# GRE Quantitative Comparison Tip #1 – Dealing with Variables

Here's 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

To help set up today's discussion, please consider the following question:

 Column A Column B 10x + 2 2x + 10

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

When confronted by QC questions involving variables, the two most common approaches are:

• Use algebra
• Plug in numbers

Each approach has its pros and cons.

Algebraic approach: For the question above, we might begin by moving the variables on one side. To accomplish this, we'll subtract 2x from both columns and subtract 2 from both columns to get:

 Column A Column B 8x 8

Then we can divide both columns by 8 to get:

 Column A Column B x 1 \blue{f_1(x)=x}
\red{f_2(x)=1}

when f_1(x) = f_2(x) \to x = 1

when f_1(x) \lt f_2(x) \to x \lt 1

when f_1(x) \gt f_2(x) \to x \gt 1

Since x can be less than 1, greater than 1 or equal to 1, it's clear that the correct answer is D.

Plug in numbers approach: We'll begin with the original question.

 Column A Column B 10x + 2 2x + 10

Then we can divide both columns by 2 to get:

 Column A Column B 5x + 1 x + 5 \blue{f_1(x) = 5x + 1}
\red{f_2(x) = x +5}

when f_1(x) = f_2(x) \to 5x + 1 = x + 5 \to x = 1

when f_1(x) \lt f_2(x) \to 5x + 1 \lt x + 5 \to x \lt 1

when f_1(x) \gt f_2(x) \to 5x + 1 \gt x + 5 \to x \gt 1

or this approach, we'll plug in different values for x and see what happens.

Now, when we plug in values for a given variable, we want to use a nice cross-section of all numbers and we want to use numbers that make it easy to evaluate each column. A nice set of numbers to choose from are: 0, 1, -1, \dfrac{1}{2}, -\dfrac{1}{2}, 100, and -100, since they represent a nice cross-section of all numbers.

Let's begin with 0. When x=0, we get:

 Column A Column B 5*(0) + 1 (0) + 5

When we evaluate this, we get:

 Column A Column B 1 5

Since Column B is greater than Column A, we know that the correct answer must be either B (Column B is always greater) or D (the relationship cannot be determined from the information given).

This is a great feature of this approach. After very little work, we can quickly whittle the number of possible answer choices down to just two.

At this point, we'll try another value of x.

Let's try x = -1. When x = -1, we get:

 Column A Column B 5*(-1) + 1 (-1) + 5

When we evaluate this, we get:

 Column A Column B -4 4

This result illustrates the main drawback of the plug-in method. When we plug in a second value for x, we see that Column B is still greater than Column A. So, perhaps it's the case that Column B is always greater than Column A, in which case the answer is B. However, we've only tried two values of x so far. Perhaps we should try another one.

 Column A Column B 5*(-10) + 1 (-10) + 5

When we evaluate this, we get:

 Column A Column B -49 -5

Since Column B is still greater than Column A, we might conclude that Column B will always be greater than Column A, in which case the correct answer is B. However, we're basing this conclusion on the results of plugging in only 3 different values of x, so we can't be absolutely certain that this is the correct answer.

In fact, unless we get two contradictory results (e.g., Column A is greater than Column B for one value of x, and then less than Column B for another value of x), we can never be 100\% certain of the answer.

When it comes to QC questions where we're comparing two algebraic expressions, we must determine which column is greater for every possible value of the given variable(s).

So, even though it appears that the correct answer here is B, we can't be certain of this, since we haven't tried every possible value of x.

In fact, it turns out the answer is not B. The answer is D.

We can see that the answer is D when we plug 1 in for x. When x=1, we get:

 Column A Column B 5*(1) + 1 (1) + 5

When we evaluate this, we get:

 Column A Column B 6 6

So, when x=1, the two columns are equal. At this point, we have contradictory results, which means we can now be certain that the correct answer is D.

