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 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·

पर्याय

• 35

• 15

• 31

• 29

Answer 1

31

Explanation:

Let the number of stones = n

The distance between every stone = 10 m

So, here an AP is formed, a = 20, d = 20

`="n"/2[2"a"+("n"-1)"d"]`

Diatance between in both side,

`⇒4800=2("n"/2 2"a"+("n"-1)"d")`

⇒ 4800 = n[2 × 20 + (n − 1)20]

∴ n = 15 (on one side)

Then, the number of stones on both side

= 2 × 15 = 30

∴ Total number of stones

= 30 + 1 (middle stone) = 31