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factor sir. as ABCD is a different stock from DCBA, we need to calculate permutations not combinations. 4 letter permutations = 26 x 26 x 26 x 26 = 26^4 5 letter permutations = 26 x 26 x 26 x 26 x 26 = 26^5 total number of possibilities: 26^4 + 26^5 = (26^4)(1 + 26) =(26^4)(27) answer is C

sure. If x=a/3+b/3^2+c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT: A. 1/27 B. 1/9 C. 4/27 D. 2/9 E. 4/9 Rewrite the equation x = a/3 + b/9 + c/27 if it helps, further we could rewrite it as x = (1/3)a + (1/9)b + (1/27)c x = (9/27)a + (3/27)b + (1/27)c a, b, c can be either 0 or 1 starting with choice A: 1/27 possible? Yes, let c=1 and a=0, b=0 choice B 1/9 or 3/27 possible? Yes, let b=1 and a=0, c=0 choice C 4/27 possible? Yes, let b=1 and c=1 and a=0 choice D 2/9 or 6/27 possible? No possible way. choice E 4/9 or 12/27 possible? Yes, let a=1 and b=1 and c=0 Choice D is not possible. to get 6/27, clearly a cannot be 1. if b=1 and c=0, we end up with 3/27 if b=1 and c=1, we end up with 4/27 if b=0 and c=1, we end up with 1/27 As far as books for math theory, I would say Manhattan GMAT books are pretty good and all you will need (for quant at least). I have written about my gmat experience here While I have not used them personally, Veritas Prep and EZ Solutions books do get good reviews on Amazon so they may be worth checking out. best.

I don't think your example works here: sqrt = square root if sqrt(n) is an integer (statement 2) then, sqrt(n) x sqrt(n) = n integer x integer = integer therefore n is an integer if the square root of n is an integer, we are basically told that n is a perfect square. hope this helps. This is Question 169 in the Official Guide GMAT: Data Sufficiency 38 | Khan Academy For Khan Academy, this questions begins at 2:35 This is Question 146 in OG 11th Edition,

sure. OG 10th Q80 Jack is now 14 years older than Bill. If in 10 years Jack will be twice as old as Bill, how old will Jack be in 5 years? a 9 b 19 c 21 d 23 e 33 j = jack's age now b = bill's age now j = 14 + b (i.e. jack is 14 years older than bill) (j+10)=2(b+10) j+10 = 2b + 20 j = 2b + 10 substitute 14 + b = 2b + 10 b = 4 bill is now 4 jack is now 18 in 5 years, jack will be 23. The answer is D. Q92 (This is question 98 in the OG Quant Review 2nd Edition - Green Book) Which of the following CANNOT be the greatest common divisor of two positive integers x and y? a 1 b x c y d x-y e x+y a) Can 1 we be the greatest common divisor of two positive integers x and y? Yes. Suppose x and y are prime. (e.g. the gcd of 2 and 3 is 1) b) yes, suppose x = y or x = 2 and y = 4 c) yes, suppose x = y or y = 2 and x = 4 d) yes, suppose x = 4 and y = 2, the GCD is 2 and x-y = 2 All of these can be the GCD so we could stop here. e) A number cannot be divisible by a larger number. Could 5 be divisible by 17. No. 5 cannot be divisible by any number larger than 5. because x and y are positive, x + y > x and x + y > y y / (x+y) is x / (x+y) is The answer is E.

another way you can do this problem is to try and find counterexamples: n(n+1)(n+2) a) even only when n is even if n=3 3 x 4 x 5 = 60 n is odd and the product is even FALSE. b) even only when n is odd if n=2 2 x 3 x 4 is clearly even FALSE c) odd whenever n is odd from a, when n is odd the product is even FALSE d) divisible by 3 only when n is odd if n=6, 6 x 7 x 8 is divisible by 6 and so also must be divisible by 3 FALSE e) divisible by 4 whenever n is even Process of elimination, only choice left.

n(n+1)(n+2) is a fancy way of saying the product of three consecutive integers if n is odd, n+1 is even, and n+2 is odd n(n+1)(n+2) = odd x even x odd = even x odd = even if n is even, n+1 is odd, and n+2 is even n(n+1)(n+2) = even x odd x even = even (if you don't understand this, you should review number properties of odd's and evens) from above, n is even regardless if n is even or odd; thus A and B are false we can eliminate them. when n is odd, n(n+1)(n+2) is even so C is false. Now we have choices D and E left. choice D: any three consecutive positive integers will have one of the numbers divisible by 3. proof: when a positive integer is divided by 3, it has remainders of 0 (i.e. divisible by 3), 1, or 2 if n has a remainder of 0, it is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3 if n has a remainder of 1, n+1 has a remainder of 2, and n+2 has a remainder of 0. n+2 is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3 if n has a remainder of 2, n+1 has a remainder of 0, and n(n+1)(n+2) is divisible by 3 Eliminate D. We could pick by process of elimination but for the sake of completion: If n is even, n can be expressed as 2k, for k an integer if n = 2k then n + 2 is also even and n + 2= 2m for m an integer n(n+1)(n+2) = 2k(n)(2m) = 4knm here knm is an integer, so 4knm and consequently n(n+1)(n+2) is divisible by 4. enjoy! p.s. for more OG alternate explanations, click here

