hi aarav

is the ans 6

i will post the solu'n once u confirm this

Very good Go ahead and post the solution.

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

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.

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.

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

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

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

@ 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.

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

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 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 mSo 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 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 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...............

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

ur advice err wish is our command....:)
aarav bhai wht is da ans of tody'z qs?

Aap logon ka answer theek hai, this was a problem where some hit and trial is required.

amar_kashyap Says

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...:)

As you wish But make sure you discuss most of the problems out there.

Aarav Says

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. You can always look back at the solutions in the original thread :-)

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...:)

Aarav Says

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. You can always look back at the solutions in the original thread :-)

ur advice err wish is our command....:)
aarav bhai wht is da ans of tody'z qs?

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..:)

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. You can always look back at the solutions in the original thread

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 mindblowin...second season thrd????

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..:)

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 okay becoz here we are dealing with numbers of chikoos, banananas and mangoes and they always come in integer values.

This can a result of my skewed thinking...

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 mindblowin...second season thrd????

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...

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 okay becoz here we are dealing with numbers of chikoos, banananas and mangoes and they always come in integer values.

This can a result of my skewed thinking...

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 mSo the given scenario is not possible
hope i make some sense..

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

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 mSo the given scenario is not possible
hope i make some sense..

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...

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 ???

:

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 mSo the given scenario is not possible
hope i make some sense..

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....?????

I also solved it this way, nothin wrong with it

_____________________
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 of chikoos and the number of mangoes is less than three-fourths the number of bananas.

How many more/less bananas did Allwin buy than mangoes?

(a) 3 (b) 6 (c) 11 (d) none of the foregoing

Soln:

Given:
b < 1/2 c m > 1/3 c
m < 3/4 b So m < 3/8 c

We know, number of mangoes is more than one third of cheekoos and less than three eighth of of chocklates. If we take number of cheekoos as 24 both the conditions cannot be satisfied. So number of mangoes has to be 23 and number of mangoes has to be 8.
The remaining number of fruits will be bananas and it will be 11

Cheekoos Bananas Mangoes
21 11 8

So Allwin bought three more bananas than mangoes.

Hence the correct answer choice would be .

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

_____________________
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 of chikoos and the number of mangoes is less than three-fourths the number of bananas.

How many more/less bananas did Allwin buy than mangoes?

(a) 3 (b) 6 (c) 11 (d) none of the foregoing

Soln:

Given:
b < 1/2 c m > 1/3 c
m < 3/4 b So m < 3/8 c

We know, number of mangoes is more than one third of cheekoos and less than three eighth of of chocklates. If we take number of cheekoos as 24 both the conditions cannot be satisfied. So number of mangoes has to be 23 and number of mangoes has to be 8.
The remaining number of fruits will be bananas and it will be 11

Cheekoos Bananas Mangoes
21 11 8

So Allwin bought three more bananas than mangoes.

Hence the correct answer choice would be .

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
hence possible vals for m,b,c are 8,11,23 to satisfy c+b+m=42
hance option 3 as 11-3=8
Not sure of itz correctness,somebody comment on ths...

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....?????

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
hence possible vals for m,b,c are 8,11,23 to satisfy c+b+m=42
hance option 3 as 11-3=8
Not sure of itz correctness,somebody comment on ths...

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 But B+M is greater than 7C/9 hence C<23.xx

=>C=23
therefore, B=11, M=8
Hecne answer is 3.

P.S - I am not sure whether my method is correct or not. Someone pls comment on this.

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

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 less than three-fourths the number of bananas.
How many more/less bananas did Allwin buy than mangoes?
(a) 3 (b) 6 (c) 11 (d) none of the foregoing



I got (a) . Chikoo = 23 banana=11 Mangoes =8.
Cheers Apple