goguma

There's a faster way to do it w/o calculating as much. Consider the 3 components of Column A: 1-(1/2)^2+(1/2)^4-(1/2)^8 Since 1 - (1/2)^2 = .75, substitute .75 back into the problem to get: .75 + (1/2)^4 - (1/2)^8 Since (1/2)^8 is a smaller number than (1/2)^4, subtracting (1/2)^8 from (1/2)^4 will result in a positive number to add to .75 Therefore, Column A (.75 + some number greater than 0) > Column B (.75)

Column A: 102 and 729 are both multiples of 3 (a number is a multiple of 3 if the sum of the digits is divisible by 3). Since the number of multiples of 3 between 102 and 720 is inclusive, it will be equal to (729-102)/3 + 1 Column B: Is B really 729 - 102/3? Or is it supposed to be (729-102)/3 ? If choice B is 729 - 102/3: 729 - 34 = 695 695 > (729-102)/3 + 1 B > A If choice B is (729-102)/3: (729-102)/3 B

Problem 1: Answer 6 The problem is looking for 3 consecutive integers (greater than 1) to multiply together. Any 3 consecutive integers will include at least one multiple of 2 (an even number) and at least one multiple of 3. Any number divisible by both the factors 2 & 3 is divisible by 6. Ex: 2*3* 4 3*4*5 4*5*6 6*7*8 7*8*9 8*9*10 10*11*12

The gist of it is the following: A right triangle has 3 sides. Of those 3 sides, two are legs (the sides opposite of the angles less than 90) and one side is the hypotenuse (the side opposite of the 90 degree angle). If the problem had stated the "largest side" it would have been referring to the hypotenuse. However, the problem stated "largest leg," indicating the leg opposite of the 60 degree angle (since the sides of a triangle are proportional to the angles of a triangle).

How I feel about my score: I'm happy. Not ecstatic, but happy. I'm looking at some of the top MPP and MIPP programs so I probably need to bump my quant score up a bit further, but this is more than satisfactory for my first attempt. I'm going to wait for the feedback from some of my past profs before I decide whether or not to take the GRE again. I think I could definitely bump up my score 50+ pts, but I'm not sure if it's worth the effort. I started studying about 3 months ago, averaging about 2.5 hrs/day. I hadn't taken any nitty gritty math classes since high school (the stats series that I took in college had more to do with distinguishing different tests than rigorous math), so I had a lot to review and relearn. Resources: (Out of the resources I have, I've probably only completed about 75% of the content overall) NOVA Math Bible: ESSENTIAL! Practice, practice, practice. Circle problems you get wrong, go back, and re-do them! Barrons: Nothing better for Verbal (especially vocabulary building). I had a pretty decent vocabulary to begin with, but I made it a point to try to learn the entire 3,500 list (I've got about 80% of it down). It helps to know the precise definition of words, rather than just a relative sense of meaning, when trying to create some of those bridges for analogies. GRE Big Book: Good source of practice. Test Magic: Excellent! A good source of problems for practice and explanations. Test magic was also a good way to keep me interested in the test. Whenever I felt like I was getting sick of studying I'd pop online and do a few problems. Kaplan GRE Advanced Math: A waste of time and money. Sure it's a source of practice problems, but the edition (09-10) I had was riddled with errors. Kaplan: Take it or leave it. PowerPrep: Good for the CAT practice, but you'll also see a number of the problems on test magic. XD 800 Score: I really liked their reading comprehension review section. It helped me get used to breaking down the passage, etc on a computer screen as opposed to on paper. If I decide to go for another round of the GRE, I'll probably spend a lot more of my time using this. Really difficult questions, and really good for time.

