CAT 2012 Kerala study group

@thevicky I have attached the fig only for the case you didn't get...

Team, its finally here, the last month before we ace the CAT !!

Its that time, of the year, when the consolidated efforts of this group will bear the max effort !!

Hence , Guys... a bit more life here will help us all in the long run !!!

IF NOT..

Initiative 1 : Post Q's from mocks which you have got wrong.
An alternative approach is just a puy away !!!

Initiative 2 : Most of us would be keeping a log of mistakes/errors committed /concepts which you have collected during your prep.


I request the team(especially the lurkers -- you know who you are) to spend 5 min in this thread daily & share a concept or a problem set a day.

Initiative 3: Team, post your doubts... you know you have them. its better to clear them here, rather than staring at the screen empty minded ...

I on my part, will try an post a concept a day!!
So if you don't see my post, you have full permission to inundate my inbox with expletives !!!


Rohan, its time to put our plan to action!!!
Rakesh, you don't have permission to inundate my box with expletives
Pradyoth....S2 is your baby !!!
Pugna.... S1 is your baby !!!!
Joe, dead_alive, seetharam7,spdsd,thevicky... Time to revitalize & rejenuvate this thread !!!

Forget 1000 posts, we're aiming for more !!!


Happy 'CAT'ting!!!









@Kevin88 Nice. Also, i suggest everyone to post every night what they studied that day. I guess if each of us try to put in more effort than the next person, we would be awesome by the time of the actual CAT.

A brilliant post about the need for a study group by @dreamer87

"I do not know what direction this thread is taking . I wanted to be a part of this team to have a group who keep pushing each other to put in more effort . You should know by now that results can be highly unpredictable when it comes to mocks or even CAT ( to an greater extent ) for that matter .. so I am never going to judge myself by my scores but yes .. there is a little thing called regret too . I am ready to fail but I do not want to have the regret that I could have done a little more hard work .. and here is where I have to generate my own motivation to make sure I am studying daily even after a bad day at the office.
You will be amazed what a little accountability can do for your motivation ..

In my opinion , the purpose of this thread is to make you not feel complacent when you get an awesome score and not feel dejected when you get a low score . We can discuss some difficult questions but there are already many threads for that very purpose . So the main thing will be to open up on this forum and support each other . I am sure all of you are brilliant and have the potential to crack CAT .. but once you start committing publicly I am sure it will help you in the execution part . I do not want you to be obsessed with this or waste time on this .. just half an hour before sleeping you come to this thread and update your previous post for progress made and create a new one for the next day . That is it .. no opening PG for the rest of the day . You have enough material/mock papers to keep you busy during the day .

So what is the advantage of writing it all down ?? Planning will help you to focus on small portions and you will not feel overburdened by the amount of study material you have . If by the end of the day you are not able to finish what you had hoped for .. do not feel disheartened .. you are either making plans which are too lofty or you need to put in a little more hard work . Just make sure that this does not take place for 2 consecutive days .. otherwise it will be difficult to break out of your old rut .
If youstill think you are better off the way you are proceeding with your prep , then no issues . I for one always used to say no to group study because I like to have my own space while studying .. so it is ok if you feel this will not work for you .

Thats enough gyaan for the day .. lets not make things too complicated here"

Concept K1 (16/09/2012)

If |x| +|y|=a
then the area of the region bounded by the given curve will be 2a^2.
This is also valid also when |x+k|+|y+k|=a

Find the area of the region bounded by the graph |x-7|+|y-9=12 (AIMCAT 1209)
Area bounded by the graph |x-7|+|y-9| = 12 is same as Area bounded by the graph |x|+y = 12 and that is
=>2*12^2=288 sq units
Carry on team!!!

Concept R1 (16/09/2012)

If N is a composite number such that N= a^p*b^q*c^r, where a,b,c are prime factors, then the PRODUCT OF THE FACTORS OF N = N^[(p+1)(q+1)(r+1) / 2 ]

Example: 12 = 2^2 * 3^1.Hence product of the factors of 12 = 12^ [ 3*2/2 ] = 12^3 = 1728


J1 17-09-12


Shortcut to find the last 2 digits

The method is different for odd and even ending bases.

Odd numbers ( i.e numbers ending in 1,3,5,7,9)

Take a simple no : 31

31^2 will have 1 as the last digit as we all know , now to find out the tens place digit , what we need to do is multiply the tens place digit with the units place of the exponent. i.e in this case it is 3*2 = 6. Thus we get the last 2 digits of 31^2 as 61 ( 31^2= 961) . This is the base on which the whole method is built.

Now in the case of numbers ending in 3,7 and 9. We know that :

3^4 = 81
7^4 = .... 21
9^2 = 81

and hence we can easily convert the given question for eg. 77^25 .

It may be written as ((77^4)^6 )*77

We can find out the last 2 digits of the bold part as shown in the below example and multiply it with 77 to get the final answer.

q. Find the last 2 digits of 23^644 ?

