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!
Not too long ago, I used something called the “balance approach” to show you how to solve a mixtures problem. But the balance technique isn’t exclusive to mixtures. In fact, the most likely time for it to come up is when a problem deals with plain ol’ averages.
Here’s an example of the kind of GRE problem I’m talking about:
At a bowling tournament in which males and females competed, the average score of the participants was 154 points. If the average score of the 8 males was 148 points, how many females were in the tournament if the average female score was 158 points?
Most of your competition is going to try to use the average formula: average equals sum divided by number, or as I prefer to write it,
Average × Number = Sum
Using the above formula gives you this:
154(8 + x) = 8(148) + x(158)
There are a couple of problems here. One, you have to do some nasty arithmetic, such as 154×8 and 8×148. And perhaps more importantly, it takes a rather impressive feat of translation to set that baby up.
Try this instead.
The overall average is 154. Each man got 148 points, which is 6 points short of the average. There were 8 men, so altogether, they dragged down the average by 8×6 = 48 points.
This means the women need to make up a 48 point deficit. Each woman scores a 158, which is 4 points above the average. If each woman contributes 4 points to overcoming the 48 point shortfall, then there need to be 48/4 = 12 of them.
That sure was a lot easier than solving the equation. And if you’re a real critical thinking wizard, you might notice that even this was too much work. The ratio of the men’s deficit (-6) to the women’s surplus (+4) is 6 to 4, or 1.5. Thus, the ratio of women to men also needs to be 1.5, and that it is: 12÷8 = 1.5.
BOB can eat a cheesecake in THREE minutes. His sister JENNY can eat a cheesecake in TEN minutes. How long will it take the two of them, eating TOGETHER, to eat three cheesecakes?
If you’re anything like I used to be, these problems drive you nuts. But I say “used to be” because after I started teaching for Kaplan, I learned that there is actually a dirt-simple plug-and-chug formula that’ll spit out the answer for you. It’s call the combined work formula, and it’s this:
A student in my class last night asked, “Why weren’t we taught this in high school?” Why, indeed. When I learned this formula was real, I wanted to go find my high school math teachers and smack ‘em.
So Bob and Jenny’s cheesecake addiction problem isn’t a hair-puller after all. Just plug their speeds into the formula…
…and remember that this is the GRE, so there’ll always be a trap. Here, remember that the question asks for the time to eat THREE cheesecakes, so multiply the above by 3:
And there’s your answer. In case you missed it in high school, don’t miss it now: the combined work formula is a mighty addition to your arsenal on Test Day!
Last night I TA-ed a “GRE Bootcamp” online event hosted by Lee Weiss. We won’t be running that particular event again for a while, so I figured I’d cover what a lot of the 400+ attendees considered the most challenging problem of the evening. Lee did an excellent job explaining it, but even so we got lots of requests in the Q&A to “go back to that mixture problem again!”
If you were at the event: hi! You might remember me as “Kaplan GRE Expert Boris.” It so happens that I’ve written before about how to solve this very type of problem, but concepts that the GRE repeats are always worth a second look.
Here was the question that confounded so many of you:
Liquid X is composed of 30% alcohol and 70% water, while Liquid Y is composed of 18% alcohol and 82% water. The two liquids are combined to form Mixture Z, which is composed of 21% alcohol and 79% water. What is the ratio in Mixture Z of Liquid X to Liquid Y?
If you’ve already read my older entry, try to solve this one on your own before you keep reading!
Now let’s solve it together. We can start by cleaning up the problem a bit. If X has 30% alcohol, it’s redundant to say that it has 70% water — the GRE just puts that there to overwhelm you with extra numbers. Same goes for Y (18% alcohol) and mixture Z (21%) alcohol. You’ll find that you can usually summarize the information much better than the GRE does!
Liquid X: 30% alcohol
Liquid Y: 18% alcohol
Mixture Z (X + Y): 21% alcohol
The question is, what’s the ratio of X to Y? Note that for simplicity’s sake, I’ve left the choices out of the picture — you can find the answer straight-up.
To solve a problem like this, you can spare yourself a heck of a headache if you use the balance approach. Imagine you’re a mad scientist, adding “parts” of X and Y to a bubbling flask until you get just the right concentration of alcohol.
Every “part” of X you add gives 9% too much alcohol to the mixture. X is 30% alcohol, but the mixture is only 21%: 30 – 21 = 9.
Every “part” of Y you add gives you 3% too little alcohol. Y is 18% alcohol, and 21-18 = 3.
You need those 3′s and 9′s to cancel out if you want just the right percentage of alcohol. To do that, of course, you need three 3′s for every 9, since 3 × 3 = 9. This means that there’s one “part” of X for every three “parts” of Y — in other words, that the ratio of X to Y is 1:3.
(Sidenote: in the problem, one of the choices was 3:1, which is the ratio of Y:X, not X:Y. Always take care, when you finish solving a GRE quantitative problem, to confirm that you answered the question the test makers actually asked! Nothing is more heartbreaking than losing a GRE point because you got the right answer to the wrong question.)
Still got questions about mixtures? Let us know in the comments below!