Along a road lie an odd number of stones placed at interval of 10 meters . These stones have to be assembled on the middle stone . A person can carry only one stone at a time. A man carried the job with one of the end stone by Carring them in succession. In carrying all the stone he covered a distance of 3 km . Find the numbers of stone.
Answers
Answer:
Let the number of stones be 2n+1 so that there is one mid stone and n stones each on either side of it.If P be the mid stone and A,B be the last stones on left and right of P respectively.
There will be n+1 stones on the left and n+1 stones on te right side of the P(P being common to both sides) or n intervals on each of 10m both on the right and left side of the mid stone.
Now,he starts from one of the end stones,picks it up goes to mid stone drops it and goes to last stone on the other side,picks it and come back to midstone.In all he travels n intervals of 10m each 3 times now from centre he will go to 2nd stone on left hand side then come back and then go to 2nd last on right hand side and again come back.Thus,he will travel n−1 intervals of 10m each 4 times.
Similarly,n−2 intervals of 10m each 4times for 3rd and so on for the last.
Hence,the total distance covered as again=3km=3000m
Or,
3×10n+4[10(n−1)+10(n−2)+..+10]=3000
=30n+40(1+2+3+...+(n−1)=3000
=30n+40( 2n - 1)
)(1+n−1)=3000
=2n
2+n−300=0
=(n−12)(2n+25)=0
=>n=12 or, n=125
As negative value is not valid
∴n=12
Hence the number of stones
=2n+1
=2×12+1
=24+1
=25
Step-by-step explanation:
Hope it helps you
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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