Step 3: Approach Strategically—Picking Numbers

Learning Objectives

In this section, you will learn how to: 

• Articulate how the Picking Numbers strategy works
• Identify situations in which it may be helpful to use Picking Numbers 
• Apply the Picking Numbers strategy to appropriate Problem Solving questions

In the previous sections, you learned about the attitude that all great GMAT test takers share: that there must be a straightforward way to solve and there’s no need to panic if you don’t find it right away. And even if you don’t use the algebra-based approach that your high school math teacher would have preferred, you will get the right answer and you will raise your score.

Identifying a more efficient alternative approach involves staying open to possibilities that might not occur to you immediately. Two such approaches—picking numbers and backsolving—are so straightforward, can eliminate so much math, and can be used so often that they deserve special mention.

Picking numbers is a powerful alternative to solving problems by brute force. Rather than trying to work with unknown variables, you pick concrete values for the variables. In essence, you’re transforming algebra or abstract math rules into basic arithmetic, giving yourself a much simpler task. Pick numbers to stand in for the variables or unknowns.

How does picking numbers work?

• Pick permissible numbers. Some problems have explicit rules such as “x is odd.” Other problems have implicit rules. For example, if a word problem says, “Betty is 5 years older than Tim,” don’t pick b = −2 and t = −7; these numbers wouldn’t work in the problem.
• Pick manageable numbers. Some numbers are simply easier to work with than others. For example, d = 2 will probably be a more successful choice than will . The problems themselves will tell you what’s manageable. For instance, d = 2 wouldn’t be a great pick if an answer choice were . (The number 12, 24, or even 120 would be much better in that case, since each is divisible by 12.)
• Once you’ve picked numbers to substitute for unknowns, approach the problem as basic arithmetic instead of algebra or number properties.
• Test every answer choice—sometimes certain numbers will yield a “false positive” in which a wrong answer looks right. If you get more than one “right answer,” just pick a new set of numbers. The truly correct answer must work for all permissible numbers.

Note that it’s just fine to pick the numbers 0 or 1 for most questions. The GMAT sometimes writes questions based on the special properties of 0 and 1, so if you aren’t considering these numbers, you won’t get the right answer to those questions. Furthermore, they are the most manageable numbers in all of math. Imagine picking numbers for the following problem with anything other than b = 0 or b = 1:

1. 7^b + 7^b + 7^b + 7^b + 7^b + 7^b + 7^b =
   1. 7^b
   2. 7^b+1
   3. 7^7b
   4. 8^b
   5. 49^b

Picking b = 1 makes short work of this exponents question:

Plugging b = 1 into the answer choices yields:

(A) 7^b = 7¹ = 7

(B) 7^b+1 = 7¹⁺¹ = 7² = 49

(C) 7^7b = 7⁷  Whatever that is, it’s a lot bigger than 49.

(D) 8^b = 8¹ = 8

(E) 49^b = 49¹ = 49

Two answer choices match the desired result of 49, so you need to pick a new number. Try b = 0. First, plug it into the expression in the question stem.

Now you only need to plug b = 0 into (B) and (E), the only two answer choices that worked when b = 1. Doing so yields:

(A) 7^b+1 = 7⁰⁺¹ = 7¹ = 7

(B) 49^b = 49⁰ = 1

The answer is (B).

Picking Numbers

There are four signals that picking numbers will be a possible approach to the problem:

• Variables in the question Stem
• Percents in the Answer choices
• Variables in the answer choices
• Must be/could be/cannot be
  

These aren’t in any particular order. You’ll see many instances of each case on Test Day. We’ll start with the first.

Picking Numbers with Variables in the Question Stem

If you see a question that has variables in the question stem or asks about a fraction of an unknown whole, pick numbers to represent the unknown(s) in the question stem and walk through the arithmetic of the problem. Look in the question stem and the answer choices for clues about what the most manageable numbers might be.

