric- ... -the-gmat/

Mike :-)
[Reveal] Spoiler: OA

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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 20 Feb 2013, 17:17
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

By the way, solution is A because we have to find the shadowed area (1) and divide by the total area (2).

How to find (2): the angle of the arc is 45º (there is a right-angled isosceles triangle), therefore the area of this part must be pi*(radius^2) divided by 8 (because we have 45º and not 360º). In this case, the area is \frac{1}{8}*pi*r^2 --> \frac{1}{8}*pi*(L*\sqrt{2})^2 --> \frac{1}{8}*pi*(L^2*2).

How to find (1): there is a right-angled isosceles triangle. Therefore the side of the radius must be L*\sqrt{2}, and the equal sides have length L. The right angle is obviously 90º and the other two angles are 45º each. Therefore the area of this triangle is \frac{1}{2}*L*L --> \frac{1}{2}*L^2. Therefore the shadowed area (1) is \frac{1}{8}*pi*(L^2*2) minus \frac{1}{2}*L^2.

Finally, we divide (1) by (2) and the solution is A.
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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 21 Feb 2013, 07:43
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wz=xz
therefore, angle xwz = angle wxz = 45 degree
let, wz = xz = k
therefore, wx = (\sqrt{2})k
therefore, area of sector wxyw = 45/360 * pi*square of (k sqrt(2) )= ( pi *square(k))/4
area of shaded portion = area of sector wxyw - area of triangle
area of triangle = 1/2 * square (k)
probability = ( shaded area )/ (total area)
={ ( pi *square(k))/4 - 1/2 * square (k) }/ ( pi *square(k))/4
= 1- 2/pi
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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 21 Feb 2013, 11:21
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johnwesley wrote:
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

I would say, this is probably at the outer limit of what could appear on the GMAT, in terms of difficulty. I would imagine the CAT would throw something like this at you only if you were getting almost every other math question right. It's true, most people cannot solve this in under 3 minutes, but most people would not ever see this question --- the CAT would have no reason to give them something this hard. Someone very skilled in math, someone acing the Quantitative Section, could see something like this and probably could dispatch this question in under a minute. It's all a matter of perspective.

BTW, I love your founder-of-the-Methodists screenname. :-)

Mike :-)
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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 21 Feb 2013, 11:55
I agree Mike, that GMAT might throw this to someone who has aced 15 questions in a row. Someone in higher rungs of Q50 and lower rungs of Q51. Still now harm in flexing your maths muscle. I was on the right track to solving this in 90 seconds but miscalculated the area of the entire arc by twice. Hence took 3 mins. The graph question on the link you shared is a bit too tough and needs very accurate graphing (one needs to verify where the line cuts the 4 sides of the square, which is not an easy task without graph).
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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 21 Feb 2013, 13:08
mikemcgarry wrote:
johnwesley wrote:
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

I would say, this is probably at the outer limit of what could appear on the GMAT, in terms of difficulty. I would imagine the CAT would throw something like this at you only if you were getting almost every other math question right. It's true, most people cannot solve this in under 3 minutes, but most people would not ever see this question --- the CAT would have no reason to give them something this hard. Someone very skilled in math, someone acing the Quantitative Section, could see something like this and probably could dispatch this question in under a minute. It's all a matter of perspective.

BTW, I love your founder-of-the-Methodists screenname. :-)

Mike :-)


Many thanks... you are right: this is a very difficult one. Many thanks for posting it.

And are you the Michael McGarry (born 17 May 1965)? The successful association footballer who frequently represented New Zealand in the 1980s and 90s? :lol:

Best!
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Re: In the diagram above, WZ = XZ, and circular arc [#permalink] New post 07 Jul 2014, 00:45
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Re: In the diagram above, WZ = XZ, and circular arc   [#permalink] 07 Jul 2014, 00:45
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