Question

Along a road lie an odd number of stones placed at intervals of $10$ metres. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried the job with one of the end stones by carrying them in succession. In carrying all the stones he covered a distance of $3$km. Find the number of stones.

Answer 1

Let the number of stones be $2n+1$ so that there is one mid-stone and n stones each on either side of it. If P be mid-stone and A, B be last stones on the left and right of P respectively.
There will be $(n+1)$ stones on the left and $(n+1)$ stones on right side of P(P being common on both sides) or n intervals each of $10$ metres both on the right and left side of mid-stone. Now he starts from one of the end stones, picks it up, goes to mid-stone, drops it and goes to last stone on the other side, picks it and comes back to mid-stone. In all he travels n intervals of $10$ metres each $3$ times. Now from centre he will go to $2$nd stone on L.H.S. then come back and then go to $2$nd last on R.H.S. and again come back. Thus he will travel $(n-1)$ intervals of $10$ metres each $4$ times. Similarly $(n-2)$ intervals of $10$ metres each $4$ times for $3$rd and so on for the last.
Hence the total distance covered as given $=3$k.m. $=3000$m.
or $3.10n+4\left[10\right(n-1)+10(n-2)+$_______$+10]$
$=30n+40[1+2+3+$________$(n+1)]=3000$
or $30n+40\left[\right(n-1\left)/2\right][1+n-1]=3000$
or $2n^2+n-300=0$
or $(n-12)(2n+25)=0$,
$\therefore n=12$.
Hence the number of stones $=2n+1=25$.