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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A certain archery target is made up of a series of tagged by: VJesus12 ##### This topic has 1 member reply ## A certain archery target is made up of a series of ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult A certain archery target is made up of a series of concentric circles creating alternating red and white scoring rings. Each successive circle has a radius 3 inches greater than the one before. The circular center region, the bulls-eye, has a radius of 3 inches, and the largest scoring ring has an area of 153π square inches. If Alex shoots an arrow that hits a random point on the target, what is the probability that Alex’s arrow hits the bulls-eye? A. 1/18 B. 1/27 C. 1/64 D. 1/81 E. 1/729 The OA is the option D. How can I know this probability? Is there a strategic way to do it? Please, I need some help. Source: Veritas Prep ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10109 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi VJesus12, We're told that the circular center region (the bull’s-eye) has a radius of 3 inches, each successive circle has a radius 3 inches greater than the one before and the largest scoring ring has an area of 153π square inches. We're asked for the probability that an arrow that hits the target hits the bull’s-eye. To start, we know that the area of the bulls-eye is 9π. To answer the question, we need to figure out the area of the ENTIRE target. Since each circle's radius is '3 more' than the immediate circle within it, we could potentially 'map out' the area of each ring until we hit 153π square inches. For example: Bull's-eye = 9π 2nd circle = radius of 6 = 36π - 9π = 27π -- area of 1st ring 3rd circle = radius of 9 = 81π - 36π = 45π -- area of 2nd ring Etc. To save some time, we should note that the area of the outer ring is 153π, so the radius of the largest circle would have to be quite a bit bigger than that of these inner circles... If you look at the area of each increasing ring, you'll notice that the area appears to increase by 18π each time.... 45π -- radius of 9 63π -- radius of 12 81π -- radius of 15 99π -- radius of 18 117π -- radius of 21 135π -- radius of 24 153π -- radius of 27 Thus, the overall area of the full target is (27^2)π = 729π and the probability of hitting a bull's-eye is 9π/729π = 1/81 Final Answer: D GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com • FREE GMAT Exam Know how you'd score today for$0

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