Math Workout for the GRE, 3rd Edition (2013)

Chapter 4. Algebra, and How to Get Rid of It

ALGEBRA

If you’re planning to attend graduate school, you’ve probably had some sort of algebraic training in the dark reaches of your past. Algebra is the fine art of determining how variable quantities relate to each other within complex functions, and it dates back more than 3,000 years to ancient Babylon.

If it seems like it’s been 3,000 years since you last studied algebra, or even if you flunked algebra last year, fear not. This chapter is devoted to the following two very important pursuits:

· refreshing your memory of the basic algebraic rules that the GRE tests, and

· removing almost all algebra from your GRE experience.

It’s impossible—and not in your best interests—to ignore algebra entirely on the GRE, because there will likely be a few questions that are best solved by using algebra. But for the most part, the best strategy is to embrace new techniques that help you look at these problems from a different perspective. Naturally, the best way to subvert the rules is to review them first.

Know the Lingo

Any letter in an algebraic term or equation is called a variable; you don’t know what its numerical value is. It varies. Until you solve an equation, a variable is an unknown quantity. Any number that’s directly in front of a variable is called a coefficient, and the coefficient is a constant multiplied by that variable. For example, 3x means “three times x,” whatever x is.

Combining Like Terms

If two terms have the same variables or series of variables in them, they’re referred to as like terms. You can combine them like this:

6a + 4a = 10a

13x − 7x = 6x

For example, if you have six apples in one hand and four in the other, you have a total of 10 apples (and a pair of humongous hands).

Solving an Equation

Whenever a variable appears in an equation and you have to find the value of the variable, you have to “isolate” it by following these steps.

· Step One: Use addition and/or subtraction to put all the terms that contain the variable on one side of the equal sign and all terms that don’t contain the variable on the other.

· Step Two: Use multiplication or division to remove whatever coefficient the variable has, until the variable is sitting there all by itself.

In order to solve an equation, you must rely on the following paramount rule of algebraic manipulation:

You can do anything you want to an equation as long as you do exactly the same thing to both sides.

Question 5 of 20

If 4x − 5 = 19, what is the value of x ?

Here’s How to Crack It

In order to get all variable terms on one side of the equal sign and all constant terms on the other, follow Step One and add 5 to both sides.

4x − 5 + 5 = 19 + 5

The 5’s on the left cancel out, and the equation becomes 4x = 24. There’s no more addition or subtraction to be done, so you’re through with Step One.

All that’s left is to deal with the coefficient: 4. Because 4x equals “4 times x,” you can simplify the equation by dividing both sides by 4:

=

x = 6

Always Check

Taking the GRE involves a lot of stress and fatigue, so it’s easy to make a careless error when you’re manipulating an algebraic equation. Therefore, it always pays to plug your answer back into the original problem to make sure it works. Let’s do it.

4(6) − 5 = 19

24 − 5 = 19

   19 = 19. Check.

Keep in mind that solutions to equations don’t always have to be integers, so don’t be concerned if your result is a fraction. As long as it works when you plug it back into the equation, you’re fine.

Question 4 of 25

If 4m + 7 = 16 − 2m, what is the value of m ?

Here’s How to Crack It

Step One

Get all the variables onto the left side of the equation by adding 2m to both sides, then move all the constants to the right side by subtracting 7 from both sides.

4m + 7 + 2m = 16 − 2m + 2m

6m + 7 = 16

6m + 7 − 7 = 16 − 7

6m = 9

Step Two

Divide both sides by 6.

=

m = =

Step Three

Check your work by plugging in.

4 + 7 = 16 − 2

+ 7 = 16 −

6 + 7 = 16 − 3

13 = 13. Check.

Take special note of Step 3 now and be prepared to use this skill a lot in the latter half of this chapter.

Solving an Equation Quick Quiz

Solve for x in each of the following:

1. 2x − 5 = 11

2. 3 − 5x = 13

3. 12x + 4 = 4x

4. 8x − 9 = x − 2

5. + 5 = -9

Explanations for Solving an Equation Quick Quiz

1. x = 8

2. x = −2

3. x = −

4. x = 1

5. x = −28

Inequalities

Inequality symbols are used to convey that one number is greater than or less than another.

The symbols used in inequalities are as follows:

> means “is greater than”

< means “is less than”

≥ means “is greater than or equal to”

≤ means “is less than or equal to”

Even though the two sides of an inequality aren’t equal, you can manipulate them in much the same way as you do the expressions in regular equations when you have to solve for a variable.

Question 13 of 20

If 5b − 3 > 2b + 9, which of the following must be true?

b > −4

b > 2

b < 2

b > 4

b < 4

Here’s How to Crack It

Adding and subtracting take place as usual, like this:

5b − 3 − 2b > 2b + 9 − 2b

3b − 3 > 9

3b − 3 + 3 > 9 + 3

3b > 12

At this point, because the coefficient of b is positive, you can divide both sides by 3 and get the final range of values for b.

To check your solution, try a number that is greater than 4 (say, 5) and see if the inequality holds true.

5(5) − 3 > 2(5) + 9

24 − 3 > 10 + 9

21 > 19. Check. The answer is (D).

Flip That Sign!

The only difference between solving equalities and inequalities is this one very important rule:

Whenever you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.

Let’s try a problem.

Question 16 of 20

If 5 − 11p ≤ 9, which of the following could be the value of p ?

Indicate all such values.

−15

−10

−2

0

1

8

24

Here’s How to Crack It

Manipulate the problem as you would a regular equality, like so:

5 − 11p ≤ 9

5 − 11p − 5 ≤ 9 − 5

−11p ≤ 4

Now that you’re dividing by −11, flip the sign and you’re done.

Any value that is greater than − is a possible value of p, so you should check the box next to 0 and every other box with a number that’s greater than zero. See how important that little rule is? If you didn’t know about it, you might have picked all the numbers that were less than −, which would be the exact opposite of the correct answers. And that would have been unfortunate.

Everybody Dance

Pretend you had two guys, Al and Bob, who wanted to each dance with two gals: Cathy and Daisy. To make sure each guy danced with each gal, we could first have Al dance with Cathy, and then have Al dance with Daisy. Now that Al’s danced with both ladies, Bob gets to dance with Cathy, and then Bob gets to dance with Daisy. Because we systematically paired Al up with each possibility, and then paired Bob up with each possibility, we’ve set up every possible pair of dance partners.

Why is this important? Some inequalities questions will ask you to find the range for two combined inequalities, and we’ll answer these questions similarly. For that type of question, you will need to find every possible result of combining the two inequalities: The least number of the first inequality will have to be paired up with the least and greatest numbers of the second inequality, and the greatest number of the first inequality will be paired up with the least and greatest numbers of the second inequality. We’ll end up with a total of four numbers: Pick the least and greatest, and that’s your combined range.

Question 12 of 20

If −5 ≤ a ≤ 12 and −10 ≤ b ≤ 25, which of the following represents all possible values of a − b ?

−30 ≤ a − b ≤ 22

−30 ≤ a − b ≤ 5

−13 ≤ a − b ≤ 5

5 ≤ a − b ≤ 22

5 ≤ a − b ≤ 30

Here’s How to Crack It

For the range of a, we have two numbers: −5 and 12. We’ll need to make sure that each of those numbers gets to dance with each of the two numbers from the range of b.

a − b

(−5) − (−10) = 5

(−5) − (25) = −30

(12) − (−10) = 22

(12) − (25) = −13

Notice how each number from a was paired up with each number from b? Our least result for a − b was −30, and our greatest was 22, so the full range is −30 ≤ a − b ≤ 22, answer (A).

Inequalities Quick Quiz

In questions 1-3, find the range of values of x.

1. 3x + 7 > 22

2. 8 − 3x < 13

3. − + 2 ≥ 5x − 9

Question 4 of 6

If y is a positive integer and 6y + 9 > 5 + 8y, then y =

Question 5 of 6

If 5f + 11 ≥ 17 + f and 7 − 4f > −13, f could equal each of the following EXCEPT:

1.6

2.4

3.9

4.1

5.3

Question 6 of 6

If −7 ≤ x ≤ 5 and −15 ≤ y ≤ 0, what is the greatest possible value of xy ?

Explanations for Inequalities Quick Quiz

1. x > 5

2. x > −

3. x ≤ 2

4. If you solve the inequality, you find that y < 2. The only positive integer less than 2 is 1, so x must equal 1.

5. Solve both inequalities. If 5f + 11 ≥ 17 + f, then f ≥ 0.5, and if 7 − 4f > −13, then f < 5. Therefore, f could be anywhere in the range of 0.5 up to but not including 5. The only answer choice that is not in this range is (E).

6. Let’s pair up our greatest and least from each inequality:

x × y

(−7) × (−15) = 105

(−7) × (0) = 0

(5) × (−15) = −75

(5) × (0) = 0

So the range is −75 ≤ xy ≤ 105 and thus the answer is 105.

Quadratics

Remember all that FOILing and its necessary factoring you did in high school algebra? Well, you probably won’t have to go through all the trial and error of that type of factoring, but there will be circumstances in which you’ll have to combine a pair of binomials. (That’s just a math term for an algebraic element that contains two terms, like “2x + y.”)

FOIL

When FOILing you combine two binomials by multiplying the First, Outside, Inside, and Last terms, and then simplify wherever possible. Let’s give it a shot.

Question 11 of 20

What is the product of (x − 3) and (2x + 7)?

2x² − 21

2x² + x − 21

2x² + 13x − 21

2x² + x + 21

2x² − 13x + 21

Here’s How to Crack It

If you’re new to FOILing, it helps to line up your products so you can keep track of what you’re doing:

So, the correct answer is choice (B).

Solving Quadratic Equations

When quadratics and equal signs come together, the result is a quadratic equation. On the GRE, quadratics are usually set equal to zero, and you’ll have find an equation’s solutions, or roots, by factoring.

Question 2 of 20

If x > 0 and x² − 5x − 6 = 0, what is the value of x ?

