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 jo

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

TestHub · Accepted Answer

Let the number of stones be 2 n + 1 . The total distance is 4800 m, so each side of the middle covers 2400 m. Distances covered to bring the stones from one side to the middle (going and coming back) are 20 , 40 , 60 , … m , n terms. 2400 = 2 n ​ [ 2 × 20 + ( n − 1 ) × 20 ] = n ( 10 n + 10 ) ⇒ n 2 + n − 240 = 0 ⇒ ( n + 16 ) ( n − 15 ) = 0 ⇒ n = 15 . Number of stones = 2 n + 1 = 31 .

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