Math ....plz help me !?

Question

A thief ran away from prison at 2:00 am towards the north with the speed of 10 kilometer\ hour ..... at 8:00 am the police officer found out about him and went after him with the speed of 210 kilometer\hour .... what will be the time when the police officer catches the thief ?? ....

Answer 1

Answer: 8:18 am

Basis:

at 8:00 am, the thief (Th) will be 60 km ahead of the police (Po). So, from that time, his total travel distance will be 60 km + 10*(time) km, where "time" is hours from 8 am.

At 8 am, Po starts at 0 km, but travels at 210 km/hour. So the total travel distance of Po - from 8 am - will be 210*(time) km.

The question is: what value of "time" satisfies the expression

210*(time) = 60 + 10*(time)

Result: (time) = [60/(210 - 10)] hours = [6/20] hours = [18/60] hours

= 18 minutes; hence, Po catches Th at 8:18 am.

  

Answer 2

The thief had a lead of 6 hours or 60km over the police officer

The police officer is going 200km/hr faster than the thief. He would catch the thief in (60/200)hr or 18 minutes

The police office will catch the thief at 8:18am

Answer 3

Notice that the thief was moving for 6hrs at 10km/hr.

So the thief has a 60km head start in front of the police officer.

The thief's distance traveled is then

60 + 10*hrs

The police officer's distance traveled is simply

210*hrs

So we want to know when they equal each other.

210*hrs = 60 + 10*hrs

200*hrs = 60

hrs = 0.3

So the police officer will catch the thief after 0.3 hours after leaving.  This would be 0.3*60 = 18 minutes.

Add that to the time the police officer left (8am)

The answer is 8:18am