Here’s How to Crack It

To answer this question, you need to solve this equation (i.e., find the equation’s roots) so you’ll have to factor the quadratic. Factoring is basically the opposite of FOILing, and it usually requires a little trial and error. Placing the two x’s in the parentheses is the easy part. The challenge lies in finding two numbers whose sum is −5 (the middle coefficient) and whose product is −6 (the last term). In this case, those numbers are −6 and 1.

x² − 5x − 6 = 0

(x )(x ) = 0

(x − 6)(x + 1) = 0

In order for the product of two numbers to be 0, one of them must be 0. So set both factors equal to 0 and solve for x.

Check your answers to make sure you didn’t make any unfortunate slip-ups.

Now, we know 6 and −1 are the roots of the equation, but the question only asks for the positive root. So the answer is 6.

Common Quadratics

As basic as FOIL is, there are a few very common multiplications of binomials that come up so frequently that you’re better off memorizing them. Standardized-test writers like them a lot, because they’re great for making easier problems seem a lot more difficult.

(x + y)² = x² + 2xy + y²

(x − y)² = x² − 2xy + y²

(x + y)(x − y) = x² − y²

Knowledge of the last formula—which is commonly referred to as a “difference of squares”—is especially useful if you come across a question that looks like this:

Question 14 of 20

If = 7, what is the value of x ?

Here’s How to Crack It

Rather than resort to cross-multiplication, here you can recognize that the fraction on the left is in the form of ; in this case, y is the constant. Because = x + y, you can rewrite the left side as = x + 3, and your math becomes easy. If x + 3 = 7, then x = 4.

Here’s another one on which a little quadratic knowledge can save you some time.

Question 19 of 20

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

If you look closely, you can see that the given term follows the format = x + y, so Quantity A must be 836 − 835, or 1. Therefore, the answer is (C). You could also use your calculator, but applying the common quadratics can be faster if you have them memorized.

Quadratics Quick Quiz

1. If x² − 5x + 6 = 0, what are the possible values of x ?

2. If x² + 7x − 18 = 0, what are the possible values of x ?

3. If x² + 8x − 15 = 0, what are the possible values of x ?

4. = 15, then x =

5. If a + b = 5 and a² + b² = 15, then ab =

Explanations for Quadratics Quick Quiz

1. Because x² − 5x + 6 can be factored to (x − 2)(x − 3), then the possible values of x are 2 and 3.

2. Because x² + 7x − 18 can be factored to (x − 2)(x + 9), the two possible values of x are 2 and 9.

3. Since x² + 8x − 15 can be factored to (x − 3)(x − 5), the two possible values of x are 3 and 5.

4. Because , the equation can be rewritten as x + 13 = 15. Therefore, x = 2.

5. This question makes you think you have to solve for a and b, but you don’t. If a + b = 5, then (a + b)² = 5², or 25. When you expand the term on the left, (a + b)² becomes a² + 2ab + b². Rewrite the equation as a² + b² + 2ab = 25; because a² + b² = 15, you can substitute 15 in the equation like this: 15 + 2ab = 25. Therefore, 2ab = 10, and ab = 5.

Simultaneous Equations

If a single equation has two variables, you can’t solve for either one. If 2x + y = 5, for example, then x = 2 and y = 1 could be one solution; but x = 1 and y = 3 could be another. But if you have two distinct equations and two variables—often referred to in high school math courses as “two equations, two unknowns”—then each variable has only one possible solution.

If a question asks you to find the value of one variable, you can usually add or subtract the equations and solve.

Question 16 of 20

If 2x − 3y = 7 and x + 3y = 8, what is the value of x ?

Here’s How to Crack It

Look at the coefficients of the y terms; if you add them together, you get zero. So line up the equations like this and add all the like terms separately.

From here, you can determine that x = 5.

When There’s Less Work Than You Think

ETS likes to build its simultaneous equation questions to look a lot more daunting and time-consuming than they actually are. Take this problem for example.

Question 9 of 20

If a + 3b = 10 and a − b = 8, what is the value of a + b ?

At first glance, you might think you have to solve for a and b individually and then add them together to get your final answer. But that isn’t the case.

Here’s How to Crack It

If you add the two equations together, you get a new equation.

a + 3b = 10

a − b = 8

2a + 2b = 18

If you divide each term in this equation by 2, you will see the answer right away.

2(a + b) = 18

a + b = 9

As it turns out, we don’t have to know what the individual values of x and y are. All we have to know is that their sum is 9.

Simultaneous Equations Quick Quiz

1. If 2x − 3y = 2 and x − 5y = −6, what is the value of y ?

2. If 3a − b = 200 and 5a + 2b = 40, what is the value of a − b ?

3. If 4m − 3n = 12 and m − 2n = 3, then m − n =

4. At a stationery store, three pens and five notebooks cost a total of $19.02 and two pens and three notebooks cost a total of $11.63. What is the price of one notebook?

Answers to Simultaneous Equations Quick Quiz

1. Because you’re solving for y, you want to make the x’s disappear. You can do this by multiplying the second equation by −2 so that it becomes −2x + 10y = 12. Now, add the equations and the x’s drop out.

2. Solve for a first by multiplying the first equation by 2: 6a − 2b = 400. Now add.

Now plug 40 into either equation to find b: 3(40) − b = 200, so b = −80. Therefore, a − b = 40 − (−80) = 120.

3. This one makes you think you have to solve for both variables individually, but you don’t if you just stack them and add them.

4. This is a simultaneous equation problem masquerading as a word problem. If three pens and five notebooks cost a total of $19.02, then 3p + 5n = 19.02. Similarly, the second equation can be rewritten as 2p + 3n = 11.63. To find the value of n, get rid of the p’s.

Each notebook costs $3.15.

All right, that was a lot of algebra to learn, especially after we said that you should use as little algebra as possible on test day. But now we’ll talk about why you should use as little algebra as possible.

WHY ALGEBRA IS NOT TO BE TRUSTED

What is x, anyway? What does it mean to have y marbles in a jar? Or z chairs at a table? Nothing, that’s what. Algebra has been the backbone of mathematics for millennia, but that doesn’t mean it will do you any good on the GRE. In fact, it’s more likely to trip you up.

Let Us Count the Ways

A basic appreciation of algebra is crucial for a good quantitative score on the GRE. But whenever you have the option, you should use arithmetic instead of algebra, for a number of reasons.

· Algebra is based on abstract unknowns, while arithmetic is a concrete system of numbers you can visualize (x apples versus 2 apples).

· You started doing arithmetic long before you even knew what algebra was, so arithmetic is far more ingrained in your brain.

· You probably haven’t done a lick of algebra in a very long time, but you perform some sort of arithmetic every day (whether you realize it or not), so arithmetic is far more familiar.

· Your calculator can help you with arithmetic calculations, but it’s useless for algebraic manipulations.

Some of you out there might think your algebra skills are passable, and that you’re not worried about making mistakes. That may well be, but we’re here to tell you that things change once you’ve left the friendly confines of preparing and you’re actually taking the GRE for real. Stress happens, and even the most brilliant math students can choke under pressure.

The other bad thing about algebra is that often when you’re performing it, you can mess something up and not even know it—because the answer you get, the wrong answer—is usually right there among the answer choices.

Do the Unexpected

If there were two roads between your house and the center of town and you knew one of them was full of landmines, which road would you choose?

This brings us to the most important reason not to use algebra. Test writers know that most students have been trained to use algebra when they see an algebraic question, and that many students make lots of careless mistakes. Therefore, most of the wrong answers test writers dream up to use in questions are based on the errors they anticipate you’ll make. If you avoid using algebra as much as possible, you’ll avoid many of the pitfalls that ETS has laid out for you.

So when you see a problem with a bunch of variables in it, don’t think the way they think you’ll think. Instead, Plug In numbers.

PLUGGING IN

Plugging In is a lot like writing your own novel; rather than wonder how many candies Phil has in his hand, assume the power to write the narrative. Decide for yourself how many he has.

When to Do It

There are two dead giveaways when you’re identifying problems that you can solve using your own numbers.

· The problems often feature the phrase “in terms of.”

· The answer choices have variables in them.

1. Recognize the Opportunity to Plug In. If you have variables in the answer choices, Plug In. If there is an unknown quantity in a problem that cannot be solved for, Plug In.

2. Set Up Your Scratch Paper. Write down letters for each answer choice and any important information from the problem.

3. Plug In an Easy Number for the Variable. Choose an easy number, such as 2, 3, 5, 10, or 100, for one of the variables. Write clearly what number you’re plugging in for which variable. If the problem gives you certain limitations for the value of the variable, such as “odd integer larger than 30,” be sure to choose a number that follows those limitations. If there are multiple variables in a problem, see if you can solve for the other variables once you’ve plugged in for the first one. If you can’t, Plug In for those variables as well.

4. Find Your Target Number. Solve the question using your numbers. Whatever is asked at the end of the question is your target number. Write it on your scratch paper and circle it.

5. Check All of the Answer Choices. Using the numbers that you plugged into the question, try out every answer choice. Put a check mark next to any answer that gives you your target number, but keep checking. You must check every answer. Most of the time, only one answer choice will work out, but if you end up with two or more answer choices that give you your target number, Plug In a new set of numbers, find a new target number, and check the remaining answer choices.

Let’s try an example:

Question 11 of 20

George is twice as old as Mary, and Mary is three years older than Juan. If Juan is j years old, then, in terms of j, what is George’s age in 10 years?

2j − 4

2j − 14

2j + 13

2j + 16

2j + 26

Trigger: Variables in the
answer choices.

Response: Plug in.

Here’s How to Crack It

Look at the answer choices first. You should recognize the opportunity to Plug In as soon as you see answer choices like those. There are variables in all of our answer choices, which means this is a perfect Plug In problem. Now that we know what we’re going to do, don’t sit and stare at the problem. We’re going to need to set up our scratch paper by writing down A B C D E. Since our variable is j, let’s Plug In an easy number for Juan’s age. Let’s say Juan is 5, so j = 5.

