a thief runs away from a police station with a uniform speed of100 m per min. after a minute a policeman runs behind him to catch. he goes at a speed of 100 m in 1^{st }min and increases his speed by 10 m each succeeding min. after how many min, the policeman catch the thief?

Let n be the number of minutes after which policeman catches the thief.

then at that instant:

Distance travelled by thief = Distance travelled by policeman

__ Distance travelled by thief__:

Since the thief moves at a constant speed of 100m per min, distance travelled by him = 100(n+1)

__ Distance travelled by policeman__:

Now the policeman increases his speed by 10m per minute after starting with an initial speed of 100m per minute.

Distance travelled by policeman in first minute: 100

Distance travelled by policeman in second minute: 110

Distance travelled by policeman in third minute: 120

Distance travelled by policeman in fourth minute: 130

As we can clearly see this is an A.P with first term 100 and common difference 10.

Hence, Total distance travelled by police in n minutes = 100 +110 + 120 +......= n/2(2*a+(n-1)d)

Now, Total distance travelled by thief in n minutes = Total distance travelled by policeman in n minutes.

- 100 (n+1) = n/2 (2*100 + (n-1)10)
- 100*n + 100 = 100*n + 5n
^{2}- 5*n - 5n
^{2}- 5*n - 100 = 0 - n
^{2}- n - 20 = 0 - n2 + 4*n - 5*n - 20 = 0
- n ( n + 4 ) - 5 ( n + 4 ) = 0
- ( n - 5 )( n + 4 ) = 0

Hence, n = 5 or n = -4.

But n > 0, ( n is the number of minutes )

Therefore n = 5.

Hence, the policeman catches the thief 5 minutes after the policeman starts running.

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