mxplusb

1. r^2-r kamakshi is correct, and that approach is very simple and elegant. I was initially thinking of it sueka's way: the whole rectangle's area is r(r+1), then you take out one row and one column, and add one for the square you took out twice. That gives you: r(r+1) - (r+1) - r +1 = r(r+1) - 2r = r^2+r-2r = r^2 - r. When you take off the outermost row and the outermost column, you're essentially taking one square out twice, the same way as you do when you take out an inner row and an inner column. 2. 1/8, or .125 It seems like everyone's good on this, but just in case: 1,000 is evenly divisible by 8, so out of the first 1,000 positive integers exactly one out of eight, or .125, are divisible by 8. Of course every number that is divisible by 8 is also divisible by 2, so that doesn't change anything. There is a certain amount of unnecessary information in these questions just put there to throw people off. 3. 1/6 This is a combinations question. Combinations = order matters, permutations = order doesn't matter. So with combinations, "2 and 6" and "6 and 2" are the same. (If you wanted to use permutations you could, but then the total number of possibilities would be 4P2=12 rather than 4C2=6.) 4. 100 I'm assuming

Agreed. Substituting real numbers is often a good approach -- don't do it in math class, except to check your work, but it works well on standardized tests.

Thank you. I try.

thepillow's explanation looks good to me. Thank you for that. I find that probability and combination/permutation problems are the most confusing problems on this sort of test. There are many different rules that can be applied, and it is often hard to figure out which one fits best. Personally I hate having to rely on an answer sheet to tell if I got the right or wrong answer -- I'd much rather be able to double-check myself. Of course on standardized tests there isn't always time. But when you're studying there is -- either you can allow yourself all the time you need to get the problem right (remembering that practicing doing a problem wrong is worse than not practicing it at all) or you can go through a practice test once while timing yourself, and then go back to the problems you're not sure of and take your time on the second untimed round, before looking up the answers. Permutation and combination formulas are shortcuts. Many problems you can test out long-hand, by actually writing down all the possibilities, and on the ones you can't you can often make a smaller version of the same problem and test that one. This way you don't have to take the strategies on trust. You can use them because you know they work, you've tested it yourself. I was thinking about this, because I've noticed the way I do test problems is often different than the way a textbook would say to do them. Textbooks tend to focus on memorizing formulas and strategies, then figuring out which one to use. I tend to not lean on formulas unless I'm very sure about them, either because I've used them so much that they're second nature or because I know how to derive them. I also rely on estimation and common sense a lot -- not relevant for this problem, but some problems will have one or a few answers that are clearly impossible. So I can cross those off without even checking.

It is amazing how quickly math knowledge can fade, if you didn't really drill it in when you first learned it. I would recommend spending a LOT of time on practice problems. The more you practice, the more it will be automatic on the test and the less time you will spend second-guessing yourself. Personally -- and your mileage may vary -- I almost never "study" formulas or attempt to memorize them. I do problems. I believe that if you do a lot of problems, with the formulas written down somewhere you can check them easily, you will probably not have to deliberately memorize them. Instead you will automatically learn the formulas by heart just from using them so much. At any rate, I've gotten A's in quite a lot of math classes with that approach.

Ah! Thank you for looking that up.

That's beautiful. I never would have looked at it that way. Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the digits are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56. In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed. And I would have gotten the correct answer very quickly, if I hadn't forgotten about the 0122 combinations. Drat.

