Along a road lies an odd number of stones

Asked by admin @ 13/08/2021 in Math viewed by 260 People

Along a road lies in odd number of stones placed at intervals of 10 metres. These stones have to be assembled around the middle Stone. A person can carry one stone at a time , A man started the job with one of the end of the stones by carrying in them succession. In carrying all the stones, he covered a distance of 3 Km. Find the number of stones.

Answered by admin @ 13/08/2021

Answer:

25 Stones

Step-by-step explanation:

Let say There are 2n + 1 stones in total

so there will be n stones each side of middle stone

Man started job from one end

so distance covered for 1st stone = n * 10 m

Distance covered for 2nd Stone = (n-1)*10 + (n-1)* 10 (as he has to go back to pick the stone And come back with stone

So covering all the stones of one side

= n* 10 + 2(n-1)*10 + 2(n-2)*10 +.................. + 2*3*10 + 2*2*10 + 2*1*10

Middle stone is already there

so for remaining n stones on other he has to do same thing but in addition he has to go another end of stone

so distance covered for other sides

= 2n*10 + 2(n-1)*10 + 2(n-2)*10 +.................. + 2*3*10 + 2*2*10 + 2*1*10

Hence total Distance covered =

3n*10 + 4(n-1)*10 + 4(n-2)*10 +.................. + 4*3*10 + 4*2*10 + 4*1*10

= 4n*10 - n*10 + 4(n-1)*10 + 4(n-2)*10 +.................. + 4*3*10 + 4*2*10 + 4*1*10

= 40 * (n + n-1 + n-2 + ......+3 +2 + 1) - 10n

= 40 n (n+1)/2 - 10n

= 20n (n+1) - 10n

= 20n² + 20n - 10n

= 20n² + 10n

= 10(2n² + n)

Distance covered = 3km = 3000m

10(2n² + n) = 3000

=> 2n² + n = 300

=> 2n² + n - 300 = 0

=> 2n² - 24n + 25n - 300 = 0

=> 2n(n-12) + 25(n-12) = 0

=> (2n+25)(n-12) = 0

n = 12 ( n = -25/2 not possible as number of stones can not be negative)

Number fo stones = 2n + 1

2 * 12 + 1 = 25

25 Stones

Along a road lies an odd number of stones

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