Question

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

Answer 1

Consider $2n+1$ stones.

Considering one side of middle stone

Distance covered the bringing $1st$ stone $=10m$

$2nd$ stone $=2\times 10\left(n-1\right)$

$3rd$ stone $=2\times 10\left(n-1\right)$ and so on....

$\begin{array}{c}Totaldistncebe{}^{\prime }{D}^{\prime }\\ ononeside,distancecovered\\ \frac{D}{2}=10n+2\times 10\left(n-1\right)+2\times 10\left(n-2\right)...2\times 10\left(1\right)\\ \frac{D}{2}=-10n+2\times 10\left(1+2+3+...+n\right)\\ \frac{D}{2}=-10n+20\left(\frac{n\left(n+1\right)}{2}\right)\\ \frac{D}{2}=10n^2\\ \therefore D=20n^2\\ Given,\\ 20n^2=3000\\ n^2=150\\ \therefore n=12.241\\ Hence,totalno.ofstone25.49\simeq 25.\end{array}$