twohundredping

A key observation to make is that despite the way the figure is drawn, the missing angle (the one on the lower left) is 90 degrees. This is because all angles of the quadrilateral must sum to be 360. From there we can do as the others have said and split the figure into two right triangles that share a hypotenuse.

Check out this post on the remainder rule: http://www.www.urch.com/forums/gre-math/133792-please-help-q-3.html#post869216 Basically when we multiply two numbers a and b together and then find their remainder divided by n we get the same thing as multiplying the remainder of a/n times the remainder of b/n and taking the remainder of this product. In a formula, we can say Remainder(a * b / n) = Remainder( Remainder(a/n) * Remainder(b/n) ). But let's consider a slightly simpler formula instead since it is basically the same: Remainder(a * b / n) = Remainder(a/n) * Remainder(b/n) From the above we can get the following general result: Remainder(x^y / n) = Remainder(x / n) ^ y We know that the remainder of 10^2 / 11 is 1. So Remainder(10^32 / 11) = Remainder( 10^(2 * 16) / 11) = Remainder((10^2)^16 / 11) = [ Remainder(10^2 / 11) ] ^16 = 1^16 = 1

All we can say about the second problem is this: Abby and Brian must have started with at least 5 chips a piece. If not, then even if Abby had 4 chips and lost them all then Brian could only have 8 chips. Abby and Brian must have started with 11 chips or fewer a piece. If not then Brian would have more than 9 chips after Abby won two games. (If Brian won any games then he would end with more than 9 chips but we assume he won 0 games to determine an upper bound). One key point. Knowing that Abby won two games doesn't mean much to the problem. It just let's us determine the upper bound of 11 chips a piece. In general Abby could win two games but Brian could win his chips right back by winning 2 games of his own. So the starting number of chips is somewhere between 5 and 11 inclusive.

The two ways I would do problem 2: First way. This path could lead you into trouble and waste time in general but I figured it wouldn't take too long. Notice that the answer choices A and B are the only expressions containing x in the problem. So let's simply substitute A and B for those expressions. A / 4 - B/8 = B/4 - A/8 (now multiply both sides by 8) 2A - B = 2B - A 3A = 3B => A = B Second way: (2x+1) / 4 - (x-1) / 8 = (x-1) / 4 - (2x+1) / 8 Rearrange the equation to get... (2x+1) / 4 + (2x+1) / 8 = (x-1) / 4 + (x-1) / 8 At this point you may (or may not) notice that 2x + 1 and x-1 must be equal to one another. If you do not notice this then of course you can go on and solve this equation. However solving the equation is error-prone and potentially highly time consuming depending on how fast you work. I personally thought of the first approach first. It was a slight risk but it seemed logical to try the substitution and I figured it couldn't waste any more than 15-20 seconds.

You probably want to post these questions in the forum section for the math section of the general GRE. This forum is located here: GRE Math I'll answer 1 and 3 since I don't really understand 2. 1) Let x be the amount of calculators in the shipment. Since $300 on x calculators then the average cost of a calculator must be 300/x. He sells x-2 or these calculators for $5 more than their average cost, i.e. 300/x +5 His revenue (which doesn't include the cost of him purchasing the items) is $420 For this problem we can simply express his revenue as the product of the items sold and the price per item. Revenue = (number of items sold) * (price per item) 420 = (x - 2) * (300/x + 5) (now solve for x) 420x = (x - 2) * (300 + 5x) 420x = 300x + 5x^2 - 600 - 10x 0 = 5x^2 - 130x - 600 0 = x^2 - 26x - 120 0 = (x - 30) (x + 4) x =-4 or x =30 Clearly we must take the positive answer so x = 30. E. 3) 80 liter solution is 25 percent alcohol. Note that 25% = 1/4. So 1/4 of the solution is alcohold and 1 - 1/4 = 3/4 of the solution is water. Originally we have 80 * 1/4 = 20 liters alcohol and 80 * 3/4 = 60 liters water. Let us first consider making a 40 percent alcohol solution. We can simply figure out how much alcohol needs to be combined with 60 liters water to yield 40 percent alcohol. And we can express 40% as 2/5 Let x represent the amount of alcohol. (amount of alcohol) / (amount of solution) = percent alcohol x / (x + 60) = 2/5 5x = 2 * (x + 60) 5x = 2x + 120 3x = 120 x = 40 So if we combine 40 liters alcohol with 60 liters water we get a solution that is 40% alcohol. Since we originally had 20 liters alcohol, 40 - 20 = 20 liters alcohol must be added. So B is an answer. Now lets consider making a 60 percent alcohol solution. Again, we can simply figure out how much alcohol needs to be combined with 60 liters water to yield 60 percent alcohol. And we can express 60% as 3/5 Let x represent the amount of alcohol. (amount of alcohol) / (amount of solution) = percent alcohol x / (x + 60) = 3/5 5x = 3 * (x + 60) 5x = 3x + 180 2x = 180 x = 90 So if we combine 90 liters alcohol with 60 liters water we get a solution that is 60% alcohol. Since we originally had 20 liters alcohol, 90 - 20 = 70 liters alcohol must be added. So E is an answer. B and E are the two answers.