Now we can work through the problem. Take it apart piece by piece. The first part of the problem states that George is twice as old as Mary, which is nice for George and Mary, but we know only how old Juan is. Let’s leave that part of the problem alone, and move on. The next part of the sentence states that Mary is three years older than Juan. Hey, now there’s something we can solve. Since Juan is 5, and Mary is three years older than Juan, Mary must be 8. Notice that since we’re using real numbers, it’s easy to make sure we’re doing the right math: Is 8-year-old Mary 3 years older than 5-year-old Juan? Definitely.

We know Mary’s age now, so we can go back to that first part of the problem that we skipped earlier. If George is twice as old as Mary, then George is 16 years old. Now we can answer the question: “If Juan is j years old, then, in terms of j, what is George’s age in 10 years?” Ignore the phrase in terms of. Whenever you see that in a plugging in problem, it just means that you Plug In. If George is 16 now, how old will he be in 10 years? He’ll be 26 years old, so that’s our target number. Write down 26 and circle it.

Now we can check all of the answer choices.

Replace j with 5 in each of the answer choices; the one that gives you an answer of 26 is the correct answer.

(A)2(5) − 4 = 6 Nope.

(B)2(5) − 14 = −4 Worse than nope, it’s impossible; no one can have a negative age.

(C)2(5) + 13 = 23 Nope.

(D)2(5) + 16 = 26 Bingo!

(E)2(5) + 26 = 36 Nope.

Since (D) is the only answer that gave us our target number of 26, that’s our answer. Your scratch paper should look something like this:

If you had solved this algebraically, you might have added j and 3, then doubled it to 2(j + 3), then distributed it to become 2j + 6, then added 10, and gotten the right answer. But say you forgot to distribute the 2, and 2(j + 3) became 2j + 3, and you added 10 to get 2j + 13. You could have chosen answer choice (C) and moved merrily along, unaware of your mistake, because ETS anticipated it.

Instead, you solved the problem using only arithmetic that you double check with your calculator if need be, and you avoided all of the traps. Let’s try another.

Question 8 of 20

If a machine working at a constant rate produces 3,000 golf balls per hour, how many golf balls can four of these machines make in y minutes?

200y

750y

Here’s How to Crack It

Woah, look at all those variables in the answer choices! Now that we’ve recognized the opportunity to Plug In, go ahead and set up your scratch paper by writing A B C D E vertically on the left side. Now choose an easy number to Plug In. There are 60 minutes in an hour, so you can make your math a little easier by choosing either a factor of 60, such as 30, or a multiple of 60. Write down y = 30.

In 60 minutes, one machine makes 3,000 golf balls. In that case, in 30 minutes one machine would make 1,500 golf balls. The question asks how many golf balls four machines can make in 30 minutes, which is 1,500 × 4 = 6,000 golf balls. Write down 6,000 golf balls and circle it. We’ve found our target number, so we now have to check all of our answer choices and see which one gives us 6,000.

Replace x with 30 in each of the answer choices; the one that gives you an answer of 6,000 is the correct answer.

(A)200(30) = 6,000 Gotcha!

(B) = 6.66 Way too small.

(C)750(30) = 22,500 Way too big.

(D) = 250 Nope.

(E) = 24,000 Nope.

The answer is (A).

You might be asking yourself, “If I got a match right away, why did I have to spend that time checking all of the others?” And that’s a good question, because your ultimate goal is to find the correct answer and scoot off to the next question as quickly as possible. Sometimes we may choose a number that works with multiple answer choices. For instance, say we had plugged in y = 60 instead. In that case, the four machines would have made 12,000 golf balls. Answer (A) still works: 200(60) = 12,000. However, look at answer (E): = 12,000. Had we picked 60, we would have had two answer choices that gave us our target number. If that happens, just pick a different number and check the remaining answers.

What to Plug In

When you plug numbers into a question, it’s perfectly fine to choose almost any number you want, as long as it doesn’t violate restrictions that the problem stipulates. Usually, the first integer that pops into your head will work just fine, although as you get better at it you’ll get a feel for picking numbers that make your math easier.

Question 13 of 20

At Markham Academy, of the m faculty members live in housing on campus grounds. Of those living on campus, own a car. Which of the following represents, in terms of m, the number of faculty members who live in campus housing and do not own a car?

Here’s How to Crack It

Since we’ve got variables in the answer choices, recognize the opportunity to Plug In and set up your scratch paper. Now let’s choose an easy number for the variable. This question has a lot of fractions in it, and fractions mean division. The fastest way to come up with a good number is to multiply the denominators: Since we know we’re going to have to divide by 3 and 4, let’s try m = 12.

Now we’ll work through the problem in bite-sized pieces. Since of the 12 faculty members live on campus, = 8 people live on campus. Write down on your scratch paper that 8 people live on campus. The problem then states that of those 8 people living on campus, own a car. = 6 people on campus own a car. The question asks how many people live on campus do not own a car, which means that out of the 8 people on campus, 2 of them don’t own a car. Now that we’ve found our target number, write down 2 on your scratch paper and circle it.

Finally, we have to check each answer choice. Replace m with 12 in each answer choice to find which one gives us our target number of 2.

(A) = 3 Next.

(B) = 2 Yes!

(C) = 1 Nope.

(D) = 8 Nope.

(E) = 9 Nope.

The answer is (B).

It Gets Easier With Practice

The more problems you work on, the better you’ll get at choosing the best numbers. If a problem involves percents, for example, you’ll probably Plug In a multiple of 100. Questions based on inches and feet might work best with a multiple of 12. If you’re dealing with units of time, you might think in multiples of 60. Numbers like these often suggest themselves when you use common sense and ask yourself, “What numbers will make this easier?” And, for that matter, “What numbers will make this harder?” Just be careful to use multiples of conversion values because using the value itself is more likely to create multiple answer choices which match the target.

What Not to Plug In

Knowing the right number to choose for a Plug In problem is useful, but it’s more important to know what numbers to avoid. These are the numbers that can have a strange effect on the algebra and skew your results.

The following chart shows the numbers that cause trouble when you plug them in for variables. These numbers aren’t forbidden, but it’s best to avoid using them for most problems.

No Variables? No Problem!

Believe it or not, you can plug into a problem that doesn’t appear to have any variables at all. In these cases, there might not be any x’s or y’s, but there will still be an unknown quantity that, if you knew it, would make the problem easier to solve. Questions like these usually have answer choices that are fractions or percentages.

Question 18 of 20

During the first month after Johanna purchased stock in Amalgamedia Inc., the value of her shares fell by 20 percent. During the following month, however, Amalgamedia’s stock price rose by 40 percent. By what percent did shares of Amalgamedia change over those two months?

4%

12%

20%

50%

68%

Here’s How to Crack It

There are no variables here, but the problem would make a little more sense if you knew the price of a share of Amalgamedia when Johanna bought into it. So make one up. The added bonus of having no variables is that there’s no plugging into the answer choices at the very end, so you can skip Step Three.

Step One: Choose a number for each variable in the problem.

Because percents are involved, let’s say the stock cost $100.

Step Two: Come up with a “target answer.”

If the stock dropped by 20% during the first month, then price dropped by × 100, or $20, and the price went from $100 to $80. After a 40% increase, the stock moved up by × 100, or $32, to $112. This represents a 12% increase from the original $100, so the answer is (B).

Plugging In Quick Quiz

Question 1 of 4

Joshua is three times as old as Kali, and Kali’s age is three years more than double Louella’s age. If Joshua is j years old, which of the following represents Louella’s age?

Question 2 of 4

Ana has a collection of books that are either fiction or non-fiction. Sixty percent of Ana’s books are fiction, and 30 percent of the non-fiction books are about politics. What fraction of the books are non-fiction books that are not about politics?

Question 3 of 4

A community park has two rectangular playgrounds—a little one for children younger than age 5, and a larger one for children aged 5 and older. The larger playground is twice as long and five times as wide as the smaller playground. If the area of the small playground is M, then the area of the larger playground is how much larger than the area of the smaller playground?

2M

5M

9M

10M

11M

Question 4 of 4

An Internet café offers Internet access at the rates of $3 per half-hour for the first two hours and $2 for every half hour after that. If Charlene used the internet at the café for x + 5 hours, how much money, in terms of x, did she have to pay?

2x + 18

2x + 15

2x + 12

4x + 24

4x + 48

Explanations for Plugging In Quick Quiz

1. Let’s start with Louella and set l = 5. If Kali is three years more than double Louella’s age, then k = 3 + (2 × 5), or 13. Joshua is three times as old as Kali, so j = 3 × 13, or 39. Louella’s age (5) is the target, and if you plug 39 in for j among all the answer choices you’ll find that (E) is the correct answer: = 5.

2. There is no variable in the problem, but knowing the number of books Ana has would be a great help. Because we’re dealing with percents, let’s say she has 100 books. If 60 percent are fiction, then the other 40 are non-fiction. Because 30 percent of those 40, or 12, are about politics, the other 28 are neither fiction nor about politics, and reduces to . The answer is (B).

3. If the length and width of the small playground are 8 feet and 10 feet, respectively, then the area of that playground is 8 × 10, or 80 square feet. Therefore, M = 80. The larger playground is twice as long (2 × 8 = 16) and five times as wide (5 × 10 = 50), so its area is 16 × 50, or 800 square feet. The difference in these areas is 800 − 80, or 720 square feet (target answer). When you Plug In M = 80 to each answer choice, you’ll see that the answer is (C). If you leapt right toward (D), you are falling for a trap answer by picking the area of the larger playground rather than the difference between the two. Slow down and make sure you understand each question completely!

4. We don’t want to choose a value of x that we’ve already seen in the question or answer choices, so let x = 6. That means Charlene worked for 11 hours. She paid $12 for the first two hours (4 half-hours at $3 each), and the remaining nine hours (18 half-hours at $2 each) cost $36. The total (and target answer) is 12 + 36, or $48. After you Plug In x = 6 to the answer choices, you’ll find the answer is (D).

“Must Be” Problems

Every so often you’ll see a question that contains variables and the phrase “must be.” Basically, you can eliminate an answer choice as soon as you find one situation when it doesn’t work.