Example:

1. Carol spends of her savings on a stereo and less than she spent on the stereo for a television. What fraction of her savings did she spend on the stereo and television?
   1. 
   2. 
   3. 
   4. 
   5. 

Lowest common denominators of fractions in the question stem are good choices for numbers to pick. So are numbers that appear frequently as denominators in the answer choices. In this case, the common denominator of the fractions in the stem is 12, so let the number 12 represent Carol’s total savings (12 dollars).

That means she spends dollars, or 3 dollars, on her stereo and dollars, or 2 dollars, on her television. That comes out to be 3 + 2 = 5 dollars; that’s how much she spent on the stereo and television combined. You’re asked what fraction of her savings she spent. Because her total savings was 12 dollars, she spent of her savings; (C) is correct. Notice how picking a common denominator for the variable (Carol’s savings) made it easy to convert each of the fractions to a simple integer.

A tricky part of this question is understanding how to determine the price of the television. The television does not cost of her savings. It costs  less than the stereo; that is, it costs as much as the stereo.

By the way, some of these answers could have been eliminated quickly using basic logic. (A) is too small; the stereo alone costs of her savings. (D) and (E) are too large. The television costs less than the stereo, so the two together must cost less than , or half, of Carol’s savings. We’ll look more at using logic to eliminate answers later in this chapter.

Applying the Kaplan Method: Picking Numbers with Variables in the Question Stem

Now let’s use the Kaplan Method on a Problem Solving question that involves picking numbers with variables in the question stem:

1. If w > x > y > z on the number line, y is halfway between x and z, and x is halfway between w and z, then
   1. 
   2. 
   3. 
   4. 
   5. 1

Step 1: Analyze the Question

This question presents a very complicated relationship among four variables. Since there are multiple variables in the problem, picking numbers is a possible approach. The answer choices are all numbers, so you won’t have to test each answer choice—you’d just plug numbers into the question. Since the phrase number line appears, you may want to consider drawing a number line to help visualize the situation:

Step 2: State the Task

The question asks you to evaluate a rather complicated fraction: . The question stem gives you no actual values for the variables, but the answer choices are numbers. This means that the value of the fractional expression must always be the same, as long as the values of the variables follow the rules described in the question stem.

Step 3: Approach Strategically

The math seems like it would be difficult and time-consuming. You’d have to translate each rule that states a relationship between variables into a separate algebraic equation and then somehow combine them to solve for the fraction in the stem. Picking numbers, on the other hand, will be much more straightforward.

The rules are pretty complicated, so pick numbers one at a time, making sure the numbers you pick are permissible (follow the rules in the stem) and manageable (easy to work with). x and z each appear twice in the rules, so starting with those two numbers would seem to make sense.

Let’s go with z = 1 and x = 3 to leave room for y.

y is halfway between 3 and 1, so y = 2.

x, or 3, is halfway between w and 1, so w = 5.

Once you’ve picked a set of permissible numbers, it’s time to plug these numbers into the expression in the stem:

The answer is (B). Picking numbers can often take you to the correct answer much more directly than algebra.

Step 4: Confirm Your Answer

Look back at the question stem and make sure you understood all the rules and picked permissible numbers. “y is halfway between x and z.” Check. “x is halfway between w and z.” Check. Another potential error would have been to make w the smallest variable instead of the largest. (Did you notice how the question stem puts w on the left of the inequality statement, even though it belongs on the right-hand side of the number line?) That’s another potential mistake you could catch in this step.

Takeaways: Picking Numbers with Variables in the Question Stem

• Most people find it easier to perform calculations with numbers than to perform calculations with variables.
• Variables may be letters (n, x, t, etc.) or unspecified values (e.g., “a factory produces some number of units of a product each month”).
• When picking numbers, be sure that the numbers are permissible and manageable.

