Question

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 1stmin and increases his speed by 10 m each succeeding min. after how many min, the policeman catch the thief?

Answer 1

$\huge{Hey Dude!!!}$

☆☞ Here is ur answer ☜☆

☆☞ 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:

☆☞ 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 + 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.

ENJOY...

GOOD BYE!!!

Answer 2

heya....

Here is your answer...

Let the police catch the thief in ‘n’ minutes Since the thief ran 1 min before the police start running.Time taken by the thief before he was caught = (n + 1) min Distance 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/minHence 100, 110, 120… are in APTotal 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) = 0n= 5 or n= -4 Hence n= 5 since n cannot be negative Therefore the time taken by the policeman to catch the thief = 5minutes.

It may help you....☺☺