I will not give a rating because I am not sure of the rubric used. But my first impressions are: solid grammar and use of vocabulary; the arguments are clearly stated and there is a good flow that is easily followed. Also the counter-arguement is acknowledged appropriately. Some potential improvements: The fourth paragraph, which addresses counterarguments, would be better placed either as the second paragraph or as next to last, so it doesn't break up the flow of the arguments that you are making. The second paragraph starts out very clear and strong, but the last sentence is confusing, in part because the word "position" is used too many times. (Perhaps "field" would be an appropriate substitute for the middle one.) Consider dedicating a full sentence in the opening paragraph to each of your three points, rather than combining two into one sentence. The word "initially" in the last sentence is, I think, inappropriate: you're talking a problem that is likely to get worse over time, rather than one that is likely to be most acute immediately after graduation. "Actually" or "in fact" would fit better. Also consider whether your main premise is better stated by your first sentence or your final one -- they are not quite the same statement. The first statement leaves unclear whether the availability of jobs should be considered at all. If your premise is that it should only be one factor of several, make it clear in the first paragraph. It is also not clear whether you think that such a factor is likely to "bolster the prospect of future employment" (but other concerns like quality of life are important too) or that the possibility of a changing job market cancels out that potential advantage, making that perception largely illusory. It seems like you're trying to use as many intermediate to advanced vocabulary words as possible; generally you use them accurately, but you may wish to use the principle that a little of that goes a long way, and tone it down a bit. (Then again that may just be my own bias speaking: I work in education, where I typically have to state concepts in a way that 10-year-olds will understand. Of course GRE applications require a different emphasis.) I think your essay has a lot going for it as it stands, and that those changes could make it an outstanding essay. Again, you've already got good grammar, good spelling, a good vocabulary, and a mostly clear, readable structure.

I like the tone of this essay -- it sounds like you're writing about something you care a lot about personally. The structure is solid: your thesis is clearly stated at the beginning and repeated in the conclusion, and each example is given an equally-weighted paragraph. The writing is also concise, without unnecessary phrases or sentences.

... unless you can use each digit more than once. In which case I will have to throw in my support for 60. OK, clearly I misunderstood the problem.

If I am understanding the question correctly, there are five digits to work with: 0,2,5,7,9 There are 5P3 = 5*4*3 = 60 ways to arrange these five digits into groups of three. One-fifth of these will start with a 0, so they are not proper three digit numbers. That leaves 48 combinations. Two-fifths of the original 60 combinations end in a 0 or 2, making the number even. Three of these combinations -- 052, 072 and 092 -- have already been eliminated for starting with 0. Take out two fifths of 60, or 24, from 48, then add back the 3 that got double-counted. That leaves 27. (I was initially skeptical of this answer, since it's not given as an option, but I have checked it in multiple ways including in the end writing out all possible combinations, so I am quite certain it is correct. Unless the question was copied wrong or I misunderstood.)

... can you come up with an example of how this might come up in a question? I feel like this question should be obvious -- but suddenly I'm unsure whether to say "the prime factorization of 1 is 1" or "1 doesn't have a prime factorization, because it's just different that way." Anyone else have any thoughts?

"further more there are only two types of right triangles one is 30-60-90 and the other is 45-45- 90" Completely wrong. As a quick example, take a piece of paper. Each corner is a right angle. Fold one corner over. Now you have a right triangle, with the folded edge as the hypotenuse. Are there only two triangles you can fold this way? Hardly.

Pythagorean equation: a^2+b^2=c^2 All the outer triangles are congruent (this is easily provable by geometry if you're not willing to just eyeball it) with one outer side 2 and one outer side 1. So the hypotenuse -- which is also the side of the inner square -- is root(2^2+1^2) = root (5) The area of a square is the square of the side. So, (root(5))^2 So the area of the square is 5.

It is not a rule, just an observation. 32=8*2*2 Therefore, any number that is divisible by 32 must also be divisible by 8 so if it is not divisible by 8 it is not divisible by 32. Also dividing by 2 twice and then checking for divisibility by 8 is exactly equivalent to checking for divisibility by 32.

