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Obra drove 160π meters along a circular track. If the area enclosed by

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Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post Updated on: 07 Jul 2016, 21:55
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Obra drove 160π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive?

A. 6.67%
B. 12.5%
C. 18.75%
D. 25%
E. 33.3%

Originally posted by EthosGMAT on 07 Jul 2016, 20:12.
Last edited by Bunuel on 07 Jul 2016, 21:55, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post 07 Jul 2016, 21:39
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area enclosed by the circular track on which she drove is 57,600π square meters

So, π(r^2) = 57,600π ---> (r^2)=57,600 ---> r=240

Circumference of the circular track = 2πr = 480π

Therefore, part of circumference covered = 160π/480π = 1/3 or 33.33%

Hence, answer is E.


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Re: Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post 08 Jul 2016, 06:25
EthosGMAT wrote:
Obra drove 160π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive?

A. 6.67%
B. 12.5%
C. 18.75%
D. 25%
E. 33.3%


pi r^2= 57600 pi

r= 240

Total distance= circumference of circle= 2pi r= 480 pi

if 480 pi is 100 %, then 160 pi= 100*160pi/480pi= 33.33% (approximately)

E is the answer
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Re: Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post 08 Jul 2016, 16:26
Thanks for the reply guys! For some reason I thought Obra drove a "part of the circular track" and that 'length of arc' which she drove is 160pi as the question says area enclosed by her on the circular track was sorta confusing wording for me.
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Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post 18 Dec 2018, 09:54
EthosGMAT wrote:
Obra drove 160π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive?

A. 6.67%
B. 12.5%
C. 18.75%
D. 25%
E. 33.3%


Area enclosed by circular track; \(\pi r^2 = 57600\pi\)

\(r^2 = 57600\)

\(r = 240\)

Circumference of the Circular track is ; \(2\pi r = 2 \pi * 240 = 480\pi\)

Obra drove \(160\pi\) of the circumference.

Hence required percentage \(= \frac{160\pi}{480\pi} * 100 = 33.3\)%

Answer E
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Re: Obra drove 160π meters along a circular track. If the area enclosed by  [#permalink]

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New post 28 Jul 2019, 10:14
EthosGMAT wrote:
Obra drove 160π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive?

A. 6.67%
B. 12.5%
C. 18.75%
D. 25%
E. 33.3%


Since the area enclosed by the circular track is 57,600π square meters, the radius of the track, r, can be determined by the equation:

πr^2 = 57,600π

r^2 = 57,600

r = √57,600 = 240

Therefore, the circumference (or length) of the track is 2πr = 2π(240) = 480π meters. So Obra drove

160π/480π = 1/3 or 33.3% of the circular track.

Answer: E
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Re: Obra drove 160π meters along a circular track. If the area enclosed by   [#permalink] 28 Jul 2019, 10:14
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