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# GRE Math: Solving Quantitative Comparisons

First off, here’s a tricky algebraic QC:

1) x > -3

If you find this puzzling, this article discusses an efficient method for handling such problems.  Before we talk about that,  a quick refresher on a few basic ideas dealing with equations and inequalities.

## Intro to Arithmetic

Suppose you have an equation, P = Q.  You probably remember that you can add, subtract, multiply, or divide one side of an equation by just about anything, as long as you remember to do the same operation to the other side.   The only restriction is — it’s illegal to divide by zero — all other adding & subtracting & multiplying & dividing is perfectly legal.

Now, suppose you have an inequality, P > Q or R < S.  It is still 100% legal to add or subtract anything you want to both sides.  If you multiply or divide both sides by a positive number, that’s perfectly legal as well.  If you multiply or divide by a negative sign, this is also allowed, but it reverses the direction of the inequality —- if P > Q, then (–5)*P < (–5)*Q.  Of course, we never can divide by zero, and it’s also problematic to multiply by zero, because that would obliterate the inequality, making both sides equal.

OK, probably all of that was review, but we need that as our ground rules.

## A powerful method for solving Quantitative Comparison

For some folks, especially folks with good mathematical intuition, what I am going to explain here will be patently obvious — probably you have already thought of it.  But, for folks for whom math is a constant struggle and not particularly intuitive, this new method might bring a landslide of insights.

You see, one way to think about an ordinary variable like x is as a “holder for a number.”  In any algebraic equation, such as 2x + 5 = 13, the variable x is holding some as yet unknown number, and of course the goal of most standard algebra is to find what number it is holding.

As you know, the GRE Quantitative Comparisons set a very different task.  We are not trying to solve for a number in the ordinary sense.  Instead, we are trying to figure out the relationship between columns A & B —- either A > B, A < B, A = B, or we can’t determine the relationship.  What is unknown is not a number but a relationship — what mathematicians call a relation.  At the outset of a typical GRE QC, the unknown relation could be < or > or =.

I will suggest introducing a “mystery symbol”, which I will denote by the arbitrary combination \$\!|\$.  This is a holder, not for an unknown number, but for an unknown relation.  We could begin any GRE QC by sticking this mystery symbol between the two columns.

What’s the point of that?  Now, we just have a funky symbol between the two columns!  How does this help us?  Well, think about it.  Whether \$\!|\$ stands for < or > or =, there are certain mathematical transactions that are 100% legal.   Here is a summary of what we can do to both sides of \$\!|\$:

a. We can add or subtract any number to both sides of \$\!|\$.

b. We can multiply or divide both sides of \$\!|\$ by any positive number.

This opens up a panoply of ways of handling the two columns.

## Example solution

I will demonstrate a solution to the QC at the top of the page using this “mystery symbol” method.   Begin with the mystery symbol between the content of the columns.

2x – 4 \$\!|\$ 5x + 7

First, because I can subtract any number from both sides, I will subtract 2x from both sides, to get all the x’s on one side.

–4 \$\!|\$ 3x + 7

Now, I will subtract 7 from both sides.

–11 \$\!|\$ 3x

I am allowed to divide by any positive number, so I will divide by 3.

–11/3 \$\!|\$ x

Well, we were told that x > –3, and –3 > –11/3, so the “value” of \$\!|\$ must be <, and column B is bigger.

## A Zen twist

Here, I made up the arbitrary symbol \$\!|\$ to demonstrate the idea of handling an unknown relationship, but of course, you don’t actually need this symbol once you understand the idea.  You can leave the space between the columns blank and simply follow guidelines (a) & (b) above, doing simultaneous operations on both columns until they are simplified to a form that allows you to decide directly.

## Summary

The problem at the top of this article was a relative easy problem — folks with good number sense might have been able to solve it by inspection.  Here’s a somewhat more challenging problem in which the same principle can be used with powerful effect.

By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.

### 16 Responses to GRE Math: Solving Quantitative Comparisons

1. shreya khastagir July 3, 2016 at 1:29 am #

Loved the way you solved the question. It just breaks down the entire expression to relating x with a value. Thanks a lot.

2. Shaha February 16, 2016 at 12:49 pm #

if you add 4 to both sides in the initial step instead of subtracting 2x would it end up with the same answer? also why does 4 remain a negative integer? shouldn’t it be written as
2x+ -4 !\$! 5x+7? I’m a little confused…

• Magoosh Test Prep Expert March 8, 2016 at 8:21 am #

Hi Shaha,

Yes, if we were to initially add 4 to both sides of the equation, we would eventually end up with the same answer. We’re allowed to add or subtract the same value from each side, just as if we were solving a normal inequality question 🙂

To answer your other question, the expressions 2x + (-4) and 2x – 4 are equal in value; in both cases, we are expressing 4 less than 2x. We can think of subtracting a number as adding the negative of that number. Does that make sense?