Actually it should be 3 and 4, respectively. I wrote it down wrong. My bad. If you divide the exponent by 4 it'll always work. For numbers with a 2-part cycle you can just divide by 2, and for numbers that always end in the same digit (anything ending in 0,1, 5, or 6) it always ends in that digit and you don't need to divide anything. I'm going to try explaining it a slightly different way. (Again, you are unlikely to need to know any of this for the test. This is just for fun.) There is a cycle that these numbers go through. 3^1 = 3 ==> units digit is 3 3^2 = 9 ==> units digit is 9 3^3 = 27 ==> units digit is 7 3^4 = 81 ==> units digit is 1 3^5 = 243 ==> units digit is 3 3^6 =729 ==> units digit is 9 and so on. The units digit will always go through this cycle. The shortcut is to look at the exponent. Find the remainder when the exponent is divided by 4. If the remainder is 1, the final number's units digit will be the same as the original number. If the remainder is 2, the final number's units digit will be the same as that of the square of the original number. If the remainder is 3, the final number's units digit will be the same as that of the original number raised to the 3rd power If the remainder is 4, the final number's units digit will be the same as that of the original number raised to the 4th power. When you divide 437 by 4, you get a remainder of 1. That is why the unit's digit of any number raised to the 437th power will be exactly the same as the units digit of the original number. 3^437 has a units digit of 3. 4^437 has a units digit of 4. I got this from a book I read on number theory, but I've also discovered a lot of cool things just by playing with a calculator or pencil and paper and looking for patterns. There's also some cool related information here: Modular Arithmetic — An Introduction

Thank you. I am glad I could help. :peach:

If I understand the second problem correctly, the answer is the earnings per share for the fourth quarter, plus the earnings per share for the first nine months if there had been 10,000,000 shares. There weren't, there were 5,000,000 shares, which each earned $7.20 per share. If earnings in the first three-quarters of the year was $7.2p per share, total earnings must have been $7.20*5 = $36 million. If there had been 10 million shares, (divide by 10 million) it would have been $3.60 per share. Adding $3.60 to $1.25 you get © $4.80. I may be misinterpreting.

For 32: well, you can first quickly check if the number is divisible by 8, because if it's not then it definitely isn't divisible by 32. If it is, well, I find dividing a number by 2 is relatively easy. I would do that twice, then check to see whether the resulting number is divisible by 8.

There is no reason to know the % change from 2006 to 2007. That's just there to throw you off. Adjusting abhishek1787's approach, Consider in 2007 dollar amount sale was 100. 2008-->8% decrease so 100 becomes 92 But you don't want the percentage of 2008 sales relative to 2007, you want the 2007 sales relative to 2008. So you take 100/92 =108.7

More information! In general, when you raise a number to a power, there's a two-part four-part cycle that the units digits go through. Except for 0,1,5, and 6, which always have the same units digit at the end. So all you need to know is where in the cycle you are. 2: 2,4,8,6 (That is, for powers of two, the units digit is 2 for 2^1, 4 for 2^2, 8 for 2^3, and 6 for 2^4. After that the cycle will start over again at 2.) 3: 3,9,7,1 4: 4,6 7: 7,9,3,1 8: 8,4,2,6 9: 9,1 So, if you have 2^437, say, you can easily figure out the units digit. If the exponent were evenly divisible by 4, the unit's digit would be the fourth number of the cycle, or 6. For remainder 3, it's the third number. For remainder 2 it's the second number. For remainder 1, the first. 436 is a multiple of 4, so 437 has a remainder of 1. Unit's digit is 2. 3^437 has a units digit of 1, and 4^437 has a unit's digit of 6, and so on. Sadly, the other digits are not so straightforward. I couldn't even tell you how many digits a number that high actually has. I doubt you will have to know any of this for the test. But it's kinda cool. There's also cool things about square numbers: for instance, square numbers can only ever end in 0,1, 4,5,6, or 9. Numbers to the fourth power can only end in 0,1, 5, or 6. So if you see a really big number ending in, say, 7, you automatically know it's not a perfect square, even if you know nothing else about it. And if it ends in 4 or 9, you know it can be a perfect square but not an integer to the fourth power. There are also patterns about which two digits a number can end in and be a perfect square (eg, numbers ending in 96 can be a perfect square, but numbers ending in 46 never are.)

