Today’s post is part one in a series of posts (Tip #2: Striving for Equality, Tip #3: Logic over Algebra, Tip #4: Comparing in Parts, Tip #5: Estimation with a Twist) devoted to Quantitative Comparison (QC) strategies.

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

Column A | Column B |
---|---|

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 2*x* from both columns and subtract 2 from both columns to get:

Column A | Column B |
---|---|

Then we can divide both columns by 8 to get:

Column A | Column B |
---|---|

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 |
---|---|

For 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, ½, -½, 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 |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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 |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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.

How about *x*= -10? Here, we get:

Column A | Column B |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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 |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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 |
---|---|

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

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 |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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 |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

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 |
---|---|

First, add 2 to both columns to get:

Column A | Column B |
---|---|

Next, factor Column A to get:

Column A | Column B |
---|---|

And now rewrite Column A as:

Column A | Column B |
---|---|

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.

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I’m confused about the final example in the article… x^2-6x+7 and -3…

When the author starts to solve Algebraically, he adds 2 to both sides… But I don’t understand where the 2 is coming from… and I also don’t understand how he factors the equation…

Maybe I’m just really out of practice…

Adding 2 to each side for x^2-6x+7 and -3 does look confusing at a glance, even if you have been getting in some good practice.

The first thing that’s important to remember is that if you add, subtract, multiply, or divide using the same exact value on both sides of an algebraic operation, then the operation is effectively unchanged; the new operation will get the same result as the old operation, so long as you do the same thing to both sides.

For example, take the very simple operation 1+x = 2. Here, x is obviously 1. 1+1 = 2. And the x value for 1+x=2 remains exactly the same if you divide both sides by–say– 4:

(1+x)/4 = 2/4 >>> (1+1)/4 = 2/4 >>> 2/4 = 2/4

…Or if you subtract 3:

(1+x)-3 = 2-3 >>> (1+1)-3 = 2-3 >>> 2-3 = 2-3

Or if you add 2. You get the idea. This also works on inequalities. 1+x>1 if x = 1; (1+x)*5 > 1*5 if x=1, and so on.

So the 2 doesn’t have to come from anywhere. It just has to be applied to both sides of the inequality, so that the results of the inequality and the relative value of both sides remains the same.

The question then, is why do we add 2? It’s certainly not just arbitrary.

In GRE algebra, applying a number to both sides is done to create a new set of values that’s easier to look at and work with, one that lends itself to certain shortcuts.

In the case of x^2-6x+7, changing it to x^2-6x+9 allows you to use your knowledge of FOIL to create a single squared value: (x-3)(x-3) or (x-3)^2. Note that this is a reverse FOIL. If you multiply the first, outside, inside, and last values of (x-3)(x-3), you get x^2-6x+9.

More to the point, you should remember the rule for the quadratic pattern x^2-Ax+B. This pattern can always be converted to a pattern of (x +or- a)(x +or- b) where the a and b values add up to A in the original pattern, and multiply for a product of B in the original pattern. In the case of x^2-6x+9, (x-3)(x-3) gets us -3 for a and -3 for b. -3 plus -3 = -6, the value that is multiplied by x in the original pattern. -3 times -3 has a product of 9, the value added to 6x in the original pattern.

Ultimately, the factored version of x^2-6x+9 is a square, (x-3)(x-3). The factored version of the original x^2-6x+7 would not be a square, and would not even be easily factorable at all. So adding 2 creates a powerful shortcut to an algebraic solution. The trick is being fluent enough with FOIL and factoring quadratics to recognize that adding something to x^2-6x+7 has the potential to get you to a quick algebraic solve.

My 2c :

It would also be helpful to use the properties of lines and parabolas and reason using them. I’ll stick to the formulas

For the 1st qn :

We are given 2 lines, y = 10x + 2 and y = 2x + 10

Either the 2 lines are parallel to each other or perhaps the same line or they must intersect

Since the coefficients are not in equivalent ratio, the 2 lines are not parallel. Nor are they the same line. So somewhere they must intersect. Now if they intersect, before the intersection (i.e. approaching from the left), one line is below the other. At the intersection both the lines have a common point. After the intersection the below line goes above. So one line can be greater than the other before the intersection and then become lesser.

Hence D

For 2nd qn :

Add +3 to both sides. By doing this we shift the whole scene up by 3 units, but the relative distances remain the same. Now that -3 has moved to 0, vertically

Now find the discriminant b^2 – 4ac = 36 – 40 = -4

This is negative and therefore doesn’t have a real square root. Meaning it’s roots are imaginary. Meaning the parabola never crosses the X axis and always stays above it. Therefore the parabola will always be greater

You can directly learn the formula for a minima/maxima of a parabola. Here the parabola has a positive coefficient for x^2. Hence it curves upwards. Hence it will have a minima.

Now, if -3 is above the minima, then the parabola will have values both above, equal and below it. If -3 is the minima, then the parabola will have values only greater and equal to -3

If the minima is above -3, then it will always be greater

excellent analysis !!

Hi,

This post is still proving useful years later!

On the first example, how do we know that “x can be less than 1, greater than 1 or equal to 1”?

Thank you!

Hi Lissa,

In QC, ‘x’ can be anything as long as there are no “constraints” above the two columns saying otherwise (x 0, something to that effect). Hope that helps!

I would graph the quadratic expression and then you can see that the parabola is greater than -3.

in the last question, if x=3 at the last equation then column b is greater. thus the answer is D

If x=3, then we get:

Column A = 3^2 – 6(3) + 7 = 9-18+7 = -2

Column B = -3

Column A is still greater than column B.

Cheers,

Brent

Great !

I take it that the tip off to using this approach is the squared variable. So if there wasn’t a squared variable and a negative # in the other column would this still work???

KLR

Exactly – the squared variable is the big tip off.

If the question were lacking the negative number in the other column, this approach wouldn’t work.

Cheers,

Brent

In the event that we try to solve x^2 – 6x + 9 = -3, is there any other way to solve it instead of adding 2 to both sides?

At the moment, I can’t think of another way to definitively solve the question. We could plug in numbers, but that isn’t very conclusive.

Cheers,

Brent