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A policeman chases a thief at a speed of 27kmph and catches him after

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A policeman chases a thief at a speed of 27kmph and catches him after  [#permalink]

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New post 18 Sep 2018, 04:09
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Question Stats:

58% (02:18) correct 43% (02:09) wrong based on 80 sessions

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A policeman chases a thief at a speed of 27kmph and catches him after 1 minute and 36 seconds. If the distance between the Policeman and the thief was 240 meters initially, What's the speed of the thief?

1) 6Kmph
2) 9Kmph
3) 12Kmph
4) 18Kmph
5) None of these

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A policeman chases a thief at a speed of 27kmph and catches him after  [#permalink]

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New post Updated on: 18 Sep 2018, 04:49
Afc0892 wrote:
A policeman chases a thief at a speed of 27kmph and catches him after 1 minute and 36 seconds. If the distance between the Policeman and the thief was 240 meters initially, What's the speed of the thief?

1) 6Kmph
2) 9Kmph
3) 12Kmph
4) 18Kmph
5) None of these


OA: D

\((V_{Policeman}-V_{Thief})* Time = Distance \quad b/w \quad Two\)

\((27-V_{Thief})* (\frac{1}{60}+\frac{36}{60*60}) = \frac{240}{1000}\)

\((27-V_{Thief})* (\frac{96}{60*60}) = \frac{240}{1000}\)

\((27-V_{Thief})* (\frac{96}{3600}) = \frac{24}{100}\)

\((27-V_{Thief})= \frac{24}{100}*\frac{3600}{96}\)

\((27-V_{Thief})= \frac{24*36}{96}=9\)

\(27-V_{Thief} = 9, V_{Thief}= 27-9= 18\) Km\hr

Originally posted by Bismarck on 18 Sep 2018, 04:46.
Last edited by Bismarck on 18 Sep 2018, 04:49, edited 1 time in total.
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Re: A policeman chases a thief at a speed of 27kmph and catches him after  [#permalink]

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New post 18 Sep 2018, 04:46
Afc0892 wrote:
A policeman chases a thief at a speed of 27kmph and catches him after 1 minute and 36 seconds. If the distance between the Policeman and the thief was 240 meters initially, What's the speed of the thief?

1) 6Kmph
2) 9Kmph
3) 12Kmph
4) 18Kmph
5) None of these


Two ways..

Speed of policeman = 27kmph=27000/3600=7.5mps
So the policeman would travel 7.5*(60+36)=96*15/2=48*15=720
But policeman had to just cover 240 m, so remaining 720-240=480 is the distance thief ran in (60+36)secs so speed =480/96=5mps=5*3600/1000=18kmph

OR

Let the speed of thief be s, so relative speed =(27-x)kmph=(27-x)*1000/3600 mps
Therefore distance =speed *time
240=96*(27-x)*10/36
\(27-x=240*\frac{36}{10*96}=\frac{36}{4}=9......x=27-9=18\)


D
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Re: A policeman chases a thief at a speed of 27kmph and catches him after  [#permalink]

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New post 19 Sep 2018, 11:12
Let the speed of thief be u, so relative speed =(27-u)kmph=(27-u)*5/18 m/s
Therefore distance =speed *time
240=96*(27-u)*5/18
u=5m/s or u=5*18/5=18kmph

so ans is D
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Re: A policeman chases a thief at a speed of 27kmph and catches him after  [#permalink]

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New post 01 Oct 2018, 03:06
Another approach:

The policemen runs 27000m in 96seconds. The distance between the two is 240m.

Therefore the relative distance between the two should be equal to \(\frac{240}{96}\) \(=\frac{5}{2}\) meter per second \(=\frac{5}{2}*3600\) \(=9000\) meters per hour.



Therefore the thief needs to be 9000meter per hour slower than the policemen. \(27000-9000=18000\) meters per second, which equals 18kmh. Hence, D.
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Re: A policeman chases a thief at a speed of 27kmph and catches him after &nbs [#permalink] 01 Oct 2018, 03:06
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