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@rengaraj • 25 Nov, 2009
My question is

A 1 KM long wire is held by 'n' poles. If one pole is removed the length of the gap increases by 5/3 meters. What is the number of poles initially ?

R.Rengaraj
{NIT 23/2007/Pg 3 (Infosys Part-1)}
{On this forum my 12th Question.
@yadavundertaker mohit • 25 Nov, 2009 length of gap with n poles = 1000/(n-1)

new length of gap = 1000/(n-2)

according to question

1000/(n-2)= 5/3 + 1000/(n-1)

1000(n-1-n+2)/(n-1)(n-2)=5/3

on solving n = 26

number of poles initially is 26.....
1000/(n-1)(n-2)= 5/3
@rengaraj • 26 Nov, 2009 Sir,
I got the question from an book but, the solution was 25. But i can't get the steps worked.
But i will try once more.
R.Rengaraj
@yadavundertaker mohit • 26 Nov, 2009 the no. of poles will be 25 after one pole is removed..

steps

length of gaps = total length in meter /(no. of ploes -1) [we minus 1 because a gap will be exist b/w two poles]
@CIVILPRINCESS • 27 Nov, 2009 yes the number of poles initially is 26.by simplifying the equation yadav got we get
n^2-3n-598=0
26 solves it.
@yadavundertaker mohit • 29 Nov, 2009 hi rengraj

did you check the answer again?....
@rengaraj • 29 Nov, 2009 Sir,
The solution is 25.
Thanks,
R.Rengaraj
@Anil Jain • 30 Nov, 2009 if n poles (n-1) gaps of each 1000/(n-1) width
if n-1 poles (n-2) gaps of each 1000/(n-2) width
so,
1000/(n-2) -1000/(n-1)=5/3
1000(1/(n-2) -1/(n-1))=5/3

so number of poles initially would be 26 only.
@rengaraj • 30 Nov, 2009 Sir,
This is an Infosys question. The given answer is 25.
R.Rengaraj
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