Because Must Be problems are looking for the answer choice that always works, we may need to try a lot of different numbers until we’re down to only one answer. The trick here is to use the numbers that most people don’t think to use. We’ll call those numbers FROZEN.

F − Fractions

R − Repeats

O − One

Z − Zero

E − Extremes

N − Negative

We won’t have to try every single FROZEN number for Must Be problems, but we may have to try several. First, however, we’ll try an easy number. It may be our usual easy numbers: 2, 3, 5, 10, or 100, but it could be any number that seems easy for the question. Don’t think too hard about finding the perfect number at this point. We’re just going to use the easy number to eliminate some or most of the answers. Once we’ve eliminated some answers, we’ll Plug In using a FROZEN number and see which other answers we can eliminate.

1. Recognize the Opportunity to Plug In. If you have variables in the answer choices and the problem says “must be,” Plug In.

2. Set Up Your Scratch Paper. Write down letters for each answer choice and any important information from the problem.

3. Plug In an Easy Number for the Variable. Choose an easy number, such as 2, 3, 5, 10, or 100, for one of the variables. If the problem gives you certain limitations for the value of the variable, such as “odd integer larger than 30,” be sure to choose a number that follows those limitations. If there are multiple variables in a problem, see if you can solve for the other variables once you’ve plugged in for the first one. If you can’t, Plug In for those variables as well.

4. Check All of the Answer Choices. You’re not looking for the right answer, but simply looking for any answers that you can eliminate.

5. Try a FROZEN Number. Check all the remaining answers using that number. If necessary, try another FROZEN number.

Trigger: Variables in the
answer; problem says
“must be.”

Response: Plug In a
simple number, and then
use FROZEN numbers.

Question 10 of 20

If |c| and d = 3c − 2, which of the following must be true?

−2 ≤ d < 10

d ≠ −2

d < 0

c < d

−14 < d

Here’s How to Crack It

First off, recognize the opportunity to Plug In. We’ve got variables in all the answer choices, and the question says “must be.” Set up your scratch paper by writing down A B C D E. Now let’s try an easy number. If c = 2, then d = 3(2) − 2 = 6 − 2 = 4. Now let’s check each answer. (A) works, because −2 ≤ 4 < 10, and (B) works, because 4 ≠ −2. Cross off (C), because 4 is not less than 0. (D) works, because 2 < 4, and (E) works, because −14 < 2. We’ve still got 4 answer choices left, so let’s try a FROZEN number. Let’s start with c = 0. If c = 0, then d = 3(0) − 2 = −2. Leave (A) because d is equal to −2. Cross off (B) because d does, in fact, equal −2. We’ve already eliminated (C), so we don’t have to check it again. We can eliminate (D), because 0 < −2 is not true. We can leave (E), because −14 < −2. Now let’s try another FROZEN number. Since there’s an absolute value in the problem, let’s try a negative number for c. If c = −3, then d = 3(−3) − 2 = −9 − 2 = −11. We’ve only got (A) and (E) left, so let’s try those. Cross off (A) because −3 is smaller than −2. The only answer left is (E), which works: −14 is less than −11. The answer is (E).

Here’s what your scratch paper should look like:

FROZEN numbers are weird numbers. These are the numbers people normally don’t think of when working on GRE math problems. However, there may be other numbers other than the FROZEN numbers that work better for certain problems. For instance, a problem about even and odd numbers may require plugging in an even number and then an odd number. We may need to try prime numbers or perfect squares on a problem about factoring. FROZEN will work most of the time, so stick with it unless you definitely see a different type of number that is necessary for that problem.

“Must Be” Quick Quiz

Question 1 of 4

If xy ≠ 0 and y is even, which of the following must also be even?

+ x

3y − 2x

+

2y − 3x

Question 2 of 4

If b and c are negative integers, which of the following must also be negative?

b³ − c³

bc² − c

bc(b − c)

b²c + bc²

Question 3 of 4

If m > 0 > n and m is odd, all of the following could be odd and positive EXCEPT

(mn)²

n³ − m³

(m − n)²

m² + n²

m(m + n)

Question 4 of 4

If positive integer a is multiplied by integer b and the result is less than a, then it must be true that b is

equal to 1

greater than 0

greater than a

greater than 1

less than 1

Explanations for “Must Be” Quick Quiz

1. If xy ≠ 0 and y is even, you can set x = 2 and y = 4 and consider the results.

= 1, which is odd. Eliminate it.

+ 2 = 4, which is even. Keep it.

3(4) − 2(2) = 8, which is even. Keep it.

+ = 2, which you can get rid of because it isn’t an integer.

2(4) − 3(2) = 2, so you can keep this as well.

Next select an odd value for x. Let’s make x = 3, leave y = 4, and check the remaining three.

+ 3 = 5, which is now odd. Get rid of it.

3(4) − 2(3) = 2, which is still even. Keep it.

2(4) − 3(3) = −1, which is odd.

You’ve eliminated all the others, so the answer is (C).

2. Plug In two negative numbers, and see what happens. Let b = −2 and c = −3.

(−2)³ − (−3)³ = −8 + 27, which equals 19. Dump it.

(−2)(3)² − (−3) = −18 + 3 = −15. Keep it.

(−2)(−3)[−2 − (−3)] = 6. Dump it.

. Dump it.

(−2)²(−3) + (−2)(−3)² = −12 + −18 = −30. Keep it.

We have two answer choices left, and we have to find a way to make one of them positive. If you let b and c both equal −1, you get what you’re looking for: (−1)(−1)² − (−1) = 0, which isn’t negative, while (−1)²(−1) + (−1)(−1)² = −2. The answer is (E).

3. Because m is odd and positive, let m = 3. Now, let’s Plug In n = −2 and see what transpires.

(3 × −2)² = 36. Positive but not odd, so keep it.

(−2)³ − 3³ = −35. Odd but not positive, so keep it.

[3 − (−2)]² = 25. Odd and positive, so get rid of it.

3² + (−2)² = 13. Also gone.

3(3 + −2) = 3. Ditto.

You have two answers left, and you want to make answer choice (A) odd or answer choice (B) positive. If you Plug In an odd number for n in (A), you can do the former: (3 × −3)² = 81, which is odd and positive. There’s no way (B) will ever be positive, so the answer is (B).

4. Think of this one from a slightly different perspective by choosing a value of a and the value after a is multiplied by b; this product ab must be less than a. But, if you multiply a positive integer by another positive integer, the product will always be the same or bigger (Try plugging in 2 for a!). So you can eliminate answer choices (A), (B), (C), and (D). But no one said that b was positive. Therefore, if a = 2 and b = −3, it’s possible for ab to equal −6, which is less than 2. The answer is (E).

Plugging In on Quant Comps

Plugging In for Quant Comp questions is similar to Plugging In on Must Be questions. We’ve got four possible answers: (A), (B), (C), and (D). As we try different numbers, we’ll eliminate answers. We’ll want to pick some easy numbers first, and then try FROZEN numbers until we’ve either eliminated all the answers but (D) or keep getting (A), (B), or (C).

Let’s review the answer choices for Quant Comp questions first.

(A) means that Quantity A is always larger than Quantity B. B is never larger than A, and they are never the same.

(B) means that Quantity B is always larger than Quantity A. A is never larger than B, and they are never the same.

(C) means the two quantities are always equal. Quantity A is never larger than B, and Quantity B is never larger than A.

(D) means we’re not sure. Sometimes Quantity A is larger, sometimes B is larger, or sometimes they’re the same.

As we try each number, we’ll check our quantities. Is Quantity A larger right now? Then eliminate answers (B) and (C), because Quantity B is not always larger than A, and the two quantities are not always the same. Is Quantity B larger right now? Then eliminate (A) and (C), because Quantity A isn’t larger, and they’re not the same. Are the two quantities the same? Then eliminate (A) and (B), because Quantity A isn’t always larger, and Quantity B isn’t always larger.

So once we know one possible value for Quantity A and B, we can cross off two answers. Already down to a 50-50 shot, just by plugging in one number! Then we’ll try to Plug In more numbers, and see if we can eliminate whichever of (A), (B), or (C) is left.

Notice that we’ll never eliminate (D) when we Plug In. Answer choice (D) is our last ditch effort. First we’re going to try as hard as possible to see if (A), (B), or (C) work. If we eliminate all those, however, then we’re going with (D).

Trigger: Quant Comp with
variables.

Response: Set-up your
scratch paper and Plug In
using FROZEN.

The Steps in Detail

· Use your scratch paper: Write down the question number, A B C D vertically on the left side of the paper, and any information given in the problem. Remember that we’re going to mostly use POE with Quant Comp Plug In questions, so writing down the answer choices is incredibly important.

· Recognize that it’s a Plug In: If it’s a Quant Comp question with variables in both the Quantities, it’s definitely a Plug In question. You should immediately know how you’re going to solve that question: by Plugging In! If one quantity has a variable in it, and the other quantity is a number, then it may be a plug in. Check to see if there’s any quantity referenced in the problem for which you can’t actually solve. If so, it’s a Plug In question.

· Try an easy number: Choose a nice, easy number to Plug In for your variable. Don’t worry about being clever here, and don’t spend time thinking of the perfect number. Just try out whatever number you can think of that is allowed by the question. 2, 3, 5, 10, or 100 are all easy numbers to try out. Work through the problem with your easy number. If you have multiple variables in the question, see if you can solve for the other variables once you’ve plugged in one of them: If not, you’ll have to Plug In for that variable as well.

· Cross off two answers: Once you’ve solved the question with your first number, compare Quantity A and Quantity B. If A is greater, cross off (B) and (C). If B is greater, cross off (A) and (C). If the quantities are the same, cross off (A) and (B).

· Try a FROZEN number: Now we’re down to two answer choices, and we want to make sure that whichever answer we have, (A), (B), or (C), always works, no matter what we throw at the problem. Once again, our FROZEN numbers are great here, although those aren’t the only possibilities. If Quantity A is greater, try any number that you think will make Quantity B greater (and vice versa).

