Question

Along a road lie an odd number of stones and distance between consecutive stones is $10m$. A person can carry only one stone at a time and his job is to assembled all the stones around the middle stone. If he starts his job from one of the end stone and in carrying all the stones, he covers a distance of $3km$. Find the number of stones.

Answer 1

Let the number of stones be $(2n+1)$ since there are odd number of stones.

Given these stones are to be kept around the middle stone.

That is there will $n$ stones kept on right side of middle stone and $n$ stones kept on the left side of the middle stone.

It is also given that there are $n$ intervals each $10$m on both the sides.

If the man collects the stones from $1st$ to $n$ stones and drops it around the middle stone.

The distance covered in collecting all the stones on the left side $=2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1\left)\right]+10\left(n\right)[\because$ the man was initially at the extreme left then $nth$ stone he will cover one way distance whereas other stones he has cover two way distance$]$

Now the person is at the middle stone. Now repeats the same process as done in collecting the stones in left side. The difference is that here he will cover distance both the way in collecting all the stones.

Hence the distance covered on the right side $=2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]$

Total distance $=2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1\left)\right]+10\left(n\right)+2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]$

$=4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1\left)\right]+30n]$

$=4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]–10n$

But the total distance covered is $3km$ or $3000m$

Hence $4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]–10n=\mathrm{3000........}\left(1\right)$

But $10\left(1\right)+10\left(2\right)+10\left(3\right)+\dots +10(n-1)+10\left(n\right)$ forms an $AP.$

Here $a=10,d=10$

Recall the sum of n terms of $AP,$ $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}\left[2\right(10)+(n-1\left)10\right]$

$=\frac{n}{2}[10n+10]$

$=5n^2+5$

Equation $\left(1\right)$ becomes,

$4[5n^2+5n]–10n=3000$

$=20n^2+20n–10n=3000$

$=20n^2+10n–3000=0$

$=2n^2+n–300=0$

$=2n^2+25n–24n–300=0$

$=n(2n+25)–12(2n+25)=0$

$=(2n+25)(n–12)=0$

$=(2n+25)=0$ or $(n–12)=0$

$\therefore n=12$ or $n=\frac{-25}{2}$

But $n$ cannot be negative or fraction

$\to n=12$

Therefore, number of stones $=2n+1=2\left(12\right)+1=25$