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06 Mar 2017, 01:48
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54% (03:07) correct
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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 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 bull’s-eye?
A. 1/18
B. 1/27
C. 1/64
D. 1/81
E. 1/729
A. 1/18
B. 1/27
C. 1/64
D. 1/81
E. 1/729
06 Mar 2017, 02:42
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The probability that Alex’s arrow hits the bull’s-eye = Area of the Bulls Eye/ Area of the entire circular Archery target
Area of the Bulls Eye = 9π square inches (Give : The circular center region, the bull’s-eye, has a radius of 3 inches)
Now we need to calculate the Area of the entire circular Archery target.
Given: the largest scoring ring has an area of 153π square inches, which means the difference between the areas of the last two consecutive rings is 153π square inches. So we need to fine the radius of the last Circle
π [(R+3)^2 - R^2] = 153π
R^2 + 9 + 6R - R^2 = 153
6R = 144
R=24
Radius of Last Circle = R+3 = 27
Area of the entire circular Archery target = 27^2 * π
The probability that Alex’s arrow hits the bull’s-eye = Area of the Bulls Eye/ Area of the entire circular Archery target
= 9π/27^2 * π
= 1/81
Answer is D. 1/81
Area of the Bulls Eye = 9π square inches (Give : The circular center region, the bull’s-eye, has a radius of 3 inches)
Now we need to calculate the Area of the entire circular Archery target.
Given: the largest scoring ring has an area of 153π square inches, which means the difference between the areas of the last two consecutive rings is 153π square inches. So we need to fine the radius of the last Circle
π [(R+3)^2 - R^2] = 153π
R^2 + 9 + 6R - R^2 = 153
6R = 144
R=24
Radius of Last Circle = R+3 = 27
Area of the entire circular Archery target = 27^2 * π
The probability that Alex’s arrow hits the bull’s-eye = Area of the Bulls Eye/ Area of the entire circular Archery target
= 9π/27^2 * π
= 1/81
Answer is D. 1/81
29 Nov 2017, 21:52
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Hi All,
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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
