Question from MockCat, SimCat, Aim Cat,...... for CAT'06

Quagmire

Quote:

Originally Posted by crazyfootballer

This is what I am saying. We need to find the nos. whose HCF with 36 is 1. So, no question of even nos. comes into picture. We are just looking for odd nos which are not multiples of 3 for our solution.

Total odd nos. = 499
Toal multiples of 3 = 333
now, out of 333 multiples of 3 we are concerned with only odd multiples of 3 which are equal to 167 (bcoz we have already discarded even nos, in other words in 499 nos. we are not talking abt even nos. at all).

Hence we need to subtract these 167 odd multiples of 3 from 499 as HCF of these nos. with 36 is 3.

Our answer thus is 499-167=332

Got it. Thanx.

ernil

Quote:

Originally Posted by warrior

Every now and then we come across some good questions in mock that we write. these questions can be
(a) the one which we are unable to solve
(b) the question seems easy on face but answer come out to be a different one ( a tricky question)
(c) a question which has been solved in some newer or shorter or different way in aimcat solutions
(d) we came across a different/newer sort of question

I have started this thread so that we can discuss questions from mocks by various institutes over here so that at the end we can have a good collection of question for revision.Also as different people join different test series we can have them all at one place.

A small contribution by all of us ( esp from those who join CL,CF which are less popular as compared to TIME,IMS) can have a big impact for all of us.

Note: Also when posting post your question in bold so that when browing through the page its easy to figure out and keep them numbered

err... can i post some easy {by yu guys standards that is..} aimcat probs here? i mean since i don't even look at the tough ones, they don't bother me much

trinity16684

[quote=Quagmire]I would have taken my chances. What I did was on intuition take x=1, which turns the eqn to a+b=2. Here only option b satisfies a+b=2.

Hi
IN PROB 1u took it 2 be1 , i think it will be 1.as product of roots =1( frm eq.)
sum of roots=4(coefficient of x^2)
so each root wil b 1
is dis approach correct

warrior

Quote:

Originally Posted by ernil

err... can i post some easy {by yu guys standards that is..} aimcat probs here? i mean since i don't even look at the tough ones, they don't bother me much

mate your wish. If you think that question is worth revising at later that and it teaches a somehwhat different or new concept then post it.

In short: agar lage question thoda hat ke hai then post it.

swethcat

despite questions... it also acceptable if anyone provide us the paper...
not here but can upload somewhere and provide us the link....

ernil

Quote:

Originally Posted by swethcat

despite questions... it also acceptable if anyone provide us the paper...
not here but can upload somewhere and provide us the link....

dude, they happen to be copy writed stuff
@ warrior. sure thing

catfever

how many 0s are there in 102! somebody please help...how to proceed??

mave

Quote:

Originally Posted by catfever

how many 0s are there in 102! somebody please help...how to proceed??

no of 0s = no of 5s in 102! = 24.

ajay_citm

Quote:

Originally Posted by warrior

Aimcat 717 was easy and it dosen't have may questions to include here but let start with this one

Problem #1: If the equation has x^4-4*x^3+a*x^2+b*x+1 = 0 has four positive roots, the ordered pair (a,b) is

(a) (-4,6) (b) (6,-4) (c) (-2,-3) (d) (-3,2)

lets assume four roots to be x,y,z,p
so x+y+z+p=4
and xyzp=1
also xy+xz+xp+yz+yp+zp=a
xyz+xyp+xzp+yzp=-b
solving only solution for x,y,z,p is
x=y=z=p=1
so a=6 and b=-4
ciao

TheCat

Quote:

Originally Posted by ajay_citm

lets assume four roots to be x,y,z,p
so x+y+z+p=4
and xyzp=1
also xy+xz+xp+yz+yp+zp=a
xyz+xyp+xzp+yzp=-b
solving only solution for x,y,z,p is
x=y=z=p=1
so a=6 and b=-4
ciao

Bhai mere..itna panga lene ka nanhi exam hall mein..if you apply Descarte's rule f sign...which says that if f(x) has N positive root, then f(x) will have N changes in sign.....this will make 6,-4 as the only option...