Quantitative questions a day 51 to 100 - The discussions

[Shashank Shekha]

@ Aarav.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.

[Shashank Shekha]

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..

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...............

[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

[warrior]

where is today's question i.e. question no 52

[Aarav]

Originally Posted by Shashank Shekha

@ Aarav.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.

Don't post in bold and big fonts. It looks bad.

Folks, todays NL has been delivered. Check your inboxes in 5 minutes.

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

Will happen from tomorrow

[Derrida]

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

[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

[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

[Aarav]

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

Really? Wake up. it's almost noon today.

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

Well, check again.

[warrior]

Originally Posted by Aarav

Really? Wake up. it's almost noon today.



Well, check again.

hi aarav

is the ans 6

i will post the solu'n once u confirm this