I have been a GRE tutor since 2012. My methods enabled me to score a 170q/167v, and my students frequently break 320 overall and have scored as high as 169q. Below are some features which help define my approach to GRE tutoring. Also see my GMAT Tutoring page (the GRE and GMAT are pretty similar).
• Emphasis on how to think quantitatively: The math syllabus for the GRE is fairly limited; the main challenge is figuring out which concepts apply to a given question. I can help you build the necessary intuitions and problem-solving techniques.
• Plenty of Verbal advice: Though the Quant section of the GRE is more amenable to organized study, I often help students improve their Verbal performance as well. Many Verbal questions ultimately deal with logic, and I can help you understand the principles.
•Flexibility and alternative explanations: There are usually multiple ways to approach GRE questions, and I will work hard to find one that makes sense to you. A good GRE tutor should help you get the most out of the tools you already have—I always strive to do this.
Visit my testimonials page to get a better idea of what working with me is like.
Looking for a challenge? Below are a couple Quant questions and solutions I have composed. They give you some insight into how I approach GRE tutoring.
Quantity A Quantity B
x2 12x2 – 72x + 116
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.
Subtract Quantity A from Quantity B to get 11x2 – 72x + 116. Now use the quadratic formula to determine that this equation has two real solutions, meaning the parabola representing (Quantity B – Quantity A) crosses the x-axis. This means (Quantity B – Quantity A) can be positive, negative, or 0; hence, Quantity B can be greater than, less than, or equal to Quantity A. If you answered B, you may have simply chosen some values of x and found that Quantity B was larger for all of them. In fact, Quantity A is larger than Quantity B only in a very limited interval near x = 3, and you might not have known to test numbers around there. This shows a potential pitfall of the “choosing numbers” strategy.
A box contains ten marbles, identical except that each is labeled with a unique integer from 1 to 10. Allen will draw three distinct marbles at random from the set of ten and then return all three to the box. Then Bill will draw one marble at random from the set of ten, return it to the box; again draw a marble at random from the set of ten, again return it to the box; and once more draw a marble at random from the set of ten and return it to the box. (In short, Allen draws three marbles without replacement, and Bill draws three marbles with replacement.)
Let X = the probability that Allen draws at least one marble with an even number and Y = the probability that Bill draws at least one marble with an even number. What is X – Y? Give your answer to the nearest .0001.
show answerIn cases like this, it is easier to compute the probability of the complement and then subtract from 1. That is, we should compute the probability that Allen or Bill doesn’t ever draw an even-numbered marble and then subtract that probability from 1. For Allen, the probability that the first marble does not have an even number is 5/10. When he draws the second marble, there are 9 available marbles, of which 4 are not even-numbered. For his last draw, there are 8 marbles, 3 of which aren’t even-numbered. So Allen’s probability of never drawing an even-numbered marble is 5/10*4/9*3/8 = .0833, and X = 1 – .0833 = .9167. Now, when Bill draws his first marble, it has a 5/10 chance of not having an even number, same as with Allen. However, since Bill returns the marble before drawing again, all 10 marbles will still be available, and the probability of not getting an even-numbered one will remain 5/10. Hence, Bill’s probability of never getting an even-numbered marble in three draws is (5/10)^3 = .125, and Y = 1 – .125 = .875. X – Y = .9167 – .8750 = .0417