At this stage of the game, it is important to spend more time REVIEWING your concerns rather than plowing through new material. I would suggest to return to your previous work and thoroughly dissect the particular problem types that are still giving you difficulty. Focus not only on content, but als...

I also agree that it is extremely unlikely that this particular issue would arise on an official GMAT. As for mathematical definitions, a square is a particular type of rectangle (just as a rectangle is a particular type of parallelogram). Thus, the figure could be a square, even if the problem type...

For a number to equal any integer to the third power, it must contain prime factors raised to the power of three. For example, a prime number raised to the third power, such as 27, will only contain 3^3 in its prime box. As for non-prime integers raised to the third power, they can always be reduced...

Van D, Though you could consider dealing with prime factors, I personally would have chosen the plugging in method as well. First, I would rephrase the questions to What is the remainder when x^2 - 1 is divided by 24. For statement (1) try the first few odd integers: -When x = 1, r=0 When x=3, r=8 I...

To me, this is an overlapping sets problem best solved with a two-circle Venn Diagram. Circle #1 is first day attendants, circle #2 is second day attendants, the overlap is attendants for both days, and Neither is the space outside the diagram. The information from the original problem plus statemen...

I agree that, putting the statement together, the prime factors of 18 can be split two ways between the two prime boxes (which already contain 5*2): Either you have 3*3*2 in one (x = 180) and y remains the same (y=10), or you have 3*3 in one (x = 90) and 2 in the other (y = 20) However, don't forget...

I approached this as a VIC problems, so I decided to plug in values for x. I'm not certain that this is the BEST approach, but it is one approach. I chose the number 2 for x and thus -1 for 1-x. Then, I tested each answer choice. a. f(x) = 1-x a. f(2) = -1; f(-1) = 2. Eliminate a. b. f(x) = 1-x^2 b....

No problem! (1/5)^m = 1/5^m, and (1/4)^18 = 1/4^18 (since 1 raised to any power = 1) Additionally, 4^18 = (2^2)^18 = 2^36. Since both sides of the equation have 1 in the their numerators, to set them = we have to set the demoninators equal to each other. The denominator of the left side is now = 5^m...

Howdy. Answer choice A indicates that Joan, for 18 years, was the former chair; it is nonsensical to think that she would be attending meetings while she was the former chair. The sentence is attempting to indicate that Joan served for 18 years as the chair, and during this time she attended meeting...

To solve this problem quickly, I "broke" several established rules of plugging in numbers. When plugging in numbers on algebraic problems (VICs), it is advisable not to choose 1, 0, or the same number more than once. Doing so will not eliminate the correct answer; however, it will potentia...

The above explanation is correct. This problem offers great practice on both rephrasing Data Sufficiency questions and deducing the easier statement to begin your work. Rephrase: Is -1 < n < 1? Since statement (2) offers no informtion about n, begin with a BD/ACE grid and eliminate BD. Stateme...

A great debate! Just the type of discourse we like to see on our forums. First, I would definitely rephrase the question to Is -1< x <1 (with x not equal to zero)? Then, start with statement (2) and a BD/ACE grid. Statement 2 simply tells us that x < 0, so clearly this is not sufficient. El...

Good work all! Though the "trial and error" approach is a good default method, I am chirping with "chirpy" on this one. Better to learn those rules of exponents so you can use them in future problems!

Though this problem is well-explained in the OG guide, it does highlight an important word: "approximate." On the GMAT, whenever the test writers use the words "approximate" or "approximately" they are in essence telling you NOT to solve explicitly. Instead, you will be...

Overlapping sets is a great approach. I also have an even quicker formula should you really feel the need for speed. In one group that is broken down into two overlapping sets, the following formula applies. Total = Group 1 + Group 2 + Neither - Both In this problem, let's make the total 100. Thus, ...