I hope this helps! Happy studying 😀

3. Carl January 4, 2016 at 11:30 am #

So I get the math logic to get to x = -11/3, x> -3, and that -11/3 is < -3. I don't understand why the "\$\!|\$" is replaced by the "<" symbol.

• Shaha February 16, 2016 at 12:52 pm #

he put those symbols instead of < in, so he would have something there until he solved for the symbol. for example before solving the expression he didn't know if it would be , or =

4. Jeffrey September 19, 2015 at 9:21 am #

I’m pretty decent student in math I would say. But this exercise certainly confused me. I was just checking other methods on how to resolve a problem. In other videos is explain that:

2x – 4 and 5x + 7, then
x=4/2 5x=-7
where: x=2 x=-7/5

This means that A is way bigger than -3 and B is just bigger than -3. So why isn’t the answer A in this case. Second, if it is the case that we have equal both equations as performed above:

2x – 4 = 5x + 7 this in order to find only one value of x, when and how I would be able to identify this. I’m a premium student of Magoosh. Thank you Mike

• Mike McGarry September 23, 2015 at 10:53 am #

Dear Jeffery,
I’m happy to respond. 🙂 This is incredibly subtle: you are blurring the distinction between an expression and an equation. An equation is mathematical statement involving an equals sign: it is the equivalent of a complete mathematical sentence. By contrast, an expression has no equal sign: it’s like a phrase (e.g. “the man of the hour,” “by the skin of your teeth“) that we may use in a complete sentence, but in and of itself, it’s not a complete sentence.
An expression: x + 2y
An equation: x + 2y = 9
We sometimes can solve an equation, or a system of equations, but we can’t “solve” an expression. If given a particular number, we plug that in and evaluate the expression, but solving it makes no sense.
This QC problem gives two expressions, (2x – 4) and (5x + 7). These will take on different values for different values of x, but there is nothing for which we can “solve.” You committed a major violation of the laws of mathematics by turning each one of these expressions into an equation. The expression (2x – 4) has no equal sign at all, but suddenly you change it to (2x = 4), something with an equal sign. From whence did the equal sign come? Essentially, you set the expression equal to zero and started solving. You actually found the two x-values that make the two individual expressions equal to zero, but this is not at all helpful in solving the problem—among other things, we have no guarantee that either one of the expressions equals zero in the first place! The basic problem is that you can take an expression you were given, stick in an equal sign wherever you like, and start freelance solving on your own. An expression is one mathematical thing, an equation is a very different mathematical think, and to change one arbitrarily into the other is a major violation of the laws of mathematics.
This is an extremely subtle point. Does all this make sense?
Mike 🙂

5. Amna Ijaz July 21, 2015 at 8:46 am #

Oh yeah! <3 I watched the two videos on QC strategies in the second module of math and solved the problem at the top the way Mike told in matching operations lesson. Got the right answer in a few seconds. I have absolutely no brains when it comes to this, so I am declaring you a very good teacher! 😀 You rock!

6. Martin August 24, 2013 at 11:18 am #

Great Post.
Just a general question on Quantit. Compar. on the GRE: Suppose we are given a term / equation with two or three variables (say x, y, z) can we always assume that x, y and z are NOT equal? Or will the question explicitly state that x, y, z are non-equal (integers)? It makes a big difference when picking numbers of course…

• Mike August 24, 2013 at 1:28 pm #

Martin,
First of all, if the question doesn’t explicitly say “integers”, never assume the variables are integers —- that’s a huge trap, one of ETS’s favorites. “Numbers” is a much much bigger category than integers. If the question doesn’t explicitly say that x & y & z are unequal (or distinct), then we have to assume that two or three of them having equal values has to be a possibility that needs to be considered. Does this make sense?
Mike 😉

7. vishal August 19, 2013 at 9:40 pm #

Hey, what if x=0 since x>-3

• Mike August 20, 2013 at 11:39 am #

Vishal,
If x = 0, then column A equals (-4) and column B equals (7), and (-4) < (7), so column B would be bigger. That fits with our conclusion.
Mike 🙂

8. Thomas July 10, 2013 at 10:15 pm #

How do we know x > –3 ? Would that be given as a question parameter?

• Mike July 11, 2013 at 2:37 pm #

Thomas,
I’m sorry — the QC question the article was discussing wasn’t visible on the page. I believe I fixed that, so now everything should be clear. Thank you for raising this question.
Mike 🙂

9. hashir July 9, 2013 at 11:30 am #

“We couldn’t find the question you were looking for” got this error when clicked on link provided in summary. Hope you can fix it. Thanks

• Mike July 9, 2013 at 2:25 pm #

Dear Hashir,
Thanks for pointing this out. I believe the problem is fixed now.
Mike 🙂

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