PG Premier League - Chennai Dream Team

[nishant_rungta]

Good news for the ones who got the answer to qn.71 as 14. TIME mods have posted on their forum:

Key Change In Q.71:

The solution given for Q.71, solves the question under the assumption that the four packets are identical, i.e., that they have an equal number of chocolates each.

However, as rightly pointed out by most of the students, the question does not explicitly mention that Qureshi bought four "identical" packets, i.e., the word identical has been missed out. Therefore, given the fact that the minimum number of daughters needs to be found, the number of chocolates in each packet will also, logically, be considered as variable (subject only to the "at least 45 chocolates" criterion mentioned) in the process of minimising the number of daughters. Hence, the answer to Q.71 becomes 14, and not 16, i.e., the packets need not have an equal number of chocolates each and thus the answer does not have to be a multiple of four.

The correct answer to Q.71 is therefore Choice (2) and not Choice (1).

This change will be taken into consideration while evaluating the answer sheets.

Regards,
Moderator.



My scores go to 48

[exile]

Originally Posted by nishant_rungta

My scores go to 48

My score comes down to 76.

[nishant_rungta]

Guys seems like there might be changes in VA keys too

DOn't know. Let's see. The TIME Mods have been replying on their forum. Here's another update from them:

Other Queries/Issues regarding the VA key will be responded to by end of day today.


Moderator

[ramaan]

Originally Posted by nishant_rungta

Good news for the ones who got the answer to qn.71 as 14. TIME mods have posted on their forum:


The correct answer to Q.71 is therefore Choice (2) and not Choice (1).

This change will be taken into consideration while evaluating the answer sheets.

Regards,
Moderator.



My scores go to 48

hiya..my scores go to 29 in QA...

[somname]

Ok,
As the strategy coach , I read some of the analyses in the last 2-3 pages, and this is what I suggest.:

VA general strategies:
1.Read an RC as if you are the author. Blank ur mind when u start the RCs. Ur individual opinion while reading/doing an RC is useless. So is you prior knowledge. The questions are asked from the viewpoint of the author, and you’ve got to answer them in that sense. Remember, for RCs, You DO NOT have any knowledge/opinion.
2.Read newspapers and enjoy them. Don’t read stupid weird articles just because you want to improve your VA. Read funny, sexy articles that you like and enjoy. It’s a much better way of improving your English than reading up on the etymology of words from thick red books.
3.Word-lists are stupid and a waste of time. They are not important for CAT. Burn your word-lists.

DI general strategies:
1.Do NOT spend more time on a set just because you’ve already spent some time on it. Wasted time does not merit more time to be wasted. If after 20 minutes of reading a set, you do not understand it, the LEAVE it!
2.Practice! Practice only from mock-cats. The BRMs are generally for ppl who are going to give cat in 2010, and I expect none of you are like that.

QA general strategies:
1.Make sure you read ALL the questions! Even if they are from pathetic topics like PnC or Probability. This way, you’ll never feel like a fool later for not having read a question completely.

General strategies:
1.Give equal importance to the three sections in terms of time. This simple rule, whenever violated, has caused havoc. Do not allot less than 40 minutes for any section, whatever be the compulsion.
2.Your primary aim in all these mocks is to clear ALL the cut-offs in every paper. Trust a 4-times cat taker on this, if you can clear ALL three cut-offs in cat, there’s a 99.99% chance of you clearing the total cut-off as well, and making it to the next stage.
3.Do not bother whether the paper is tough or not. Your job is to make sure that by the end of the paper you have read all the questions and feel happy that you’ve given it your best shot.
4.Remember, the Real competition is with yourself and your past scores. The zillion other people writing the mock are not your problem. It’s their problem that you are writing the mock as well.


I have read the analysis of Ramaan, hantan, exile, elusive_lady, monster, ravi and rungta and have given my analysis on the basis of reading these analyses.

Ya, one more thing guys… this PGPL thing is a practice series, and that is what it should be taken as.. The real thing’s on 16th Nov, and that’s when the underdogs shall have their day!

Soham

[Greenspan]

Originally Posted by somname

Read funny, sexy articles that you like and enjoy.

This could include your local papers' Page 3 news with all the hotties prancing around in their skimpy dresses too.. Show this post by somname to your parents (if you are at home), and maybe you will get away with watching whatever you like.... and soham you forgot to mention FTV music.... that channel has trance like music that helps increase your concentration levels!

Forgot to mention the serious stuff.... gems such as that of soham above and that of other stalwarts around in other threads like harshad and raven are not to be forgotten. It is very easy to acknowledge them by pressing the "thank" button, but they would be much happier and so would you, if you were to remove a print out of the same and hang it onto whatever you look at the first thing in the morning! I am serious!

