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 #5: Estimation with a Twist

what happened to the denominator? Shouldnt it still have 14x-42/35 on Column A and

15-5x/35?

Hi Trevor,

Sure! The first thing we do in this question is create a common denominator (35) for each of the terms. This means that we multiply the term in column A by 7, and the term in column B by 5. Once we have a common denominator, we can basically ignore the denominator and focus on the numerator! Both of the numerators will be divided by the same number, so we don’t have to worry about the denominator in this question. You can include 35 in this question, but really we are comparing the two numerators to determine the answer here. Does that make sense?

To tackle this problem I used the algebraic approach as follows:

18y/(7y-x)=3. So, this is an equation and I can cross multiply:

18y=3(7y-x)

18y=21y-3x If I subtract 18y from both sides:

0=3y-3x Factor out 3

0=3(y-x)

So, if y-x=0, then y=x

This is the way I looked at the problem. Thanks for the tips!

Thank you Yoxarys.

To me, it seems you magically carried the 3 over the = sign and multiplied it by (7y-x) which breaks some rule in my head that says “you must do that on both sides” if you can do that at all.

Why does that procedure work? What is it called?

Hi Myles,

From what I see, Yoxarys hasn’t done anything strange or ‘illegal’ here. The original equation is 18y/(7y-x) = 3, so the ‘3’ started out on the right side of the equation. He then multiplied both sides by (7y-x). The (7y-x) in the denominator of the left side of the equation canceled out, and he multiplied 3 by (7y-x). This follows the “do things on both sides” rule. I’m happy to explain further if there’s another point of confusion!

Are there any specific Clemmonsdogpark problems you could direct me to in order to practice this principle?

Hi, Claire

Unfortunately (or fortunately?), there are a lot of Clemmonsdogpark practice questions for Quantitative Comparison, and it’s difficult to pin down exactly which ones you should use logic over algebra. While there is always the recommended approach, you should always work out problems in a way that makes sense to you, individually, and take the explanations as a guide.

If you’d like practice on QC problems in general, I’d recommend filtering in Practice for Quantitative Comparison under Math and going through all of them, and trying out different strategies to see which seems most intuitive to you. Sorry I can’t be of more help!

Best,

Margarette

Thanks Brent and Margarette. The explanation and the suggestion is a great deal for QC questions.