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Re: Quantitative questions a day 51 to 100 - The discussions -
18-04-2006, 05:24 PM
@ Aarav.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.
Last edited by Shashank Shekha; 18-04-2006 at 05:47 PM.
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Re: Quantitative questions a day 51 to 100 - The discussions -
18-04-2006, 05:24 PM
Quote:
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Originally Posted by amar_kashyap
I am getting (d)..ie none of the foregoing..
b=c/2-k becoz bananas are less than half the chikoos, similarly
m=c/3+p
m=3b/4-x=3c/8-3k/4-x
Now c , m and b have to be integer. c must be of the form 24n.So c must be=24
Now m+b=18 becoz c+m+b=42
Now b must be 4,8,12,or 16.But as b is less than half of c it has to be less than 12 ie b=4 resulting into m=14 which is not possible becoz as indicated from third eqauation m<b
So the given scenario is not possible
hope i make some sense.. 
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Bandhu, am not able to understand that how u concluded that c/2 and c/3 must be integers. According to question b<c/2 and m<c/3. This implies that c/2 and c/3 need not be integers. I do approve the point that c,m and b must be integral values but that doesn't mean that c must be of form 24n.
A much easier solution to the prashna may be this one and this is not a hit and trial.
ATQ:
b= GIF{1/2c} ; GIF:- Greatest Integer Factor
m= LIF{1/3c} ; LIF:- Least Integer Factor
Also, m= GIF {3/4b}...........................(a)
Since, c+b+m= 42
Hence:
c+GIF{1/2c}+LIF{1/3c}= 42...............(b)
=> c= LIF{42*6/11}
Hence, c= 23.
The value of m and b is found to be 8 and 11 respectively by using above formed equations. The values satisfy equation (a).
The method is not very conventional and may be wrong but itz not kinda hit and trial.
Chiao.............
Shashank...............
Last edited by Shashank Shekha; 18-04-2006 at 05:45 PM.
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 09:02 AM
Hey Aarav Bhai,
I would request u to post daily question at 12:00 midnight or 6 a.m.Bcoz most of the junta here is in college and we have to attend classes.
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:18 AM
where is today's question i.e. question no 52
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:27 AM
Quote:
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Originally Posted by Shashank Shekha
@ Aarav.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.
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Don't post in bold and big fonts. It looks bad.
Folks, todays NL has been delivered. Check your inboxes in 5 minutes.
Quote:
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Originally Posted by Apple
Hey Aarav Bhai,
I would request u to post daily question at 12:00 midnight or 6 a.m.Bcoz most of the junta here is in college and we have to attend classes.
Apple 
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Will happen from tomorrow :-)
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:30 AM
Is the answer (d) ? f(a) + f(b) + f(c) + f(d) = 0. will post the solution in afternoon once I get todayz work done
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:40 AM
answer is zero
let g(x)= x^4-x^3-x^2-1
and f(x) = x.g(x)
now g(x) is zero at a b c d
pretty straight forward
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:45 AM
f(x) is not x.g(x)
Actually f(x) = x^2 * g(x) + 1/x
Thus, f(a) + f(b) + f(c) + f(d) = 1/a + 1/b + 1/c + 1/d = 0
as 1/a, 1/b, 1/c, 1/d are roots of the eqn. g(1/x) = 0 i.e. x^4 + x^2 + x - 1 = 0
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 11:47 AM
Quote:
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Originally Posted by warrior
answer is zero
let g(x)= x^4-x^3-x^2-1
and f(x) = x.g(x)
now g(x) is zero at a b c d
pretty straight forward
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Really? Wake up. it's almost noon today.
Quote:
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Originally Posted by Derrida
f(x) is not x.g(x)
Actually f(x) = x^2 * g(x) + 1/x
Thus, f(a) + f(b) + f(c) + f(d) = 1/a + 1/b + 1/c + 1/d = 0
as 1/a, 1/b, 1/c, 1/d are roots of the eqn. g(1/x) = 0 i.e. x^4 + x^2 + x - 1 = 0
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Well, check again.
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Re: Quantitative questions a day 51 to 100 - The discussions -
19-04-2006, 12:11 PM
Quote:
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Originally Posted by Aarav
Really? Wake up. it's almost noon today.
Well, check again.
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hi aarav
is the ans 6
i will post the solu'n once u confirm this
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