b) A thief, after committing a theft, runs at a uniform speed of 50 m/

1.	b) A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policemanHe goes 60 m in first minute and increases his speed by Sm/minute every succeeding131minute. After how many minutes, the policeman will catch the thief ?
Answer» Thief has a lead of 50*2 = 100m Relative distance covered by cop in 1st min = 60-50 = 10 Relative distance covered by cop in 2nd min= 65-50 = 15 Relative distance covered by cop in 3rd min = 70 -50 = 20 10, 15, 20 ... is an AP whose sum is 100a= 10 , d = 5 Sum = 100 100 = n/2(2a + (n-1)d) = n/2(20 +(n-1)5)100= n/2(20 + 5n -5)200 = n(15 + 5n)divide by 540 = n(n + 3)n^2 + 8n - 5n - 40 =0n(n+8) - 5(n+8)= 0 >> n = 5 as it cant be negative Therefore cop catches after 5 min + 2min (when cop was at rest) = 7min

b) A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policemanHe goes 60 m in first minute and increases his speed by Sm/minute every succeeding131minute. After how many minutes, the policeman will catch the thief ?

Answer»

Thief has a lead of 50*2 = 100m

Relative distance covered by cop in 1st min = 60-50 = 10

Relative distance covered by cop in 2nd min= 65-50 = 15

Relative distance covered by cop in 3rd min = 70 -50 = 20

10, 15, 20 ... is an AP whose sum is 100a= 10 , d = 5

Sum = 100

100 = n/2(2a + (n-1)d) = n/2(20 +(n-1)5)100= n/2(20 + 5n -5)200 = n(15 + 5n)divide by 540 = n(n + 3)n^2 + 8n - 5n - 40 =0n(n+8) - 5(n+8)= 0

>> n = 5 as it cant be negative

Therefore cop catches after 5 min + 2min (when cop was at rest) = 7min