this is much tougher to answer than to answer "should india persist with its fab four batsmen" [ if u didnt get it, dont fume over it...it reflects my bad taste at humour] considering that u r in for Fin. then it is indeed a bad score in Quants but when u say that ur UG results in maths pu...

Congrats!! 3 weeks and a 710, not bad huh!

I myself got a decent score and committed the same mistake..under- estimated it or say over-estimated myself...both are not good signs.

Anyways!! wishing you all the very best for ur future.
Amen!

Regards,
chetan

Thanks Ozzy, All, Here i'll just give a quick debrief about me so that when any of you feel to give ur advice, u have sm idea on my profile and thus, u cn clear it out a little as to what expectations are VALID for me: the score is a reality check ( but as 45 days have given me a decent score, it's ...

the reason for 2^x - 2^(x-2) written as 2^x ( 1- 1/4) is:
{ plugging the values }

2^4 - 2^2 = 16- 4 = 12 or
2^4 ( 1- 1/4) = 16 ( 3/4) = 12
or
2^5 - 2^3 = 2^5 ( 1- 1/4) = 32 (3/4) = 24

i hope, it is clear now

good luck!

if D^2 = 20 => L^2 + W^2 = 20
Also, L+W = 6 => L^2 + W^2 + 2L*W = 36 i.e. L*W = 8
so, we have L* ( 6 - L) = 8 or L=2/4

now, the question says clearly that L is the length, so for me,
L = 4 is the correct answer.

What is the OA?

answer is (A)
-------------------------------

from (1) => x/2 is an integer, thus, x is an even integer for sure.
from (2) => 2*x is an integer, but 2*0.5 is also an integer, as is 2*1, so with (2) we aren't sure if the number x is integer for sure.

thus, only first stmt is sufficient

(E) ? logic: Is xy > x^2y^2? (1) 14x^2 = 3 (2) y^2 = 1 ------------------------------- RHS => x^2y^2 = (x^2)^y^2 = (3/14)^sq(y) = 3/14 =.21 now, xy = x or -x (depending upon value of y) now RHS is always greater than -x i.e. LHS < RHS thus, LHS when, x =sqrt(3/14) =.41 i.e. LHS > RHS thus, answer is...

y(u-c) = 0 & j(u-k) =0 also, c<k => c-k <0 .....this is the hint here. but, the only link in 2 eqns is u, so we need to get a value for it. therefore, in y(u-c)=0 ,assuming that y!= 0 => u =c now => in j(u-k) = j(c-k)=0 => j =0 [my original answer :( ] also, y can as well be 0 (contradicting our...