Practice Set: Picking Numbers with Variables in the Question Stem

6. If , then
   1. −1
   2. 1
   3. 2
   4. 3
   5. 4
7. DeShawn set aside some of his savings to make charitable donations. If he gives  of the money to Charity A and  of the remainder to Charity B, what fraction of the original amount that he set aside does DeShawn have left to distribute to other charities?
   1. 
   2. 
   3. 
   4. 
   5. 
8. During a sale, a store sells 20 percent of its remaining stock each day, without replenishment. After 4 days, what fraction of its original stock has it sold?
   1. 
   2. 
   3. 
   4. 
   5. 

Picking Numbers with Percents in the Answer Choices

Picking numbers also works well on Problem Solving questions for which the answer choices are given in percents. When the answers are in percents, 100 will almost always be the most manageable number to pick. Using 100 not only makes your calculations easier, but it also simplifies the task of expressing the final value as a percent of the original.

Example:

1. The manufacturer of Sleep-EZ mattresses is offering a 10 percent discount on the price of its king-size mattress. Some retailers are offering additional discounts. If a retailer offers an additional 20 percent discount, then what is the total discount available at that retailer?
   1. 10%
   2. 25%
   3. 28%
   4. 30%
   5. 35%

Since the answers are in percents, pick $100 as the original price of the mattress. (Remember, realism is irrelevant—only permissibility and manageability matter.) The manufacturer offers a 10% discount: 10% of $100 is $10. So now the mattress costs $90.

Then the retailer offers an additional 20% discount. Since the price has fallen to $90, that 20% is taken off the $90 price. A 20% discount is now a reduction of $18. The final price of the mattress is $90 − $18, or $72.

The mattress has been reduced to $72 from an original price of $100. That’s a $28 reduction. Since you started with $100, you can easily calculate that $28 is 28% of the original price. Choice (C) is correct.

Notice that choice (D) commits the error of simply adding the two percents given in the question stem. An answer choice like (D) will never be correct on a question that gives you information about multiple percent changes.

Applying the Kaplan Method: Picking Numbers with Percents in the Answer Choices

Now let’s use the Kaplan Method on a Problem Solving question that involves picking numbers with percents in the answer choices:

1. In 1998, the profits of Company N were 10 percent of revenues. In 1999, the revenues of Company N fell by 20 percent, but profits were 15 percent of revenues. The profits in 1999 were what percent of the profits in 1998?
   1. 80%
   2. 105%
   3. 120%
   4. 124.2%
   5. 138%

Step 1: Analyze the Question

Profits and revenues are both changing over time. Since the answer choices are percents, picking 100 is a good idea.

Step 2: State the Task

The profits in 1999 were what percent of the profits in 1998?

You’ll have to compare the profits of the two years. The year 1998 is the basis of the comparison. You’re not solving for the increase or decrease, just the relative amount.

In other words, you want this fraction, , expressed as a percent.

Step 3: Approach Strategically

You should start with the original profits, since the question presents its 1999 information as a change from the 1998 information. You read this about 1998 profits: “In 1998, the profits of Company N were 10 percent of revenues.” So you’ll know profits if you know revenues. You’re given no information about revenue at all, so just pick a number. Given the percents in the answer choices, pick $100. That makes profits in 1998 equal to $10.

Now, what about 1999 profits? You read, “In 1999 . . . profits were 15 percent of revenues.” So you have to know revenues in 1999 to know profits. What does the question say about 1999’s revenues? “In 1999, the revenues of Company N fell by 20 percent.” Revenues were $100, so they fell $20 to $80. Profits, then, were 15% of $80, or 0.15($80) = $12.

By the way, you don’t have to calculate that directly. You can use Critical Thinking and say:

Plugging your results back into the question yields .

Just multiply by 100% to convert the decimal to a percent: 1.2 × 100% = 120%. Choice (C) is correct.

Step 4: Confirm Your Answer

Make sure to reread the question stem to confirm that you interpreted the relationships correctly.

Takeaways: Picking Numbers with Percents in the Answer Choices

• On a question involving multiple changes in percent, one of the trap answers will most likely involve simply adding or subtracting the percents.
• When you see a percent question with unspecified values, pick 100 for the unknown value. Doing so will make the calculations much easier.

