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 a distance of 3 km. Find the number of stones?

My answer is 25.I don't know whether it is right or not.

I have been stuck for hrs in this question.kindly plz help me.

Take a look at the distances for each stone, numbering them outward. I'm assuming that we only include the distances actually carrying stones. The actual distance walked would be twice that.

# d
1 10
2 20
3 30
...
n 10n

So, after collecting n stones on each side, the distance walked with stone in hand would be

2*Sn = 2 * n/2 (20+(n-1)*10) = 3000
n^2+n=300

Hmm. Not an integer solution. But it is close to 17, not 25.

Maybe I have misinterpreted the problem. How did you arrive at 25? It seems like a reasonable number.

Let there be n stones to the right of P and n stones to the left of P.'P' means person. Therefore total no.of stones=n+n+1=2n+1.

A man starts from A goes to P drop the stones and come back.Again,he goes upto S(n-1) and picks it up and comes back to P.This process is repeated till the person collects all the stones on the left.The same procedure is repeated for stones on the right side also.

Distance covered by the person to the left hand side=10*n+2*10(n-1)+2*10(n-2)+...+2*10*2+2*10*1]
=10n+20[(n-1)+(n-2)+...+2+1]
Similarly,Distance covered by the person to the rights and side:
So,we get 20[n+(n-1)+(n+2)+....+2+1]
Total distance=20[n+(n-1)+(n-2)+....+2+1]+10n+20[(n-1)+(n-2)+...+2+1]-10n

=20[n+(n-1)+...+2+1)+20n+20[(n-1)+(n-2)+....+2+1]-10n
=20[n+(n-1)+...+2+1)+20[n+(n-1)+....2+1]-10n
=40[n+(n-1)+....+2+1]-10n
=40[1+2+....+(n-1)+n]-10n
=40[n/2(1+n)]-10n
=20n(1+n)-10n=20n+20n^2-10n
=20n^2+10n

Given that the total distance covered is 3 km,i.e;3000m.
Therefore,20n^2+10n=3000
20n^2+10n-3000=0
2n^2+n-300=0
(n-12)(2n+25)=0
Therefore,n=12 or n=-25/2
n is a natural number.
Therefore,n=12
Hence,number of stones =2n+1
=2*12+1=25

Similarly,distance covered by the person to the right hand side

Sorry.I wrote ,rights and side:-it's wrong:)

To solve this problem, we need to calculate the number of stones by considering the distance covered.

Given:
- The person carried the stones from one end to the other, covering a distance of 3 km.
- The stones are placed at intervals of 10 meters.
- The number of stones is odd.

First, let's convert the distance covered to meters:
1 kilometer = 1000 meters
So, the person covered 3 km = 3 * 1000 = 3000 meters.

Next, let's calculate the number of intervals between the stones:
Since the stones are placed at intervals of 10 meters, we can find the number of intervals by dividing the total distance covered by 10:
Number of intervals = distance covered / interval length = 3000 / 10 = 300.

However, we need to keep in mind that the number of stones is odd and we are considering carrying from one end to the other.

Since the person starts carrying the stones from one end, they will pick up the end stone and walk only halfway to the middle stone. At this point, the person has carried (300 / 2) = 150 intervals.

Therefore, the number of stones assembled around the middle stone will be:
Number of stones = 150 intervals + 1 middle stone = 151 stones.

So, there are 151 stones along the road.