The probability of what? For any given situation, all probabilities must add up to 1. (Or 100%, if you're using percentages.) P(A and B) = P(A)*P(B), assuming independence P(A or B) = P(A) + P(B) - P(A and B) ~P(A) = 1 - P(A) Seriously, though, don't take a formula-based approach. That'll mess you up, because it's too easy to use the wrong formula if you don't know what's going on. Study practice problems instead, so you can recognize the type of problem you're looking at. I did a short search for a website or PDF with what you're looking for, but all I found were formulas that were far more advanced than what you need. This page here has a very short explanation of four types of probability problems and what approach to use for each (it's geared at the GMAT, but I think the GRE is similar enough for the tips to be valid.) See if it helps: GMAT Hacks: GMAT Probability In Perspective

Oo, am I the first one to answer this? Solving the hard way: I'm going to get all algebraic here, because that's the way I roll. m=6k + 2, where k is an integer n= 6j +3, where j is an integer (This is just the definition of remainder. I believe it intuitively makes sense as well.) m-n = 6k + 2 - (6j +3) = 6(k-j) -1 k and j are both integers, so k-j is an integer too. (Also greater than zero, but I don't think this is actually relevant.) Now, we're looking for something in the form 6 * (an integer) + (a number between 0 and 5). -1 doesn't cut it. But, what if you do this: = 6(k-j) -1 +6 -6 Then you can get = 6(k-j-1)+5 (note, k-j-1 is an integer since k, j, and 1 are all integers.) And your remainder is 5. (This would still hold if we didn't know m>n, because negative numbers have remainders according to the same formula. -7/6 is -2 remainder 5.) Easy alternative approach: Of course, since "cannot be determined" is not an answer, you can also just substitute numbers with the appropriate remainders for m and n, such as 8 for m and 3 for n. I hope this helps!

I was going to explain how I solved the problem, then I realized Fahad Ali did it better. (And arq's approach is basically the same, just without the table.) Samuel, if you are concerned about the time it will take to solve a problem, skip ahead and come back to it later if you have time, or see if you can eliminate unlikely answers and take a guess. You're absolutely right that it's important not to spend too long on any one problem if it means you don't get around to answering other ones. But if there is a shorter way to solve this particular problem, I don't see it. The numbers seem very improbable: hardly any of the under-20 students were susceptible (less than 2%), but 20% of the over-20 students were? Seems unlikely. Oh well, I guess the point is showing you can think abstractly, not showing that you have any common sense in regards to numbers. Sigh. chrismagoosh, I believe your interpretation of the numbers on the first problem is incorrect. 40%= (susceptible and over 20)/(total susceptible), not (susceptible and over 20)/(total students). (As far as I can tell, your responses to the other 3 questions are both accurate and concisely explained, so thank you for that. Personally I strive for a high level of accuracy, but I tend to get ridiculously long-winded in my explanations.)

Generally if there is not reason to believe the events in a probability problem are dependent, you can assume they are independent. . Independent means that the outcome of one does not affect the outcome of the other. Two subsequent coin flips are independent. But, if you're drawing from a deck of cards (or pulling marbles out of a jar or socks from a drawer or whatever) and not putting them back in between, the events are not independent. For instance if you draw a card from a full deck it has a 1/2 chance of being red, a 1/4 chance of being a heart, and a 1/52 chance of being the King of Hearts. But if it is the King of Hearts and you draw again, the next card has a 25/51 chance of being red, a 12/51 chance of being a heart, and a zero chance of being the King of Hearts. The second drawing is dependent on the outcome of the first. Probability formulas for independent events: Both: P(A and B) = P(A)*P(B) Either: P(A or B)= P(A) + P(B) - P(A)*P(B) You can also double-check by realizing P(A or B) = 1 - (1-P(A))*(1-P(B)) -- in plain English, the probability that either happens is 1 minus the probability that both don't happen. That's actually the way I solved the problem, even though it looks more complicated. Basically, lacosteaef is spot on. arq is confusing independent and mutually exclusive, I think. (Mutually exclusive means they can't both happen, like one card being a spade and the King of Hearts at the same time.) Thus possibly demonstrating the general Internet principle that the people who seem most sure of themselves are often the least accurate ;)