Practice Set: Picking Numbers with Percents in the Answer Choices

9. If a bicyclist in motion increases his speed by 30 percent and then increases this speed by 10 percent, what percent of the original speed is the total increase in speed?
   1. 10%
   2. 40%
   3. 43%
   4. 64%
   5. 140%
10. A driver delivered food from a restaurant to a customer’s home by way of a certain route. If the driver then completed 20 percent of the return trip by way of the same route before the delivery vehicle broke down, what percent of the round trip did the driver complete?
    1. 10%
    2. 20%
    3. 60%
    4. 70%
    5. 80%
11. Sofia runs a business delivering boxes of frequently used household supplies to monthly subscribers. After hiring a workflow manager, she delivered 20 percent more boxes per month while working 20 percent fewer hours herself each month. Hiring a workflow manager increased Sofia’s personal productivity per hour by what percent?
    1. 40%
    2. 50%
    3. 60%
    4. 140%
    5. 150%

Picking Numbers with Variables in the Answer Choices

Whenever the answer choices contain variables, you should consider picking numbers. The correct answer choice is the one that yields the result that you got when you plugged the same number(s) into the question stem. Make sure that you test each answer choice, just in case more than one answer choice produces the desired result. In such a case, you will need to pick a new set of numbers and repeat the process only for the answer choices that worked out the first time.

Example:

1. If a > 1, which of the following is equal to ?
   1. a
   2. a + 3
   3. 
   4. 
   5. 

The question says that a > 1, so the most manageable permissible number will probably be 2. Then .

Now substitute 2 for a in each answer choice, looking for choices that equal 2 when a = 2. Eliminate choices that do not equal 2 when a = 2.

Choice (A): a = 2. Choice (A) is possibly correct.

Choice (B): a + 3 = 2 + 3 = 5. This is not 2. Discard.

Choice (C): . Possibly correct.

Choice (D): . This is not 2. Discard.

Choice (E): . This is not 2. Discard.

You’re down to (A) and (C). When more than one answer choice remains, you must pick another number. Try a = 3. Then .

Now work with the remaining answer choices.

Choice (A): a = 3. This is not 1. Discard.

Now that all four incorrect answer choices have been eliminated, you know that (C) must be correct. Check to see whether it equals 1 when a = 3.

Choice (C): . Choice (C) does equal 1 when a = 3.

This approach to picking numbers also applies to many confusing word problems on the GMAT. Picking numbers can resolve a lot of that confusion. The key to picking numbers in word problems is to reread the question stem after you’ve picked numbers, substituting the numbers in place of the variables.

Example:

1. A car rental company charges for mileage as follows: x dollars per mile for the first n miles and x + 1 dollars per mile for each mile over n miles. How much will the mileage charge be, in dollars, for a journey of d miles, where d > n?
   1. d(x + 1) − n
   2. xn + d
   3. xn + d(x + 1)
   4. x(n + d) − d
   5. (x + 1)(d − n)

Suppose that you pick x = 4, n = 3, and d = 5. Now the problem would read like this:

A car rental company charges for mileage as follows: $4 per mile for each of the first 3 miles and $5 per mile for each mile over 3 miles. How much will the mileage charge be, in dollars, for a journey of 5 miles?

All of a sudden, the problem has gotten much more straightforward. The first 3 miles are charged at $4/mile. So that’s $4 + $4 + $4, or $12. There are 2 miles remaining, and each one costs $5. So that’s $5 + $5, for a total of $10. If the first 3 miles cost $12 and the next 2 cost $10, then the total charge is $12 + $10, which is $22.

Plugging x = 4, n = 3, and d = 5 into the answer choices, you get

(A) d(x + 1) − n = 5(4 + 1) − 3 = 22

(B) xn + d = 4 × 3 + 5 = 17

(C) xn + d(x + 1) = 4 × 3 + 5(4 + 1) = 37

(D) x(n + d) − d = 4(3 + 5) − 5 = 27

(E) (x + 1)(d − n) = (4 + 1)(5 − 3) = 10

Only (A) yields the same number you got when you plugged these numbers into the question stem, so (A) is the answer. No need for algebra at all.

