A long a road lie an odd number of stones placed at intervals of 10 meters. These stones have to be assembled around the middle stone. A person can carry only one stone ar a time. A man started the job with one of the end stones by carrying them in succession. In carrying all the stones, the man covered a total distance of 3 kilometers. Then the total number of stones is

Explanation

Let there be (2n+1) stones, one in the middle and n on each side of it in a row.
Let A and B be the end stones on the left and right of middle stone P, respectively.
Clearly, there are n intervals each of length 10 metres on both the sides of P.
Now, suppose the man starts from A.
He picks up the end stone and goes to the mid-stone, drops it and goes to(n-1)th stone, picks it up, goes to the mid-stone and drops it.
This process is repeated till he collects all stones from the left side at P.
Distance covered in collecting stones between A and P.
`=10xxn+2[10xx(n-1)+10xx(n-2)+..+10xx2+10xx1]`
After collecting all the stones from left side, the man goes to the stone B from P, picks it up, goes to P and drops it.
Then he goes to (n-1)th stone on the right and the process is repeated till he collects all stones at P.
Distance covered in collecting stones between B and P
`=2[10xxn+10xx(n-1)+10xx(n-2)+...+10xx2+10xx1]`
`therefore` Total distance covered
`=10xxn+2[10xx(n-1)+10xx(n-2)+..+10xx2+10xx1]+2[10xxn+10xx(n-1)+...+10xx2+10xx1]`
`=4[10xxn+10xx(n-1)+..+10xx2+10xx1]-10xxn`
`=40[1+2+3+...+n]-10n`
`=40[n/2(1+n)]-10n`
`=20n(n+1)-10n`
`=20n(n+1)-10n`
`=20n^(2)+10n`
But the total distance covered is 3 km=3000 m
`therefore20n^(2)+10n=3000`
`rArr2n^(2)+n-300=0`
`rArr(n-12)(2n+25)=0`
`rArrn=12`
Hence, the number of stones = 2n+1=25