arjmen

Vote E Interior angle (in degrees) of a regular n-gon is given by the expression (n-2)*180/n (n-2)*180 will be divisible by n if: 180 is divisible by n; or in a few cases where n is even and (and consequently so is n-2) and (n-2)*180 is divisible by n although 180 is not divisible by n. 180 = (2^2)*(3^2)*5 Number of factors of 180 = (2+1)*(2+1)*(1+1) = 18 When n is even but not a factor of 180, then n can take [(3+1)*(2+1)*(1+1)] - [(2+1)*(2+1)*(1+1)] = 6 further values So the number of different values n can take for which the interior angle of an n-gon is an integer = 22 (since n cannot equal 1 or 2 - a polygon is a closed figure)

This is a partition problem with some interesting aspects: we have distinct individuals to be composed into un-named teams. Denote the 6 distinct students as: A,B,C,D,E and F A B C D E F Denote a partition with a "|". There will have to be 2 of these to partition the 6 individuals into 3 teams. The partitioning symbol "|" can be placed in any of 5 spaces (since a team needs atleast one member). So number of ways to partition into 3 un-named teams = 5C2 = 10 (Note: You shouldn't permute the individuals or you'd get similar/redundant team permutations since we have un-named teams here) The number of ways to permute 3 distinct things among 3 groups = 3P3 = 3! Total team-task permutations possible = 10*3! = 60

Was previously discussed here. The question is also better written on the linked post.

I see two issues here: 1. You seem to be seeing the table as a free circle (and hence (9-1)!/2, etc.). A table is however a fixed circle. 2. Subtracting (b) from (a) gives you a solution that also includes permutations where atleast 2 women (but not all 4) are seated beside each other; however, the question clearly expects you to preclude such permutations. On a general note, this question has been covered in the quant forums (GMAT and GRE) under various guises. And this page provides solutions to many similar questions including one with cats, rats and cockroaches!

You could also use alligation. Initially: a/o = (60-56)/(56-40) = 4/16 = 2/8 (notice that 2+8=10) Finally: a/o = (60-52)/(52-40) = 8/12 = 2/3 So Mary needs to lose 5 oranges (compare the denominators)

I agree with Chix. First seat the men --> 4! ways (I'm not trying to discriminate here - this just makes the solution simpler) Then arrange the 4 women in the 5 spots available --> 5P4 = 5! ways Total ways to seat = 4!5! Sruthi07, The solution would have been different in the linear case. Ways to seat 5 men and 4 women in a row = 5!*6P4

"Lattice Path" as addressed by Mathworld: http://mathworld.wolfram.com/LatticePath.html

Agreed with D. Alligation x/y = (5-2)/(10-5) = 3/5 OR x/(z-x) = 3/5 I: Suff II:Suff Haven't done this (GMAT quant) in a long while.

2 to 1 might not be ideal (not for us blokes anyway), but it's not as bad as my last degree. MEng Aeronautics, where the ratio was more like 10 to 1; a prime example of a sausage fest.

When you square both sides of the equation in statement 1, you have, (ly-xl + ly-zl)^2 = (z-x)^2 and NOT (y-x)^2 + (y-z)^2 = (z-x)^2

Thousand's - 3 choices (4, 5 and 6); 2 Even and 1 Odd Hundred's - 10 choices (0, 1, 2 ..., 9); 5 Even and 5 Odd Ten's - 10 choices (0, 1, 2 ..., 9); 5 Even and 5 Odd Unit's - 5 choices (0, 2, 4, 6, 8) Th-Hu-Te-U even-even-even-even ---> 2*4*3*2 = 48 even-even-odd-even ---> 2*4*5*3 = 120 even-odd-even-even ---> 2*5*4*3 = 120 even-odd-odd-even ----> 2*5*4*4 = 160 odd-even-even-even ----> 1*5*4*3 = 60 odd-even-odd-even -----> 1*5*4*4 = 80 odd-odd-even-even -----> 1*4*5*4 = 80 odd-odd-odd-even ------> 1*4*3*5 = 60 Sum = 728

Stem: For |x| I: x x > 1 or -1 Insuff II: x Insuff I&II: X Sufficient

It might be helpful if you could show us step-by-step how you reduce ly-xl+ly-zl=lz-xl to 2yz = 0. As I said before, I cannot see any reason why y or z or both need to be zero for the equation ly-xl+ly-zl=lz-xl to be true. E.g., x=1, y =2, z=3 ly-xl = 1 ly-zl = 1 lz-xl = 2

Why can you not answer this question using just the number line method? Why bother squaring both sides? The stem asks: Is "distance(xy) > distance(xz)" on the number line. Neither of the statements talk about the distances we want to know. Clearly, the answer is E. The answer would have been C had the 2nd statement been "x=0" And mukkimouse, can you please stop saying "The Panacea For Inequality Is To Square Them." Firstly, you're not squaring inequalities, you're squaring absolute values. Secondly, the panacea (if there is one) for solving absolute value problems is to understand the idea of absolute values, not just square them willy-nilly.

How do you get 2yz = 0 from ly-xl+ly-zl=lz-xl because I don't see anything in the latter equation that suggests that either y or z or both are zero?

You've probably heard of this: "The ratio of the areas of 2 similar triangles is equal to the ratio of the square of their corresponding sides." This is what I meant by "Area is proportional to the square of the sides", although I must say that is not precise.

1) The triangles are similar. Area is proportional to the square of the sides. Therefore, S = sqrt(2)*s 2) Let, Sqrt(9+sqrt(80)) = a Sqrt(9-sqrt(80)) = b a^2 = 9+sqrt(80) b^2 = 9-sqrt(80) 2ab = 2*sqrt(81-80) = 2 (a+b)^2 = 20

Previously discussed here

dhiru, although I agree with the OA, I do not agree with your explanation. If x > y^2 > z^4, x > y > z for say x = 100, y = 8, z = 2 z > y > x for say x = 1/4, y = 1/3, z = 1/2 x > z > y for say x = 1, y = -1/3, z = 1/2

The interest has to be calculated on a reducing balance. If monthly repayment = x At the end of the 3 month period, 1.1*[1.1*{1.1*(1000)-x}-x]-x = 0 => 3.31x = 1331 => x ~ 402

I: This equation can be stated as "On the number line, the sum of the distance between x and y, and the distance between y and z is equal to the distance between x and z." x Insuff II: Insuff I&II: x Insuff Hence E Had the stem said that the 3 numbers were all different, then C would've been correct.

Another solution would be: (8C1*6C1*4C1)/3! = 32

Most mathematicians consider zero even; the evenness of zero is not an idea relegated to GMAT land. By definition an even number is an integer of the form 2n, where n is an integer. Zero is therefore Even by definition.

Repeated post.

I'd love to know how you'd solve this question without assuming that the runners run at their respective constant speeds (or atleast that their respective average speeds are constant). The answer to the question as is cannot be 16,800m because that is the minimum distance run by Q before the three runners meet. P would have run a longer distance and S a shorter distance than Q. The only way the runners can meet if they're running at different speeds round a track is if the fastest runner laps the second fastest and the slowest at the same instant. The three must therefore have run different distances (the difference in the distances run have to be multiples of 2100m (the length of the track). Had the question asked "What is the distance run by Q before the three runners meet for the first time after starting?", then the answer would be 16,800m. But that is not what the question asks. In this question the speeds are such that the 3 will meet only at the start line everytime they meet.