@mk.... In your first post of this thread, keep a list of the permalinks of such posts, for a quick reference for puys around.

[ramaan]

The BRMs are generally for ppl who are going to give cat in 2010, and I expect none of you are like that.

oops..whats this BRM meant..???????....

[nishant_rungta]

Originally Posted by ramaan

The BRMs are generally for ppl who are going to give cat in 2010, and I expect none of you are like that.

oops..whats this BRM meant..???????....

Well it's basically the Basic Reference Material that an institute gives u to prepare for cat. Their course material

Agree to Soham's points word by word. High time I start reading articles more seriously and religiously.

[monsterkartik]

Originally Posted by somname

The BRMs are generally for ppl who are going to give cat in 2010, and I expect none of you are like that.

Soham, you are wrong. I havent even started off with my BRM's. Will start off and finish them asap. Never thought that its expected that we should be thorough with our basics

Basics of Geometry, PnC, probabaility, Algebra, set theory, number systems (remainder theorem, divisibility rules) are pending for me

Cheers

MK

[nishant_rungta]

Funda 3

Today's concept...few tricky questions on divisibility


concepts...


a^n - b^n is always divisible by a-b

a^n - b^n is divisible by a+b when n is even.

a^n + b ^n is divisible by a+b when n is odd.

a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.

when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.





Answers to yesterday's questions.

1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt----667

2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.----such a no. doesnt exist as 5 rem with aint possible.

3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.----872

4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?----80

5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.----50

6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11? ----3

7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.---cannot be detrmnd

kudos to rockeezee who got most of them right!



questions for today...

1. 32^23 + 17^23 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


2. 32^23 - 17^23 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


3. 32^232 + 17^232 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


4. 32^232 - 17^232 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....

a. 40 b. 20 c. 80 d. all of these. e. none of these.


what is the remainder when 42527152653425416242624272427215287 is divided by :

6. 16

7. 32

8. 64



regards
maxximus


Today's concept...finding factors of a no. n sorting relevant questions.


concepts...

#1. to find total no. of factors of a number.

=>first of all...express it in terms of prime numbers

e.g 1260^4.

express it as (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.

=>now add 1 to the powers of every prime no. n multiply them all...u get the total no. of factors of 1260^4.

i.e (8+1)*(8+1)*(4+1)*(4+1) = 2025 = total no. of factors of 1260^4.




#2 to find no. of odd factors of a number.

leave power of 2 and multiply powers of all other prime nos after adding one to them.

i.e. (8 + 1)*(4+1)*(4+1) = 225 = no. of odd factors of 1260^4



#3 to find no. of even factors of a number.

take the difference of total factors and odd factors.

i.e. 2025 - 225 = 1800. = no. of even factors of 1260^4



#4 to find in how many ways can a given no. be represented as product of two relatively prime factors.

in this case, the power of prime no. becomes irrelevant as all the powers shud lie with the same factor else the two factors wont be relatively prime.

i.e in the above case 1260 = (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.

now powers 8,8,4,4 have no importance...whats important is how many prime nos. are there...they are four...viz 2,3,5,7. hence 4.

now v have to form nos. using these 4 primr nos. its a like a question askin u...if u have 4 sweets...in how many ways can u eat them? the answer is 4C0..when u eat none... + 4C1 when eat any one...+4C2 + ...4C4...when u eat all.

similarly here, answer wud be 4C0 + 4C1 + 4C2 ...4C4 = 2^4 = 16.

but the factors have to exist in pairs...hence 2^4/2 = 8 factors are possible.

on a general note...a no. formed of n prime nos has 2^(n-1) pair of co-prime factors


Answers to yesterday's questions


1. 32^23 + 17^23 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


2. 32^23 - 17^23 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


3. 32^232 + 17^232 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.


4. 32^232 - 17^232 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.



5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....

a. 40 b. 20 c. 80 d. all of these. e. none of these.



what is the remainder when 42527152653425416242624272427215287 is divided by :

6. 16----7

7. 32----23

8. 64----55

kudos to irevani, gk who got all correct...so too to deep, rockeezee...who got jus 1 wrong....bhaiyon...if a no. is divisible by 80...its obviously divisible by 40 n 20.




questions for today...

on the very simple fundas above, very tricky questions can be framed.



Find the smallest no. that has exactly.....

1. 16 factors

2. 12 factors

3. 60 factors

4. how many factors of 27000 are perfect cubes?

how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...

5. 3 terms

6. 4 terms

7. 5 terms

8. 6 terms

How many values of a are possible if x^2 + ax + 2400 has...

9. integral roots

10. roots which are natural nos.



regards
maxximus

Kudos to Maxximus for this