Posts and Comments

@catdog
79

Posted 22 Jan '11

> anant_jain8 Says
>
> Why the hell they want to blame us for the past wat we did in 10th and 12th.... Just 40% weightage to CAT score.... !! .... They are just not interested to know how we are performing now.... :disappointed: ... :(....
IIM I during our times(2006-200 :: used to be ...

anant_jain8 SaysWhy the hell they want to blame us for the past wat we did in 10th and 12th.... Just 40% weightage to CAT score.... !! .... They are just not interested to know how we are performing now....... :(....

IIM I during our times(2006-200 used to be the only IIM which used to give lot of weightage on CAT score and less to other subjective parameters like board results, work ex etc. I feel that was right thing to do. Now these IIMs have started aping foreign universties and give more importance to parametere like board examsm, work ex, NGO work ex and other nonsense. Result of this is there is no objectivity in CAT results now...If a candidate with 99.5 +%tile is not getting a single call then there is some serious flaw with CAT methodoogy for 2 very basic reasons

1. Lot of subjectivity creeps in to the evaluation process

2. Candidates are being punished for their past performance without actually ascertaing the reason for that performance

Just Imagine brilliant guy who scores 99.8 and has scored low in board becuase of inadequate support at home... and this is very reason why IIMs can't compared to IITs

- 7 Likes

@catdog
79

Posted 21 Jul '09

> *registration procedure......????*
i'm trying to enroll for dec 2009 l1..... can anyone tell me the procedure... i mean i want to enroll before the second deadline on august 21st... my doubt is can i make an online transfer or do i have to use a credit card only and after payment do i have...

registration procedure......????

i'm trying to enroll for dec 2009 l1..... can anyone tell me the procedure... i mean i want to enroll before the second deadline on august 21st... my doubt is can i make an online transfer or do i have to use a credit card only and after payment do i have to send any paperwork to cfa institute or the process is over and i am done with my responsibility...????

now will they send me the study material...???

just visit CFA institute website click on enroll rest is self explanatory

You can make payment using credit card (VISA or MASTER) thats most convenient (you will have to pay 3.5% for internation payments)

No paper work required ..entire process is paperless ..just make sure you notedown your CFA account user is and passowrd . In November you will get your exam tickets online. Just carry its print out and along with government issued I Card ..

Yes you will get set of CFA books at your postal address ( so make sure you enter the right postal address). Normally people get books within a week.

- 2 Likes

@catdog
79

Posted 14 Jul '09

> Hello Sir,
Can you plzzz elaborate it further..?
x1 + x2 + x3 =12
where x1= correct answers (x1>=1)
x2 = incorrect answers
x3 is not attempted
now x2 is always of form 4n(where n=0 1 2 etc)
this can be modelled as co. of polynomial model we can state same prob...

Hello Sir,

Can you plzzz elaborate it further..?

x1 + x2 + x3 =12

where x1= correct answers (x1>=1)

x2 = incorrect answers

x3 is not attempted

now x2 is always of form 4n(where n=0 1 2 etc)

this can be modelled as co. of polynomial model we can state same problem as

Find co. of x^12 in following polynomial

------A------- --------B-----|------C---------

(x+x^2+x^3...)(1+x^4+x^8....)(1+x+x^2+x^3...)

where as you must have noticed

A corresponds to x1 , B to x2 and C to x3

as you notice x is common in A and can be taken out

hence you need to find out Co. of x^11 in

------A------- |--------B-----------C---------

(1+x^1+x^2...)(1+x^4+x^8....)(1+x+x^2+x^3...)

A is expansion of (1-x)^-1

B is expansion of (1-x)^-1

C is expansion of (1-x^4)-1

So you need to find Co. of x^11 in

=(1-x)^-2 (1-x^4)-1

Which means finding out Co of x^11 X^7 x^3 in (1-x)^-2

which is ( note Co of x^k in series (1-x)^-n is (n+k-1)C k)

=12*1+8*1+4*1

=24

@catdog
79

Posted 14 Jul '09

> Try this
In a test of 12 questions, a candidate gets 1 mark for every right answer, 0 for unanswered questions, and a negative 0.25 for every wrong answer. What is the total number of ways in
which a particular candidate may score a positive integral value of net score ?
Source:ww(dot)qua...

Try this

In a test of 12 questions, a candidate gets 1 mark for every right answer, 0 for unanswered questions, and a negative 0.25 for every wrong answer. What is the total number of ways in

which a particular candidate may score a positive integral value of net score ?