Applying the Kaplan Method: Picking Numbers with Variables in the Answer Choices

Now let’s use the Kaplan Method on a Problem Solving question that involves picking numbers with variables in the answer choices:

1. Cindy paddles her kayak upstream at m kilometers per hour, and then returns downstream the same distance at n kilometers per hour. How many kilometers upstream did she travel if she spent a total of p hours for the round trip?
   1. mnp
   2. 
   3. 
   4. 
   5. 

Step 1: Analyze the Question

Cindy is going upstream and then the same distance back downstream. Notice that the answer choices contain variables. Whenever the answer choices have variable(s), consider picking numbers as an approach.

Step 2: State the Task

The question asks, “How many kilometers upstream did she travel if she spent a total of p hours for the round trip?” You’re solving not for the total distance of her round trip, just one leg of it.

Step 3: Approach Strategically

It will be hard to pick all the variables at once, so don’t try to do so. You can always go one variable at a time. It’s often easiest to start with the variable that appears most often, as that value often has the greatest influence over what numbers will be manageable. In this problem, that’s the distance upstream (it appears twice, since she goes the same distance downstream). So if you picked that distance to be, say, 6 kilometers, you could then easily pick manageable numbers for m and n.

But for the sake of argument, let’s say you didn’t see that you needed to pick distance. That’s OK. If you start by picking two manageable numbers for m and n—for example, m = 2 and n = 3—then the question becomes this:

Cindy paddles her kayak upstream at 2 kilometers per hour, and then returns downstream the same distance at 3 kilometers per hour. How many kilometers upstream did she travel if she spent a total of p hours for the round trip?

To figure out the time, you’d need the distance. As none is given, you can pick one. If the speeds are 2 kilometers per hour and 3 kilometers per hour, a distance of 6 kilometers works nicely. Now the question is this:

Cindy paddles her kayak 6 kilometers upstream at 2 kilometers per hour, and then returns downstream 6 kilometers at 3 kilometers per hour. How many kilometers upstream did she travel if she spent a total of p hours for the round trip?

Now the time is very straightforward to calculate. Six kilometers upstream at 2 kilometers per hour means a total of 3 hours. Six kilometers downstream at 3 kilometers per hour means a total of 2 hours. That’s 3 + 2 = 5 hours for the round trip.

Now plug m = 2, n = 3, and p = 5 into the answer choices to see which yields 6 for the number of kilometers traveled upstream.

(A) mnp = 2 × 3 × 5 = 30. Eliminate.

(B)  . Eliminate.

(C)  . Eliminate.

(D)  . (D) could be right, but you need to test (E) to make sure.

(E)  . Eliminate.

Only choice (D) remains, so it is correct.

Step 4: Confirm Your Answer

Make sure to reread the question stem to confirm that you interpreted the information correctly.

Picking numbers can make word problems much easier to understand, so it’s important that you don’t hastily start writing down equations. Were you to immediately start writing down algebraic equations for this question, you’d wind up with six variables: speed upstream, speed downstream, time upstream, time downstream, distance for each leg of the trip, and total time. That’s a lot to deal with. See how picking numbers simplifies things.

Takeaways: Picking Numbers with Variables in the Answer Choices

• Questions with variables in the answer choices comprise the majority of situations in which test takers should consider picking numbers.
• After picking numbers, reread the question stem, substituting your number(s) for the variable(s).
• When picking numbers with variables in the answer choices, check each answer choice to make sure the numbers you picked don’t happen to work for more than one choice.
  • Exception: For “could be” questions (e.g., “Which of the following could be odd?”), you can safely choose the first answer choice that works.

Practice Set: Picking Numbers with Variables in the Answer Choices

12. If , ,  , and   and if  , then 
    1. 
       
    2. 
       
