rezbipul

congratulations for reaching so far.

wow! a post after 4 years!!!

Yeah you are right.

Here is your solution: y must be one because y is the carry from the addition of the digits at tens position. That means z = 2. Now the result of the addition is 122. We get this equation: 2 * (10x+1) = 122; [xy means 10*x+1] => x = 3 So x+y+z = 3+1+2 = 6

http://www.www.urch.com/forums/gre-math/94632-math-problems-collection-post614911.html#post614911

Here are the collections of math problems collected from this forum (until September, 2007) on topics like probability, permutations/combinations, series etc. Math Problems Disclaimer: If you think they are an overkill of your time, don't blame me.

I have similar lists of problems on permutation-combination, percentage, probability, mixture etc. I was willing to donate these lists and other materials for people here. But I felt like I over-prepared for the actual test and didn't want to burden others.

Its too easy c You can verify by real data choose the first number divisible by both 5 and 7. Its 35 so it must be divisible by 35 and not 70.

Is it 8? if so it should be easy. if M is A's superset then M would be like {2,3,4} {2,34,6} {2,3,4,7} {2,3,4,6,7} etc i mean for 6,7 and something chosen from B, each one can be either present or absent.

Hi, The John Hopkins degree is MS IT not MS CS. JHU MSIT is nowhere near ambitious. It would be if it had been MS CS. You can easily get thru hopefully. Carnegie melon info sec institute (INI) give MS ISM (Information Sec Management) and it gives 23k$ award to almost everybody and it seems to have more waiting. I have seen people with lower profile getting easily into that program. There is another program in info sec (again not CS) in purdue. The institute is CERIAS. Not sure about funding but it is another place for info sec FYI.

HI Did you check the schools' requirement? Here is a useful link: CS Graduate Acceptance Estimator: List of Departments

a = 2b a+2 = 2b + 2 = 2 * (b+1) (a+2) = 2*(b+1)/2 as b is even b+1 is odd hence d is the right answer.

4. 0.66 for neither A 0.32 for neither B greatest intersection is 0.32 38. a = 64 d= 256 equal distance = (256-64)/3 = 64 hence h = 64+64*7 = 8*64 = 512 gcf question: the numbers could be 60 and 120 then gcf would be 60. among the choices 20 is the greatest divisor of 60.

p (neither heart, nor king) = 1 - p (either heart or king) p (either heart or king) = p (heart) + p (king) - p (heart and king) = 13/52 + 4/52 - 1/52 = 16/52 p (neither heart, nor king) = 1 - 16/52 = 36/52 = 9/13

P (king or heart or red) = p(k) + p(h) + p® - p(k and h) - p(h and r) - p(r and k) + p(k and h and r) = 4/52 + 13/52 + 26/52 - 1/52 - 13/52 - 2/52 + 1/52 = 28/52 = 7/13