Question

Along a road lie an odd number of stones placed at intervals of 10metres.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 3km.Find the number of stones?

Answer 1

first see the solution in pic

Then
$10 {n}^{2} + 10 {n}^{2} + 10n = 3000 \\ 2{n}^{2} + n - 300 = 0 \\ n = 12 \\ z \\ stones \: = 2n + 1 = 25$

Answer 2

Answer:

Answer

Step-by-step explanation:

Let the number of stones be 2n+1 so that there is one mid stone and n stones each on either side of it.If P be the mid stone and A,B be the last stones on left and right of P respectively.

There will be n+1 stones on the left and n+1 stones on te right side of the P(P being common to both sides) or n intervals on each of 10m both on the right and left side of the mid stone.

Now,he starts from one of the end stones,picks it up goes to mid stone drops it and goes to last stone on the other side,picks it and come back to midstone.In all he travels n intervals of 10m each 3 times now from centre he will go to 2nd stone on left hand side then come back and then go to 2nd last on right hand side and again come back.Thus,he will travel n−1 intervals of 10m each 4 times.

Similarly,n−2 intervals of 10m each 4times for 3rd and so on for the last.

Hence,the total distance covered as again=3km=3000m

Or,

3×10n+4[10(n−1)+10(n−2)+..+10]=3000

=30n+40(1+2+3+...+(n−1)=3000

=30n+40(

2

n−1

)(1+n−1)=3000

=2n

2

+n−300=0

=(n−12)(2n+25)=0

=>n=12 or, n=125

As negative value is not valid

∴n=12

Hence the number of stones

=2n+1

=2×12+1

=24+1

=25