· If you can’t eliminate (A), (B), or (C), choose that. You may have to try several different numbers until you can prove to yourself that the answer is always (A), (B), or (C). If none of the FROZEN numbers look like they’ll change the quantities much, and there aren’t any other numbers you can try, then go ahead and pick whichever answer, (A), (B), or (C), you keep getting.

· If different numbers gave you different answers, pick (D). Crossed off (A), (B), and (C)? Then and only then do you pick (D).

Question 6 of 20

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

At first glance, you might think instinctively that Quantity B must always be greater, because 3 is greater than 2. The GRE is engineered to take advantage of these instincts, however. Follow our steps and you’ll see what we mean.

Start by setting up your scratch paper. Write down A B C D, and copy down Quantity A and Quantity B. Any scratch work we have to do, such as working out expressions for each quantity, will be done on the right side of the paper.

Choose an easy number. Let’s try x = 3. Write down x = 3 between the two columns. Now let’s find the values of each quantity. Quantity A is 2(3) + 1 = 6 + 1 = 7. Quantity B is 3(3) + 1 = 9 + 1 = 10. Right now, B is greater than A, so we can eliminate (A) and (C).

Now let’s see if we can pick a number that makes it so that Quantity B is no longer greater. Check your FROZEN numbers: How about zero? Quantity A is 2(0) + 1 = 0 + 1 = 1. Quantity B is 3(0) + 1 = 0 + 1 = 1. Now Quantity A equals Quantity B, which means B isn’t always larger. Cross off (B). Different numbers gave different answers so choice (D) is correct.

Your scratch paper should look something like this:

See how this works? Rather than guess, you can use a simple, methodical approach to eliminating the three wrong choices and choosing the one that remains. Let’s try another:

Question 5 of 20

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

Here is another situation in which instinct might lead you down the wrong road. Look how big Quantity A looks! How could Quantity B possibly measure up?

As always, get your hand moving by setting up your scratch paper. Write down A B C D, and copy down Quantity A and Quantity B. Now we’re ready to choose an easy number. Let’s try x = 2. Quantity A is 1,000x + 10, so it’s 2,000 + 10 = 2,010. Quantity B is 2² = 4. Since Quantity A is definitely greater than Quantity B, cross off answers (B) and (C).

We’re left with (A) and (D), so now let’s try FROZEN. We don’t have to try every single number, but let’s look to see if any of them look like they’ll make Quantity B greater than A. We could try zero again, but this time it just confirms what we already know: If x = 0, Quantity A would equal 0 + 10 = 10 and Quantity B would be 0, so Quantity A is still greater.

We’ve tried two sets of numbers now, but we’re still not done. Keep looking over FROZEN to see if any other numbers look promising. How about negative numbers? Let’s try x = −5. Quantity A is 1,000(−5) + 10 = −5,000 + 10 = −4,990. Quantity B is (−5)² = 25. Since A is negative and B is positive, Quantity B is now greater, and we can eliminate (A). Our answer is (D).

Here’s what your scratch paper should look like for this problem:

Sometimes, we’ll have limitations on what we can Plug In for certain variables. Whenever we Plug In, we’ll have to check to make sure that the numbers we plug in are allowed by the problem.

Question 2 of 20

42 < m < 49

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

We’ve got a lot more pieces of information to deal with on this problem, so rather than sitting and staring at this problem, get that hand moving by setting up your scratch paper. Write down A B C D, and copy down the inequality and Quantity A and Quantity B.

When we choose our easy number, we’ll have to keep in mind that the problem has limited the numbers we can choose: 42 < m < 49. So let’s choose an easy number somewhere between 42 and 49, such as 45. Let’s check each quantity. Quantity A is ≈ 0.803. Since that’s greater than 0.75, we can cross off (B) and (C).

Now it’s time to try FROZEN, but you may have already noticed something about our FROZEN numbers: We can’t use all of them for this problem. Because m has to be greater than 42 and less than 49, we can’t use zero, one, or any negative numbers. Looks like we’re stuck with extreme numbers and fractions. No problem. We’ll use a number either as extremely large or extremely small as is possible within our limited range for m, and we’ll use some non-integers.

Before we choose another number, however, let’s look at what we want to get. Right now Quantity A is greater than B, and we want to see if we can make it less. To make Quantity A less, we’ll want to choose as small a number as possible for m. Since we’re limited by 42 < m < 49, let’s choose the smallest number we can for m, such as m = 42.001. We could get smaller, but let’s start there for now. ≈ 0.7500179. Close, but still a little greater than B, so we can’t eliminate any answers. Let’s go even smaller then, and try m = 42.00001, which is about as much as we can enter into the on-screen calculator. ≈ 0.7500002. A is still greater than B and nothing we pick seems to change that. We’re down to (A) and (D), and since we couldn’t eliminate (A), that’s our answer.

As usual, here’s an example of what your scratch paper should look like. Notice that for each fraction entered into the calculator, we wrote down the fraction and the result of the calculation on the right side of our scratch paper.

If we have multiple variables, then we’ll only Plug In for one variable at a time. If we can’t solve for any of the other variables, then we’ll Plug In a different number for the next variable, and see if we can solve for any of the other unknowns.

Question 4 of 20

4a = 12b

2b = 10c

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

Variables in both Quantities mean you probably noticed that we can Plug In. Pick up your pen and start setting up your scratch paper. Write down A B C D and copy down the two equations, Quantity A, and Quantity B. We’ve got three variables in this problem, which means we’ll still choose an easy number, but we’ll then have to see if we can solve for the other variables. If we can’t, then we’ll make up numbers for those variables as well.

First off, look at the equations. We’re going to have to divide by 12 to find b and by 10 to find c, so let’s pick an easy number divisible by both 12 and 10, such as 120. We could actually pick any number, but 120 will keep us from having to deal with too many fractions.

If a = 120, then 4(120) = 12b, and 480 = 12b, and b = 40. Now use b to find c, so 2(40) = 10c, 80 = 10c, and c = 8. Quantity A is therefore 120, and Quantity B is 15(8) = 120. Since the two quantities are the same, eliminate (A) and (B).

Now let’s try FROZEN. We could try a = 0, but that would make b = 0 and c = 0, so our answer is still (C). How about negative numbers? If a = −2, then 4(−2) = 12b, and b = . Looks like we’re going to end up trying fractions with this set of numbers. Since 2 = 10c, and c = . Quantity A is −2, and Quantity B is 15 = −2, our answer is still (C). Since we can’t find any numbers to eliminate (C), that’s our answer.

Scratch paper:

As we Plug In for multiple variables, Plug In different numbers for each variable on your first attempt.

Question 5 of 20

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

Pick up your pen and set up your scratch paper: We’ve got variables in our answers, which means this is a Plug In problem. Once you’ve done that, we need to figure out two different easy numbers: Let’s say a = 2. Does that tell us what b is? Nope, we don’t have any way of figuring out b, so let’s plug in for b as well, and say b = 3. Quantity A is 2 × 3 = 6. Quantity B is 2 + 3 = 5. Quantity A is greater, so cross off (B) and (C). Do any of our FROZEN numbers seem promising? We could try negative numbers, but if we try a = −2 and b = −3, then Quantity A is (−2)(−3) = 6 and B is (−2) + (−3) = −5, and A is still greater. But who says we have to make both numbers negative? Let’s try a = −4 and b = 5. Now Quantity A is (−4)(5) = −20, and Quantity B is (−4) + (5) = 1, and Quantity B is greater, so we can cross off answer (A). Our only answer left is (D).

Be Vigilant

When you’re focusing on the two quantities, it’s easy to miss the text immediately above. You wouldn’t think it would happen often, but it does. So keep a watchful eye and make sure that everything you Plug In to the Quant Comp problem is nice and legal.

Quant Comp Plugging In Quick Quiz

Question 1 of 6

k is a negative integer.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 2 of 6

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 3 of 6

h is a positive integer.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 4 of 6

w, x, y, and z are consecutive, even integers, and 0 < w < x < y < z

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 5 of 6

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 6 of 6

Albert is a inches tall, Beatrice is b inches tall, and Charlene is c inches tall, and c is the average (arithmetic mean) of a and b.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Explanations for Plugging In Quick Quiz

1. We’ve got variables in our answer choices, which means this is a Plug In question. Set up your scratch paper, and let’s try some numbers. Let’s start with an easy negative number: k = −2. In that case, Quantity A is 5 − 2(−2) = 5 + 4 = 9. Quantity B is 2 − 5(−2) = 2 + 10 = 12, so eliminate (A) and (C). Let’s try a FROZEN number. We’re stuck with negative numbers, so how about k = −1? Quantity (A) is 5 − 2(−1) = 5 + 2 = 7, and Quantity B is 2 − 5(−1) = 2 + 5 = 7, so both quantities are equal. Eliminate (B), and the answer is (D).

2. Set up your scratch paper with A B C D, Quantity A, and Quantity B. Start with something easy: b = 5. Quantity A is (5 − 15)(15 + 5) = (−10)(20) = −200. Quantity B is 225 − (5)² = 225 − 25 = 200. Quantity B is greater, so eliminate (A) and (C). Since we want Quantity B to be smaller, we want b to be greater, so let’s try an extremely large number, such as 100. Quantity A is (100 − 15)(100 + 15) = (85)(115) = 9,775. Quantity B is 225 − 100² = 225 − 10,000 = −9,775. Now Quantity A is greater, so we can eliminate (B), and the answer is (D).

3. Variables in both Quantities? Pick up your pen and set up your scratch paper. Now that your hand is moving, let’s pick an easy number for h. If h = 2, then Quantity A is (2)² − 16 = 4 − 16 = −12. Quantity B is 2² + 2 − 12 = 4 + 2 − 12 = 6 − 12 = −6. Quantity B is therefore greater (remember that negative numbers closer to zero are greater than negative numbers farther from zero), so eliminate (A) and (C). Now let’s try a FROZEN number. Since h has to be positive, we can’t pick negative numbers or zero, so let’s try an extremely large number. If h = 100, then Quantity A is 100² − 16 = 10,000 − 16 = 9,984. Quantity B is 100² + 100 − 12 = 10,000 + 100 − 12 = 10,100 − 12 = 10,088. Quantity B is greater, and nothing seems to change that, so (B) is our answer.

