A thief runs away from a police station with a uniform speed of 100m/minutes

Let the police catch the thief in ‘n’ minutes

Since the thief ran 1 min before the police start running.

Time taken by the thief before he was caught = (n + 1) min

Distance travelled by the thief in (n+1) min = 100(n+1)m

Given speed of policeman increased by 10m per minute.

Speed of police in the 1st min = 100m/min

Speed of police in the 2nd min = 110m/min

Speed of police in the 3rd min = 120m/min

Hence 100, 110, 120… are in AP

Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]

= (n/2)[2 x 100 +(n – 1)10]

After the thief was caught by the police,

Distance traveled by the thief = distance travelled by the police

100(n+1)= (n/2)[2 x 100 +(n-1)10]

200(n + 1) = n[200 + 10n – 10]

200n + 200= 200n + n(10n – 10 )

200 = n(n – 1)10

n(n – 1) – 20 = 0

n2 – n– 20 = 0

n2 – 5n + 4n – 20 = 0

n(n – 5) + 4(n – 5) = 0

(n – 5) (n+4) = 0

(n – 5) = 0 or (n + 4) = 0

n= 5 or n= -4

Hence n= 5 since n cannot be negative

Therefore the time taken by the policeman to catch the thief = 5minutes