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 3 km. find the number of stones.

Answer 1

Hint: In the solution, first we have to calculate the total distance covered by that person. In the problem, it is given that the person covers a total of 3000 meter. Thus, we have to compare his calculated distance with the given value. We get the value of the total number of an interval of the stones. Once we get the total number of intervals of the stone, we can calculate the total number of stones present.

Complete step by step solution:
Let us consider the number of stones to be $2n + 1$.
Considering, C be the mid stone and A, B be the last stones on the left side and right side of C respectively.
Therefore the number of stones on the left side is $n + 1$.

Similarly, the number of stones on the right side is $n + 1$.
Given that the interval of each stone is 10 meter.
Thus, for $2n + 1$ number of stones there are $n$ intervals on the left side and $n$ intervals on the right side of the mid stone.
Given that the man carries one stone at a time and covers a total distance of 3 km.
Considering the man started carrying the stone from the right side.

Now, he moves from one of the end stones, picks it up and drops it to the midst of stone and goes to the last stone on the other side, picks it up and returns to the midst of stone. In all, he now goes n intervals of 10 m every 3 times from the middle to the 2nd stone on the left side and then moves to the 2nd last on the right side and returns again.

Hence, the total distance covered $ = 3 \times 10n + 4[10\left( {n - 1} \right) + 10\left( {n - 2} \right) + \ldots + 10]$ (i)
It is given that distance he cover is 3 km
Converting 3 km to meter
$3\;{\rm{km}} = 3000\;{\rm{m}}$ (ii)

Now, comparing both (i) and (ii), we get
$\begin{array}{l}3 \times 10n + 4[10\left( {n - 1} \right) + 10\left( {n - 2} \right) + \ldots + 10] = 3000\\ \Rightarrow 30n + 40\left( {1 + 2 + 3 + \ldots + \left( {n - 1} \right)} \right) = 3000\\ \Rightarrow 30n + 40\left( {2n - 1} \right)\left( {1 + n - 1} \right) = 3000\\ \Rightarrow 2{n^2} + n - 300 = 0\\ \Rightarrow \left( {n - 12} \right)\left( {2n + 25} \right) = 0\\ \Rightarrow n = 12, - 12.5\end{array}$

Since, a negative value is not possible, we have to consider only$n = 12$.
Now, we have to calculate the number of stones.
Substituting the value of $n$, we get
Number of stones
$\begin{array}{l} = 2n + 1\\ = 2 \times 12 + 1\\ = 25\end{array}$

Hence, the total number of stone is 25.

Note: Here we have to determine the total number of stones present for the given data. In the problem the total distance and the interval of each stone are given, so we can easily determine the total number of stones.