Question

Along a road lies an odd number of stones placed at intervals of $10$ m. These stones have to be assembled around the middle stone.A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle ,carrying stones in succession thereby covering a distance of $4.8$ km. Then the number of stones is?

Answer 1

Since man picks up stones successively, so

Total distance covered $=10m+20m+30m+40m+....$

So, Since this is an $AP$

Sum $=\left[2\right(10)+(n-1\left)\right(10\left)\right]\frac{n}{2}=4.8km=4800m(n+1)n=960n^2+n-960=0n\approx 30$

So, $30$ stones plus one one middle. Thus, $31$ stones are there in total