Source:ww(dot)quantexpert(dot)in

x1 +x2+x3=12

x2=0,4,8,12

(x+x^2+x^3...)(1+x^4+x^8+x^12....)(1+x+....

find coeffcient of 0 4 8

11 7 3

=(1-x)^-2 *(1-x^4)-1

=11 + 7 + 3

=21

@catdog
79

Posted 13 Jul '09

> solution:
suppose 'A' goes to another person. he takes one pebble and takes 11 of his pebbles. again A goes to another person and gives him one of both the types of pebbles, he has and takes 11 pebbles from him.
same way he meets 11 persons and does the same. at last A and eleventh pers...

solution:

suppose 'A' goes to another person. he takes one pebble and takes 11 of his pebbles. again A goes to another person and gives him one of both the types of pebbles, he has and takes 11 pebbles from him.

same way he meets 11 persons and does the same. at last A and eleventh person have all types of pebbles and now other 10 need another pebble for which A need to meet them once apart from initial 11 meetings.

so total 11+10=21

why not divide and conquer method

say

1-2 3-4 5-6 7-8 9-10 11-12 ->6 meetings

then

1-3 2-4 5-7 6-8 9-11 10-12 -> 6 meetins

then

1-5 1-9 3-7 3-11 2-6 2-10 4-8 4-12 -> 8 meetings

total 20 meetings

@catdog
79

Posted 13 Jul '09

> question: in a group of 12 people, each one has 12 pebbles of a single type. no two people have a single pebble of the same type. in a meeting 2 people can meet and exchange any no. of hands pebbles. what is the minimum no. of meetings required so that every person will have pebbles of all type...

question: in a group of 12 people, each one has 12 pebbles of a single type. no two people have a single pebble of the same type. in a meeting 2 people can meet and exchange any no. of hands pebbles. what is the minimum no. of meetings required so that every person will have pebbles of all types?

22

11

21

33

66

First

1-2 3-4 5-6 7-8 9-10 11-12 meet each other total 6 meetings

now

1-3 2-4 5-7 6-8 9-11 10-12 total 6 meetings

now we have groups

1234 5678 9101112

1-5 1-9 2-6 2-10 3-7 3-11 4-8 4-12 8 meetings

total 20 meetings

- 1 Like

@catdog
79

Posted 13 Jul '09

> The method is bit lengthy but hope u get it:
1\. First 1003 WHOLE nos. are 0 to 1002 (not till 1003)
2\. The natural no. thus formed will be 012345...10011002
3.Lets consider the the above no. till 999 i.e. 0123...998999 for this arrangement the addition of the nos. will be given by
*...

The method is bit lengthy but hope u get it:

1. First 1003 WHOLE nos. are 0 to 1002 (not till 1003)

2. The natural no. thus formed will be 012345...10011002

3.Lets consider the the above no. till 999 i.e. 0123...998999 for this arrangement the addition of the nos. will be given bySum(0-999)= 1000/2 {0+(1000-1)1)= 500*999=499500

This no. is divisible by 9 (4+9+9+5+0+0=27)

so now we have to consider the remaining part i.e 100010011002

This no. is not divisible by 9 (1+1+1+1+2=6) and gives a remainder 6.

So we can conclude that for the natural no. 012345...100010011002 the remainder is 6.

Hope this has solved ur doubt.. i am very bad at explaining but even then i hope u have understood...

I not convinced with bold part..I dont think its sum of digits

For any number of form X X X

Number of times 1 occurs = 100 + 100 + 100 =300

Same for 0 2 3 4 ...9

so sum of digits for number of form 123456891011...999 is

=300*9*5 which is divisible by 9

approach after this remains same

@catdog
79

Posted 12 Jul '09

> hi guys,
m currently working in wealth management profile. planning to do cfa as i have heard and seen in this thread thats its an good option. do u think my work exp in the same profile will help me for the work exp required for getting cfa charter. and do u think cfa can be beneficial fo...

hi guys,

m currently working in wealth management profile. planning to do cfa as i have heard and seen in this thread thats its an good option. do u think my work exp in the same profile will help me for the work exp required for getting cfa charter. and do u think cfa can be beneficial for my career.

that will be so great if anyone can suggest me on this. m following this thread from so long...

thanks in advance

Hi pawan,

It depends on your desgination and JD as wel,l but since wealth planning is very much related to CFA course curriculum i think your wrk expereince should be counted. You can also mail CFA your detailed JD designation company etc and they wud respond on that.

- 1 Like

@catdog
79

Posted 11 Jul '09

> g(n) = GCD
=GCD
=GCD
=GCD
For max(g(n)) , GCD should be 2n+1 , so 22-n = 0--> n = 22
So *max(g(n) = 2*22+1 = 45* ::
Cool thats awsome approach..

g(n) = GCD

=GCD

=GCD

=GCD

For max(g(n)) , GCD should be 2n+1 , so 22-n = 0--> n = 22

Somax(g(n) = 2*22+1 = 45

Cool thats awsome approach..

- 1 Like

@catdog
79

Posted 10 Jul '09

> prateek7563 Says
>
> i didnt understand the bold part. y divisible by 8 and not 24.
m*(m+1)*(2m+1)/6 should be divisible by 4
m*(m+1)*(2m+1)=24k
now out of m m+1 and m+(m+1) at least one will be divisible by 3
(if m=3n+1 then 2m+1=6n+3 is divisible by 3
m=3n+2 then m+1 is di...

prateek7563 Saysi didnt understand the bold part. y divisible by 8 and not 24.

m*(m+1)*(2m+1)/6 should be divisible by 4

m*(m+1)*(2m+1)=24k

now out of m m+1 and m+(m+1) at least one will be divisible by 3

(if m=3n+1 then 2m+1=6n+3 is divisible by 3

m=3n+2 then m+1 is divisible by 3)

so now what remains is to prove that m m+1 and 2m+1 is also divisible by 8

0Commentsnew comments