    3. 
       
    4. 
       
    5. j + k + m
13. In 15 minutes, Jyoti can read p pages of her economics book. How many minutes will it take her to read r pages of her marketing book, if she reads both books at the same rate?
    1. 
       
    2. 
       
    3. 
       
    4. 
       
    5. 
       
14. To raise money for a charity, v volunteers each pledge to contribute an equal portion of the total goal of d dollars. If r more volunteers join the fund-raising campaign and agree with the original volunteers to contribute equally to the goal, how much less must each original donor give? 
    1. 
    2. 
    3. 
    4. 
    5. 

Picking Numbers on Must Be/Could Be/Cannot Be Questions

This is a slightly different style of question that asks things like “Which of the following must be an even integer?” or “Which of the following CANNOT be true?” These questions are usually based not on algebra or arithmetic but rather on the properties of the numbers themselves. Some of these questions can be very abstract, so picking numbers really helps. Just as with word problems, making a number properties question concrete helps you to understand it.

Example:

1. If a and b are odd integers, which of the following must be an even integer?
   1. a(b − 2)
   2. ab + 4
   3. (a + 2)(b − 4)
   4. 3a + 5b
   5. a(a + 6)

You can run through the answer choices quite quickly using the picking numbers strategy. Try a = 1, b = 3. Only (D), 3 + 15 = 18, is even.

Another approach that works well for many “must be/could be/cannot be” problems is to pick different numbers for each answer choice, looking to eliminate the wrong answers. The principle of focusing on what’s asked is particularly important here, as you need a clear idea of what you’re looking for in a wrong answer. Always characterize what you’re looking for among the answer choices before you dive in and start testing them. For instance, in the example you just saw, any expression that comes back even may be the correct answer, but any expression that comes back odd must certainly be incorrect and should be eliminated.

Knowing how to test the answer choices on “must be/could be/cannot be” questions is essential to your success on these questions. As you just saw, on a “must be” question, it’s not enough to stop when you reach the first answer choice that works with the numbers you picked. If an answer choice works, that only proves that it could be true, not that it must be true every time. To be sure you’re choosing the right answer to a “must be” or “cannot be” question, pick numbers to eliminate all four incorrect answers. However, for a “could be” question (e.g., “Which of the following could be odd?”), you can safely choose the first answer choice that works. If none of the answer choices work using the numbers you picked, you will need to pick another set of numbers and test the choices again. Always think critically about what the question stem is asking and how to pick numbers to find the answer most efficiently.

Roman numeral questions often feature must be/could be/cannot be language in their question stems. Let’s look at how to handle one of these questions.

Example:

1. If integers x and y are distinct factors of 24, then which of the following CANNOT be a factor of 24?
   1. (x + y)²
   2. x² − y²
   3. xy + y²
   1. I only
   2. I and II
   3. II and III
   4. II only
   5. III only

Roman numeral questions are rare but can be big time wasters if you don’t work strategically. Evaluate statements one at a time, eliminating answer choices as you go. Usually you should start with the statement that appears most frequently in the answer choices.

This question asks you to figure out which statement or statements can never be a factor of 24. That means you can eliminate a statement if it could ever possibly equal a factor of 24. If you pick some numbers that don’t yield a factor of 24, all that means is that the statement might be part of the right answer. But if you pick numbers that do yield a factor of 24, you know that any answer choice that includes that statement can be eliminated. It’s much more straightforward, therefore, to prove answers wrong rather than to prove them right.

Since the question involves factors of 24, it’s a good idea to list these factors out. Anytime a question deals with a small set of specific numbers, write them down. You’ll find keeping track of the options to be much easier:

1, 2, 3, 4, 6, 8, 12, 24

Statement II appears three times among the answer choices, so if you can eliminate it, you’ll be down to two choices right away. Squaring something large like 24 or 12 is going to produce a large number, which wouldn’t be a factor of 24. So choose smaller numbers instead: x = 4 and y = 2, perhaps. Then x² − y² would equal 16 − 4, or 12. Because 12 is a factor of 24, you can eliminate any answer choice containing Statement II. That leaves only (A) and (E).

