Many quantitative questions have variables in both columns. While your first instinct may be to work algebraically, this strategy is not always best. Often the fastest way to a solution is by plugging in different values to see which column is greater.

Developing a sense of when to plug in and when to solve algebraically takes practice. Here are a few helpful guidelines when trying to determine which approach to use.

## Work algebraically if the question is a polynomial

If you are dealing with a polynomial, simplify. Plugging in may require too much calculation. Instead, work with the familiar algebraic forms shown below:

Now let’s take a look at a question.

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

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

In Quantitative Comparison we can make it so each side is equal. Then we can balance the equation, adding and subtracting, multiplying and dividing, where necessary.

First, note that

At this point we have to be careful. While algebra tells us to divide both sides by (x – 2), we need to be aware of the following: if x is between 0 and 2, (x-2) yields a negative. However, (x + 2) yields a positive, meaning (x – 2)(x + 2) gives us a negative. In this case (A) would be bigger.

However, if we divide each side by (x – 2) and solve we get the following:

DIVIDE EACH SIDE BY (x – 2)

SUBTRACT ‘X’ FROM BOTH SIDES

This hardly looks like a solution (in fact it looks like I forgot to go to grade school!). However, what this yields is the important insight: Column B is now 4 greater than Column A. Therefore, the answer is (D). Alternatively you could pick numbers, such as ‘0’ and ‘4’. Each gives us different answers, leading to the same conclusion: Answer (D).

## If the question has variables, but there no polynomials, plug in values.

This advice pertains to variables that are not in polynomial form, as seen above. Here come up with easy numbers to plug in to see which values the columns yield.

0 > x > y > z > -1

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

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

It would be very nice if we could just stop here and choose answer (B). However, things are not so simple. When plugging in one set of numbers we will inevitably come up with one outcome. Once we’ve plugged in and come up with one answer, whether it is (A), (B), or (C), our job is to disprove that answer.

Can we make Column A larger than Column B? Well what if we plug in a values for x and y that are very close to one another.

**Takeaway**

You should be adept at both algebra and plugging-in to efficiently—and accurately—answer a quantitative comparison question that contains variables. Typically it is best to simply check for polynomials before plugging in.

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For the first question you say: “However, what this yields is the important insight: Column B is now 4 greater than Column A. Therefore, the answer is (D)”. Wouldn’t the answer be “The quantity in Column B is greater”; answer B? You said column B would be 4 greater.

Hi Ryan,

The key to this explanation is the word “now.” In this question, we have one example where Column A is greater than Column B: “While algebra tells us to divide both sides by (x – 2), we need to be aware of the following: if x is between 0 and 2, (x-2) yields a negative. However, (x + 2) yields a positive, meaning (x – 2)(x + 2) gives us a negative. In this case (A) would be bigger.”

Then, we continue to solve this equation by dividing by (x-2), which eventually yields a situation in which Column B is greater than Column A. This means that there are viable solutions for x in which Column A is greater, and other viable solutions in which Column B is greater. This means that we can’t determine whether Column A or Column B is greater without more information about ‘x’. In this situation, we must choose answer choice “D”. I hope this helps 🙂

Hello!

I’ve been getting stuck on questions like:

x> IyI> z

Quantity A: x + y Quantity B: IyI + z

I automatically assumed the answer was B because x is greater than IyI which means y could be positive or negative. Therefore the absolute value of y would have to be greater than the non-absolute value. I have a feeling like I’m missing something?

Hi Sarah,

Happy to help! 🙂

You’re thinking about this the right way but making a small mistake. The problem is that you do not know if x and z are positive or negative, either. So if x and z are positive, then B would be larger, but if x and z are negative, then it is possible that y could be positive and make x + y larger than |y| + z. Therefore, it is not possible to determine. Does that make sense? I hope so!

doesn’t x > |y| > z imply that x must be positive as |y| must be positive?

Hi Kai,

You’re completely right, which means that the explanation above isn’t quite right. If x > |y|, then x must be positive. On the other hand, y doesn’t have to be positive, since if y=0, then |y| = 0, and 0 is not positive or negative. With that being said, in the original question, we have

x + y vs. |y| + z.We know that x > |y| and that x > z. However, we don’t know the relationship between y and z or y and |y|. Here are a couple of possibilities:

1. y > zIn this case, x > |y| and y > z. This means that the two variables on the left are greater than the two variables on the right, so Quantity A is greater.

2. y = 0 = |y|If y = |y| = 0, then we know that x > z and y = |y|, so Quantity A will be greater.

3. |y| > z, y < zIf, for example, y = -1 and z = 0.5, then |y| > z but y z but |y| > y. Unlike in the other two cases, in this situation, we cannot definitely say which quantity is greater. This is because the variables on one side are not both larger than the variables on the other side.

So, we’re unable to determine whether Quantity A or Quantity B is larger.

why cant we use the formula for (x-2)2 instead of writting it as simply (x-2)(x-2)

same question… did you got the answer ?

Hi Chris,

For question number one, instead of setting the polynomials equal to each other, can we set each polynomial equal to zero? So (x-2)(x-2) = 0 & (x-2) (x+2) = 0, in which case we’d get x = 2 for A and X = 2 or -2 for B, and so D would still be the correct answer.

Now that I think about it, if they’re both equal to zero they’re both equal to each other, right? So this should work for all similar equations? Thanks in advance!

