Question

Along a road lies an odd number of stones of weight 10 kg each, placed at intervals of 10 metres. These stoneshave to be assembled around the middle stone. Nirvah, a stone loader can carry only one stone of 10 kg at atime. He started the job with one of the end stones by carrying them in succession. In carrying all the stones,he covered a distance of 3 km. Find the number of stones,​

Answer 1

The total number stone is 25.

Given

To find the number of stones.

Let "( 2n + 1 )" be the number of stones.

From the figure,

Consider "A" and "B" be the end stones on left-end and right-end.

Here, "P" is the middle stone.

         "n" is the intervals where each stones placed at intervals of length 10 m  

          on both the sides of "P".

Weight of each stone is 10 Kg.

Nirvah, carries the stone from "A" and moves to mid stone "P" and Nirvah drops the stone at "P" -----> ( n - 1 )    

Distance covered from A ⇄ P is,

              = (10×n) + 2 [ 10×(n-1) + 10×(n-2) + ....... + 10×2 +10×1 ]

Now, Nirvah carries the stone from "B" and moves to mid stone "P" and Nirvah drops the stone at "P"

Distance covered from B ⇄ P is,

              = 2 × (10×n) + 2 [ 10×(n-1) + 10×(n-2) + ....... + 10×2 +10×1 ]

Total distance covered is,

              = (10×n) + 2 [ 10×(n-1) + 10×(n-2) + ....... + 10×2 +10×1 ] + 2 × (10×n) + 2 [  10×(n-1) + 10×(n-2) + ....... + 10×2 +10×1 ]

              = 4 [ (10×n) + 10×(n-1) + 10×(n-2) + ........ + (10×2) + (10×1) ] - (10×n)

              = 40 [ 1 + 2 + 3 + ....... + n ] - 10n

             [ 1 + 2 + 3 + ....... + n ] = $\frac{n}{2}$ × (1+n)

              = 40 [ $\frac{n}{2}$ × (1+n) ] - 10n

              = $\frac{40}{2}$ × [ n×(1+n) ] - 10n

              = 20 × [ n×(1+n) ] - 10n

              = 20n (n+1) - 10n

              = 20$n^{2}$ + 20n - 10n

              =  20$n^{2}$ + 10n

Nirvah covered a distance of 3 km.

Converting Km to m, multiply by 1000

             20$n^{2}$ + 10n = 3 × 1000

             20$n^{2}$ + 10n = 3000

             20$n^{2}$ + 10n - 3000 = 0

Divide the above equation by 10,

              2$n^{2}$ + n - 30 = 0

              ( n-12 ) ( 2n+25 ) = 0    ( n = negative values are omitted )

Where, n = 12

Therefore, total number of stones = ( 2n+1 ) = ( (2×12) + 1 ) = ( 24+1 ) = 25.

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