DWarrior

approximately 9.25million was distributed in September, and there are total 38 charitable organizations. So 9.25 mil divided among 38 will be somewhere between 200k and 300k.

a-b is the same as a+(-b) which is the same as a+(-1)(b) So 15 - (6-4)(-2) = 15 + (-1)(6+(-4))(-2) then distribute the -1 inside the parentheses: 15 + ((-1)*6+(-1)*(-4))(-2) = 15 + (-6+4)(-2) = 15 + (-2)(-2) EDIT: and yeah, see the recommended site above.

You're right, our sample space is restricted to only people who've done something, so it's basically P(HA|chose option), which is the value that = P(HA|yoga)+P(HA|meds). This would be a tricky question even for a probability course imo.

|a| Hence, |x+2| -y so -y+2 But clearly, -(y+2)=-y-2 so -(y+2) which we can rewrite in modulus notation: |x| But this doesn't tell us the relationship between |x| and y, since |x|=1, y=2 work, as does |x|=2, y=1 D

summation P(Ei)P(A|Ei ) for any i = 1,2,3,...n is the same as P(A) You can prove Baye's Theorem from: P(A and B) = P(A)*P(B|A) = P(B)*P(A|B) the theorem is then just algebra: P(A)*P(B|A) = P(B)*P(A|B) P(B|A) = P(B)*P(A|B)/P(A) The problem is with this part: Prob of person having heart attack = .4 P(Yoga or Meditaion/Reduce HA) = .3 => P(Yoga or Meditaion/has HA) = .7 The two events aren't "reduce HA" and "has HA". It's "doesn't have HA" and "has HA", the yoga reduces risk of having a heart attack by 30%, so yoga people will have a heart attack 70% as often as regular people, or .4*.7 The same thing for P(drugs | has HA) and the other parts.

I guess I'll spoiler this because it's pretty interesting

Only possible if both numbers chosen are odd (since even*anything=even), and you're picking with replacement (since numbers can be the same). The odd numbers between 1 and 20, inclusive, are {1,3,5,7,9,11,13,15,17,19}, 10 in all, and there is a total of 20 numbers to choose from (20-1+1). So the probability is: (10/20)*(10/20)=1/4

The question specifically says "two different numbers are chosen," so you can't get doubles. It wants P(a+b=8 | a=/=b) I suppose that means "with replacement" is not really possible since if you happen to redraw it, you'd just have to put it back in and go again.

Just came back from the test. I'm a Math major planning to do math in grad school, so verbal isn't of any significance afaik. People told me to just forget about it and do my best on test day. I didn't care at all what I got on the Verbal, but I read enough about it to know what question types to expect. I was positively surprised to find I got a 660. I didn't get the essay scores yet, but I barely finished the Issue one (no time to proof-read/improve) and didn't get to finish writing the introduction for my Analysis part. I'm guessing they'll be around 4-5. I'm generally slow to write essays, and I didn't bother practicing because I don't think math programs will care. I'm saddened by the Quant score though, as I expected to get 800. From what I've heard, math grad schools prefer their applicants get 800Q, but hopefully 790 will be high enough to not be an issue. I didn't study much for it, just did the problems and a couple of the math sections from the practice tests on PowerPrep, got 800 on both with about 10 minutes to spare both times, and I did some Kaplan Math Workbook problems. I had about 12 minutes left going into the final 4 problems of the Quant section, so I triple checked them all. Finished with 6 minutes left to spare. I probably should have double checked the problems, but they all seemed easy, and I didn't know how difficult they would keep getting so I didn't want to waste time. I guess I should have double checked all of them regardless, I probably made a couple of dumb mistakes that took me out of 800. It didn't help that I had an exam on special relativity an hour before the GRE that I had to study for yesterday. Anyway, I know the Math Subject GRE is the more important one for what I want to do, which I took a few weeks ago. I didn't get the score for it yet, but I know I didn't do well on it. :(

We're not subtracting one digit from itself, we're subtracting the first digit from the last digit. So the number could be 54321 and the difference will be 5-1=4, or the number could be 24321 and the difference will be 2-1=1. Same idea for divisibility by 5.

indeed, thanks.

http://img208.imageshack.us/img208/1560/myquestion.jpg Darkened circle is the correct answer. The question just asks to multiply data from the two graphs together. In 1980, households received average of $320 benefits in actual dollars, and there were ~4.1 million (~4,100,000) such households. That means the households in 1980 received a total of $1,312,000,000 in actual-dollar benefits, which is not close to the expected answer. It seems either PowerPrep or I have lapsed on a factor of 10 somewhere.

