Please tell the solution of question 40.

Solution:

Let the number of stones be 2n+1 . [since number of stones is odd]
let us write the mid-point stone as P and M and N are the left and right end stones respectively.
therefore n number of stones are on the right side of P and n number of stones are on the left side of P.
the distance PM = 10n , since the stones are placed at the interval of 10m.
and M is the nth stone from P.

Let the man starts with the end stone M.
therefore the distance covered in collecting all the left stones.
$$\begin{gathered} =10n+2*\left[10\right(n-1)+10(n-2)+.........+10] \\ =10n+20*\left[\right(n-1)+(n-2)+.......+1] \\ =10n+20*\frac{(n-1)n}{2} \\ =10n+10n^2-10n \\ =10n^2 \end{gathered}$$
here we multiplied by 2, since all the distances are covered twice , except PM.
now the distance covered in collecting all the right stones: [here all the distances are covered twice PN also]
$$\begin{gathered} =2*[10n+10(n-1)+10(n-2)+........+10*2+10] \\ =20*[n+(n-1)+(n-2)+........+2+1] \\ =20*\frac{n(n+1)}{2} \\ =10n^2+10n \end{gathered}$$
therefore total distance covered:
$$\begin{gathered} 10n^2+10n^2+10n=3*1000 \\ 20n^2+10n-3000=0 \\ 2n^2+n-300=0 \\ 2n^2+25n-24n-300=0 \\ n(2n+25)-12(2n+25)=0 \\ (2n+25)(n-12)=0 \\ \Rightarrow n=-25/2 or n =12 \end{gathered}$$
since n is a positive integer. n =12
total number of stones = 2 × 12 + 1 = 25 stones