4. Plug In values for the four variables according to the directions. If we let w = 2, x = 4, y = 6, and z = 8, then Quantity A equals 2(2 + 4) + 2, or 14, and Quantity B equals 6 + 8 − 2, or 12. Quantity A is greater, so eliminate answer choices (B) and (C). Now try greater numbers, such as 20, 22, 24, and 26. Quantity A is 2(20 + 22) + 2, or 86, while Quantity B is 24 + 26 − 2, or 48. Quantity A has remained greater, so the answer is (A).

5. You might think the answer to this question is (D), because you don’t know the values of m and n. Plugging in numbers, however, reveals a different picture. Say m = 4 and n = 8; Quantity A is the price of 4 books at $5 each, or $20, while Quantity B is the price of 8 books at $6, or $48. Because B is greater, you can eliminate (A) and (C). The only numbers that can make Quantity A greater are negative numbers, and you can’t Plug In negatives or zero because we’re dealing with an actual number of items, and you can’t have a negative number of books or a negative price. The answer is (B).

6. Don’t just Plug In three values for a, b, and c, because the numbers are related. You can pick whatever you want for a and b (such as a = 36 and b = 48), but c has to be the average of these values. Therefore, c = = 42. The combined height of all three people is 36 + 48 + 42, or 126 inches. To convert to feet, divide by 12; 126 ÷ 12 = 10.5. Because Quantity B is , or 10.5, the answer is (C).

PLUGGING IN THE ANSWER CHOICES

It’s great when we can supply our own numbers to the Plug-In party, but sometimes we don’t even have to exert that much effort. Often, the GRE is courteous enough to supply five numbers that we can use to Plug In. Awfully sporting, don’t you think?

This isn’t the limit of the GRE’s largesse, because as we mentioned before, the answer choices are always listed in numerical order. In order to eliminate the incorrect answer choices most efficiently, you should “Plug In the Answers,” (PITA), starting with answer choice (C), because it’s in the middle.

The steps for PITA are:

1. Recognize the opportunity to PITA. Questions which ask for the value of something (how much, how many, what is the value, et cetera) are PITA questions. If the answer choices are numbers, you can probably PITA.

2. Write out the answer choices on your scratch paper. Write the numbers for each answer choice as well as the usual vertical A B C D E.

3. Label your answer choices. If the question asks for the number of hats David has, label the column of answer choices as “David’s hats.” If it’s asking for the value of x, label the column of answer choices with an x on top.

4. Use answer choice (C). Pretend that the answer to the question is whatever number answer choice (C) is. If you have more or fewer than five answer choices, use the middle answer choice.

5. Step through the problem. Use (C) to work through the problem in bite-size pieces. As you come up with values for each step, make a new column next to your answer choices.

6. Check your answer. If you end up with numbers that don’t match up, then (C) is incorrect. If (C) was too small, cross off any answers less than (C), usually (A) and (B). If (C) was too big, cross off any answers greater than (C), usually (D) and (E). Check the remaining answers.

If you’re not sure whether your answer was too big or too small, then cross off (C) and pick another answer choice. Once you’ve tried, say, (B), you’ll notice that (B), if it was incorrect, was either closer to the answer you wanted than (C) was, in which case (A) is the answer, or farther away from the answer you wanted, in which case try (D) or (E).

Trigger: “How much,”
“How many,” “What is the
value,” numbers in the
answer choices.

Response: Plug In the
answers.

Question 16 of 20

When three consecutive integers are multiplied together, the result is 17,550. What is the greatest integer of these three integers?

24

25

26

27

28

Here’s How to Crack It

It seems like there is some algebra we could do to solve this problem, but it may not be obvious what that algebra is. That’s fine. In fact, we should avoid using algebra on this problem. It’s exactly what ETS knows most people will do, and most people will subsequently make algebraic mistakes.

This question wants to know the greatest number, and we’ve got numbers in the answers. Recognize the opportunity to PITA and write down your answer choices. Since the question is asking for the greatest integer, label the column of answer choices “Greatest Integer.”

Let’s use answer choice (C) first. If the greatest of the three integers is 26, then what are the other numbers? Now it’s time to step through the problem. The problem states that we have three consecutive integers. If the greatest integer is 26, then the other two must be 24 and 25.

To check our answer, we’ll have to use the last part of the problem: When the numbers are multiplied together, the result is 17,550. Pull up the calculator and multiply together our three numbers. Remember to do one step at a time with the calculator, and write down the result at each stage. 24 × 25 × 26 = 600 × 26 = 15,600. We wanted our result to be 17,550, so the answer we checked was too small. Therefore, we can cross off (C), because it’s wrong, and also (A) and (B): If (C) was too small, then (A) and (B) must be too small as well.

We’ve got (D) and (E) left. Which one should we check? It doesn’t matter. If we check (D) and it’s wrong, the answer is (E). If we check (D) and it’s correct, then we’re done. The same is true of (E). Let’s try (D). If the greatest integer is 27, then the other two consecutive integers must be 25 and 26. 25 × 26 × 27 = 650 × 27 = 17,550, which is exactly what we wanted. The answer is (D).

Here’s our scratch paper for this problem:

Let’s try another one:

Question 12 of 20

Dale and Kelly collect rare coins and Dale has twice as many rare coins as Kelly. If Dale gives Kelly 6 coins leaving Dale with 10 more coins than Kelly, how many coins did Kelly have initially?

20

22

28

38

44

Here’s How to Crack It

Again, start with (C). If Kelly had 28 rare coins initially, then Dale had twice as many, or 56. Dale then gave Kelly 6 coins and had 50 left; Kelly now has 6 additional coins, or 34 coins. The difference between 50 and 34 is 16, which is greater than 10. So 28 is out (choice (C) can be eliminated), and because it’s too big, we can eliminate (D) and (E) as well.

If Kelly started with 20 rare coins, then Dale had 40, and after the transaction, Kelly had 26 and Dale had 34. Those numbers are only 8 apart, so answer choice (A) is too small. You’ve eliminated everything else, so there’s nothing left to do. The answer is (B).

PITA Quick Quiz

Question 1 of 4

If 3x ² + 5x − 2 = 0, what is a possible value of x ?

−2

−1

0

1

2

Question 2 of 4

If 3^x = 9^{x − 4}, then x =

2

4

5

8

16

Question 3 of 4

Margarita bought x identical sweaters for $300. If each sweater had cost $7.50 less, she could have bought x + 2 sweaters for the same amount of money. How many sweaters did she buy?

6

8

10

12

14

Question 4 of 4

In the inventory at a certain car dealership, of the cars are four-door sedans and of the rest are SUVs. If the remaining 24 vehicles are pickup trucks, how many vehicles are in the inventory at the car dealership?

60

90

120

180

240

Explanations for PITA Quick Quiz

1. Because 3(−2)² + 5(−2) − 2 = 0, the correct answer is (A).

2. You’ll need your calculator for this one. Because 3⁸ = 6,561 and 9^{(8 − 4)} = 9⁴ = 6,561, so the answer is (D).

3. Try (C) first. If Margarita bought 10 identical sweaters for $300, then they cost $30 each. Two sweaters more would equal 12 sweaters, each of which would cost , or $25. Because $30 and $25 are not $7.50 apart, you can eliminate (C) and move on. But which way? You want the difference between prices to be greater, so try a smaller number. If she bought 8 sweaters, then they cost $37.50 each. An extra 2 sweaters would make 10, which we know would cost $30 each. The difference is $7.50, and you’re work is complete. The answer is (B).

4. If the car dealership has 120 vehicles, then of those 120, or 80, are four-door sedans. Of the remaining 40, are SUVs. Because × 40 is 16, the number of cars that remain to be categorized is 40 − 16, or 24. Bingo! The answer is (C).

PITA also works on the other question types. With All That Apply questions, we’ll typically be looking for the greatest and least answers that work, and crossing off any answers greater than the greatest answer that works and less than the least answer that works.

Question 14 of 20

Last year, Company X spent between and of its yearly advertising budget on print ads. If Company X spent $31,120 on print ads, which of the following could have been Company X’s advertising budget last year?

Indicate all such statements.

$106,720

$115,880

$121,960

$126,400

$138,240

$145,000

$157,100

Here’s How to Crack It

With All That Apply questions, the GRE will often say “which of the following could be.” That, and the presence of numbers in our answer choices, tells us that this is a PITA question. Write down A B C D E F G vertically on your scratch paper. Write down the answers and label them as “ad budget.” For each answer, we’ll calculate and of each ad budget, and see if $31,120 lies between those two numbers.

Let’s start with (D), our middle answer. If the ad budget was $126,400, then they must have spent between of $126,400 = $31,600 and of $126,400 = $25,280 on print ads. The problem states that Company X spent $31,120 on ads, and $25,280 < $31,120 < $31,600, so $126,400 could have been the amount in Company X’s advertising budget. Put a check mark next to (D).

We know that (D) works, but we need to know which other answers work. We’ll work our way through the lesser answer choices until they don’t work anymore, and then through the greater answer choices until those don’t work. Let’s try (C) next. If the ad budget was $121,960, then the company must have spent between of $121,960 = $30,490 and of $121,960 = $24,392 on print ads. Company X, however, spent $31,120 on advertising, which is more than that $30,490 allowed by answer (C). Since now our budget is too small, cross off answer (A), (B), and (C).

Now let’s try the greater answer choices. If the ad budget was $138,240, then the company must have spent between of $138,240 = $34,560 and of $138,240 = $27,648 on print ads. Since $27,648 < $31,120 < $34,560, the company’s total budget could have been $138,240. Put a check mark next to (E), and let’s try a greater number.

For answer choice (F), if the total advertising budget was $145,000, then the company spent between $29,000 and $36,250 on print ads, which matches our answer. Put a check mark next to (F).