Nope, see :Mathwords: Leg of a Right Triangle

Column A: 8*z = 8z Column B: 1. Since the triangle is 30-60-90 (with ratios of x, x(sqrt3), 2x), and the largest leg is 6z, we know that it is the leg opposite of the 60 degree angle. Set 6z equal to the ratio of x(sqrt3). 6z = x(sqrt3) 6z/(sqrt3) = x 6z(sqrt3)/(sqrt3)(sqrt3) = x 6z(sqrt3)/3 = x 2z(sqrt3) = x 2. The perimeter of a 30-60-90 triangle = x + x(sqrt3) + 2x. 2z(sqrt3) + (largest leg) + 2[2z(sqrt3)] = perimeter 2z(sqrt3) + 6z + 4z(sqrt3) = perimeter 6z + 6z(sqrt3) = perimeter 3. Since (sqrt3) is approximately equal to 1.7, 6z + 6z(sqrt3) is greater than 8z. B > A NOTE: In a right triangle, the largest leg is not the hypotenuse. Also, keep in mind that a 30-60-90 right triangle has the ratios of 1:(sqrt3):2, and not 30:60:90.

Stilted also means "stiff or formal." It's 'unnatural' in the sense that it is affected (fake, pretentious, etc).

Answer: D The problem as it is stated gives you no reference as to what the probability that Linda is 85 years old is. Assuming that the probabilities are independent, if the probability that Linda will be 85 years old is 0, then .6*0 = 0, and B is greater. However, if the probability that Linda will be 85 years old is 1, then .6*1 =.6, and A is greater.

Answer A: A prodigal person by definition does not preserve. Vexed means discussed or debated at length, and sulk means to be moodily silent. Thus, a vexed person by definition does not sulk.

The question is asking you to determine by what % the new price must be decreased in order to match the old price Step 1: Determine the absolute difference between the old price and the new price Old Price = New Price - X Old Price = P New Price = P(1+.2) = 1.2P P = 1.2P - .2P Step 2: Determine what % X is of the new price x/(New Price) = .2/1.2 = .1666 = 16.66%

Hypotenuse = r(sqrt2) It's a right triangle with two equal legs (leg 1 = r, leg 2 = r) so you know that it's a 45,45,90 degree triangle with the ratios x,x,(x)(sqrt2). Substituting 'r' as the variable in place of 'x,' this gives us the side lengths of r,r,®(sqrt2) where the hyp is r(sqrt2).

First imagine that you have two circles, one inside the other. The flowerbed is the inner circle and the walkway is the remaining area. The information you are given is that there is a flowerbed (circle) with a radius r. Surrounding that flowerbed is a walkway with a width of r/2. So if you were to draw a line from the center of the flowerbed, to the outer edge of the walkway, it would equal r+(r/2). Effectively, r+(r/2) is the radius of the larger circle (flowerbed + walkway). We'll call this new radius R to avoid confusion. R = r+(r/2) The area of the larger circle (flowerbed + walkway) = (pi)R^2 = (pi)[r+(r/2)]^2 = (pi)(3r/2)^2 = (pi)(9/4)r^2

I would actually recommend doing the reverse. As someone who hadn't taken real math (a series of statistics 3 years ago discounted) since high school (approx 5 years ago), and I found the Nova Math Bible's explanations more than adequate. The trick is that you have to make sure to practice problems that you've gotten wrong repeatedly after reading the explanations. The book breaks up types of math problems into different sections and is quite effective in grounding you in GRE math. As for other resources, go through the scores posted forum and look at what the high scorers have written. Many of them post their study regimes, resources used, and ratings of those resources.

(area of the flower bed) = (pi)r^2 (area of the walk + flowerbed) = (pi)[r+(r/2)]^2 [note: the radius of the (area of the flowerbed + walk) = (radius of the flowerbed) + (width of the walk)] (area of the walk) = (area of the flower bed) - (area of the walk + flowerbed) (area of the walk) = (pi)[r+(r/2)]^2 - (pi)(r^2) = (pi)(3r/2)^2 - (pi)(r^2) = (pi)(9/4)(r^2) - (pi)(r^2) = (pi)(r^2)[(9/4) - 1] = (pi)(r^2)(5/4) Therefore: (pi)(r^2)(5/4) > (pi)(r^2) (area of the walk) > (area of the flowerbed) B > A