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M14-06

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Bunuel
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M14-06 [#permalink] New post   16 Sep 2014, 00:51
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69% (01:57) correct 31% (02:18) wrong based on 80 sessions
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If an athlete ran the first half of a 200-meter race in 10 seconds and the second half in 12 seconds, how much faster was his average speed during the first half compared to the second?

A. 2.0 km/h
B. 3.0 km/h
C. 3.6 km/h
D. 6.0 km/h
E. 7.2 km/h
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D
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Bunuel
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Re M14-06 [#permalink] New post   16 Sep 2014, 00:51
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Official Solution:

If an athlete ran the first half of a 200-meter race in 10 seconds and the second half in 12 seconds, how much faster was his average speed during the first half compared to the second?

A. 2.0 km/h
B. 3.0 km/h
C. 3.6 km/h
D. 6.0 km/h
E. 7.2 km/h


The runner's average speed for the first 100 meters was \(\frac{100}{10}=10\) m/s, which is equivalent to \(3600*\frac{10}{1000} = 36\) km/h.

For the second 100 meters, the average speed was \(\frac{100}{12}\) m/s or \(3600*\frac{100}{12 *1000} = 30\) km/h.

The difference, therefore, is 6 km/h.


Answer: D
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Bunuel
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M14-06 [#permalink] New post   08 Dec 2016, 13:48
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Check other Conversion problems to practice from our Special Questions Directory.
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Re: M14-06 [#permalink] New post   09 Oct 2017, 19:15
Could you please explain this part 3600∗[10][/1000]=36 kmh
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Re: M14-06 [#permalink] New post   09 Oct 2017, 19:45
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Voksu wrote:
Could you please explain this part 3600∗[10][/1000]=36 kmh



Hi...
You have to convert m into km and sec into h...
So 10m/s
1m= 1/1000 km ..... 10m=10/1000
1hour=60minutes..
And 1minute=60sec, so 1h =60*60 sec=3600sec
Therefore 1sec=1/3600 h...
Substitute in 10m/s....
(10/1000)/(1/3600)=3600*10/1000 kmph..
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Re: M14-06 [#permalink] New post   11 Jun 2020, 05:17
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I think this is a high-quality question and I agree with explanation.
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Re: M14-06 [#permalink] New post   05 Oct 2023, 12:04
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M14-06 [#permalink]
05 Oct 2023, 12:04
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