[vipul88]

DISCLAIMER: Everything written in this post is from a heart that loves DI, vipul88 is not responsible for anything that this post might cause

2-cents on DI section from my side:

The first few minutes we spend on any section have a huge bearing on how we are going to perform in that section. So u should have complete familiarity with the set that u attempt first. Never venture into unknown waters initially. Now it can't be that u are not having any idea about all the sets asked in exam. In that case, the paper is really testing or u should be prepared to give the test next year ( sorry if this sounds harsh, but it's true).
So the best option would be select a DI set first. The chances of cracking a DI set are always more than cracking an obscure LR set. Again if a set based on tournament is asked and u hv the confidence to crack any tournament set, u may start with tht as well.

But to all those who are weak in DI, i'll suggest u to start with a DI set. So my suggestion to all of you is to start working on speed maths a little bit. The time u save while doing DI can be allotted to a tough LR set. Develop good calculation speed and try to attempt all DI sets, and may be 1 LR set and this should be enough to secure a 97-98%ile. ( I'm assuming tht there would be atleast 2 DI sets).

Now those who want 99.5+ in DI, start loving DI and LR. If u can't enjoy doing DI/LR it would be tough to get 99.5+. I know it's all about marks and we not giving CAT for fun, but i personally believe tht unnecessary pressure doesn't help in this section.
You see what kind of a person u are, also determines how u'll fare in this section. If u are by nature a bundle of nerves and allow even small occasions to get to u, chances are tht u'll find going tough in this section.
The reason is that compared to other sections the stakes are very high in this section. Other sections do have single questions but DI section generally has questions in sets. At a time 4-5 questions are at stake. That translates into around 15-20 marks which will be the difference between a 94%ile and a 99%ile. So at the back of the mind there is always this thought that if I'm not able to crack the set, then those 15-20 mins tht i'hv invested in the set would become a liability. This pressure does not help ur cause really. The best way to avoid this mental trap is to start loving those sets. Just don't think abt cut-off and %iles and other things while attempting DI and specially LR. I know what i'm saying is tough to follow but trust me the day u clear these demons from ur mind, u'll start doing exceedingly well.

Further Gyan (only if u want it) after some time.

[saurav_:)]

Originally Posted by naga25french

1) |x|+|y|+|z| = 15

for any value like x=1,y=2 and z = 12
there are 8 ways for representing the nos. with + or -
so total = 17C2*8 = 1088
in this we will subtract 15*4*3 + 6 = 186 considering the 0's taken...
hence total values = 1088 - 186 = 902

This could have been done in this way also..
Total Positive integral solutions== 14C2*8 =728
and the zeroes solutions= 186 as derived..
so total solutions= 728+186=914...
plz xplain where am i wrong..!!

[saurav_:)]

Originally Posted by naga25french

2) a + b + c + d + e + f + g = 6

whole number soln = 12C6 = 924

required answer = 924 - 1 = 923

plz xplain this.. i hve forgotten the concepts..

[Love_CAT]

Originally Posted by saurav_:)

This could have been done in this way also..
Total Positive integral solutions== 14C2*8 =728
and the zeroes solutions= 186 as derived..
so total solutions= 728+186=914...
plz xplain where am i wrong..!!

Another way of solving |x|+|y|+|z|=15

(i)All are non zero integers
So total number of soln: 14C2*8=728

(ii)One zero two non zero integers
So total number of soln: 14C1*4*3=56*3=168

(iii)One non zero integer
So total number of solution: 6

Total number of solutions of the equation are
728+168+6=902

[saurav_:)]

Originally Posted by Love_CAT

Another way of solving |x|+|y|+|z|=15

(i)All are non zero integers
So total number of soln: 14C2*8=728

(ii)One zero two non zero integers
So total number of soln: 14C1*4*3=56*3=168

(iii)One non zero integer
So total number of solution: 6

Total number of solutions of the equation are
728+168+6=902

this solution seems alright..

17c2*8 has an error as it includes the Both the +ve n -ve signs of zeroes... along with non zeroes terms... which is very difficult to count..
plz correct if i am wrong..

[Love_CAT]

Originally Posted by saurav_:)

this solution seems alright..

15c2*8 has an error as it includes the Both the +ve n -ve signs of zeroes... along with non zeroes terms... which is very difficult to count..
plz correct if i am wrong..

No,
x+y+z=15 has (15-1)c(3-1)=14c2 positive integral solutions..so for each x,y,z we will be getting 2 numbers (positive and negative)..
a total of 14c2*8=728..

Hope it helps..

[first_timer]

Wrote unproc 10 today ( from my friend's account )....had an okayish perfomance....

QA - 12C 6W = 42 (93.7)....should have been much,much better
DI - 14C 2W = 54 ( 97.4)
EU/RC - 14C 10W = 46 (89.3)....thought I had done well,but percentiles indicate otherwise
OA - 40C 18W = 142 ( 98.57)

Btw, the overall cut-off was set at 110.....yet at 142 I am getting 98.5....

either the number of test-takers is extremely low or the cut-off is set too low....

[Intekhab_ali]

Originally Posted by naga25french

2) a + b + c + d + e + f + g = 6

whole number soln = 12C6 = 924

required answer = 924 - 1 = 923

In dis prob...m proceeding by findin d number of integral solutions of a+b+c+d+e+f=6;a+b+c+d+e+f=5 n so on till 1...m getting d correct answer 923 as d case of 0 will give a whole number....
plz explain ur approach bro as it is way 2 short....y dat extra g...thanks

[naga25french]

Originally Posted by saurav_:)

plz xplain this.. i hve forgotten the concepts..

Originally Posted by Intekhab_ali

In dis prob...m proceeding by findin d number of integral solutions of a+b+c+d+e+f=6;a+b+c+d+e+f=5 n so on till 1...m getting d correct answer 923 as d case of 0 will give a whole number....
plz explain ur approach bro as it is way 2 short....y dat extra g...thanks

we have the eqn as a+b+c+d+e+ f <= 6

to remove < sign , add one more variable .. this is commonly used method

so we get a+b+c+d+e+f+g = 6

12C6 = 924

but there is case where a = 0 , to remove that subtract by 1

924 - 1 =923

hope this helps

[first_timer]

Exam season kick starts from tomorrow.....

All the best to people appearing for IRMA