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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·
Options
35
15
31
29
Solution
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
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