Look at Statement III next. Not only are you squaring and multiplying, but you’re also adding, so this number is going to get big fast. To keep it in your target range, pick the smallest numbers on your list, x = 1 and y = 2. Then xy + y² would equal 1(2) + 4, or 6. That’s a factor of 24, so (E) is eliminated. You know that (A) is the right answer without ever having to evaluate Statement I.

Applying the Kaplan Method: Picking Numbers on Must Be/Could Be/Cannot Be Questions

Now let’s use the Kaplan Method on a Problem Solving question that involves picking numbers with “must be/could be/cannot be” questions:

1. If j and k are integers, and 2j + k = 15, which of the following must be true?
   1. j + k is odd.
   2. j + k is even.
   3. j is odd.
   4. k is odd.
   5. k > j

Step 1: Analyze the Question

You’re given the equation 2j + k = 15 and told that j and k are integers. Those two variables also show up in the answer choices. Whenever the answer choices contain variables, consider picking numbers.

Step 2: State the Task

You’re looking for an answer that MUST be true. Since you know to characterize the answer choices on “must be/could be/cannot be” problems, you know to remind yourself that if an answer choice could be false, even in one case, it can be eliminated.

Step 3: Approach Strategically

Pretend for a moment that you weren’t able to confidently pick different numbers, trying to see whether each answer could be false and thus eliminated. You’d have to pick one set of numbers and plug them into all the choices.

Let’s say that you start with j = 4 and k = 7.

(A) 4 + 7 is odd. True. Can’t be eliminated.

(B) 4 + 7 is even. False. Eliminate.

(C) 4 is odd. False. Eliminate.

(D) 7 is odd. True. Can’t be eliminated.

(E) 7 > 4. True. Can’t be eliminated.

Now you’d have to try a different set of numbers, hoping to eliminate some more. (E) looks like the easiest one to target, as you just have to think of a j that’s greater than or equal to k. You could choose j = 5 and k = 5. There’s no need to test (C) or (B), as they’ve already been eliminated.

(A) 5 + 5 is odd. False. Eliminate.

(D) 5 is odd. True. Can’t be eliminated.

(E) 5 > 5. False. Eliminate.

Only (D) wasn’t eliminated, so it must be the correct answer.

Step 4: Confirm Your Answer

Make sure to reread the question stem to confirm that you interpreted the stem correctly.

Takeaways: Picking Numbers on Must Be/Could Be/Cannot Be Questions

• On these questions, you can pick numbers and plug them into every answer choice . . .
• . . . or you can pick different numbers for each answer choice, trying either to eliminate it or to confirm it.
• Characterizing what you’re looking for in the answer choices will help you to pick numbers.
• On Roman numeral questions, start eliminating answer choices by either testing the options that appear most often in the choices or testing the options that are easiest to evaluate.

Practice Set: Picking Numbers on Must Be/Could Be/Cannot Be Questions

15. If integers a and b are distinct factors of 30, which of the following CANNOT be a factor of 30?
    1. ab + b²
    2. (a + b)²
    3. a + b
    1. I only
    2. II only
    3. III only
    4. I and II only
    5. I, II, and III
16. The integer a is less than −1, and the integer d is greater than 1. If −1 < b < 0 and 0 < c < 1, then which of the following must be the greatest negative number?
    1. b − a
    2. b − c
    3. b − d
    4. c − a
    5. a − c
17. In the set S of consecutive integers from 1 to 10, inclusive, D is the subset of all odd numbers, M is the subset of all prime numbers, and Q is the subset of all numbers whose positive square root is also a member of set S. If d is an element of D, m is an element of M, and q is an element of Q, which of the following could be true?
    1. 
       
    2. 
       
    3. m = q
    4. 
       
    5.