CAT 2007: QQAD, Practice Test Discussions

[gaurav.r.arya]

Originally Posted by gopi_siri

I think the answer is 2/9. As the expressions which need to be maximised and the constraint expression have equal weightage to p and q, p and q must be equal at the point of maxima. This gives me 2/9 when both p and q are 1/sqrt(3).

take pq=.5 ........
you'll get 1/4 as the answer which is greater than 2/9.

[Aarav]

Originally Posted by gaurav.r.arya

take pq=.5 ........
you'll get 1/4 as the answer which is greater than 2/9.

The doubt keeps occuring on this question.

Let's have step by step solution to this now.

p^2+pq+q^2 = 1, but p^2+q^2 >= 2pq => pq <= 1/3.
We need to maximize pq(1-pq) or 1/4*(1-(1-2pq)^2), thus pq has to be max i.e. pq=1/3 and pq(1-pq) = 2/9.

1/4 comes when you ignore the fact in bold and get imaginary p and q.

[Aarav]

Originally Posted by gaurav.r.arya

Q) In how many ways can 8 persons sit at a round dining table in such a manner that all will not have the same neighbours in any 2 arrangements?

ans> i'm getting 2880... out of 5040 possible cases.

Quite sure its wrong!!!

AARAV kindly enlighten me on how you would approach this......
Probability, perms and combs..... - > never really got a hang of it!!!

Thanks

Gaurav, post this on Combinatorics thread and let this remain for QQAD Test discussions only. BTW, 2880 is too large an answer. Check for smaller values n=4 and n=5.

[gopi_siri]

pq=0.5 and p=q doesn't satisfy the constraint expression. I couldn't find any values for p and q which can satisfy pq=0.5 and p2+q2=0.5. So, pq can't be 0.5.

[Aarav]

Originally Posted by MBA_007

just one request ,is baar please no errata,bahut irritation hoti hai nd time waste hota hia
i know u deliberately do so as CAT did last year but i want just for once neat and clean paper

I wish I were able to lie and say the erratas in the paper were caused deliberately. We promise a clean and flawless Practice Test 5.

[KOOL030]

Originally Posted by Aarav

The doubt keeps occuring on this question.

Let's have step by step solution to this now.

p^2+pq+q^2 = 1, but p^2+q^2 >= 2pq => pq <= 1/3.
We need to maximize pq(1-pq) or 1/4*(1-(1-2pq)^2), thus pq has to be max i.e. pq=1/3 and pq(1-pq) = 2/9.

1/4 comes when you ignore the fact in bold and get imaginary p and q.

aarav,i think the inequality in bold above has emerged from A.M>=G.M. taking values p^2 and q^2.
however, if we take A.M>=G.M for (p^2+q^2) and (pq) we have
p^2+q^2+pq/2>=sq. root[pq(p^2+q^2)]
=)1/2>=sq. root[pq(p^2+q^2)]
=)pq(p^2+q^2)<=1/4.
so max. is 1/4.

i think there are certain assumptions for using A.M>= G.M.please throw some light!!!

[Aarav]

Originally Posted by KOOL030

aarav,i think the inequality in bold above has emerged from A.M>=G.M. taking values p^2 and q^2.
however, if we take A.M>=G.M for (p^2+q^2) and (pq) we have
p^2+q^2+pq/2>=sq. root[pq(p^2+q^2)]
=)1/2>=sq. root[pq(p^2+q^2)]
=)pq(p^2+q^2)<=1/4.
so max. is 1/4.

i think there are certain assumptions for using A.M>= G.M.please throw some light!!!

When the numbers are positive then AM >= GM. The equality occurs when all the numbers are equal. So by your logic p^2 + q^2 = pq which is never true for any real p, q.
Caveat: Don't use AM/GM/HM rule blindly especially when the numbers are not independent. Here p^2 + q^2 and pq are dependent.

[chidvi]

HI guys where can i get this 4test papaer to download .. help me..

regards
chidvi

[Aarav]

Originally Posted by chidvi

HI guys where can i get this 4test papaer to download .. help me..

regards
chidvi

Post #499 on this thread.

[MITS@BITS]

Hello Aarav sir,
I have one doubt in usin AM>=GM, I did not find proper ans for this. In some kind of questions, there are no of variables---some are dependent & some are independent.So, how can we decide, where we use this and where not.
plzz give me reply