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20 Feb 2013, 17:48
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (03:11) correct
29%
(02:58)
wrong
based on 214
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
right isosceles triangle with sector.JPG [ 14.66 KiB | Viewed 8220 times ]
(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:
https://magoosh.com/gmat/2013/geometric- ... -the-gmat/
Mike
20 Feb 2013, 18:17
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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.
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.
21 Feb 2013, 08:43
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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
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
