Question

a thief runs away from a police station with the uniform speed of 100 m/minute.after one minute the police man runs behind the thief to catch him he goes with speed of 100m/minute in the first minute and increases his speed by 10m/minute each succeeding minute.after how many minutes ,the police will catch the thief?

Answer 1

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 100 m 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 the total distance travelled by police in n minutes 
= 100 + 110 + 120 +... = n/2(2*a + (n - 1)d) 
Now the 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 + 5n2 - 5*n 
5n2 - 5*n - 100 = 0 
n2 - 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.