coyote

Statement 1 : X > Y^2 Since y^2 > 0 for all Y, we can conclude that X > 0 We do not know whether X and Y are integers or fractions. Hence we cannot compare X and Y. Statement 2 : X and Y are positive integers. Obviously insufficient. Combining both, X > |Y| and X>0 Thus sufficient to see that X>Y IMHO it is C.

Merci, Bucky! I'll make it point to study the Princeton strategies well. :)

That's a fantastic score, buckykatt! Congratulations. Could you please tell us a little more about how you prepared for the verbal section? Did you take any special efforts for verbal before the 2nd test besides the practice tests?

25 and 26 do not have any common factors. So their LCM is 25*26 Hence the LCM of 2*25 and 2*26 is 25*26*2

:2cents: If M is a positive integer, for sqrt(M) > 25 we need M>25*25 Statement 1 : M is divisible by 50 M >= 50 Insufficient. Statement 2 : M is divisible by 52 M >= 52 Insufficient. Statement 1 + Statement 2 : M is divisible by 51 and 52. M >= LCM (50,52) M >= (2*25*26) i.e. M >= (25*52) Hence we can conclude that M > 25*25 IMHO it's C.

:2cents:IMHO it's B. 25^sqrt(2) = square of 5^sqrt(2) = [5^sqrt(2)]^2 = [5^(2*sqrt2)] = [5^2] ^ sqrt(2) Discussing other options, 5^2 = 25 [A = C] D. 25^2sqrt(2) = {[5^sqrt(2)]^2}^2 E. 5^sqrt(2)^2 = 5^[sqrt(2)^2] = 5 raised to square of sqrt(2)

The population increase is a GP. a,ar,ar^2.... where r=2 Hence on 6th day, population = 3*2^5 Since no.of days IMHO the answer is 96.

Of course we should. as Olm rightly said, 0 is divisible by any integer. Let's see it the other way, 1/2 = 0.5 = 0 + 5/10 = 0+(1/2) = (0+1) / 2 i.e remainder 1.

Welcome. Glad I could be of some help. :)

12^10 = (3*4)^10 Thus 12^10 = 3^10 * (2^2)^10 = 3^10 * 2^(10*2) = 3^10 * 2^20. [ (a^b)^c = a^(b*c) ] Since 3^10 will not contain any even factor , the maximum possible power of 2 in 12^10 will be 2^20.

Just a glimpse will tell you that 7/22 = 0.3xxx As we know 22/7 = pi = 3.142 Thus 22/7 - 7/22 = 3.142 - 0.3xxx The nearest integer is 3.

Choose 1 even integer from a set of 5 even integers --> 5C1 Choose 1 odd integer from a set of 5 odd integers --> 5C1 Total selections = 5C1 * 5C1 = 25

An invaluable piece of advice there. Heartiest Congratulations to one of the best contributors in TM! Good Luck! Keep us posted about your admission process!

A detailed analysis in the debrief! [clap] Applause for the same. Your score of course, was not unexpected. Great show mate! We'll miss you in TM! Best wishes for your applications!

:2cents:IMHO it's B Since the strip fits around the rug, length of strip = perimeter of rug. Say length and breadth of rug are L and B. 2(L+B) = 30 LB = 54 Solve to get L=9 and B=6 Thus col A = 9 Col B = 10