The Official CAT 2007 Quant Thread

[the_fence_crosser]

I've had an extremely short endeavour at pagalguy, as indicated by the number of times I've posted. However short it may have been, this connection has been infinitely memorable for me....nobody knows how many among us would finally make it to the IIM's, but what really matters is the way in which all of you guys have come together in helping others with problems, irrespective of their difficulty level. While most people I know have been all quite with their own secret ways of preparation, you guys have come out with whatever armoury you've got, and whats better than sharing that extra bit of knowledge with everyone. Kudos.

At this juncture, it is of little actual relevance and value that I post further questions for you all to solve. With only 3 days to go for the big day, whats more important to share would be a few valuable tips/ checklists that should be kept in mind while giving the test. I've come across way too many tips from various sources, but a few really caught my eye as being of utmost importance.
1.) Do NOT go to the test expecting a set pattern. NO. Instead, go for the test with an extremely open mind and a cool head.
2.) Keep an exit strategy in place, not giving more than 3-4 minutes for a problem.
3.)READ ALL QUESTIONS. Yes, it's wiser to read all questions and then select the ones you are comfortable with, rather than randomly selecting questions to solve.
4.) Read the questions properly, keeping in mind that each and every word in the question carries a significance.
5.) Have fun solving questions in the CAT. It's not everyday that you get such a chance.

It's a very small list, though quite potent. That apart, remember that the CAT is just another test, and by no means is it the be all and end all of our career. So remember to have fun, and irrespective of how you'll perform on the big day, remember to visit this thread and post your experiences on CAT 2007.
All the best!!!!!

[joel1702]

Originally Posted by andy_jaan

As there are 7 places (sessions) to be filled, with every subject getting atleast 1 session the remaining 2 additional sessions can be given to any 1 subject or 2 different subjects.
Therefore in first case no. choose no. of ways will be 5!*(7!/3!) and in second case no. of ways will be 5C2*{7!/(2!*2!)}
so the answer should be 2100 (I have done orally so please check this last calculation )



Look under as by basic knowledge here A(N) should be like n(2n+1)/6, Therefore start checking from the least and try breaking the no.*6 in two factors of n and (2n+1) kind.
With 121, its not possible
with 196 we can see that 196*6=24*(2*24+1)

Tata

Andy

no the Answer is not 2100. it is 16800.

any idea how to get that>

[andy_jaan]

Originally Posted by joel1702

no the Answer is not 2100. it is 16800.

any idea how to get that>

Oh sorry, it was a calculation mistake (what I was afraid of ), Please calculate your self (the last part) and you'll get 16800 as answer.
BTW solution (procedure) is correct and valid.

i.e 5*7!/3! + 5C2*[7!/(2!*2!)} = 4200 + 12600 = 16800

Andy
P.S. Not being a aspirant myself I often tend to be casual but otherwise I myself support to do all calculation on paper else silly mistake like this one can cause serious damage in real exam.

[parikhdhruv]

help wid dis 1.......
CAT 2005 question..
p=1!+(2*2!)+(3*3!)..........+(10*10!)
then p+2 divided by 11 gives a remainder of?

[deep_agrawal]

Originally Posted by parikhdhruv

help wid dis 1.......
CAT 2005 question..
p=1!+(2*2!)+(3*3!)..........+(10*10!)
then p+2 divided by 11 gives a remainder of?

It is divided by 11! and not 11.

Lets do this way:

Start with p = 1! which can be written as 1*1!
Hence p = 1
p + 2 = 3
p + 2 when divided by 2! is 1

Now check with p = 1!+2*2!
This results in p = 1+ 4 = 5
now p + 2 = 7 which when divided by 3! gives 1

Similarly for p = 1! + 2*2! + 3*3!
p = 23
p+2 = 25
and p+2 when divided by 4! is 1 again.

Hence the general case can be observed. Hence the answer is one

[the_fence_crosser]

Originally Posted by parikhdhruv

help wid dis 1.......
CAT 2005 question..
p=1!+(2*2!)+(3*3!)..........+(10*10!)
then p+2 divided by 11 gives a remainder of?

Note that p = 1! + (3!-2!) + (4!-3!) + ...+(10!-9!) + (11!-10!)
= 1! - 2! + 11!

p+2 = 1 + 11!
therefore p+2 / 11! would give remainder 1

[medulla]

Hi,

What is the strategy in following types of series questions:

Q Find the sum of the series 1/3 + 1/15 + 1/35 + 1/63 + ... up to 99 terms

1] 99/199
2] 198/199
3] 98/99
4] 9/199
5] Cannot be determined

--- Medulla

[gaurishankar]

Originally Posted by medulla

Hi,

What is the strategy in following types of series questions:

Q Find the sum of the series 1/3 + 1/15 + 1/35 + 1/63 + ... up to 99 terms

1] 99/199
2] 198/199
3] 98/99
4] 9/199
5] Cannot be determined

--- Medulla

T99=1/197*199
1/3=1/2(1-1/3)
1/15=1/2(1/3-1/5)
1/35=1/2(1/5-1/7)
.............

............
1/197*199=1/2(1/197-1/199)
Adding all=1/2(1-1/199)=99/199
hence 1

[medulla]

Originally Posted by gaurishankar

T99=1/197*199
1/3=1/2(1-1/3)
1/15=1/2(1/3-1/5)
1/35=1/2(1/5-1/7)
.............

............
1/197*199=1/2(1/197-199)
Adding all=1/2(1-1/199)=99/199
hence 1

Thanks gaurishankar,

What i want to know what made you think that i have to take 1/2 common..
Are their any specific patterns for series questions..
plz let me know..
Medulla

[gouravklr]

there r 5 red balls...6 green balls...7 blue balls....8 yellow balls...in how many ways can u select 10 balls...??
well..i kno it will be the coeffcnt of x^10..is
(1+x+x^2+.....x^5)(1+x+x^2+.....x^6)(1+x+x^2+..... .x^7)(1+x+x^2+...x^
but i dunno how to calculate that....
please anybdy throw some light......



best of luck to everybody...
it's now...OR....now...