@rengaraj • 25 Nov, 2009
Sir / madam,
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 ?
Advance Thanks,
R.Rengaraj
{NIT 23/2007/Pg 3 (Infosys Part-1)}
{On this forum my 12th Question.
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 ?
Advance Thanks,
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
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
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]
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.
n^2-3n-598=0
26 solves it.
@yadavundertaker mohit • 29 Nov, 2009
hi rengraj
did you check the answer again?....
did you check the answer again?....
@rengaraj • 29 Nov, 2009
Sir,
The solution is 25.
Thanks,
R.Rengaraj
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.
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
This is an Infosys question. The given answer is 25.
R.Rengaraj
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