I just did the practice test containing this question. Yes, the OP did not post the question completely: http://img816.imageshack.us/img816/1885/yourquestion.jpg And then the answer is unambiguous. If you interpret it as "drawing without replacement, order matters", then the solution is posted by eoliv001 above, result is 1/6 If order doesn't matter, then you only have one combination {2,6} and 2C4=6 choices, so result is 1/6 still If you interpret it as "drawing with replacement", then your first pick has 2/4 choices, but for the second pick, you need the other specific number, and it must be different from your first choice (since the problem states the numbers picked are different) so you have 3 total choices for second. Hence, (2/4)*(1/3)=1/6 still. So no matter how you interpret it, the answer is 1/6. It's only if you miss the "different numbers" part that you start getting wrong answers.

C is correct assuming B reads "Number of different triangles that can be drawn taking any three points as three fixed vertex points." otherwise the question is nonsensical.

You can rule out 2 almost immediately because it relies on last digit of the difference being even, so that would imply (first digit) - (last digit) = even, which does not have to be the case. This fact rules out 4 and 6. You can similarly rule out 5, since it also relies on the last digit of the difference being 5 or 0. This leaves 9 as the only answer. In fact, if you want to prove it, represent the 5 digit number as abcde where each letter is between 0 and 9, inclusive (well, a isn't 0) Then its reflection is edcba, and their difference is: 10000a+1000b+100c+10d+e-(10000e+1000d+100c+10b+a) group terms by letter: (10000-1)a+(1000-10)b+(100-100)c+(10-1000)d+(1-10000)e 9999a+990b+0-990d-9999e 9*(1111a+110b-110d-1111e) technically this would all be in absolute value, but the fact that it's divisible by 9 remains the same.

Agree on Q2, 7

Do the modular arithmetic questions imply positive integers by default? Because if negatives are allowed, then yeah you have an infinite number of counterexamples, like -6*8+1 btw, -25 actually leaves remainder -1 when divided by 2, 4, 6, or 8 (which corresponds to 1, 3, 5, 7 respectively). You'd want -23

My solution is the same as above: The outer diameter is 33+4, since it's 2cm more on each end, so outer radius radius is 37/2 cm. The inner radius is 33/2 cm Then just subtract volume of larger from the smaller: 70*pi*(37/2)^2 - 70*pi*(33/2)^2=(37^2-33^2)*70pi/4

This problem has come up before, here is how I solved it: 100210*90021 = (100000+210)*(90000+21) = big number + 21*100000 + 210*90000 + small number = big number + small number + 21*(100000+900000) = big number + small number + 21*(1,000,000) 100021*90210 = (100000+21)*(90000+210) = big number + 210*100000 + 21*90000 + small number = big number + small number + 21*(1000000+90000) = big number + small number + 21*(1,090,000) where: big number = 100000*90000 small number = 21*210 This just exploits (a+b)*(c+d)=ac+ad+bc+bd. It's somewhat cumbersome to write out, but fairly easy to do on paper as you don't have to write most of it down. The idea is that both "big number" and "small number" will show up in both products, so you can just omit them.