It is important to note that the question states that x and y are positive integers so you shouldn't (without proper reasoning) plug in 0 for x or y. Now because divisibility works the way it does, it doesn't matter if you plug in positive numbers, negative numbers, or 0. Anyway, I just came here to post a systematic approach. Start with the equation in the premise: 3x + 7y is divisible by 11. I will work this problem modulo 11. We can always add or subtract a multiple of 11 modulo 11 and get an equivalent expression. 3x + 7y = 0 (mod 11) (Now I want to eliminate the coefficient on x; alternatively, I could have done this for y) 4 * (3x + 7y) = 0 (mod 11) x + 6y = 0 (mod 11) Now the idea is to multiply this equation by the coefficient on x in each answer and see if I can get the same equation as in the answer. A) 4 *(x + 6y) = 4 * 0 (mod 11) 4x + 2y = 0 (mod 11) Since there is no way to manipulate 2y mod 11 to make it 6y mod 11 we can eliminate this answer choice. Alternatively, we could have eliminated this answer choice at the beginning since the 6y's matched and there is no way to turn the x mod 11 into 4x mod 11. B) x + 6y = 0 (mod 11) There is no way to manipulate 6y mod 11 to make it y + 5 mod 11 so we can eliminate this choice C) 9 * ( x + 6y) = 9 * 0 (mod 11) 9x + 10y = 0 (mod 11) There is no way to manipulate 10y mod 11 into 4y mod 11 D) 4 * (x + 6y) = 4 * 0 (mod 11) 4x + 2y = 0 (mod 11) (Now notice that the coefficient on y in the answer is negative) 4x + 2y - 11y = 0 (mod 11) 4x - 9y = 0 (mod 11) Since we started with the equation we knew to be true and obtained the same equation as D then we know that answer choice D is correct.

Yes that is exactly how I always think of it. A little note that I'm sure you are aware of though is that it is actually 4! * 2! . So if we had to arrange n people around a table such that one group of k people sits together there are (n-k+1)! * k! possible arrangements. And Mike its not the unique neighbors the question is asking for. It's each person being in a particular chair. So ABCDEF is different from BCDEFA.

The first powerprep test I took required me to know how many quarts were in a gallon. Actually though, I think you just needed to know that a quart was smaller than a gallon. On the actual GRE I took in July, I didn't have any units questions. Refer to post 4 in the following thread to see the only conversions I imagine you will have to know: http://www.www.urch.com/forums/gre-math/132301-conversion-factors.html

I'm pretty sure everything here is correct but I cannot guarantee it as I have not done normal distributions in awhile. The area between the 15th and 30th percentiles must be the same as the area between the 30th and 45th percentiles. Because of the shape of the curve, the left side of the curve specifically, the area between the 15th and 30th percentiles will accumulate slower than the area between the 30th and 45th percentiles. This is because the normal distribution is an increasing function on the left side of the curve. Therefore, the 30th percentile must correspond to an x-value greater than 450. Since the kth percentile corresponds with the x-value of 450, k must be less than 30. The powerprep explanation is given here in posts 19 and 20: http://www.www.urch.com/forums/gre-math/133932-what-do-you-think.html

