Question

To be solved as a question of ARITHMETIC PROGRESSION...

A thief runs from police-station with a uniform speed of 100 m/min. After 1 min, a policeman runs behind the thief to catch him. He goes with a speed of 100 m/min in the first min and increases his speed by 10 m each succeeding min. After how many minutes will the policeman catch the thief.

Answer 1

11 minutes

Answer 2

remember to click good answer :p

Answer 3

where is the solution?????? I want detailed solution by the Arithmetic Progression method....

Answer 4

The distance traveled by the thief after \(n \text{ min }\) can be represented by \(a(n)=100n\). The distance of the policeman is represented by \(b(n)=100(n-1)+10+20+..+10(n-2)=100(n-1)+10\frac{(n-2)(n-1)}{2}\)
\(=100(n-1)+5(n-2)(n-1)\). Solve \(a(n)=b(n)\).

Answer 5

Thanks Anwar, can u pls explain b(n).....

Answer 6

That would give \(n=6\). So, the policeman will catch the thief after 6 minutes.

Answer 7

Answer I have been given is 5 min !!!

Answer 8

yeah, 5 minutes is true calculating from the time when the policeman started running.

Answer 9

Yeah that makes sense....
But pls explain b(n)......

Answer 10

You can see that I define a(n) and b(n) so that at n=1the thief would run 100m and the policeman will start running at n=2.

Answer 11

that i understood...... what about rest....

Answer 12

Okay. Let me make a schedule that can clear things for you:
n (min) distance traveled by thief a(n) distance traveled by policeman b(n)
1 100 0
2 100+100 100
3 100+100+100 100+110=200+10
4 100+100+100+100 100+110+120=300+10+20
n 100n 100(n-1)+10+20+30+..+10(n-2)

Answer 13

I hope the schedule above makes it clearer. I would like to add that \(10+20+..+10(n-2)=10(1+2+..+(n-2)=10\frac{(n-2)(n-1)}{2}\).

Answer 14

Thanks a ton Anwar !!!! You explained it very lucidly......... I also had made the same comparison but cud not "see" the last step.....

Wish I could give you more medals...☺☻

Answer 15

One is enough! Thanks! :D