A thief runs away from a police station with a uniform speed of 100m/minute. After 1 minute, a policeman runs behind the thief to catch him. He runs at a speed of 100m/min in the first minute and increases his speed by 10m/min in each succeeding minute. At the end of which minute, will the policeman be able to catch the thief?

The thief runs at constant speed: $100$ m/min with a $1$-minute head start.

The policeman starts chasing after $1$ minute with increasing speed: $100$ m/min initially, then $+10$ m/min each succeeding minute.

---

The policeman starts at $t = 1$ minute with speeds:

• 1st minute of chase: $100$ m/min
• 2nd minute of chase: $110$ m/min
• 3rd minute of chase: $120$ m/min
• And so on...

---

Calculating distances minute by minute:

Global Time (min)	Thief Position	Policeman Speed	Policeman Distance	Policeman Position	Minutes Since Police Ran
$1$	$100$ m	$0$	$0$	$0$ m	0
$2$	$200$ m	$100$	$100$	$100$ m	1
$3$	$300$ m	$110$	$100 + 110 = 210$	$210$ m	2
$4$	$400$ m	$120$	$210 + 120 = 330$	$330$ m	3
$5$	$500$ m	$130$	$330 + 130 = 460$	$460$ m	4
$6$	$600$ m	$140$	$460 + 140 = 600$	$600$ m	5

---

The policeman catches the thief at the end of the 6th minute (total journey), but that is the 5th minute since the police started the chase.

The question says: "He runs at a speed of 100m/min in the first minute", so, we're looking at minutes since the police started running (i.e. at the end of the 5th minute).

However, this question is ambiguous, and it was contested (but the answer-key remained the same).