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# The vertices of rectangle ABCD are (12,5), (-12,5), (-12,-5)

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Manager
Joined: 04 Jan 2008
Posts: 117
The vertices of rectangle ABCD are (12,5), (-12,5), (-12,-5) [#permalink]

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11 Sep 2008, 05:20
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Question Stats:

100% (01:40) correct 0% (00:00) wrong based on 1 sessions

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The vertices of rectangle ABCD are (12,5), (-12,5), (-12,-5) and (12,-5). What is the approximate probability that the coordinate of a random point within the rectangle is such that $$x^2+y^2$$< 25?

12.5%
25%
33%
40%
50%

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SVP
Joined: 17 Jun 2008
Posts: 1502

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11 Sep 2008, 11:04
It will be the ratio of area of square enclosed within (5,0), (0,5), (-5,0), (0,-5) and the original rectangle and will be approx 25% (~22%).
Senior Manager
Joined: 04 Jan 2006
Posts: 276

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11 Sep 2008, 11:44
scthakur wrote:
It will be the ratio of area of square enclosed within (5,0), (0,5), (-5,0), (0,-5) and the original rectangle and will be approx 25% (~22%).

I think I should be more like "the ratio of area of circle enclosed with in radius 5 and the original rectangle 24 x 10"
Area of rectangle = |12 -(-12)| x |5 -(-5)| = 24 x 10 = 240
Area of any point in circle = Pi*5^2 = 25Pi
25Pi is approximately 25 * 3

Prob[Circle in Rectangle] = 25 * 3/ 24 * 10 = 25 / 80 = around 30 - 34%

Choice C is the correct answer.
SVP
Joined: 17 Jun 2008
Posts: 1502

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12 Sep 2008, 05:37
devilmirror wrote:
scthakur wrote:
It will be the ratio of area of square enclosed within (5,0), (0,5), (-5,0), (0,-5) and the original rectangle and will be approx 25% (~22%).

I think I should be more like "the ratio of area of circle enclosed with in radius 5 and the original rectangle 24 x 10"
Area of rectangle = |12 -(-12)| x |5 -(-5)| = 24 x 10 = 240
Area of any point in circle = Pi*5^2 = 25Pi
25Pi is approximately 25 * 3

Prob[Circle in Rectangle] = 25 * 3/ 24 * 10 = 25 / 80 = around 30 - 34%

Choice C is the correct answer.

Thanks devilmirror for correcting.....committed another silly mistake.....hope do not do during the first 15-20 questions in the actual test.

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Zumit PS 012   [#permalink] 12 Sep 2008, 05:37
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