Hi Amira,

Yes, I think it is dangerous to work backwards that way, by assuming they are the same, which is answer (C). Instead, plug in different values to see whether the relationship between the two columns is constant. If the relationship is not constant, then the answer is (D).

Novas gre bible has strategies such as only needing to plug in three times in a question, eliminate the greatest or least answer choice, if it is asking for that, eliminate numbers that merely repeat from the question, eliminate ones that are to easily derived by elementary operations, and eliminate answer choices that state not enough information. The information that is in here is great, but is it truthful? Thanks.

In order to answer your question, I’d have to see a specific problem and see if that advice really holds up. I could see some of that advice actually back-firing, so I would take it with a grain of salt :).

Hello Chris

For the question like x/y vs y/z shown above, is there any trick by which we can make sure that we have tried all the possible outcomes. I tried two values and got A greater for both values which would have given me the wrong answer.

Hi Nitish,

Conditions permitting, make sure you try -1, 1, 0, 1/2, -1/2, 2, -2, and any big number.

If at that point, you can’t come up with answer (D), then you can be pretty confident the answer not (D).

I have been using wordsmart and my dictionary and theatrrus. Good idea?

Yes Craig,

Word Smart is book I highly recommend. A thesaurus can be helpful but make sure you learn the nuances between words. To do so, just look in your trusty dictionary :).

sounds good. Thanks so much.

Chris, my buddy is worried about the vocab. 1000 words and so forth. I told him to use you as a guide, because you can determine how each of the books differ in their approaches. I have been printing out your blog pages for myself and him. Are the books that state these are the most used words for example, the right books to use? Thanks.

Craig,

That’s a good question. I think in general the books that state ‘most used words’ are pretty good words to learn. By no means should you limit yourself to these books, nor for that matter think that they are better than books that do not make this vaunted claim. For instance, a book that claims it has the 15o most current words will not be a more effective study tool than the Barron’s 1100 book or Princeton Review Word Smart. Test day the chances of you seeing words from the Barron’s and PR are much higher – there are simply more words in those books.

Hope that helps!

Chris, Root list, top gre words in context, word groups. All of these are from my Kaplan gre book i had for a while. Useful to use?

Hi Craig,

The Kaplan roots are a waste of time (its more roots fault than Kaplan’s). The word grouping is downright dangerous and leads one to lump words together that can be very different. I wouldn’t recommend the book.

Quick clarification for question one….at the end you say “Alternatively you could pick numbers, such as ’0′ and ’4′” …. do u mean just plug in 0 into both column 1 and 2 and compare….and/or plug in 4 into both column 1 and 2 and compare and reach the final conclusion?

Thanks

My approach was: From the start I plugged in and tried 0, -1, and 2 in both columns. I got 4 > -4, 9>-3, and 0 = 0 respectively …Is this a bad way to go about the problem?

Sure, plugging in numbers and comparing columns is a very effective technique on QC. Once you’ve established a relationship using one set of numbers use another set of numbers to see if you can disprove the relationship. As long as it is allowed by the problem, using ‘0’, ‘1’ and fractions is typically effective.

Chris, What is the best way to use QC Strategies on Geometry? I use manhattans and it is good, but it is very very in depth. How can I be sure for example that i can take my shapes to extremes?

And also, for inscribed figures, can you have special right triplets in them? Ex: 3,4,5? Or is it usually 45,45,90 (30,60,90)? Thanks.

HI Craig,

Good question – as long as there are no parameters to a QC geometry question (right angles, x > y, etc) you can manipulate the figure at will.

For your other question, you can have 3:4:5 inscribed in circles. For that matter you can have any right triangle set of triplets inscribed in a circle as long as the length of the diameter is the length of the hypotenuse.

Hope that helps!

I think the 1st example is incorrect!

The answer must be D.

At the step dividing both sides by (x-2) is unable as you don’t know which does x is stand for (For instance if x = 2 the equation would be indefinite)

Try: if x = 1 then A = 1 > B = -3

if x = 3 then A = 1 < B = 5

Excellent point!

Yes, you are absolutely right :). That is definitely a mistake on my part. Thanks for catching that!

i think answer to first question should be D because for some value of x ( for x= -1) quantity in column A is greater and for some other value of x (for x=3) column B is greater and also for x=0 both column have equal value.

Yes, you are right. My error :).

In example one can we cancel out (x-2) even if we don’t know the sign on the variable?

Actually, that is my mistake. I’ve change the text to reflect the correct approach. Sorry for any confusion :).

Hi Chris,

I have a point to raise regarding your first example involving the polynomial comparison. When treating the two polynomials as functions and graphing, we see there are values in which the function in column A is either greater, less, or equal to the function in column B; for x=0, column A yields a value of 4 while column B yields a value of -4; for x = 3, column A yields a value of 1 while column B yields a value of 5; for x=2, columns A and B both yield a value of 0. Ergo, I believe the answer should be (D).

Yes, you are absolutely correct. That is my oversight. Thanks for catching that :).

I appreciate the second example, I was convinced it was B but then I realized if you try to plug in different numbers you end up with A. So delusory!

No problem – it’s a tricky question :).

This is great chris. I always see the posts but then u don’t know what to do. When u say compare, make like another is that an algebra based strategy? And also can u make one like this for the other question types? Thanks.

Yes, I hope that is helpful :). I think the best approach is knowing that you can use a mixture of algebra and plugging in values to arrive at the answer.