A thief is capable of running 20kms a day.

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?
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Questions by srividhya athreya

Showing Answers 1 - 7 of 7 Answers

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.

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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.
                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. 


  • Jun 10th, 2009

39 days...
let the no. of days be 'n'..after n days thief runs n*20 kms.. police runs 1+2+...+n kms..

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  • Jul 25th, 2009

Let the no. of days needed is x.

In x days, the theif runs 20x kms.

On day 1, police runs 1km.
On day 2, he runs 2 kms.
Similarly, on the xth day, he runs x kms.

In the entire x days, police runs (1+2+3+......+x) kms i.e  x(x+1)/2 kms.

Equating 20x and x(x+1)/2, we get




Hence, it takes 39 days to catch the theif.


  • Nov 5th, 2009

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

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