A decent explanation I made back in the day. Not my best but its all I got. http://www.www.urch.com/forums/gre-math/132462-salary-new-gre-question.html It is important that you understand the concept of weighted means here as well. This is always true but I will show a particular example. Suppose we have the two numbers, one 3 and one 7. The average of the two numbers is 5. However, now consider that I have x 3's and y 7's. If x > y, then the mean will now be closer to 3 (i.e. less than 5) but no less than 3. This is because y 3's and y 7's would average out to 5. Since I still have more 3's left, the mean will get smaller. The mean cannot get any smaller than 3 even if I add an infinite amount of 3's. You can see this logically or by doing a infinite limit like in calculus. Alternatively, if y > x, then the mean will now be closer to 7 (i.e. greater than 5) but no more than 7. This is because x 3's and x 7's would average out to 5. Since I still have more 7's left, the mean will get larger. The mean cannot get any larger than 7 even if I add an infinite amount of 7's. You can see this logically or by doing a infinite limit like in calculus. Using the concept of weighted means in the original problem we can see that any average salary S such that 25000

First question answered here: http://www.www.urch.com/forums/gre-math/134741-points.html 2) Way to determine maximum number of groups: 3 + 2(7) + 3 = 20. So we can have 9 groups with 2 groups of 3 and 7 groups of 2. Basically here we want as many groups with 2 people as possible. Remember that at least one group must contain 3 people. So I subtracted multiples of 3 from 20 and used the greatest number divisible by 2 to determine the number of groups with 2 people. Way to determine minimum number of groups: 3 + 4(2) + 3(3) = 20. So we can have 6 groups with 4 groups of 3 and 2 groups of 4. Basically here we want as many groups with 4 people as possible. Remember that at least one group must contain 3 people. So I subtracted multiples of 3 from 20 and used the greatest number divisible by 4 to determine the number of groups with 4 people.

Since S is not on the line PQR, PQS will always be able to form a triangle. From the triangle rule, PS Your visualization is incorrect because from the conditions you established PQR is a right triangle but PSR doesn't have to be a right triangle. In fact, PSR will never be a right triangle unless PQ = QR. Also you may want to avoid this visualization because it is just one particular example (although this is a popular approach on many GRE problems). Here is a visualization of all the points S can be. Visualize or draw a line PQR. Now draw a circle with center Q and radius QR. Any point S on that circle and not on line PQR would satisfy the conditions of S not on the line and QS = QR. And lastly, I want to say that I don't discourage the one particular case approach in general. Sometimes your one particular case can be expanded for all cases but in this problem it isn't very useful.

The first approach I thought of was an inequality approach. Since P lies below line l, P satisfies y 3 We can quickly determine that the 4th answer choice is NEVER true. In case you are interested in proving every other case then here are the methods: Since P lies inside the first quadrant, rt > 0 Since P lies above line m, P satisfies y > 1/2x which gives t > 1/2r which rearranges to 2t > r Since P lies above line n, P satisfies y > -x + 1 which gives t > -r + 1 which rearranges to t + r > 1 Since P lies below line l, P satisfies y 3 Since 3r - 3 > t and r > 0, 4r > 3r - 3 > t

From the question given, I shouldn't have assumed that this was exponential growth. I think I arbitrarily chose exponential growth just because it made the problem more interesting. You are correct that this problem is supposed to have population x in 1990 and 1.30 * x in 1998. On the actual GRE, the determining factor that would have made this problem exponential growth is if it instead said "The population of a town is increases at a constant rate of 30% every year from 1990 to 1998."

The fact that you have a prime number of extra outcomes is inconsequential. This is because the valid approach uses addition and the invalid approach uses multiplication. If you want to go through exactly which cases are doubles to show that there are 53 invalid choices then I can help.