Question

a thief run with a uniform speed 100 m/minute in first minute. After 1 minute police man run after the thief to catch him. He goes with a speed of 100 m/minute in first minute and increases his speed by 10 m/minute every succeeding minute. After how many minute the police man will catch the thief ?

Answer 1

Answer.

Let the police catch the thief in ‘n’ minutesSince the thief ran 1 min before the police start running. Time taken by the thief before he was caught = (n + 1) minDistance 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