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GRE Quantitative Comparison Tip #3 – Logic over Algebra

In previous posts (Tip #1 – Dealing with Variables, Tip #2: Striving for Equality), I have discussed two approaches when tackling Quantitative Comparison (QC) questions involving variables. Those approaches are:

1) Apply algebraic techniques

2) Plug in numbers

In those posts, I noted that the algebraic approach is typically the faster and more reliable approach.

In today’s post, we’ll examine a third strategy that can sometimes be the fastest and easiest approach. We’ll call this the “logical approach.”

To set things up, please consider the following QC question: A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

The algebraic approach to this question looks something like this:

First multiply both sides by 35 (the least common multiple of 5 and 7) to get: Then add 5x to both sides to get:  Then add 42 to both sides to get: And, finally, divide both sides by 19 to get: So, now we’re comparing x and 3, and the given information tells us that x is greater than 3.

This means the correct answer must be A.

Now let’s take the original question and use logic to solve it (in about 5 seconds). Column A: If x > 3, then 2x > 6, which means that 2x-6 must be positive.

Column B: If x > 3, then 3-x must be negative.

So, the two columns can be rewritten as: From here, we can see that Column A is always positive and Column B is always negative. As such, Column A will always be greater than Column B. So, the correct answer is A.

Let’s try another one. See if you can solve it in your head. A. The quantity in Column A is greater

B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

For this question, I’ll leave the algebraic approach to you.

Let’s apply some logic.

First, we’re told that . In order to apply some logic, let’s refer to the denominator as “something.” In other words, 18y divided by “something” equals 3. Well, we know that 18y divided by 6y equals 3, so that “something” must equal 6y.

In other words, it must be the case that 7y-x = 6y

Now consider the fact that 7y-x = 6y. If we now refer to “x” as “something,” we can see that 7y minus “something” equals 6y. Since we know that 7y-y=6y, we can see that “something” must equal y.

In other words x = y.

Now that we have concluded that x=y, we’ll return to the original question: If x=y, we can see that the answer here must be C.

So, although the algebraic approach is typically the superior approach for quantitative comparison questions involving variables, be sure to take a moment to see whether the problem can be solved by applying a little logic.

Heres’ the whole series of QC tips:

Tip #1: Dealing with Variables

Tip #2: Striving for Equality

Tip #3: Logic over Algebra

Tip #4: Comparing in Parts

Tip #5: Estimation with a Twist

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