Powerscore sentence correction and critical reasoning are good. I have bought only one EZ Solutions books - a workbook - the book was fine but a lot of the questions did not have multiple choice answers which I didnt like. but the ez solutions do get very favorable reviews on amazon so they should be helpful for your studies i have blogged more about gmat resources here. best wishes.

Bobogee: These are hard questions so feel free to reply if something in these explanations are not clear. h(100) = 2 x 4 x 6 x 8 x .... x 98 x 100 h(100) = (2^50)(1 x 2 x 3 x 4 x .... x 49 x 50) by factoring out a 2 out of each number from above, it follows that h(100) is divisible by 1, 2, 3, 4, 5, 6, 7, ... , 49, 50. Since h(100) is divisible by all the numbers from 2 to 50, h(100) + 1 is not divisible by any of the numbers from to 2 to 50. In other words, since h(100) is divisible by all numbers from 2 to 50, we know that when h(100) + 1 is divided by any number from 2 to 50, the remainder is 1 (and hence not divisible) By law, every whole number must have a unique prime factorization. None of the numbers from 2 to 50 are factors of h(100) + 1. Thus h(100) + 1 must be prime (in which case its prime factor is greater than 40) or have a prime factor greater than 50 (in which case its prime factor is greater than 40) Thus, the answer is E. Question 2: Inversely proportional function y = k/x Directly proportional to the square function y = kx^2 where k is a constant If B increases by 100% (or doubles), the speed is halved. We need the reaction speed to double again. for x^2 to double so we need to multiply x by square root 2. rt 2 is approx 1.41..... so increase 40% note: i dont think this question is that good as it assumes k is equal in both equations.

Dear Mello: I have written about my experience taking the GMAT and what resources to use. I scored a 750 while studying for 6 weeks. They can be found here My advice is after you familiarize yourself with the GMAT Format, try a few practice questions and briefly review the topics. Soon after, take the practice test. I recommend taking the first practice test from GMATPrep software that is free on mba.com (these are from the people that make the gmat). Take the practice test (you can skip the essays) and from your score, see where to go from there.

@vgp: from the question, the coefficient on the x2 term is 1. Thus your second example does not work.

If you are looking for templates, I have written guides for both the AWA Argument and Issue Essays that walk you through both types of essays: They can be accessed here: Issue Template: How to Write an Analysis of Issue AWA | GMAT Hints Argument Template: How to Write an Analysis of Argument AWA | GMAT Hints Based on my experience (and I scored a 6 on the AWA), on the actual test day you will end up adjusting your template based on the actual prompt you are given that day.

To clarify my point about STATEMENT 2. Try to draw a parabola pointing upwards with the point (0, c) where c is a negative number that does not have one negative zero (i.e. root) and one positive zero (i.e. root). IMO, you cant do it. Therefore 2 is SUFFICIENT.

I think the answer is B. Here's why. First let's start with what we know: x^2 + bx + c = 0 This tells us the parabola is pointing upward. Since rs Think about it graphically: statement 1: b Proof by example: b is negative: (x-3)(x-4) = x^2 - 7x + 12, roots are 3 and 4, rs > 0 b is negative again: (x-3)(x+1) = x^2 - 2x - 3, roots are 3 and -1, rs statement 2: SUFFICIENT. c Think about it graphically: Whenever our parabola is below the x-axis. There is one zero on the left and one zero on the right. At x=0, we are on the y axis at some negative number (that is, c). Because we are below the x axis, there is one zero to the left and one zero to the right** **If you get what I am talking about, please explain it as I don't think I am explaining it properly.

The question has been discussed here .

I self-studied for the GMAT and scored a 750. In my opinion, the Official Guides are absolutely necessary for good GMATpreparation as they have best questions. But sometimes, the answer choices are not the best: As a result, I have written shortcut and alternative solutions for some of the questions in the OG 12th Edition book: Alternative Solutions for the OG 12the Edition I have also written shortcuts for some of the Quant 2nd Edition Book (Green Book). They can be found here. In my opinion, the 2nd Edition Quant Review book problems are not very hard. If you are scoring 80% or above there is really no point in getting this book. As far as Verbal: The OG questions and answers are both pretty good. The Verbal Review (Blue Workbook) is not easier like the quant one.