Question

Along a road lie an odd number of stones placed at intervals of 10 m/ These stones have to be assembled around the middle-stone.A person with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km.Then. the number of stones is

• A. 35
• B. 15
• C. 31
• D. 29

Answer 1

The correct option is C 31

Let the number of stones =n
The distance between every stone= 10 m
So, here an AP is formed , a=20, d=20
$=\frac{n}{2}[2a+(n-1\left)d\right]$
Distance between in both side,
$\Rightarrow 4800=2(\frac{n}{2}2a+(n-1\left)d\right)$
$\Rightarrow =n[2\times 20+(n-1\left)20\right]$
$\therefore$ n=15 (on one side)
Then, number of stones on both side
=2 X 15 = 30
$\therefore$ Total number of stones
=30+1 (middle stone)=31