Let me know if this is of any help: Definition: P(A|B) means "probability of A occurring, given that B has occurred." It is read as "the conditional probability of A, given B" For example, you roll a 6-sided die and you got an odd number, what's the probability that you rolled a 5? In this notation, P(it's a 5 | it's an odd number). There are 3 odd numbers possible: {1,3,5} and only one of them is a 5, so P(5|odd)=1/3. side remark: The following trivialities should be immediate: 1. P(A|A)=1 because if A has occurred, A obviously occurred. 2. If P(B)=1, then P(A|B)=P(A), since the probability of A occurring given that anything has occurred doesn't really give us any more info and so probability of A occurring remains the same. For example, probability that a student is a female given a student is a person; knowing that our student is a person doesn't give us any more information whatsoever, since all students are people (disregarding comic absurdity) 3. If P(B)=0, P(A|B) is undefined. This is because if B never occurs, it doesn't make sense to speak of what happens once B occurs. If you deem B to be impossible, and it happens, that means your original estimate for probability of B was incorrect and so everything else is nonsensical. In general, P(A and B) = P(A)*P(B|A), Which says that the probability of having A and B occur together is the same as looking how often A occurs and once A occurs (aka, "given A") how often B occurs. Hence, you multiply "how often A occurs" and "how often B occurs given A has occurred". side remark: P(A and B) = P(A)*P(B|A) = P(B)*P(A|B). So you can look at it as A occurs and once it has occurred, how often B occurs. And equivalently, how often B occurs and once B has occurred, how often A will occur. To continue with the die rolling example, what's the probability of rolling a 5 and an odd number on a single roll? This should be pretty obvious intuitively, if you roll a 5 then you've obviously rolled an odd number. In math terms, P(rolling 5 and rolling odd number) = P(odd)*P(5|odd) = 1/2 * 1/3 = 1/6, which is the same as P(5), which is exactly what our intuition told us. Definition: Independence of A and B means P(A|B)=P(A) and P(B|A)=P(B) You should have the intuitive understanding that independence of two events means one event does not influence the other. That's exactly what the definition says. Given that B has occurred, it has no bearing on the probability of A occurring. Similarly, A occurring has no bearing on the probability of B occurring. The notation just summarizes that. Theorem: if A and B are independent, P(A and B)=P(A)*P(B) Proof: We know in general, P(A and B)=P(A)*P(B|A). But, since A and B are independent, P(B|A)=P(B). So, P(A and B)=P(A)*P(B|A)=P(A)*P(B) So you can only apply the easier formula when A and B are independent events, and now you should know why. In case of cards, suits and ranks are independent events. You can verify this by looking at their probabilities. P(Spade) = 13/52=1/4. P(Spade | it's an Ace)=1/4 (since there are 4 aces and only one of them is a spade). Similarly, P(Ace)=4/52=1/13. P(Ace | Spade)=1/13, since there are 13 spades each of different rank, and only 1 is an Ace. If this made sense, then you should be able to understand why P(sophomores and female)=1/4

It appears to imply "inclusive" by default. Sources: http://www.www.urch.com/forums/gmat-problem-solving/90662-average.html#post591541 GMAT Number Properties 2: Number Rules

I haven't taken the GRE myself yet, and I've only looked at one book, but what was the reason you didn't do well on the GRE? I think if your math is up to par, the difficulty comes primarily from having to work fast and not because the problems are difficult. It's probably better to do the easy stuff super fast than to know how to solve every problem but require a lot of time. Are you making sure you're doing the practices timed?

The first way that comes to mind: Since 24 is even, we can instantly rule out C Furthermore, it's always possible to pick X>24 that fits all that criteria (for example, 6*8+1), so it's also immediate that B is not an option It remains to see whether we can find a fitting number Look at X=1, it leaves remainder 1 when divided by any other natural number, so the answer is D If you miss X=1 you will get the wrong answer. I guess the trick is to not forget to look at the remainder itself when dealing with such remainder problems.

You're adding 2.1million each year for 2 years and want to know percentages. Let N be the sales in previous year, In 1996, this increase represents 2.1mil / N * 100 percent In 1997, this increase represents 2.1mil / (N+2.1mil) * 100 percent So in 1997, the numerator stays the same, but denominator increases, hence the percentage decreases. A

If you have a bunch of fractions, you will probably have other information available to help your sorting. For example, you can pretty much immediately group fractions into >.5, You can then compare fractions where either numerators or denominators are equal. Given equal denominator, higher numerator is greater (example, 7/9 > 5/9). But given equal numerator, lower denominator is greater (example, 5/8 > 5/9). Then look at fractions whose denominators are multiples of the others, for example, 5/12 and 11/24 are easy to compare. Also, adding equal amount to numerator and denominator increases the fraction. For example, compare 21/23 and 24/26. 24/26=(21+3)/(23+3), so we know 24/26 is greater. I think by this time you should have enough comparisons down to be able to use the cross-multiplying trick on remaining fractions. You can also use it for more than two numbers at a time if the numbers being multiplied aren't too big.