We’ve only got one answer choice left. Remember, with All That Apply questions you must click on every answer that applies, which sometimes means checking many different answers. If the company had $157,100 in its advertising budget, then it spent between $31,420 and $39,275 on print ads. That’s too big of a budget, so we can cross off answer (G).

The answers are therefore (D), (E), and (F).

PITA Works for Quant Comp, Too

Sometimes you can Plug In some of the information that they give you in one of the Quantities in order to solve a problem.

Question 11 of 20

On Monday, Dave’s Car Emporium had h used cars in its inventory. Dave sold of the cars to his other dealership on Tuesday, and on Wednesday 7 people brought in their used cars as trade-ins. This brought the total number of used cars on the lot to 67.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Here’s How to Crack It

Is it possible that h = 84? Find out by plugging 84 into the problem: If Dave’s had 84 cars on Monday and sold of them, then he would have had 63 cars left ( × 84 = 21, and 84 − 21 = 63). If he acquired 7 more on Tuesday, this would have brought his total up to 70. This doesn’t match the 67 figure in the question; in fact, it’s too big. Therefore, the number h must be less than 84, and the answer is (B).

Quant Comp PITA Quick Quiz

Question 1 of 3

Mendel purchased three pairs of dress pants at a men’s store. One pair cost $65, a second pair cost $85, and the average (arithmetic mean) price for all three is $73.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 2 of 3

A rectangular swimming pool with a uniform depth is 20 feet long and 40 feet wide and holds a total of 9,600 cubic feet of water.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 3 of 3

A local movie theater normally sells individual tickets for $9 each, but it offers 20% discounts to all members of groups of 10 or more.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Answers to Quant Comp PITA Quick Quiz

1. Use PITA and assume that the third pair of pants cost $72. If that were the case, then the total cost of all three pairs of pants would be 65 + 85 + 72, or 222, and the average price would be 222 ÷ 3, or $74. Because the problem tells you the average price was $73, the average you calculated is too large, which means the $72 that you plugged into the problem is too large. Therefore, $72 must be greater than the price of the third pair of pants, and the answer is (B).

2. If the dimensions of the swimming pool are 20 by 40 by 12, the volume of the pool is 20 × 40 × 12, or 9,600 cubic feet. Because this matches the volume mentioned in the question, the answer is (C).

3. First, it’s important to recognize that the group is large enough to qualify for the discount. If the group spent $112 at the movies, then each spent $112 ÷ 16, or $7. This is too little, because the discounted price of a ticket is 80% of $9, or $7.20. Therefore, they must have spent more than $112, and the answer is (A).

We’ve run through a number of examples in this chapter, and now it’s your turn. As you practice these questions, train yourself to recognize patterns and determine the numbers that will either undermine or confirm your initial results.

Plug In Early, Plug In Often

If you finish this book and take away just one new skill, make this the one. Plugging In is a million-dollar idea that can bring about solid score improvements right away, because of how efficiently it (1) makes problems more accessible and (2) reduces the chance that you’ll make careless errors. You should practice Plugging In as much as you can until it becomes instinctive.

Algebra Drill

Question 1 of 35

ac < bc

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 2 of 35

A video game goes on sale for 15% off the original price and a customer can save an additional 10% on the sale price if she has a coupon. If Bailey buys the video game at the sale price and uses a coupon, what percent of the original price does she pay?

12.5%

25%

75%

76.5%

83.5%

Question 3 of 35

A merchant sold an equal number of 5-cent and 10-cent screws. If the total cost of the screws was $3.00, what was the total number of screws sold?

25

30

40

44

50

Question 4 of 35

At a constant rate of 12,000 rotations per hour, how many rotations does a spinning top make in p minutes?

12,000p

200p

Question 5 of 35

If 3x = 12, then 8 ÷ x =

Question 6 of 35

st = −6

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 7 of 35

A contest winner receives of his winnings in cash and also receives four prizes, each worth of the balance. If total value of the cash and one of the prizes is $35,000, what is the total value of his winnings?

$70,000

$75,000

$80,000

$95,000

$140,000

Question 8 of 35

Connie has more marbles than Joey, and Joey has fewer marbles than Mark.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 9 of 35

If 2 < r < 8 and 1 < s < , which of the following expresses all possible values of rs ?

1 < rs < 5

2 < rs < 20

< rs < 8

< rs < 20

5 < rs < 10

Question 10 of 35

Set A = {−1, 0, 1}

Set B = {−2, −1, 0, 1}

If a is a member of Set A and b is a member of Set B, what is the least possible value of a − b² ?

2

0

−2

−5

−9

Question 11 of 35

12m² − 8m − 64 =

4(3m + 8)(m − 2)

4(3m − 8)(m + 2)

4(3m − 2)(m + 8)

4m² − 64

4m − 64

Question 12 of 35

If a + b − 2c = 12, and 3a + 3b + c = 22, what is the value of c ?

−2

0

10

17

34

Question 13 of 35

Kevin decided to consecutively number the T-shirts in his closet. He wrote one number on each of his T-shirts, starting with 1 on the first T-shirt. When he was finished numbering he had written a total of 59 digits.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 14 of 35

If the sum of x distinct, positive integers is less than 75, what is the greatest possible value of x ?

8

9

10

11

12

Question 15 of 35

An office supply store sells staplers for $5 each, and boxes of staples for $2 each. On Monday, the store sold a total of 22 staplers and boxes of staples combined, for which it collected a total of $74. How many staplers did the store sell?

6

10

11

12

44

Question 16 of 35

Carmen has 12 collectibles, and t, the value of Carmen’s favorite collectible, is between 8 and 9 dollars.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 17 of 35

Which of the following is equivalent to 8a + (2ab − 4a)b − 4ab ?

2a(b − 2)²

2a(b² − 2b + 2)

4a(b² − 2b + b)

4ab(1 − 2b)

a²(2b + 2b²)b

Question 18 of 35

Timmie can buy his favorite pens in $10 packs that contain p pens, or he can buy the same pens singly at a cost of $1.12 each.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 19 of 35

If x = and x ≠ 0, then =

Question 20 of 35

If x is an integer less than or equal to −2 and y is an integer with an absolute value greater than or equal to 5, which of the following statements must be true of xy?

Indicate all such statements.

xy is an integer

xy is negative

xy ≤ −10

xy ≥ 10

xy = −10

|xy| ≥ 10

Question 21 of 35

Six years ago, Jim’s age was four times Carol’s. Jim is now j years old, and Carol is now c years old.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 22 of 35

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 23 of 35

= a² − b² = 32

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 24 of 35

If a > 12 and b < 7, which of the following must be true?

Indicate all such values.

a + b > 12

a − b > 5

a + b < 19

a − b < 12

ab > 84

ab < 84

Question 25 of 35

x + y = 14

y + 4 = 10

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 26 of 35

If x ≠ 0, which of the following must be true?

x < x²

< x

x² < x³

1 − x < x

x < x + 2

Question 27 of 35

If x > 0, y > 0, and z > 0, then

Question 28 of 35

2x² + 3xy − 2y² = 0

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 29 of 35

Pat has 4 teapots more than Judi, but 3 teapots fewer than Rudy. If Pat has y teapots, which of the following is an expression for the total number of teapots that Jodi and Rudy have?

2y − 7

2y − 1

2y + 9

y + 7

y + 9

Question 30 of 35

If x ≠ 0 and one-half of x is equal to four times x², then x =

Question 31 of 35

q ≠ 0

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 32 of 35

For all non-zero numbers x and y, if 3x = 5y, then =

Question 33 of 35

Terrence has as many cards as Phillip. If Terrence were to win one card from Phillip, he would have as many cards as Phillip. If Terrence wins an even number of cards from Phillip, which of the following could be the number of cards that Phillip has left?

Indicate all such values.

24

20

19

18

17

14

Question 34 of 35

40 < t < 50

10 < s < 12

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined from the information given.

Question 35 of 35

If x and n are integers and x(^{n + 2)} = 16x², what is the least possible value of 16 + x?

EXPLANATIONS FOR ALGEBRA DRILL

1. D

Plug In. Let a = 2, b = 5, and c = 1. Quantity B can be greater than Quantity A, so eliminate answer choices (A) and (C). Now change the sign of c. Let a = 5, b = 2, and c = −1. Quantity A can be greater than Quantity B, so eliminate answer choice (B). Only choice (D) remains.

2. D

Plug In. $100 for the original price. The sale price is $100 − $15 = $85. The coupon reduces the price another $8.50, to $76.50. This is 76.5% of the original price, making choice (D) correct.

3. C

Plug In the answer choices, starting with choice (C). If there are 40 total screws, then there are 20 5-cent screws and 20 10-cent screws. The 5-cent screws cost 20 × $0.05 = $1.00, and the 10-cent screws cost 20 × $0.10 = $2.00. The total cost of the screws is $3.00. This information matches all of the information given in the problem, so choice (C) is correct.

4. B

Plug In, and let p = 30 minutes. Because the top rotates 12,000 times per hour, it must rotate 6,000 times in 30 minutes, which is one half-hour. Plug 30 for p into the answer choices to find the target answer of 6,000. You’ll find it only in choice (B): 200 × 30 = 6,000.

5. 2

First, solve for x by dividing both sides of the equation by 3. x = 4. Next, answer the question: 8 ÷ 4 = 2.

6. D

There are variables in the columns, so Plug In. Try s = 2 and t = −3. Quantity A is greater, so cross off choices (B) and (C). Now switch the numbers around and try s = −3 and t = 2. This time Quantity B is greater, so the correct answer is choice (D).

7. C

Plug In the answers, starting with choice (C). If the winnings were $80,000, the cash is × $80,000 = $20,000. Then, take of the remainder of the prize, $60,000: × $60,000 = $15,000. The total of the cash and the first payment is $20,000 + $15,000 = $35,000. Because choice (C) is correct, there is no need to check the other answers.

8. D

There are unknowns in the columns, so set up your scratch paper and Plug In more than once. Try 10 marbles for Connie, 5 marbles for Joey, and 7 marbles for Mark. Quantity B is greater, so cross off choices (A) and (C). Now try 10 marbles for Connie, 5 marbles for Joey, and 10 marbles for Mark. This time the quantities are equal, and you can eliminate choice (B). The correct answer is choice (D).

