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# M09-25

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Math Expert
Joined: 02 Sep 2009
Posts: 54496

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16 Sep 2014, 00:40
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Difficulty:

25% (medium)

Question Stats:

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

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 54496

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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 $$2\pi(R + \frac{\pi}{2}) - 2 \pi R = 2 \pi \frac{\pi}{2} = \pi^2$$

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Intern
Joined: 13 May 2015
Posts: 18
Concentration: Finance, General Management
GMAT 1: 330 Q17 V12
GPA: 3.39

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30 Mar 2016, 19:20
Intern
Joined: 15 Jun 2016
Posts: 1

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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.
Intern
Joined: 29 Sep 2016
Posts: 16
Location: United States
Concentration: Finance, Economics
GPA: 3.01

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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.
Intern
Joined: 07 Feb 2016
Posts: 20
GMAT 1: 650 Q47 V34
GMAT 2: 710 Q48 V39

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

$$r1=π$$
$$c1=2rπ=2π^2$$

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

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

Moderators: chetan2u, Bunuel

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