Question

Along a road lies an odd number of stones of weight 10 kg placed at intervals of 10m. These stones have to be assembled around the middle stone. A stone loader can carry only 1 stone of 10 kg at a time. He started the job with one of the end stones by carrying all the stones in succession. In carrying all the stones. He covered a distance of 3 km. Find the number of stones.

Answer 1

Hint: We will start solving this question by letting the total number of stones. After it, we will find the distance covered in both sides of the middle stone and equate the distance covered with the total distance given. We will make a series according to the given question.

Complete step-by-step answer:
The total number of stones lying on the road are odd according to the question. So, let the total number of stones = 2n+1
Now, we will first find the distance covered from one side of the middle stone. Now, as we let the total stone be 2n+1, there are no stones on both sides of the middle stone.
Now, as the loader carry the stone, the distance covered in carrying the first stone = 10(2)
Again, distance covered in carrying 2nd stone = 10(2)(2)
Distance covered in carrying the 3rd stone = 10(3)(2)
So, total distance covered in carrying (n-1) stones = 10(2) + 10(2)(2) + 10(3)(2) + …. 10(n-1) (2)
So, distance = 2[10 + 10(2) + 10(3) + ……… 10(n-1)]
Also, distance covered in carrying nth stone = 10n
So, total distance covered in carrying stones on one side = 2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n
As both sides have equal numbers of stones.
So, distance covered in carrying stones on other side = 2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n
So, total distance covered carrying all the stone = 2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n + 2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n
According to the question, the total distance = 3km = 3000 m.
So, we get
2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n + 2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n = 3000
2{2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n} = 3000
2[10 + 10(2) + 10(3) + ……… 10(n-1)] + 10n = 1500
Now, we can see that on the left-hand side, there is a series of arithmetic progression of (n-1) terms. So, we will use the formula of sum to find the number of stones. Also, the above expression can be written as,
2[10 + 10(2) + 10(3) + ……… 10(n-1) + 10n] - 10n = 1500 … (1)
We have added 10n and subtract 10n in the left-hand side, so we get the sum of n terms.
Now, the sum of n terms with first term a and common difference d = $\dfrac{n}{2}(2a + (n - 1)d)$
So, the sum of n terms with a = 10 and d = 10 is S = $\dfrac{n}{2}(2(10) + (n - 1)10)$ = $5{n^2} + 5n$
So, from equation (1), we get
\[2(5{n^2} + 5n) - 10n = 1500\]
\[10{n^2} + 10n - 10n = 1500\]
\[10{n^2} = 1500\]
${n^2} = 150$
n = 12.24
As, n is an integer. Therefore, n = 12
So, total number of stones = 2(12) +1 = 24+1=25
So, total number of stones lying on the road are 25


Note: Whenever we come up with such types of questions, we will first make the series according to the question. Then we will use the properties and formulas of the series formed. For example, if a geometric progression is formed, we will use the formulas of GP, like in this question, where there is a series of AP and its formulas are used. In such questions, the final answer may come as a floating value instead of an integer, so we round off the final result to get the integer value.