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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Monday, November 17, 2008
Sunday, November 16, 2008
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 !!
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 !!
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