Articles published in November 2013

Free GRE Events This Week: December 1- December 7

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Here are the free GRE events we’re holding this week (All times local unless otherwise specified):

Remember, our Black Friday special ends December 7th. Receive $150 off all December GRE classes with the code BLACKFRIDAY150.

free

12/3/13– New York, NY- Free Trial Class- 6:30PM-9:30PM

12/7/13– New York, NY- Free Trial Class- 10:00AM – 1:00PM

Looking for more free events? Check out our Free Events Listing Page.

Manhattan Prep’s Black Friday Deals

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BlackFriday150GRE

Happy Black Friday! In case you’re too full of turkey and stuffing to make your way out to the stores today, we’re serving up something extra special. Today through December 7th, we’re offering $150 off all of our December GMATLSAT, and GRE courses*! This deal includes all Complete Courses– in-person as well as Live-Online. To receive this limited-time discount, register for a course that starts in December and enter the code “BLACKFRIDAY150” at checkout.

This is only the beginning of the holiday season, which means we have many more amazing things coming your way. Be sure to check back next week when we unlock our most student-focused holiday campaign. You can also follow us on Facebook and Twitter to keep up with everything happening at Manhattan Prep. Oh what fun this is going to be!

*Offer is valid for courses starting in the month of December only. Not valid for students currently registered for courses, or with any additional offers. Offer expires 12/07/2013 for GRE courses.

The Math Beast Challenge Problem of the Week – November 25, 2013

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Math Beast
Each week, we post a new GRE Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for two free Manhattan Prep GRE Strategy Guides.

If n is a positive integer greater than 1, then the function p(n) represents the product of all the prime numbers less than or equal to n. Which of the following is the second smallest prime factor of p(11) + 12?

 

To see this week’s answer choices and to submit your pick, visit our Challenge Problem page.

Manhattan Prep’s GRE Social Venture Scholar Program

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Manhattan Prep is offering special full tuition scholarships for up to 4 individuals per year (1 per quarter) who will be selected as part of Manhattan Prep’s GRE Social Venture Scholars program. This program provides the selected scholars with free admission into one of Manhattan Prep’s GRE live online Complete Courses (an $890 value).

These competitive scholarships are offered to individuals who (1) currently work full-time in an organization that promotes positive social change, (2) plan to use their master’s degree to work in a public, not-for-profit, or other venture with a social-change oriented mission, and (3) demonstrate clear financial need. The Social Venture Scholars can enroll in any live online preparation course taught by one of Manhattan Prep’s expert instructors within one year of winning the scholarship.

The deadline our next application period is December 6, 2013.

More details and how you can apply can be found here.

GRE Quantitative Comparison: Don’t Be a Zero, Be a Hero

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GRE-ZERO-HERO-BLOGWhen it comes to quantitative comparison questions, zero is a pretty important number, because it’s a weird number. It reacts differently from other numbers when placed in some of the situations. And zero isn’t the only weirdo out there.

Most of us equate “number” with “positive integer”, and for good reason. Most of the numbers we think about and use in daily life are positive integers. Most of our math rules were learned, at least at first, with positive integers.

The GRE knows this, and takes advantage of our assumption. That’s why it’s important to remember all the “other” numbers out there. In particular, when testing numbers to determine the possible values of a variable, there are a few categories of numbers you want to keep in mind.

If I’m going to think about picking numbers, I want to pick numbers that are as different as possible. I try to choose my numbers from a mixture of seven categories, which can be remembered with the word FROZEN:

FR: fractions (both positive and negative)
O: one and negative none
ZE: zero
N: negatives

So we’ve got positive and negative integers (the bigger the absolute value, the better), positive and negative one, positive and negative fractions, and zero. Don’t forget, zero is an integer too!

There are other categories of numbers to think about, particularly if they are mentioned in the problem: odd versus even, prime versus non-prime, etc. But the seven groups listed above account for most of the different ways that numbers behave when you “do math” to them. Because of that fact, picking numbers from different categories can be a fast way to understand the limits of a problem.

To illustrate my point, let’s think about the value of x raised to the power of y. What happens to the value of that expression as y gets bigger? Let’s simplify our lives even further by stipulating that y is a positive integer.

What first comes to mind is the idea that as we increase the value of the exponent, we increase the value of the expression. Well, if x is a positive integer, that’s true: the expression gets exponentially bigger as y increases. Unless x is the positive integer 1, in which case the expression stays the same size, regardless of the value of y. The same is true if x is equal to 0. If x is a positive proper fraction, the expression gets smaller as the value of y increases.
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The Math Beast Challenge Problem of the Week – November 18, 2013

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Math Beast
Each week, we post a new GRE Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for two free Manhattan Prep GRE Strategy Guides.

Orren drove from his home to work in the morning at x miles per hour, and returned home in the evening at y miles per hour. His average speed for the round trip was 60 miles per hour.

                     Quantity A                                                              Quantity B

                          x                                                                               30

  • Quantity A is greater.
  • Quantity B is greater.
  • The two quantities are equal.
  • The relationship cannot be determined from the information given.

 

To see this week’s answer choices and to submit your pick, visit our Challenge Problem page.

The Math Beast Challenge Problem of the Week – November 11, 2013

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Math Beast
Each week, we post a new GRE Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for two free Manhattan Prep GRE Strategy Guides.

gre challenge

  • Quantity A is greater.
  • Quantity B is greater.
  • The two quantities are equal.
  • The relationship cannot be determined from the information given.

 

To see this week’s answer choices and to submit your pick, visit our Challenge Problem page.

Free GRE Events This Week: November 11- November 17

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Here are the free GRE events we’re holding this week (All times local unless otherwise specified):

free

11/11/13– Online- Mondays with Jen– 7:00PM – 8:30PM (EDT)

11/17/13– Online- Free Trial Class- 2:00PM – 5:00PM

Looking for more free events? Check out our Free Events Listing Page.

The Math Beast Challenge Problem of the Week – November 4, 2013

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Math Beast
Each week, we post a new GRE Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for two free Manhattan Prep GRE Strategy Guides.

Positive integer n leaves a remainder of 2 after division by 6 and a remainder of 4 after division by 5. What is the remainder when n is divided by 30?

 

To see this week’s answer choices and to submit your pick, visit our Challenge Problem page.

Manhattan GRE’s Halloween Challenge Problem

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We hope everyone had a happy Halloween! Yesterday we asked our friends on our Manhattan GRE Facebook page to attempt this Trick-or-Treat Halloween Challenge Problem. As promised, today we are sharing the answer and explanation to the problem:

Halloween

Let’s use x for the number of bags produced by the original recipe, and for the weight of each of the bags. Given those variables, our first equation is simply xy = 600. We also need to create an equation that represents the new recipe. Since the number of bags produced has increased by 30, and the weight of each bag has decreased by 1, the new equation is (x + 30)(y – 1) = 600. Remember, the total weight is still 600 ounces. Foiling this equation yields xy – x + 30y – 30 = 600.

We now have two equations with two variables. There are several different paths we can go down here, but all involve substitution of one of the variables, and all will yield a quadratic. The simplest path is to recognize that since xy = 600, we can substitute for xy in the second equation to get 600 – x + 30y – 30 = 600. Subtracting the 600 from both sides, and adding an x to each side gives us 30y – 30 = x. We can now substitute for x in the first equation.

wordbeast

 

 
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