A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.
Let us assume total time be n minutes
Also, total distance covered by thief in n minutes =Speed × Time
= 100 × n
= 100 n m
And, Total distance covered by policeman = 1st min. + 2nd min + …… (n – 1) terms
Here, we have:
a = 110
d = 110 – 100 = 10
‘n’ = n – 1
We know that, 
200n = (n – 1) [200 + 10n – 20]
10n2 + 180n – 10n – 180 – 200n = 0
10n2 – 30n – 180 = 0
10 (n2 – 3n – 18) = 0
n2 – 3n – 18 = 0
n (n – 6) + 3 (n – 6) =0
(n + 3) (n – 6) = 0
∴ n + 3 = 0
n = - 3
Or n – 6 = 0
n = 6
As the value of n i.e. time cannot be negative therefore n = 6
∴ Time taken by the thief to catch the thief = n – 1
= 6 – 1
= 1 minutes
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