Question

A theif runs away from a police station with a uniform speed of 100m/min.. After one minute , a police man runs behind the theif to catch him. He goes at a speed of 100m/ min.in the first minute and increases his speed by 10m/min. In each succeeding minute. After how many minutes will the police man catch the theif?

Answer 1

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
 

Answer 2

IT'S LONG BUT U Will UNDERSTAND

Let time taken by thief = n minutes
speed of thief = 100m/min
So the distance travelled by thief is 100n

Let time taken by policeman= (n-1) minutes
distance travelled by police in A.P.
100n , 110n , 120n........... 100(n-1)

Therefore
100n = (n-1)/2{2(100) + (n-1-1)10}
200n = (n-1){200 + (n-2)10}
200n =(n-1){200 + 10n -20}
200n = (n-1)(180 + 10n)

200n = 180n + 10n²- 10n-180
200n = 10n² + 170n - 180
10n² -30n - 180 = 0
n² -3n -18 = 0
n² -6n +3n - 18 =0
n(n-6) +3(n-6) =0
n = 6

So time taken by policeman is 6 -1 = 5 min