Ultimate Quant Marathon Blog For IIM CAT

I could not post due to some engagements. Here are a bunch of problems to compensate

Question 1)

A + B + C + D = D + E + F + G = G + H + I = 17 where each letter represent a number from 1 to 9. Find out number of ordered pairs (D,G) if letter A = 4.
a) 0                       b) 1                 c)2                        d) 3                 e) none of these

Question 2)

The sequence 1, 3, 4, 9, 10, 12….. includes all numbers that are a sum of one or more distinct powers of 3. Then the 50th term of the sequence is
a. 252                    b. 283                     c. 327                      d. 360                  e) none of these

Question 3)

Given that g(h(x)) = 2x² + 3x and h(g(x)) = x² + 4x − 4 for all
real x. WHich of the following could be the value of g(-4)?
a)1                     b) -1                          c) 2                 d) -2                   e) -3

Question 4)

If a, x, b and y are real numbers and ax+by = 4 and ax² +by² = 2 and
ax³ + by³= −3 then find (2x − 1)(2y − 1)
a)4                      b) 3                    c) 5                 d) -3          e) cannot be determined.

Question 5)

K1,K2,K3…K30 are thirty toffees. A child places these toffees on a circle, such that there are exactly n ( n is a positive integer) toffees placed between Ki and Ki+1 and no two toffees overlap each other. Find n
a)4                        b) 5                     c) 9                 d) 12                       e) 13

Question 6)
For the n found in previous  question, which of the two toffees are adjacently
placed on the circle? ( All other conditions remaining same)
a) K11 and K13                    b) K6 and K23                   c) K2 and K10              d) K11 and K18
e) K20 and K28

In a triangle ABC, perpendiculars BD and CE are drawn to the sides AC and AB. Points  D and E are joined, then the ratio of the area of ADE to the area of ABC is:

1) Cos²A 2) Sin²A 3) Cot²A 4) Tan²A  5) None of these

Hi People,

I am sure quite a few of you who visit this blog took CAT 2008, as did I. And on 9th Jan 2009, came the results after about two months and the result was according to expectation at least for me, and I hope it was same for you.

My Scores

Quant     64 ( 16 Correct, 0 incorrect)  99.88%ile

Logic and DI  44 ( 12 correct, 4 incorrect) 98.43%ile

Verbal     76 ( 21 Correct, 8 Wrong)  99.49%ile

Overall  184  ( 49 Correct, 12 Wrong) 99.97%ile

Calls:

IIM Ahemdabad, Calcutta, Indore, Kozhikode, Lucknow!

Wish me luck for the next stage!!

Thanks a lot for being with me !

Best of Luck everyone!

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 !!

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 !!