Question

Along a road lie an odd number of stones placed at intervals of 10 metres These stones have to be assembled around he 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

The correct option is C 25

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 $10$m 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 $10$m each $3$ times now from centre he will go to $2$nd stone on left hand side then come back and then go to $2$nd last on right hand side and again come back.Thus,he will travel $n-1$ intervals of $10$m each $4$ times.
Similarly,$n-2$ intervals of $10$m each $4$times for $3$rd and so on for the last.
Hence,the total distance covered as again=$3$km=$3000$m
Or,
$3\times 10n+4\left[10\right(n-1)+10(n-2)+..+10]=3000$
$=30n+40(1+2+3+...+(n-1)=3000$
$=30n+40\left(\frac{n-1}{2}\right)(1+n-1)=3000$
$=2n^2+n-300=0$
$=(n-12)(2n+25)=0$
$=>n=12$ or, $n=125$
As negative value is not valid
$\therefore n=12$
Hence the number of stones
=$2n+1$
=$2\times 12+1$
=$24+1$
=$25$