Step 1 - write the no: in such a way that a number ending in 1 is inside the brackets.
( 23^4)^161 ( since 3^4 ends in 1)

Step 2:
(23^2 * 23^2) ^161

= > (529 * 529)^161

Taking only the last 2 digits ( ..29 * ..29 = ...41)
Step 3:
(....41)^161

= > ..41 is the answer


This method can also be used to find out the remainders . The pre requisite here is that you should know all the squares from 1 to 30 ๐Ÿ˜ƒ and the short cut to calculate squares upto 100

Wil explain the method for even numbers another day ๐Ÿ˜›
Dheee poyi dhaaa vannu...!!!!
Concept Rkd 1(17-09-2012)
Consider a m*n matrix. where A and B are two points on the two diametrically opposite corners.
Then the number of shortest paths that can be taken by a person to travel from A to B is
m+nCm or m+nCn

Please not that there was a mistake in the example that i posted above , it has been corrected.


Thank you Kevin for pointing it out :)
@joethaliath No probs

K2 (17/9/12) : CALENDARS : Points to Remember !!

1. 1st January 0001 was a Monday.
2. Calendar repeats after every 400 years.
3. Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
4. Odd days- remainder obtained when no. of days is divided by 7. Normal year has 1 odd day and leap year has 2 odd days.
5. Calendar moves ahead by number of odd days.
6. While checking leap year just analyze whether February falls in that period or not.
7. Century has 5 odd days and leap century has 6 odd days.
8. Take out net odd days.( add all the odd days and again divide by 7)
9.In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
10. 1st January 1901 was Tuesday.
11. We calculate odd days on the basis of the previous month.


K (2.1) : Concept in a concept (@raku1989 dhaaa vannu... Dheee poyi !! )

Reposting...(related to Calendar's)

Zeller's Rule Formula:

F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C


K = Date => for 25/3/2009, we take 25
In Zellers rule months start from march.
M = Month no. => Starts from March.
March = 1, April = 2, May = 3
Nov. = 9, Dec = 10, Jan = 11
Feb. = 12
D = Last two digits of the year => for 2009 = 09
C = The first two digits of century => for 2009 = 20

Example: 25/03/2009
F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 - (2 x 20)
= 25 + 12/5 + 09 + 09/4 + 20/4 - 2x20
=25+2+09+2+5-40
[ We will just consider the integral value and ignore the value after decimal]
= 43 - 40 = 3

Replace the number with the day using the information given below.

1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
7 = Sunday
So it's Wednesday on 25th march, 2009.

If the number is more than 7, divide the no. by 7. The remainder will give you the day.
Prep Destresser for the day !!

Before joining Kerala Study Group :

(a+b)^n = Use Binomial Theorem

After joining Kerala Study Group : Same as Peter


Apt analogy for my errors in 1305...

Team not only concepts, pls post relevant examples /Previous Cat Q's /Doubts /Alternative approaches too...

Hope no one bans me for excessive SPAMMING here !!!



@Kevin88 said:Prep Destresser for the day !!(a+b)^n = Use Binomial Theorem Same as Peter
Hope no one bans me for excessive SPAMMING here !!!
All (except @Kevin88)
Nadeshaa Kollanda!! Paavam payyana....Arivilaymakonda
Amongst 200 housewives who watch atleast one of 4 TV channels in the afternoon, 150 watch Sony, 155 watch Zee, 160 watch Star, 165 watch Imagine

Q1. Minimum number of housewives who watch all four channels
Q2. Maximum number of housewives who watch all 4 channels

Trying doing this without options!
@raku1989 Max no is it 143 ??

I used trial and error
Max no had to be less than 150. So tried backwards from 150.

Common to 4 = 143
only sony = 7
only zee = 12
only star = 17
only imag = 22
total = 201
@seetharam7 Yup, its 143...
@raku1989 Gud sum..

A U B U C U D = 630.
Excess of housewives = 630-200 = 430.
For max number of housewives who watch all 4 channels = 430/3 =143.xx=143.

For min no of housewives = 200 - {(200-150)+(200-160)+(200-155)+(200-165)
= 30.


@Kevin88 For max no, why did you divide by 3 ?
@seetharam7
When you add a quantity to the intersection of the 4 sets, you will get an excess quantity of 3

Eg: If you add 5 here to the intersection of all 4 sets, then (Assume initially a=b=c=d=0)
==> A+B+C+D=20, i.e 15 is additional (3*5)


What I did today
Went through the basics of geometry, coordinate geometry and quadratic equations. Wrote the sectional test on quadratic. Had taken Aimcat1319 on saturday and had screwed it up badly (32+33). Analysed it and took Aimcat 1318 today. Scored 35+30. Just went through the PG thread for that mock and found that it is a decent score and would fetch me 95%ile and clear both cut offs. Sad about the score. Happy with the %ile.
@Kevin88 What would the answer be if the question was max no who watched atleast 3 channels ?
Concept S1

For questions related to chess boards
Assume 8 X 8 board

No of AXA squares = (9-A)^2
Eg: no of 1X1 sqaures = (9-1)^2 = 8^2 = 64
no of 7X7 squares = (9-7)^2 = 2^2 = 4

Total no of squares = 1^2 + 2^2................+8^2 = n(n+1)(2n+1)/6 = 204
Total no of rectangles = 1^3 + 2^3............+8^3

Also, this formula can used for any square board. For N X N board, substitute N+1 instead of 9.
Eg: 4 X 4 board, we need no of 2X2 squares, its (5-2)^2 = 9