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Updated on: 26 Jun 2013, 03:26
Originally posted by guerrero25 on 26 Jun 2013, 03:11.
Last edited by guerrero25 on 26 Jun 2013, 03:26, edited 1 time in total.
Last edited by guerrero25 on 26 Jun 2013, 03:26, edited 1 time in total.
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E
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Difficulty:
25%
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Question Stats:
75% (01:29) correct
25%
(01:50)
wrong
based on 151
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Circles X and Y are concentric. If the radius of circle X is three times that of circle Y, what is the probability that a point selected inside circle X at random will be outside circle Y?
(A) 13
(B) \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{2}\)
(D) 5/6
(E) 8/9
(A) 13
(B) \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{2}\)
(D) 5/6
(E) 8/9
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26 Jun 2013, 03:18
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guerrero25
Say the radius of the smaller circle (Y) is 1, then the radius of the larger circle would be 3.
The area of X = \(\pi{R^2}=9\pi\);
The area of Y = \(\pi{r^2}=\pi\).
The probability = (Favorable)/(Total) = \(\frac{9\pi-\pi}{9\pi}=\frac{8}{9}\).
Answer: E.
14 Oct 2013, 01:54
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Small circle radius = a, so area = PI a ^ 2
Large circle radius = 3a, so area = 9. PI a ^ 2
Area of shaded region = 9PI a^2 - PI a^2 = 8 PI a^2
So, probability = ( 8 PI a^2 ) / (9 PI a^2 )
= 8/9 = Answer = E
Large circle radius = 3a, so area = 9. PI a ^ 2
Area of shaded region = 9PI a^2 - PI a^2 = 8 PI a^2
So, probability = ( 8 PI a^2 ) / (9 PI a^2 )
= 8/9 = Answer = E
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