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 Along a road lie an odd number of sto...
 Entrance Exam
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Along a road lie an odd number of stones placed at intervals of 10metres.These stones have to be assembled around the middle stone.A person can carry only one stone at a time.A man carried the job with one of the end stones by carrying them in succession.In carrying all the stones he covered a distance of 3km.Find the number of stones?
Let there be (2n + 1) stones.
let P be the middle stone, A the extreme most stone on the left of P, and B the extreme most stone on the right of P.
There are n intervals each of 10m on both the sides.
Suppose man starts from point A, picks up stone A, goes to P, drops it, then goes to (n1)th stone on left, picks it, goes to P, drops it. This process is repeated, till he collects all the stones on the left hand side.
Therefore, the distance coverd in collecting all the stones on the left side, will be
10 X (n) + 2[ 10(n1) + 10(n2) + ................ + 10 (2) + 10 (1) ]
The part after the first positive sign will be multiplied by 2 (i 've already done it) as that distance is being covered twice, whereas the distance AP will be covered only nce.
After putting the closest stone on the left hand side to the centre, his position will be at the centre. So, now he will go to point B, pick up the stone, drop it at P, and the same process is repeated all over again. The only difference will be, that while the distance AP was being covered nly once, the distance BP will also be covered twice.
therefore, the distance covered on right side, will be,
2 { 10 X (n) + 2[ 10(n1) + 10(n2) + ................ + 10 (2) + 10 (1) ] }
therefore total distance covered will be
= 10 X (n) + 2[ 10(n1) + 10(n2) + ................ + 10 (2) + 10 (1) ] + 2 { 10 X (n) + 2[ 10(n1) + 10(n2) + ................ + 10 (2) + 10 (1) ] } = 4 { 10n + 2 [ 10(n1) + 10(n2) + ............ + 10(2) + 10(1) ] }  10n >> Because 10n is being taken within the bracket to be multipied by 2
= 4 X 10 {1+2+3+............+n}  10n
= 40 {1+2+3+............+n}  10n
= 40 [ n/2 X (1+n) ] >> Applying the formula S_{n} = n/2 (a + a_{n})
= 20n(1+n)  10n
= 20n + 20n^{2 }  10n
= 20n^{2} + 10n
But that is equal to 3 km, i.e. 3000m
Therefore,
3000 = 20n^{2} + 10n
2n^{2} + n  300 = 0
n(2n + 25) 12(2n + 25) = 0
n = 12, or, n = 25/2, which cannot be.
Therefore, n = 12
Therefore, number of stones, will be, 2n+1 = 2n+1, as assumed.
therefore, number of stones = 25