A certain archery target is made up of a series...

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

A certain archery target is made up of a series...

by BTGmoderatorLU » Sat Oct 21, 2017 7:10 pm

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 bull's-eye, has a radius of 3 inches, and the largest scoring ring has an area of 153Ï€ 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 bull's-eye?

A. 1/18

B. 1/27

C. 1/64

D. 1/81

E. 1/729

The OA is D.

Can any expert explain this PS question please? How can I solve it? Thanks.

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Source: Beat The GMAT — Problem Solving | Sat Oct 21, 2017 7:10 pm

GMAT/MBA Expert

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by [email protected] » Sun Oct 22, 2017 10:33 am

Hi LUANDATO,

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 153Ï€ 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 9Ï€. 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 153Ï€ square inches. For example:
Bull's-eye = 9Ï€
2nd circle = radius of 6 = 36Ï€ - 9Ï€ = 27Ï€ -- area of 1st ring
3rd circle = radius of 9 = 81Ï€ - 36Ï€ = 45Ï€ -- area of 2nd ring
Etc.

To save some time, we should note that the area of the outer ring is 153Ï€, 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 18Ï€ each time....

45Ï€ -- radius of 9
63Ï€ -- radius of 12
81Ï€ -- radius of 15
99Ï€ -- radius of 18
117Ï€ -- radius of 21
135Ï€ -- radius of 24
153Ï€ -- radius of 27

Thus, the overall area of the full target is (27^2)Ï€ = 729Ï€ and the probability of hitting a bull's-eye is 9Ï€/729Ï€ = 1/81

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Wed Nov 20, 2019 6:00 pm

BTGmoderatorLU wrote: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 bull's-eye, has a radius of 3 inches, and the largest scoring ring has an area of 153Ï€ 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 bull's-eye?

A. 1/18

B. 1/27

C. 1/64

D. 1/81

E. 1/729

The OA is D.

Can any expert explain this PS question please? How can I solve it? Thanks.

The area of the largest ring is the difference between the area of the largest circle and the second largest circle. We can let x = the radius of the second largest circle. Thus, we have:

Ï€(x + 3)^2 - Ï€x^2 = 153Ï€

(x + 3)^2 - x^2 = 153

x^2 + 6x + 9 - x^2 = 153

6x + 9 = 153

6x = 144

x = 24

Since the entire target is the area of the largest circle, the area of the target is Ï€(24 + 3)^2 = Ï€(27)^2. Since the area of the bull's eye is Ï€(3)^2, the probability of randomly hitting the bull's eye is Ï€(3)^2/[Ï€(27)^2] = (3/27)^2 = (1/9)^2 = 1/81.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]