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A spaceship that is traveling 3,600 kilometers per hour...

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by AAPL » Tue Jan 23, 2018 6:46 am
A spaceship that is traveling 3,600 kilometers per hour can make a circular orbit around a spherical planet in 4 minutes. Approximately how far is the spaceship from the center of the planet?

A. 24,000 kilometers
B. 1,200 kilometers
C. 240 kilometers
D. 120 kilometers
E. 40 kilometers

The OA is E.

I don't have clear this PS question, I appreciate if any expert explain it for me. I solve it of the following way,

I need to determine the distance that the spaceship cover around of the planet, right? Then the distance will be,
$$D=Speed\ \cdot Time=3,600\cdot\left(\frac{4}{60}\right)=240\ kilometers$$
Then I know that the orbit will be spherical, that's mean that I can say that the spaceship is traveling over a circumference with lenght 240 kilometers, right?

Therefore, I need to determine the radius of this circumference and it will be,
$$L_{circumference}=2\cdot\pi\cdot r\ ,\ then\ r=\frac{L}{2\cdot\pi}$$
$$Finally,\ r=\frac{240}{2\cdot\pi}=38.19\approx40\ kilometers$$
Thank you so much.
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Re:

by DavidG@VeritasPrep » Tue Jan 23, 2018 7:12 am
AAPL wrote:A spaceship that is traveling 3,600 kilometers per hour can make a circular orbit around a spherical planet in 4 minutes. Approximately how far is the spaceship from the center of the planet?

A. 24,000 kilometers
B. 1,200 kilometers
C. 240 kilometers
D. 120 kilometers
E. 40 kilometers

The OA is E.

I don't have clear this PS question, I appreciate if any expert explain it for me. I solve it of the following way,

I need to determine the distance that the spaceship cover around of the planet, right? Then the distance will be,
$$D=Speed\ \cdot Time=3,600\cdot\left(\frac{4}{60}\right)=240\ kilometers$$
Then I know that the orbit will be spherical, that's mean that I can say that the spaceship is traveling over a circumference with lenght 240 kilometers, right?



Therefore, I need to determine the radius of this circumference and it will be,
$$L_{circumference}=2\cdot\pi\cdot r\ ,\ then\ r=\frac{L}{2\cdot\pi}$$
$$Finally,\ r=\frac{240}{2\cdot\pi}=38.19\approx40\ kilometers$$
Thank you so much.
Sure. Your solution is valid. You could also, once you've determined that the circumference is 240km, use the answer choices and a little logic. We want the radius. Obviously the radius must be less than the circumference. That kills A, B, and C.

Now, if you recognize that the circumference is the diameter times Pi, you can see that D can't possible work, as the diameter is itself 240 kilometers, and to get the circumference, you'll be multiplying that by a value slightly greater than 3. So the answer has to be E. Not much math required.
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by [email protected] » Tue Jan 23, 2018 11:46 am
Is there a way to know what level these question's' fall into on BEAT THE GMAT, sorry am a new user seeking to know more
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by Brent@GMATPrepNow » Tue Jan 23, 2018 11:58 am
[email protected] wrote:Is there a way to know what level these question's' fall into on BEAT THE GMAT, sorry am a new user seeking to know more
No, BTG doesn't have that functionality (yet).
I'd say the question falls in the 500-600 range

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by DavidG@VeritasPrep » Tue Jan 23, 2018 12:09 pm
Brent@GMATPrepNow wrote:
[email protected] wrote:Is there a way to know what level these question's' fall into on BEAT THE GMAT, sorry am a new user seeking to know more
No, BTG doesn't have that functionality (yet).
I'd say the question falls in the 500-600 range

Cheers,
Brent
I'll both second Brent's assessment of the question's difficulty level and caution you against devoting too much energy to assessing how the algorithm might rate a particular problem. Better to use your practice tests to determine which content areas that give you difficulty, and then address those areas before taking your subsequent exam. What's difficult for someone else might not be difficult for you and vice versa.
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by Jeff@TargetTestPrep » Wed Jan 24, 2018 9:43 am
AAPL wrote:A spaceship that is traveling 3,600 kilometers per hour can make a circular orbit around a spherical planet in 4 minutes. Approximately how far is the spaceship from the center of the planet?

A. 24,000 kilometers
B. 1,200 kilometers
C. 240 kilometers
D. 120 kilometers
E. 40 kilometers
Let's first determine the circumference C of the circular path of the spaceship around the planet:

C = 3600 km/hr x 4 min

C = 3600 km/hr x 4/60 hr

C = (3600 x 4/60) km

C = 240 km

Now we have to determine the radius of the orbit since that is the distance the spaceship is from the center of the plant:

240 = 2Ï€r

240/(2Ï€) = r

Since π ≈ 3, then r ≈ 240/6 = 40 km.

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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