Question

A thief away from a police station with a uniform speed of 100 metre per minutes. After 1 minute a police man runs behind the thief to catch him. He goes at a speed of 100 M per minute in first minute and increases the speed 10 metre per minute on each succeeding minutes. After how many minutes the police man catches the thief.

Answer 1

Suppose the police man catches the thief after t minutes.

Uniform speed of the thief = 100m/min

∴ Distance covered by thief in (t + 1) minutes = 100 m/min × (t + 1) minutes = 100 (t + 1) m

Distance covered by police man in t minutes

= Sum of t terms of an AP with first term 100 and common difference 10



When the policeman catches the thief, we have



∴ t = 5  (t cannot be negative)

Thus, the policeman catches the thief after 5 minutes.

Answer 2

Let the police catch the thief in n minutes

Time taken by thief=n+1
Distance by thief =100(n+1)
Speed of police = 100,110,120......
Distance travelled by thief =distance travelled by police
100(n+1)=n/2(200+10n-10)
200(n+1)= n(200+10n-10)
200n+200=200n+n(10n-10)
200=10×n(n-1)
20=n^2-n
n^2 - n - 20=0
n^2-5n+4n-20 =0
n(n-5)+4(n-5)=0
(n+4)(n-5)=0
n= 5 or -4
n=5