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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:

1. 35
2. 15
3. 29
4. 31

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It can be shown that there is only one strategy to handle the given problem hence no question of optimal strategy . Thats why they have asked for "number of stones" and not "minimum or maximum number of stones"..

Let we have (2n+1) number of stones so we have 'n' stones on one side of the middle stone and 'n' on other side

Distance between middle stone and right to it is 10 m..So to bring it to middle 20 m is covered..

Similarly for 2nd next one , we need distance  =  40 m

and so on till nth next which is last from right , we need distance = 20n

Hence total distance     =     20(1+2+3...+n)  =   20n(n+1)/2  =  10n(n+1)

Likewise to bring the left stones we need total distance  =  10n(n+1) m

Thus total distance  =  20n(n+1) m which is equal to 4800 m as mentioned in the question.

Hence ,

20n(n+1)   =   4800

==>  n(n+1)      =   240

==>  n             =   15

Hence number of stones  =  2n + 1  =  2(15) + 1  =  31

Hence D) should be the correct option.

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