@seetharam7
That took some time !! I'm getting it as 200 itself. But i dont know the individual distributions.
Method : (HW- House Wife)
To maximise the no who watched atleast 3 =>max(HW watching 3) + Min of (HW watching 4)
= 170 + 30 =200.
Now i worked backwards.
Excess HW from Q = 630-200=430 cases.
Max(HW watching 3) accommodates 170*2=340 cases.
Min of (HW watching 4) accommodates 30*3=90 cases
U can thus find total but individual distributions among channels would be thru trial & error i guess !!!
From where are practicing both sections from ?
@Kevin88 I am going through the basic material of TIME. Not going topic wise. I write a mock, analyse, find a weak area, work on it. 
That's it ! Haven't worked on Sec II for a while. Have some practice exercises of TIME. Hope to go through it.
That's it ! Haven't worked on Sec II for a while. Have some practice exercises of TIME. Hope to go through it.
This thread is really on a roll now 
.Keep up the good work guyz...... 
Hi all!! I jus wanted to know; how much time on an average, do u spend to solve each question in DI part? General case, ok....
After how much time do u give up on the problem, deeming it 'unsolvable'???
Any source for DI caselets? Offline downloadable material appreciated too.....
Concept J2 - Last 2 digits of numbers ending in even numbers(2,4,6,8) 18-09-12
Use the pattern of the number 1024 =2^10 i.e.
*2^10 raised to even power ends with 76 and
* 2^10raised to odd power ends with 24
e.g.) 2^(788) = ((2^10)^ 78 )* (2^8) = (....76 )*(256) = ...56
@Ash27Winz
Well personally i stick to a thumb rule of giving DI - 20-25 Minutes max.
However, it depends on the difficulty level of the set. A max of 2.5 min per Q; else bookmark
It always better to leave the unsolvable ones to the end; if the method does not click on a couple of reads, skip it- max 1 min.
All of these are thumb rules, but you need to be flexible with them.
Browse through the DI thread, prep section for caselets
Well personally i stick to a thumb rule of giving DI - 20-25 Minutes max.
However, it depends on the difficulty level of the set. A max of 2.5 min per Q; else bookmark
It always better to leave the unsolvable ones to the end; if the method does not click on a couple of reads, skip it- max 1 min.
All of these are thumb rules, but you need to be flexible with them.
Browse through the DI thread, prep section for caselets
V1. To find the last 2 digits of the square of any number.
x^2, (50-x)^2, (50+x)^2, (100-x)^2, (100+x)^2, (150-x)^2 etc ends with x^2.
All you need to know is last 2 digits of squares till 25^2.
TCYonline is offering free mock tests. Please go through the site if interested. I haven't given it yet but would like to. Has anyone given any of 'em ?
Team.. Whats with the silence & deadlock here !!! 
Was busy with my Internal Audit the last 2 days & was unable to log in 
Anyway an easy one from my end.
Q : For how many intergral k does the inequality log 2 + log (2x^2 + 2x + 7/2) >= log (kx^2 + k) possesses at least one solution?
2)Let C be mini 4X4 chessboard . In how many ways is it possible to select two squares of C such that the midpoint of the segment joining the centres of the two squares should also be the centre of a square?
Could anyone explain the underlying logic if the Q was a NXN square ?
anybody from trivendram here?
@Pratishruti I believe 
Where is everyone ???
3) The total number of 7 digit positive integers whose digits are in increasing order (not necessarily strictly) is ?
Aimcat 1304
QA/DI 26A 24C 2W == 70
VA/LR 28A 15C 13W == 32
OA 54A 39C 15W == 102
It was the easiest mock of the season. Just went through the thread, everyone is scoring above 100.
Someone help me with section 2 !
@Kevin88 said: Where is everyone ???3) The total number of 7 digit positive integers whose digits are in increasing order (not necessarily strictly) is ?
Can the digits repeat or is it distict digits ?
Ans for distinct digits
1st digit = 9 possibilites (except 0)
2nd digit = 9 possibilites
3rd digit = 8
4th digit = 7
5th digit = 6
6th digit = 5
7th digit = 4
total possibilites = 9*9*8*7*6*5*4 = 544320
in each possibility exactly one arrangement will have numbers in increasing order
So = 544320/ 7! = 108
@Kevin88 said: Team.. Whats with the silence & deadlock here !!! Was busy with my Internal Audit the last 2 days & was unable to log in Anyway an easy one from my end.Q : For how many intergral k does the inequality log 2 + log (2x^2 + 2x + 7/2) >= log (kx^2 + k) possesses at least one solution?
I finally got
(4-k)x^2 + 4x + (7-k) >=0
Trial and error
k=0,1,2,3,4,5,6,7....8 values..
when k=8, we get
-4x^2 + 4x -1 >=0....so no value of x will satisfy the inequality...
@Kevin88 said: 2)Let C be mini 4X4 chessboard . In how many ways is it possible to select two squares of C such that the midpoint of the segment joining the centres of the two squares should also be the centre of a square?Could anyone explain the underlying logic if the Q was a NXN square ?
Absolutely no idea 
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