Question

Along a road lie an odd number of stones placed at intervals of 10 m. These stones have to be assembled around the middle stone. 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. Then how many stones were there?

Answer 1

Hint- Here, we will proceed by finding the pattern separately between the distance travelled by the man on either side on the middle stone in order to pick up each of these stones and place them around the middle stone.

Complete step-by-step solution:
Let the total number of stones be (2n+1) stones because the total number of stones are odd in number.
As we know that the middle term of n odd terms is given by ${\left( {\dfrac{{n + 1}}{2}} \right)^{{\text{th}}}}$
By replacing n in the above formula by (2n+1), the middle stone along the road is given by
${\left( {\dfrac{{\left( {2n + 1} \right) + 1}}{2}} \right)^{{\text{th}}}} = {\left( {\dfrac{{2n + 2}}{2}} \right)^{{\text{th}}}} = {\left( {n + 1} \right)^{{\text{th}}}}$ stone
From ${1^{{\text{st}}}}$ stone to the middle stone i.e., ${\left( {n + 1} \right)^{{\text{th}}}}$ stone, there are total n number of stones placed in between them. Since, the distance between any two adjacent stones is 10 m.
So, Distance between the first stone and the middle stone = 10(Number of stones occurring in between)
$ \Rightarrow $Distance between the first stone and the middle stone = 10(n) = 10n

The man is carrying the first stone to place it around the middle stone after travelling a distance of 10n metres. Now, he will move to the second stone from the middle stone. As, there are total (n-1) stones in between the middle stone and the second stone. So, the next distance covered by the man will be 10(n-1) metres. Now, he will be picking up the second stone and will be placing it at the middle stone after covering a distance of 10(n-1) metres. Similarly, for the third, fourth and rest of the first n stones. When he has to move from the middle stone i.e., ${\left( {n + 1} \right)^{{\text{th}}}}$ stone to the ${n^{{\text{th}}}}$ stone in order to pick the ${n^{{\text{th}}}}$ stone, he will be covering a distance of 10 metres and then for placing this stone around the middle stone he will again be covering 10 metres distance.
We will be adding all the above distances and the whole sum will give the distance travelled by the man only one side of the middle stone.
Total distance travelled by the man on one side = $10+10+20+20+30+30+…..+10(n-1)+10(n-1)+10n$
$ \Rightarrow $ Total distance travelled by the man on one side = $ 2[10+20+30+…..+10(n-1)]+10n $
For any AP having first term a and the common difference as d,
Sum of first n terms of an AP = $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Since, $[10+20+30+…..+10(n-1)]$ represents the sum of first (n-1) terms of an AP having first term as 10 and common difference as 10.
$ \Rightarrow $ Total distance travelled by the man on one side = $2\left[ {\dfrac{{\left( {n - 1} \right)}}{2}\left[ {2 \times 10 + \left( {n - 1 - 1} \right)10} \right]} \right] + 10n$
$ \Rightarrow $ Total distance travelled by the man on one side = $2\left[ {\dfrac{{\left( {n - 1} \right)}}{2}\left[ {20 + \left( {n - 2} \right)10} \right]} \right] + 10n$

$ \Rightarrow $ Total distance travelled by the man on one side = $\left( {n - 1} \right)\left( {10n} \right) + 10n = 10{n^2} - 10n + 10n = 10{n^2}$ metres.

Now, the man is at the middle stone, he has to travel to the last stone i.e., ${\left( {2n + 1} \right)^{{\text{th}}}}$ stone after travelling a distance of 10n meters to pick that stone and then again he will travel 10n meters to place it around the middle stone and rest will be same as considered for the first end side.
Total distance travelled by the man on other side = $10+10+20+20+30+30+…..+10(n-1)+10(n-1)+10n+10n$
$ \Rightarrow $ Total distance travelled by the man on other side = $ 2[10+20+30+…..+10(n-1)+10n] $

For any AP having first term a and the common difference as d,
Sum of first n terms of an AP = $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Since, $[10+20+30+…..+10(n-1)+10n]$ represents the sum of first n terms of an AP having first term as 10 and common difference as 10.
$ \Rightarrow $ Total distance travelled by the man on other side = \[2\left[ {\dfrac{n}{2}\left[ {2 \times 10 + \left( {n - 1} \right)10} \right]} \right]\]
$ \Rightarrow $ Total distance travelled by the man on other side = \[2\left[ {\dfrac{n}{2}\left[ {20 + 10n - 10} \right]} \right] = n\left[ {10 + 10n} \right]\]

$ \Rightarrow $ Total distance travelled by the man on other side = \[n\left[ {10 + 10n} \right] = 10{n^2} + 10n\] metres
Total distance travelled by the man = Total distance travelled by the man on one side + Total distance travelled by the man on other side
$ \Rightarrow $Total distance travelled by the man = \[10{n^2} + 10{n^2} + 10n = 20{n^2} + 10n\] metres
It is given that the total distance travelled by the man is 3 km = 3000 metres
\[
   \Rightarrow 3000 = 20{n^2} + 10n \\
   \Rightarrow 10\left( {2{n^2} + n} \right) = 3000 \\
   \Rightarrow 2{n^2} + n = 300 \\
   \Rightarrow 2{n^2} + n - 300 = 0 \\
 \]
Solving the above quadratic equation by splitting the middle term method, we get
\[
   \Rightarrow 2{n^2} - 24n + 25n - 300 = 0 \\
   \Rightarrow 2n\left( {n - 12} \right) + 25\left( {n - 12} \right) = 0 \\
   \Rightarrow \left( {2n + 25} \right)\left( {n - 12} \right) = 0 \\
 \]
Either \[
  2n + 25 = 0 \\
   \Rightarrow 2n = - 25 \\
   \Rightarrow n = - \dfrac{{25}}{2} \\
 \] or \[
  n - 12 = 0 \\
   \Rightarrow n = 12 \\
 \]
Since, the number of stones can never be negative and should always be an integer. So, \[n = - \dfrac{{25}}{2}\] is neglected.
So, n = 12 is the correct value of n.
Therefore, Total number of stones = 2n+1 = 2(12) + 1 = 24+1 = 25 stones

Note- In this particular problem, the only difference between the pattern of the distances covered by the man on either side of the middle stone is that for one side, the man is starting to pick from the end stone of that side whereas for other side, he is starting from the middle stone and then we will be going to the end stone of that side in order to pick it up.