This week we have something special for our readers.
Whether you’ve seen us in class, in a free event, on this blog, or elsewhere, Kaplan always presents a unified front, as though we agree on everything. In fact, the smart, opinionated people who work for Kaplan disagree about lots of things, and today we’re offering a glimpse of our intellectual disputes.
Kaplan’s GRE bloggers Boris and Teresa take to their chairs to debate a common topic: does a GRE calculator make the test easier, or harder? Hear their arguments, pro and con, in the video below!
As always, let us know what you think in the comments below!
I hear this from a lot of you. Unfortunately, as I explained recently, having more time on the GRE wouldn’t actually help you get a higher score, since the GRE is a scaled test. So let’s leave the complaining to our competition, shall we? Instead of moaning about the clock, strive be as awesome as you can at solving problems. If you’re great, you’ll also be fast. Here’s a quantitative comparison that’s pretty simple, but also a nice illustratration of the fact that speed isn’t something that comes independently of problem solving skill.
This problem, like several I’ve been looking at recently, comes from our GRE Bootcamp event:
Quantity A: The sum of all integers from 9 to 29, inclusive
Quantity B: The sum of all integers from 12 to 30, inclusive
At a glance, the “math” way to solve this problem is time-consuming but direct: add up the sums in both columns, then compare. Since there’s an on-screen calculator on the GRE, some of your competition will solve the problem this way. And boy does it take a long time.
Let me be very clear: directly totaling both columns isn’t just a slow way to solve the problem. It’s a BAD way. Someone who solves the problem in this head-on, brute force fashion, then says to themselves, “I’m fine with the problems, it’s the timing that kills me,” is being dishonest with themselves. They are not ”fine with the problems.” They are very much unfine!
Instead, when you have to compare two quantities, start by eliminating what they have in common. If a quantity appears in both columns, then it isn’t helping either one to be bigger than the other.
Here, both columns include the range of numbers 12-29. Thus, totaling that range would be a waste of time. Ignore it and look instead at what’s different:
Quantity A: The sum of all integers from 9 to 11, inclusive
Quantity B: The of all integers from … never mind, it’s just 30!
And since 9 + 10 + 11 clearly equals 30, you can click choice (C) — “The two quantities are equal” — in under 10 seconds and score the point. That’s the beauty of the GRE: if you’re awesome, speed comes for free. Practice will get you there!
When a GRE quantitative problem features multiple ratios, many of you suffer headaches. This is because the “math” way of solving the problem is brutal, and students who don’t use logic will dive head-first into a morass of ugly substitutions, mistakenly assuming that the GRE is a math test. Here’s the kind of problem I’m talking about:
In a particular mixed candy bag, the ratio of Skittles to M&M’s is 4 to 5, while the ratio of Reese’s Pieces to M&M’s is 9 to 7. What is the ratio of Skittles to Reese’s Pieces?
The “math” way to do this problem is to set up two equations, solve one for M&M’s, and plug that value into the other one. If that sounds painful, that’s because it is. Don’t do this. Make a simple table instead:
S | M | R
4 : 5
7 : 9
Take a moment to confirm that you understand where the numbers above are coming from. They’re just a translation of the information in the word problem.
The question asks for the ratio of S to R. Can you just say it’s 4 to 9? No way. The value connecting them — the M — is different. It’s 5 in one ratio and 7 in the other. So, rewrite the ratios to make the M term the same in both, creating a kind of “bridge.”
Multiply the first ratio by 7: 7×(4:5) = 28:35
Multiply the second ratio by 5: 5×(7:9) = 35:45
Next, check out your new table:
S | M | R
28 : 35
35 : 45
Now you can just “walk across the bridge,” as it were — the ratio of S to R is simply 28:45. Try this technique on your next multiple-ratios problem and let us know how it goes!
In honor of the upcoming weekend, I decided to devote this entry to some great math-themed movies that you can watch the next time you need a break from your GRE studies. Giving yourself time off is important, and there’s no reason that you can’t use that time to let some of these classic (and not-so-well-known) stories characters inspire you to knock your next study session out of the park.
In no particular order, here are my recommendations:
- 21: Several MIT students are trained to count cards, and go to Las Vegas to win millions playing blackjack. I’m always a fan of Kevin Spacey, who places a professor, and hey – you may come away from this movie with a completely new, very valuable skill set that you can take on your next trip to Vegas.
- Moneyball: While everyone who saw this movie focused on Brad Pitt (understandably so), there was a lot of fun discussion about how to use math to cut through bias and human, error-prone perception. If you’re a sports fan, you’ll enjoy this baseball-themed movie. If you’re not a sports fan, don’t worry – it has Brad Pitt and Jonah Hill in it, and is very well-written.
- Fermat’s Room: This Spanish movie is about several mathematicians who are trapped together in a room and forced to solve “enigmas” or risk being killed as the room becomes increasingly small, all while trying to determine who is trying to kill them – it’s a combination of “Clue” and the scene from Star Wars in which Luke Skywalker & company are trapped in the Death Star’s garbage compactor. At the very least, it’ll make you feel better about the fact that the GRE testing room won’t close on in you if you get an incorrect answer!
- Good Will Hunting: Matt Damon as a math prodigy who, with the guidance of mentor –therapist Robin Williams, must decide what to make of his life. Certainly a good pick-me-up for anyone who is at a crossroads (as many who are applying to grad school are!), but be warned: You might come out of this movie speaking with a Boston accent.
These are just a sampling of some of the great movies that you can use to motivate you in your GRE prep – what are your favorite math-themed movies? Let us know in the comments!
For anyone who ate too much chocolate last week and is looking for a sweet treat that won’t add to his or her waistline, look no further: How would you like a quick way to solve rate & speed problems that have lots of variables? It probably doesn’t sound quite as good as a box of truffles does, but truffles unfortunately aren’t going to get you into grad school.
The beautiful thing about rate problem is that they virtually always are written as a fraction. Think about it: Any rate – miles/gallon, cookies/jar, heartbeats/minute – is just one term divided by another term. So when you’re setting up a rate problem in which you have variables without real numbers, figuring out where each variable goes in the fraction – the numerator or the denominator – will often get you directly to the correct answer. Let’s look at an example to demonstrate this:
Sweetheart Candies sells boxes holding p pieces of chocolate each. The boxes are shipped in crates, each holding b boxes. What is the price charged per piece of chocolate, in cents, if Sweetheart Candies charges m dollars for each crate?
What’s this question asking us for? We’re looking for the price per piece of chocolate, in cents. The keyword “per”, along with the fact that the answer choices are all fractions, clue us into the fact that this is a rate problem. We have three variables, so let’s see where we can put them in the fraction.
Anything that comes before the “per” will go in the fraction’s numerator. In this case, that’s the price. m is the only variable that has anything to do with money, so it must be on top. If we glance back up at the answer choices right now, which answer choices can we eliminate because they don’t correctly place m? That’s right: (A), (C), and (E) are now gone.
The only difference between (B) and (D), the two remaining choices, is that (B) multiplies m by 100. Is that extra 100 a necessary component of the rate that we’ve been asked to set up? Let’s think: The problem asks for the price, in cents, but m represents a number of dollars. So yes, we absolutely need to multiply m by 100 to get from dollars to cents. And voila – choice (D) is out, and we’ve determined that (B ) is the correct answer.
The next time that you’re confronted with a tough rate problem that involves multiple variables, don’t despair – just figure out whether each variable belongs on the top of the bottom of the fraction. On Test Day, you’ll have nailed points and moved on before your competition is even done reading the problem!