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A motorcycle stunts man belonging to a fair, rides over

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by BTGmoderatorLU » Mon Apr 16, 2018 2:56 pm

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A motorcycle stunts man belonging to a fair, rides over the vertical walls of a circular well at an average speed of 54 kph for 5 minutes. If the radius of the well is 5 meters then the distance traveled is:

A. 2.5 km
B. 3.5 km
C. 4.5 km
D. 5.5 km
E. None of the above

The OA is C.

Please, can anyone help me to solve this PS question? I'm confused. Thanks!
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Source: — Problem Solving |

by [email protected] » Tue Apr 17, 2018 7:10 pm
Hi All,

We're told that a motorcycle stuntman rides over the vertical walls of a circular well at an average speed of 54 kph for 5 minutes (in simple terms, he's riding his motorcycle in a circle) and the radius of the well is 5 meters. We're asked for the distance the stuntman travels. While this question is a bit awkwardly-worded, it's essentially just a rate question.

Since we know the motorcycle's speed and time traveled, the fact that it's going in a circle is irrelevant (and by extension, so is the radius of the circle).

5 minutes = 5/60 = 1/12 of an hour

Distance = (54 Km/hour)(1/12 hour)
Distance = 54/12 = 4 6/12 = 4.5 km

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Jeff@TargetTestPrep » Thu Apr 19, 2018 4:41 pm
BTGmoderatorLU wrote:A motorcycle stunts man belonging to a fair, rides over the vertical walls of a circular well at an average speed of 54 kph for 5 minutes. If the radius of the well is 5 meters then the distance traveled is:

A. 2.5 km
B. 3.5 km
C. 4.5 km
D. 5.5 km
E. None of the above
Since distance = rate x time and 5 minutes = 5/60 = 1/12 hour, we have:

Distance = 54 x 1/12 = 54/12 = 9/2 = 4.5 km

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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by deloitte247 » Sun Apr 22, 2018 10:14 am
Speed= 54kph
Time= 5 minutes
Radius= 5 meters
Distance traveled= ??

Average Speed = distance/time taken
Distance= average speed * time taken
$$dis\tan ce=\ 54\cdot\frac{5}{60}\left(to\ convert\ to\ \sec onds\right)$$
$$dis\tan ce=\ 54\cdot\frac{1}{12}=\frac{54}{12}=4.5kms$$
hence the answer is option B
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