This question illustrates the need to plug in a variety of numbers. The first 3 numbers I plugged in were 0, -1 and -10. It wasn't until I plugged in a positive value that we had contradictory results, which allowed us to see that the correct answer is D.

## Pros and Cons

The main drawback of the plug-in approach is that, unless we get two contradictory results, we can never be certain of the correct answer.

The algebraic approach, on the other hand, will almost always allow us to determine the correct answer with absolute certainty.

Given all of this, it seems that the algebraic approach is the best approach. This is true to a certain extent. The problem with the algebraic approach is that it's often the more difficult of the two approaches.

So, even though the algebraic approach may be the superior approach, we should also remember that plugging in numbers can often be the faster approach, and it's an approach that can used if you don't recognize how to solve a QC question algebraically.

For example, consider this question:

 Column A Column B x^2 - 6x + 7 -3

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 \blue{f_1(x) = x^2 - 6x + 7}
\red{f_2(x) = -3}

when f_1(x) = f_2(x)

\to x^2 - 6x + 7 = -3

\to x^2 - 6x + 10 = 0

\to x^2 - 6x + 9 + 1 = 0

\to (x + 3)^2 + 1 = 0

because (x + 3)^2 \geqslant 0 and 1 \gt 0, (x + 3)^2 + 1 \gt 0

any value of x does not meet (x + 3)^2 + 1 = 0.

when f_1(x) \gt f_2(x)

\to x^2 - 6x + 7 \gt -3

\to x^2 - 6x + 10 \gt 0

\to x^2 - 6x + 9 + 1 \gt 0

\to (x + 3)^2 + 1 \gt 0

any value of x meet (x + 3)^2 + 1 \gt 0.

when f_1(x) \lt f_2(x)

\to x^2 - 6x + 7 \lt -3

\to x^2 - 6x + 10 \lt 0

\to x^2 - 6x + 9 + 1 \lt 0

\to (x + 3)^2 + 1 \lt 0

because (x + 3)^2 \geqslant 0 and 1 \gt 0, (x + 3)^2 + 1 \gt 0

any value of x does not meet (x + 3)^2 + 1 \lt 0.

Plug in numbers approach:

Can you see the algebraic solution? In a test situation, you can give yourself some time to solve the question algebraically, but if you don't readily see a useful approach, you should start plugging in numbers. Let's do that.

Let's begin with x = 0. When x = 0, we get:

 Column A Column B (0)^2 - 6*(0) + 7 -3

When we evaluate this, we get:

 Column A Column B 7 -3

Since Column A is greater, we know that the correct answer is either A or D.

Let's try plugging in another number (if we're lucky, it will yield contradictory results).

Let's try x = 1. When x = 1, we get:

 Column A Column B (1)^2 - 6*(1) + 7 -3

When we evaluate this, we get:

 Column A Column B 2 -3

Once again Column A is bigger, so perhaps the answer is A.

Should we keep plugging in more numbers? Maybe, maybe not. There are many factors at play here. Are you ahead on your timing or behind? Do you have a strong feeling about this question (don't discount intuition on QC questions).

Now, if we keep plugging in different values for x, we will keep getting the same results. That is, Column A is bigger than Column B. Given the inevitable absence of contradictory results, you will be forced, at some point, to choose between A and D. That's the reality of the plugging in numbers approach, so everyone has to get used to that at some point.

Okay, now let's look at one way to solve this question algebraically.

 Column A Column B x^2 - 6*x + 7 -3

When we evaluate this, we get:

 Column A Column B x^2 - 6*x + 9 -1

Next, factor Column A to get:

 Column A Column B (x-3)*(x-3) -1

And now rewrite Column A as:

 Column A Column B (x-3)^2 -1

Since the square of any value is always greater than or equal to zero, we know that Column A will always be greater than -1, which means the correct answer is A.

The big takeaway of this post is that you have at least 2 possible approaches at your disposal when you encounter QC questions involving variables. So, be sure to consider both approaches.

In the next post, we'll look at a useful strategy for plugging in numbers.