CAT 2008 Quant anlysis

The monster of all exams, IIM-CAT was held on 16th Nov. But, I must say the section 1, i.e quant, was not that monster which it was in 2007. But as we say when you expect a monster, even a mouse will appear as a rabbit.

The paper was well set, with a nice blend of sitters, easy questions and tough questions.

Any well prepared student should eb able to solve 12-14 questions with 90% accuracy, in around 40-45 mins

The geometry problems were so easy, it looked like, sitting and solving ncert book exercises of class 9.

The real gem where the function problems and the series involving roots.

The number theory problems can all be classified as mock type, as all of them have appeared in mocks

For a while, I was in the notion as if I was taking a mock which was a mix of cl ims and time. neither too tough or too easy, a blend..

My take on cutoff 35+-2

Good Luck !!

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CAT 2008 Key and Solutions

Hi,

I will be posting my own solutions and keys for quant.

Any error on my part is totally unintended and I would not be liable to any damage claims

You are free to use these though

Rahul

Set 444

Q1.

The number of terms in (a+b+c)^20

it is same as number of solutions of a+b+c=20 which is 22c2=231 option 1

Q4)

Seed(n)=9 if the number is a multiple of 9

Hence 9,18, 495

55 numbers in total

5)

any two number is replaced by a+b-1

so basically we will end up with sum of all 40 numbers -39

40*41/2-39=781

Question 9

10/4=x/r

x=2.5r

A=2pir(r+h)=2pier(10-1.5r)

r=10/3

gives Area =100pie/3

Q10

obtuse angles are possible if

15 is the largest side or x is the largest side

take case 1

x+8>15 =>x>7

also x^2+8^2<15^2

x<13

so we get x=8,9,10,11,12

similarly we will get

18,19,20,21,22

hence total 10 values for x

Q11

m+(m+1)^2+(m+2)^3=9(m+1)^2

m^3-3m-2m^2=0

m=0,-2,3

Hence option 1

Q12

4 digit integers <=4000

first check upto 3999

we will get 3*5*5*5=375 and add 1

we get 376

Question 13

root(1+1/1^2+1/2^2)=3/2=2-1/2

hence we can write as 2008-1/2008 option 1

alternate

3/2=1+1/1.2

7/6=1+1/2.3

..

we add

1+(1-1/2)+1+(1/2-1/3)..+1+(1/2007-1/2008)=

2007+1-1/2008=2008-1/2008

Question 14

a/sinA=2r

17.5/3/9=2r

r=26.25

option 5

Question 15

f(x)f(y)=f(xy)

f(0)^2=f(0)

and f(1)^2=f(1)

consistent solution is f(1)=1

so f(2).f(1/2)=f(1)

4.f(1/2)=1

f(1/2)=1/4

Question 16)

7^2008

last 2 digits of 7 have a cycle of 07,49,43,01

Hence 2008=4k

hence last two digits is 01

Alternate is

7^2008=(7^4)^502=(2401)^502=(2400+1)^502=01

option 3

Question17)

[(2a)^2-2a^2/root(3)]/2a^2/root(3)=2root(3)-1

option 5

18)

let the roots be k-1, k, k+1

then 3k=a

k(k-1)+k^2-1+k(k+1)=b

3k^2-1=b

so min value of b is -1

Question 19 and 20

9a+3b+c=0

25a+5b+c=-3(4a+2b+c)

37a+3b+4c=0

gives a=b

hence

ax^2+ax+c=p(x-3)(x-q)

compare coefficients gives

q=-4

Hence 19 option  2

20 is option 5 cannot be determined

as we do not know a


we can just say a+b+c=2a+c=-10a

Question 25

first common term is 21


the common terms will form an AP of CD=lcm(4*5)=20

21+19.20=401 <417

21+20.20=421>417

Hence 20 terms !!

Mini Concept: Multinomial Theorem

Some Trickery of Multinomial Theorem

x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2

if we extend this to r variables then the formula becomes (n+r-1)C(r-1)

if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)

Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips

Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.

so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182


Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?

Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D

now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways

case 2) we leave 0 and pick two positive and one negative

then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2

case 3 it will be same as case 2
we get 50C2

hence total is 2.50C2+50=2500

Problem Of The Week 53

How many ways can a size k + 1 subset with maximum element m + 1 can be created from the given set S={1,2,3,.....,n+1} ?

Problem Of The Week 52

Suppose that and are positive numbers for which

,

what is the value of ?

Problem Of The Week 51

Let P(x)=ax^4 +bx^3+cx^2+dx+e be a polynomial with all integer coefficients and a=1. If√2+√5 is one of the roots of P(x)=0 , which of the following can be the value of |(b+c+d+e)|?

1) 103

2) 89

3) 63

4) 23

5) 5