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M14-06

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M14-06 [#permalink]

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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, by how much was his average speed over the first half larger than that over the second?

A. 2.0 kmh
B. 3.0 kmh
C. 3.6 kmh
D. 6.0 kmh
E. 7.2 kmh
[Reveal] Spoiler: OA

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New post 15 Sep 2014, 23:51
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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, by how much was his average speed over the first half larger than that over the second?

A. 2.0 kmh
B. 3.0 kmh
C. 3.6 kmh
D. 6.0 kmh
E. 7.2 kmh

The runner's average speed over the first 100 meters was 10 m/s or \(3600*\frac{10}{1000} = 36\) kmh. The runner's average speed over the second 100 meters was \(\frac{100}{12}\) m/s or \(3600*\frac{100}{12*1000} = 30\) kmh. The difference is 6 kmh.

Answer: D
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Re: M14-06 [#permalink]

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I don't like how you don't post the conversion formulas. On the test the formulas are provided thus your questions without them are unrealistic and unnecessarily difficult.

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New post 08 Dec 2016, 12:48
Mermaidk wrote:
I don't like how you don't post the conversion formulas. On the test the formulas are provided thus your questions without them are unrealistic and unnecessarily difficult.


This type of questions are called conversion problems and they never provide with any formula.

Check other Conversion problems to practice from our Special Questions Directory.
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Collection of Questions:
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Re: M14-06 [#permalink]

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New post 09 Oct 2017, 18:15
Could you please explain this part 3600∗[10][/1000]=36 kmh

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New post 09 Oct 2017, 18:45
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] 09 Oct 2017, 18:45
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