Question

A thief , after committing a theft , runs at a speed of 50 m/min . After 2 mins , a policeman runs to catch him. He goes 60m in the first min and increases his speed by 5 m/min every succeeding minute. After how many minutes,the policeman will catch the thief?

Answer 1

Let the police catch the thief in ‘n’ minutes

Since the thief ran 2 min before the police start running.

Time taken by the thief before he was caught = (n + 2) min

Distance travelled by the thief in (n+2) min = 50(n+2)m

Given speed of policeman increased by 5m per minute.

Speed of police in the 1st min = 60m/min

Speed of police in the 2nd min = 65m/min

Speed of police in the 3rd min = 70m/min

Hence 60, 65, 70… are in AP

Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]
= (n/2)[2 x 60+(n – 1)5]

After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police

50(n+2)= (n/2)[2 x 60+(n-1)5]

50(n + 2) = n[60 + (n-1)5/2]

50n + 100= 60n + n(n – 1 )5/2

100 = 10n+n(n – 1)5/2
​​​​​
n(n – 1)5/2+10n=100

Multiply each term with 2,

n(n-1)5+20n=200

divide each term with 5,

n(n-1)+4n=40
n^2-n+4n=40
n^2+3n-40=0
(n+8)(n-5)=0

n=-8 ot n=5

Hence n= 5
since n cannot be negative

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