9. B

Break this problem into two parts, and use POE. Because r > 2 and s > 1, the product of rs must be greater than the product of 2 × 1. The only answer choice in which 2 < rs is choice (B). In addition, because r < 8 and s < , the product of rs must be less than 8 = 20. Choice (B) is also the only answer choice in which rs < 20.

10. D

To make the value of a − b² as small as possible, make a as small as possible and b² as large as possible. The smallest number is Set A is −1, so that’s the value for a. Use −2 for b because that makes b² = 4: a − b² = −1 − 4 = −5.

11. B

Plug In, and let m = 3. The equation is (12 × 3²) − (8 × 3) − 64 = 108 − 24 − 64 = 20, and 20 is the target. Plug 3 for m into the answer choices, and match the target of 20. Only choice (B) matches the target: 4(3 × 3 − 8)(3 + 2) = 4(9 − 8)(5) = 4(1)(5) = 20.

12. A

Since you’re looking for c, try to make a and b cancel out of the equations. Multiply the first equation by −3, giving you −3a − 3b + 6c = −36. Now add the equations. The a and b values cancel out, giving you 7c = −14, so c = −2.

13. A

There is an unknown in one column and a number in the other, so PITA. If there were 35 T-shirts in Kevin’s closet that would be 9 digits for the first 9 T-shirts and 2 × 26 = 52 for the next 26 T-shirts. This a total of 61 digits, which is more than 59 mentioned in the question, so the number of T-shirts must have been less than 35. The correct answer is choice (A).

14. D

Write down the integers starting with 1 and keep adding: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66. If you added 12 to the total, you would reach 78. Thus, there are 11 distinct positive integers that have a sum less than 75.

15. B

Label the number of staplers sold x, and the number of boxes of staples sold y; now you can write the equation x + y = 22. Additionally, you know that $5x + $2y = $74. You have two equations and two unknowns, so stack the equations in order to add or subtract them. Since you’re solving for x, multiply the first equation by 2 to get 2x + 2y = 44; now you have the same number of y’s, so they’ll cancel out if you subtract one equation from the other. Subtracting gives you 3x = 30, so x = 10.

16. D

Once you’ve set up your scratch paper to Plug In on a Quant Comp problem, start with t = 8.1. The value in Quantity A is 0.675; Quantity B is greater, so eliminate choices (A) and (C). Next, try t = 8.9. Now the value in Quantity A is about 0.74; this time, Quantity A is greater, so eliminate choice (B) and select choice (D).

17. A

Plug In. If a = 2 and b = 3, then 8a + (2ab − 4a)b − 4ab = 8(2) + (2(2)(3) − 4(2))(3) − 4(2)(3) = 16 + (12 − 8)(3) − 24 = 16 + 12 − 24 = 4. Your target is 4. Plug a = 2 and b = 3 into all of the answers. Only answer choice (A) yields 4: 2(2)(3 − 2)² = 4(1) = 4.

18. A

There’s an unknown in one column and a number in the other, so Plug In the answers. If p is 9, Timmie pays 9 × $1.12 = $10.08 to buy the pens singly; that’s more than it would cost him to buy the pens in the pack, but it’s supposed to be cheaper to buy the pens singly. Therefore, p must be less than 9, and the correct answer is choice (A).

19. D

Plug In. If y = 8, x = = 18. The problem is asks for , which is . Choice (D) is the only match.

20. A and F

The product of two integers must be an integer, so answer choice (A) is correct. The requirement that |y| ≥ 5 is equivalent to the following conditions: y ≥ 5 or y ≤ −5. If y ≥ 5 and x ≤ −2, then xy ≤ −10; if y ≤ −5 and x ≤ −2, then xy ≥ 10. Hence, xy ≤ −10 or xy ≥ 10, which is the same thing as saying |xy| ≥ 10, so choice (F) is correct. Answer choice (B) is only true of some values of xy; choices (C) and (D) are partial answers along the way to choice (F), but are incomplete and hence incorrect. Of course, you can simplify this problem greatly by plugging in for x and y. First, make x = −2 and y = 5: xy = −10, so eliminate answer choice (D). Next, leave x = −2, but make y = −5: Now xy = 10, so eliminate answer choices (B), (C), and (E). No matter what value you Plug In—as long as you meet the requirements—answer choices (A) and (F) will always work, so they must be true.

21. A

Plug In a few different rounds of numbers. The question states j − 6 = 4(c − 6). If you choose c = 7, then j = 10. In that case, 7 > 6.5 and Quantity A is greater, so you can eliminate choices (B) and (C). To decide between answers (A) and (D), choose a couple of very different numbers to plug in, but recall that negative numbers and zeros don’t work for ages. Let c = 66 to check the high end of the range. For c = 66, j = 246, and 66 > 65.5, and Quantity A is still 0.5 greater. Choose one more number to be sure. If c = 10, j = 22. Once again Quantity A is 0.5 greater than Quantity B because 10 > 9.5, so you can confidently select choice (A).

22. C

If you remember your exponent rules, you know that Quantity A reduces to x. However, you could also Plug In on this QC question, and when you’re dealing with exponents or absolute values, you should Plug In negative numbers. Go through FROZEN. If x = 0, then 0 = 0 and the expressions are equal: Eliminate choices (A) and (B). If x = 1, then 1 = 1. If x = −2, then −2 = −2. If x = 100, then 100 = 100. If x = , = At this point you can be confident that choice (C) is the correct choice.

23. A

Recognize the common quadratic and this becomes easier. You are initially given a + b and a² − b², and this may remind you of the common quadratic (a + b)(a − b) = a² − b². From the other given information, solve for a + b, which equals 16. Substituting 16 for (a + b) and 32 for (a² − b²) in the quadratic, you can solve for (a − b) = 2. Now Quantity A is 16 and Quantity B is 2² = 4, making Quantity A greater.

24. B

Solve this Must Be problem by plugging in. Let a = 13 and b = 6. These numbers eliminate choices (C) and (E). Then try some extremes, such as a = 100 and b = −100. These new numbers eliminate choices (A) and (D). Try a = 100 and b = 1, which eliminates choice (F). Only choice (B) always works, because the values of a and b must be more than 5 apart.

25. A

Solve the second equation and you’ll find that y = 6. Plug that in to the first equation, and you’ll find that x = 8. Therefore, Quantity A is greater.

26. E

Plug In different numbers until there is only one answer choice left. If you try a simple number such as x = 2, you’ll find that you can’t eliminate any answer choices. That means it’s time to start thinking about different kinds of numbers. If x = , choices (A), (B), (C), and (D) are eliminated. Only choice (E) works.

27. E

Plug In x = 2, y = 3, and z = 4. Put these values into the equation: = = . Now go through the answer choices and find . The only one that works is answer choice (E).

28. D

You can effectively Plug In answer choice (C) by assuming for a moment that 2x = y. When you substitute this into the quadratic equation, it satisfies it and validates choice (C). Eliminate choices (A) and (B), neither of which can be the final answer if choice (C) is ever correct. Factor the quadratic equation: (2x − y)(x + 2y) = 0. This means 2x = y (which you proved above) or x = −2y. Plug In to this second scenario, which you have yet to examine. When x = 0, then y = 0 and the columns are still equal, validating choice (C). But when x = 1, y = −0.5, making Quantity (A) greater. So choice (D) must be the correct answer.

29. B

Plug In. If y = 6, then Pat has 6 teapots, Judi has 2 teapots, and Rudy has 9 teapots. The target answer, the number of teapots that Judi and Rudy have combined, is 11. Go to the answer choices, and Plug In 6 for y. Answer choice (B) is the only answer choice that matches your target of 11.

30.

Translate the words in the question into an equation: x = 4x². Because the question states that x ≠ 0, you can divide both sides of the equation by x to get = 4x. Divide both sides by 4 to find that x = .

31. D

Notice that the trap answer is choice (C). There are variables in the columns, so Plug In. Try q = 7. Quantity A is 0 and Quantity B is 7. Quantity B is greater, so cross off choices (A) and (C). Now try q = −7. This time both columns are 7. Different numbers gave different answers, so the correct answer is choice (D).

32.

Plugging In is the easiest way to solve this question. Let x = 5 and y = 3. Plug those numbers into the expression: .

33. C and E

The first sentence tells you that T = P, and the second tells you that (T + 1) = (P −1). Simplify the second equation to get: T = P − . Because both equations express the value of T, set them equal to one another: P = P − . Multiply by 12 to get rid of the fractions, and you’re left with 8P = 9P − 21; P, therefore, equals 21. If Phillip has 21 cards and loses an even number of them, he could have 21 − 2 = 19 cards or 21 − 4 = 17 cards left. Answer choices (C) and (E) are correct.

34. D

If you simply subtract the second inequality from the first, it can appear as if choice (B) should be the answer. Be careful of merely doing the obvious on a question like this. Because there are variables, set up to Plug In on a Quant Comp. Choose something easy to start with, like t = 45 and s = 11. t − s = 45 − 11 = 34. Eliminate choices (A) and (C). Your task is to now attempt to make the value of Quantity A greater than (or equal to) 39. This appears difficult at first, because of the limited amount of range you’re given. FROZEN isn’t particularly helpful: You can’t Plug In 0, 1, or negatives for this problem, but you can try fractions. Let t = 49 and s = 10.5 so 49 − 10.5 = 38.5. You’re closer, but still not quite there. This time, let t = 49.5 and keep s = 10.5 so 49.5 − 10.5 = 39. Eliminate choice (B), and select choice (D).

35. 12

To find the least possible value of 16 + x, first find the least possible value of x. Rewrite the equation as (xⁿ)(x²) = 16x². Divide both sides by x² to get xⁿ = 16. If both x and n are integers, then there are only a very few values for x and n. Minimize 16 + x by using a negative value for x. Because, (−4)² = 16, x = −4 is the least possible value for x. The least possible value of 16 + x is 16 − 4 = 12.