Question

Along a road lie an odd number of stones placed at intervals of 10 metres. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. 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. Find the number of stones
(a) 26
(b) 23
(c) 25
(d) 20

Answer 1

Hint: Take the total number of stones as ‘2n + 1’ where take centre as 1 and ‘n’ on left and ‘n’ on right side.
Find the sum of distance covered by them and then equate it with 3 km or 3000 m.

Complete step-by-step answer:
In the question we are told that along a road lie an odd number of stones and distance between consecutive stones in 10 m. A person can carry only stone at a time and his job is to assemble all the stones around the middle stone. If he starts his job from one of the end stones and carries all the stones, he will cover a distance of 3 km. So, we have to find the total number of stones.
Now as we know that number of stones is odd so we can write as 2n + 1.
We are given that these stones are to be kept around the middle stone.
That is there are ‘n’ stones to be kept on right side and ‘n’ stones on the left side
Now it is also said that n intervals of each 10m on both sides.
If the man collects the stones from first to n stones and drops it around the middle stones on the left side = 2[10(1) + 10(2) + 10(3) + …. +10(n – 1)] + 10(n). (i.e. the man was initially at the extreme left then ${{n}^{th}}$ 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 on the left side. The difference is that here he will cover distance both the way in collecting all stones.
Hence the distance covered on the right side = 2[10(1) + 10(2) + 10(3) + ……. + 10(n – 1) + 10(n)]
So total distance $\begin{aligned} & =2[10(1)+10(2)+10(3)+.........+10(n-2)+10(n-1)]+10(n) \\\ & \text{ }+2[10(1)+10(2)+10(3)+.........+10(n-1)+10(n)] \\\ & =4[10(1)+10(2)+10(3)+.........+10(n-2)+10(n-1)]+30(n) \\\ \end{aligned}$
which can be further written as,
$\text{Total Distance = }4[10(1)+10(2)+10(3)+.......+10(n-1)+10(n)]-10(n)$
In the above expression of total distance,
10(1) + 10(2) + 10(3) + ……+10(n) form an A.P or arithmetic progression where a = 10 and d = 10, here represents the first term and d be a common difference.
So, the sum will be $=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Where n is the number of terms.
So, sum of the A.P term is $\begin{aligned} & =\dfrac{n}{2}\left( 2\left( 10+10\left( n-1 \right) \right) \right) \\\ & =\dfrac{n}{2}\left( 10n+10 \right) \\\ \end{aligned}$
So, the sum of the AP term is $5{{n}^{2}}+5n$
Hence the total sum is $4\left[ 5{{n}^{2}}+5n \right]-10n$ which can be further simplified as $20{{n}^{2}}+10n$.
Total distance is known to us which is 3Km or 3000m
So, $20{{n}^{2}}+10n=3000$
which can be further simplified and written as,
$2{{n}^{2}}+n-300=0$
Now, by using middle term factor we get,
$2{{n}^{2}}-24n+25n-300=0$
Hence, by factorization we can say that,
(2n + 25)(n – 12) = 0
So, values of n of which equation satisfies is $-\dfrac{25}{2}$ and 12
Here n represents the number of stones. So, n can’t be negative. Hence n is 12.
So, the total number of stones is 2 x 12 + 1 = 25
Hence, the total stones are 25.

Note: Students should be careful about the value of n. From the given information we should be able to analyse Arithmetic Progression is formed. Apply the formula of the sum of the nth term of A.P to get the result.