Question

A theief , after committing a theft runs at a uniform speed of 50m/min. After 2 min a policeman runs to catch him.He goes 60m in first min and increase his speed by 5m/min every suceeding min.After how many mins the policeman will catch the thief?
Plzz answer quickly,it is an AP Question.

Answer 1

Let the policeman catch the thief in n minutes.

Given that uniform speed of the thief = 50m/min.

Given that After 2 minutes a policeman runs to catch him at the speed of 60 in first minutes.

(n + 2) minutes = 50(n + 2) minutes.

Given that speed of policeman increase by 5m/min.

The speed of police in 1 minute = 60m/min.

The speed of police in 2 minutes = 65m/min.

The speed of police in 3 minutes = 70m/min.

The speed of police in 4 minutes = 75m/min.


Hence 60,65,70,75 are in AP.

Let a be the first term and d be the common difference.

a = 60,d = 65 - 60 = 5.


We know that sum of n terms of an ap = n/2(2a + (n - 1) * d)

                                                                 = n/2(2(60) + (n - 1) * 5)   ------ (1)



The distance traveled by the thief = Distance traveled by the police

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

100n + 200 = n(120 + 5n - 5)

100n + 200 = 120n + 5n^2 - 5n

100n + 200 = 5n^2 + 115n

5n^2 + 115n - 200 = 100n

5n^2 + 15n - 200 = 0

n^2 + 3n - 40 = 0

n^2 - 5n + 8n  - 40 = 0

n(n - 5) + 8(n - 5) = 0

(n - 5)(n + 8) = 0

n = 5 (or) n = -8.

n cannot be negative.


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


Hope this helps!