A thief is capable of running 20kms a day. A policeman who is to catch the thief can run 1km on the 1st day, 2kms on the 2nd day and so on. Then how long will it take the policeman to catch the thief?
Let police can caught thief in n days. and since police can run 1km in one day 2km in two days...and so on..... therefore polic can caught thief in n days by running following kms 1+2+3+......n kms n(n+1)/2......(1) and thief can run 21n kms in n days.....(2) now from eqns (1) and (2) we have: n(n+1)/2 21n > n^2+n 42n > n^2-41n 0 > n(n-41) 0 > n 0 n 41. since days can't be 0. hence in 41 days police will caught thief.
let us assume it takes `x` no of days to catch the thief for police man. so by then the thief would run (20*x) km and the police will run (1+2+3+........+x)km. hence 20x x*(x+1)/2 therefore x 39
It takes 39 days for police man to catch the thief.
Let 'n' be the number of days after which police catches thief. Then police had run '20n' kms. The thief runs 1km for 1st day '2' for 2nd & so on & 'n'kms on nth day. So the total distance he runs is [1+2+3+......+n (n*(n+1)/2)]. 20n (n*(n+1)/2) implies n 39. It takes 39 days for the police to catch the thief.
A thief can run 20 kms a day. Let thief be catched in x days so total km is 20*x km run by policeman 1+2+3+4.........+x is an A.P is x(x+1)/2 equating both 20*x x(x+1)/2 x 39
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