Question

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Answer 1

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 100
a= 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 5
40 = n(n + 3)
n^2 + 8n - 5n - 40 =0
n(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



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