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M09-25

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M09-25  [#permalink]

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New post 16 Sep 2014, 00:40
1
1
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

76% (01:08) correct 24% (01:03) wrong based on 95 sessions

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Re M09-25  [#permalink]

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New post 16 Sep 2014, 00:40
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Official Solution:

If the Earth's orbit around the Sun is a circle, by how much will the length of the Earth's orbit increase if the radius of this orbit grows by \(\frac{\pi}{2}\) meters?

A. 1 meter
B. 2 meters
C. \(\pi\) meters
D. \(2\pi\) meters
E. \(\pi^2\) meters

Let \(R\) be the radius of the orbit. The difference in length between the current orbit and the hypothetical one will be \(2\pi(R + \frac{\pi}{2}) - 2 \pi R = 2 \pi \frac{\pi}{2} = \pi^2\)

Answer: E
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Re: M09-25  [#permalink]

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New post 30 Mar 2016, 19:20
i am lost..............please help
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Re: M09-25  [#permalink]

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New post 28 Oct 2016, 05:53
I don't quite follow the explanation given either. However, I think a simpler way to solve may be just to realize that orbit=circumference. So if the radius increases by r/2 how much does the circumference increase. And since circumference= pi*di if the radius increases by pi/2 multiply by 2 to get diameter=pi, and then plug into circumference equation you get pi*pi or pi^2.
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Re: M09-25  [#permalink]

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New post 08 Nov 2016, 13:25
Details of Bunuel's method:

The length of the orbit is asking for the circumference.

If the new orbit (circle) is increased by π/2 then the circumference of this would be 2π(R + π/2). The question asks for the increase (or the difference) of the new circumference to the original circumference. Thus 2π(R + π/2) - 2πR. When using distributives it would = 2πR + 2π^2 / 2 - 2πR. The 2π^2 /2 cancels out to be π^2 and 2πR - 2πR cancels out. Thus π^2 becomes the answer.

Hope this helps.
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Re: M09-25  [#permalink]

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New post 08 Jul 2017, 05:43
1
You can also pick smart numbers:

\(r1=π\)
\(c1=2rπ=2π^2\)

\(r2=\frac{3π}{2}\)
\(c2=2rπ=2\frac{3π}{2}π=3π^2\)

\(diff=3π^2-2π^2=π^2\)
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Re: M09-25   [#permalink] 08 Jul 2017, 05:43
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