A thief,after committing a theft runs …

Let the time taken by police to catch the thief be n minutes.
Since, the thief ran 2 minutes before the police started running.
Time taken by thief before he was caught = (n + 2) mins
Distance travelled by the thief in (n + 2) mins = 50(n + 2) m
Given that speed of the police increased by 5 m/min.
Speed of police in the 1st min = 60 m/min
Speed of police in the 2nd min = 65 m/min
Speed of police in the 3rd min = 70 m/min
Now, this forms an A.P. with a = 60 and d = 5
$\therefore$ Total distance travelled by the police in n minutes
$= \frac { n } { 2 } [ 2 \times 60 + ( n - 1 ) \times 5 ]$
$= \frac { n } { 2 } [ 120 + 5 n - 5 ]$
$= \frac { n } { 2 } [ 115 + 5 n ]$
$= \frac { 5 n } { 2 } [ 23 + n ]$
After the thief was caught by the police,
Distance travelled by the thief = Distance travelled by the police
$\Rightarrow 50 ( n + 2 ) = \frac { 5 n } { 2 } [ 23 + n ]$}
$\Rightarrow 10 ( n + 2 ) = \frac { n } { 2 } [ 23 + n ]$
$\Rightarrow$20n + 40 = 23 + n²
$\Rightarrow$n² + 8n - 5n - 40 = 0
$\Rightarrow$n(n + 8) - 5(n + 8) = 0
$\Rightarrow$(n + 8)(n - 5) = 0
$\Rightarrow$n + 8 = 0 or n - 5 = 0
$\Rightarrow$n = -8 or n = 5
Thus,time is always postive, n=5
Thus, the time taken by police to catch the thief is 5 minutes.