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?
5 answers
I have been stuck for hrs in this question.kindly plz help me.
My answer is 25.I don't know whether it is right or not.
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.
# 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
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:)
Sorry.I wrote ,rights and side:-it's wrong:)
13 answers