Happy solving :-)
I'm starting a new thread as the discussions will happen on this new thread for next 50 questions and the sectional test :-)
Happy solving :-)
Happy solving :-)
Very good Go ahead and post the solution.
hi aarav
is the ans 6
i will post the solu'n once u confirm this
is the ans 6
i will post the solu'n once u confirm this
Really? Wake up. it's almost noon today.
Well, check again.
Well, check again.
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
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
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
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
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
Don't post in bold and big fonts. It looks bad.
Folks, todays NL has been delivered. Check your inboxes in 5 minutes.
Will happen from tomorrow
Folks, todays NL has been delivered. Check your inboxes in 5 minutes.
Will happen from tomorrow
where is today's question i.e. question no 52
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
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
Bandhu, am not able to understand that how u concluded that c/2 and c/3 must be integers. According to question b2 and m3. 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 ...
A much ...
@ Aarav.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.
Bhaiya i dont think we need to do any sort of hit and trial for todays question.
Aap logon ka answer theek hai, this was a problem where some hit and trial is required.
As you wish But make sure you discuss most of the problems out there.
As you wish But make sure you discuss most of the problems out there.
Thik hai fir..ulti ginti se suru karte hai. I think it will be okay if we move backwards from the 5oth page to first one of previous MB thread. ..Hope tht will be okay...:)
ur advice err wish is our command....:)
aarav bhai wht is da ans of tody'z qs?
aarav bhai wht is da ans of tody'z qs?
My advice here would be that you guys go thru each and every problem discussed in the mind-blowing section. It helped us a lot in CAT 2005. Almost 10 mark questions were from the very questions we discussed there. There are I guess 200 problems there, pick them up in order and start discussing. Y...
Not necessarily. But i am trying to pick up those questions of previous MB thread which were not solved, and some new problems from other sources too...:)
Hoping for your active participation out there..:)
Hoping for your active participation out there..:)
haan ,i think practically it has 2 b int ,actually first i had strtd workin the way u hav done,then i thaught if we adhere 2 pure mathmtcl way thy can b both frctn n int...:)
Letz c wht aarav n othr junta say...
BTW i hav ths doc of urz havin 100 odd qs,r u discussin the same in da mindbl...
Letz c wht aarav n othr junta say...
BTW i hav ths doc of urz havin 100 odd qs,r u discussin the same in da mindbl...
Tht's what i am doubting. See when we say bananas are less than half of chikoos, it has to be less by integer no only naa.It can be 1or 2 or3 or4 ...less the half of chikoos. Being less than half of chikoos by a fraction quantity doesn't make sense practically, though in equation form it looks ok...
All is well till the assumption in RED.. remember x and kneednt have to be integers
My soln
take chikkoo number as K
K + ( 1/3K + 1) + = 42
16/9 k + 10/3 ~ 42
k~ 38 *9 /16
k ~ 2.4 *9
k~22.mm
trial and error around 22
chik ban man
23 11 8
Ans a)3
My soln
take chikkoo number as K
K + ( 1/3K + 1) + = 42
16/9 k + 10/3 ~ 42
k~ 38 *9 /16
k ~ 2.4 *9
k~22.mm
trial and error around 22
chik ban man
23 11 8
Ans a)3
amar bhai i couldn't understand y u r takin c to be of the form 24 n
put c=23,b=11,m=8 thn also ur eqns are satisfied only i think the k,p,x vals will be decimals,not pure integers.....
if we take 2b=c-k then k has to be an int,
thatz what i think,may b i m incorrect...
put c=23,b=11,m=8 thn also ur eqns are satisfied only i think the k,p,x vals will be decimals,not pure integers.....
if we take 2b=c-k then k has to be an int,
thatz what i think,may b i m incorrect...
c + m + b = 42
c + 3/4 * 1/2 * c + 1/2 * c > 42
c > 22.4 => c >=23
m>=8
b >=5
c + m + b = 42
2b + 2/3 * b + b < 42
b < 11.4 => b <= 11
m<=8
c<=23
m=8, c= 23, b=11
Ans = (a) - 3 more banana's.
i need to do this faster how ???
:
c + 3/4 * 1/2 * c + 1/2 * c > 42
c > 22.4 => c >=23
m>=8
b >=5
c + m + b = 42
2b + 2/3 * b + b < 42
b < 11.4 => b <= 11
m<=8
c<=23
m=8, c= 23, b=11
Ans = (a) - 3 more banana's.
i need to do this faster how ???
:
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 i...
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 i...
I also solved it this way, nothin wrong with it
I am also getting 23 11 and 8 as solution so ans is A -3
Hey i dont receive the questions.
Please tell me how to subscribe to question list so that i also get questions along with u guys.
Rani
Hey i dont receive the questions.
Please tell me how to subscribe to question list so that i also get questions along with u guys.
Rani
_____________________
Quant Question # 51
_____________________
Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number ...
Quant Question # 51
_____________________
Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number ...
i also solved it this way...
when i got c>22.2...i assumed it to be 23...and it worked fine for the other case
is there any better method....?????
when i got c>22.2...i assumed it to be 23...and it worked fine for the other case
is there any better method....?????
given c+b+m=42--------1)
and b<1/3*c--------------2)
m>1/3*c-----------3)
m<3/4*b-----------4)
from 1)
m=42-b-c>42-c/2-c (from eqn 2)i.e 42-3c/2
i.e 2m>84-3c
i.e 2m>84-9m(from eqn 3) i.e m>7.6
similarly
c=42-b-m
>42-5c/6
i.e c>22.2
similarly
b>10.5
hen...
and b<1/3*c--------------2)
m>1/3*c-----------3)
m<3/4*b-----------4)
from 1)
m=42-b-c>42-c/2-c (from eqn 2)i.e 42-3c/2
i.e 2m>84-3c
i.e 2m>84-9m(from eqn 3) i.e m>7.6
similarly
c=42-b-m
>42-5c/6
i.e c>22.2
similarly
b>10.5
hen...
C+M+B=42
B2 M<34B M>C/3
C/3 < M < 3C/8 1)
4C/9 < B < C/2 2)
Adding 1) and 2)
7C//9 < B+M < 7C/8
C + 7C/8 = 42 (taking extreme)
=> C = 22.xx But B+M is less than 7C/8 hence C>22.xx
C + 7C/9 = 42 (taking other extreme)
=> C = 23.xx ...
B2 M<34B M>C/3
C/3 < M < 3C/8 1)
4C/9 < B < C/2 2)
Adding 1) and 2)
7C//9 < B+M < 7C/8
C + 7C/8 = 42 (taking extreme)
=> C = 22.xx But B+M is less than 7C/8 hence C>22.xx
C + 7C/9 = 42 (taking other extreme)
=> C = 23.xx ...
my answer to the qn # 51 :
c=23 , b=11 , m =8
the eqn comes to c/2>3/4b>m>c/3
only soln is above one ; so option (a) 3 is the correct option
c=23 , b=11 , m =8
the eqn comes to c/2>3/4b>m>c/3
only soln is above one ; so option (a) 3 is the correct option
Question:51
Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number
of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of
chikoos and the number of mangoes is le...
Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number
of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of
chikoos and the number of mangoes is le...
