## 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 !!

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

dude i will continue form here, f(1) = 1…

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

f(1/2) {1 -f(1)} = 0

now f(1/2) can be anything….1/4 or 0 or..so it should be cannot be determined.

let me know if m wrong anywhere…thanx in advance…

manishNovember 17, 2008 at 7:19 am

yes your are

when we are specifically given f(2) to come down to one case

we should do that !!

Hence f(1/2)=1/4

outtimedNovember 17, 2008 at 8:08 am

abt qn 5)

i’am hell confused by this problem while many say its easy….

how come only the first 40 are added?

after the 1st iteration a+b-1 is added and a and b are erased

so in the next iteration a can be the one that was written before which is greater than 40…is’nt it? or i’am i missing something

SuhasNovember 18, 2008 at 11:31 am

thats pretty obvious man

just think of a smaller set if you cant understand

for example

1,2,3

after first iteration we will be left with 1+2-1=2

and after second iteration 2+3-1=4

which is also equal to 1+2+3-1-1

outtimedNovember 19, 2008 at 7:22 am