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.

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!

Today we wrap up the series that began here: a look at how the GRE test makers play possum and try to hide correct answers in plain sight. We’ll close on a simple but powerful note.

The test makers like for there to be as few negative signs as possible in their fractions. Take a look at this one:

This would never be the correct answer choice, since there’s a way to cut a negative sign. If you ever get a fraction with variables as your answer and don’t see it in the answer choices, don’t panic. Chances are, you just need to fiddle with the negative signs a bit. Multiply the top and bottom of the fraction by -1:

This is good to know in general: multiplying (a - b) by a negative sign causes the variables to flip around and make (b - a). You can tweak the order of terms around a minus sign to your heart’s content with negative multiplication.

If you’ve encountered other tricky answer choice situations, let us know in the comments!

In my last entry, I showed you how to rejigger radical fractions to make them look the way the test makers want them to look. The example in that entry was a pretty easy one, though — all you had to do was multiply the top and bottom of the fraction by a radical. This time we’re going to crank it up a notch.

Imagine you worked out the math to a GRE quant problem and got this:

There’s a radical in the bottom of that fraction, so it will never be the credited answer to a GRE quantitative problem. As we saw last time, you’ve got to get the radical out from underneath the fraction. Some of you might perhaps be tempted to multiply the fraction through by the square root of 5, similar to what we did last time. Notice, though, what’ll happen to the bottom of the fraction if you do that:

Whoops! We got rid of the radical all right, but another one popped up in its place. What you have to do instead is multiply the top and bottom of the fraction by the complement of the denominator: that is, copy the denominator, but flip the sign. In this case, there’s a + to begin with, so switch it to a – and multiply by . Why do this? Well, you’re about to see!

Since this one’s more complicated, let’s do the two operations separately. First, on top, we’ve got:

Easy enough. The bottom is this:

And here’s why multiplying by the complement is so effective: this is a difference of squares. You absolutely have to know this property on Test Day:

So,

Notice how this gets rid of the radicals cleanly, without making any new ones! All together, then:

Finally, every term in that fraction is a multiple of 4, so divide through by 4 to get the answer:

Click, confirm, and another seemingly impossible GRE challenge bites it. Got any questions or points of confusion about manipulating radicals? Let us know in the comments!

Last week, I showed you one trick that the test makers use to “hide” right answers. Here’s a second:

By mathematical convention, square roots shouldn’t appear in the denominators of fractions. The GRE adheres to this convention, which means that the answer you come up with will sometimes look radically (*cough*) different from the credited answer. For example, you might solve a problem and get:

But when you look at the answer choices, that fraction won’t appear anywhere. The correct answer would actually be .

This is a scary moment for many of you, since apart from the radical, those two numbers don’t look anything alike! As always, don’t panic. When you get an answer that looks nothing like the choices and there are radicals involved, make sure you don’t have any radicals on the bottom of the fraction. If you do, great! That’s your problem, and it’s an easy one to fix. Multiply the top and bottom of the fraction by any radicals on the bottom — in this case, by the square root of 3 — to “clean up” the fraction:

Remember, when you multiply a radical by itself, the number doesn’t change — all that happens is the radical “pops off.” And now, the 6 on top cancels with the 3 on the bottom to get , the answer you can calmly and happily click on Test Day.