Part 4. QUANTITATIVE ABILITY: TACTICS, STRATEGIES, PRACTICE, AND REVIEW

Chapter 11. Discrete Quantitative Questions

Of the 28 questions in the quantitative section, 10 are called discrete quantitative questions, which is just a fancy name for standard multiple-choice questions, similar to those that you have encountered on many other standardized tests, such as the PSAT and SAT I.

In this chapter you will learn all of the important strategies to help you answer multiple-choice questions on the GRE. However, as invaluable as these tactics are, use them only when you need them. If you know how to do a problem and are confident that you can do it accurately and reasonably quickly, JUST DO IT!

Before the first discrete quantitative question appears on the screen, you will see the following instructions:

Select the best of the answer choices given.

This one-sentence set of directions is quite simple. Unfortunately, it is also quite useless. First of all, it is misleading. It is never the case that two of the answers are bad and three are good, and you then need to choose the bestanswer from among the three good ones. Every quantitative multiple-choice question has exactly one correct answer and four incorrect ones. Second, the implication is that you should just solve each problem and then look at the five choices to see which one is best (i.e., correct). As you will learn in this chapter, that is not always the best strategy.

As we have done throughout this book, we will continue to label the five answer choices A, B, C, D, and E and to refer to them as such. Of course, when you take the GRE, these letters will not appear—there will simply be a blank oval in front of each of the answer choices. When we refer to Choice C—as we do, for example, in TACTIC 1 (on page 326)—we are simply referring to the third answer choice among the five presented.

HELPFUL HINT

When you take the GRE, dismiss the instructions for these questions instantly— do not spend even one second reading them— and certainly never accept their offer of clicking on “HELP” to return to them during the test.

Testing Tactics

 Test the Choices, Starting with C

TACTIC 1, often called backsolving, is useful when you are asked to solve for an unknown and you understand what needs to be done to answer the question, but you want to avoid doing the algebra. The idea is simple: test the various choices to see which one is correct.

NOTE: On the GRE the answers to virtually all numerical multiple-choice questions are listed in either increasing or decreasing order. Consequently, C is the middle value, and in applying TACTIC 1, you should always start with C. For example, assume that choices A, B, C, D, and E are given in increasing order. Try C. If it works, you’ve found the answer. If C doesn’t work, you should know whether you need to test a larger number or a smaller one, and that permits you to eliminate two more choices. If C is too small, you need a larger number, and so A and B are out; if C is too large, eliminate D and E, which are even larger.

Examples 1 and 2 illustrate the proper use of TACTIC 1.

If the average (arithmetic mean) of 5, 6, 7, and w is 10, what is the value of w?

(A) 8     (B) 13     (C) 18     (D) 22     (E) 28

SOLUTION. Use TACTIC 1. Test Choice C: w = 18.

·        Is the average of 5, 6, 7, and 18 equal to 10?

·        No:  =  = 9, which is too small.

·        Eliminate C, and, since for the average to be 10, w must be greater than 18, eliminate A and B, as well.

·        Try D: w = 22. Is the average of 5, 6, 7, and 22 equal to 10?

·        Yes:  =  = 10. The answer is D.

Every problem that can be solved using TACTIC 1 can be solved directly, usually in less time. So we stress: if you are confident that you can solve a problem quickly and accurately, just do so.

Here are two direct methods for solving Example 1, each of which is faster than backsolving. (See Section 14-E on averages.) If you know either method you should use it, and save TACTIC 1 for those problems which you can’t easily solve directly.

DIRECT SOLUTION 1. If the average of four numbers is 10, their sum is 40. So, 5 + 6 + 7 + w = 40 ⇒ 18 + w = 40 ⇒ w = 22.

DIRECT SOLUTION 2. Since 5 is 5 less than 10, 6 is 4 less than 10, and 7 is 3 less than 10, to compensate, w must be 5 + 4 + 3 = 12 more than 10. So, w = 10 + 12 = 22.

Judy is now twice as old as Adam, but 6 years ago, she was 5 times as old as he was. How old is Judy now?

(A) 10     (B) 16     (C) 20     (D) 24     (E) 32

SOLUTION. Use TACTIC 1: backsolve starting with C. If Judy is now 20, Adam is 10, and 6 years ago, they would have been 14 and 4. Since Judy would have been less than 5 times as old as Adam, eliminate C, D, and E, and try a smaller value. If Judy is now 16, Adam is 8; 6 years ago, they would have been 10 and 2. That’s it; 10 is 5 times 2. The answer is B.

(See Section 14-H on word problems for the correct algebraic solution.)

Some tactics allow you to eliminate a few choices so you can make an educated guess. On those problems where it can be used, TACTIC 1 always gets you the right answer. The only reason not to use it on a particular problem is that you can easily solve the problem directly.

If 3x = 2(5 – 2x), then x =

HELPFUL HINT

Don’t start with C if some of the other choices are much easier to work with. If you start with B and it is too small, you may only get to eliminate two choices (A and B), instead of three, but it will save time if plugging in Choice C would be messy.

SOLUTION. Since plugging in 0 is so much easier than plugging in , start with B: then the left-hand side of the equation is 0 and the right-hand side is 10. The left-hand side is much too small. Eliminate A and B and try something bigger — D, of course; it will be much easier to deal with 1 than with  or . Now the left-hand side is 3 and the right-hand side is 6. We’re closer, but not there. The answer must be E. Notice that we got the right answer without ever plugging in one of those unpleasant fractions. Are you uncomfortable choosing E without checking it? Don’t be. If you know that the answer is greater than 1, and only one choice is greater than 1, that choice has to be right.

Again, we emphasize that, no matter what the choices are, you backsolve only if you can’t easily do the algebra. Most students would probably do this problem directly:

3x = 2(5 – 2x) ⇒ 3x = 10 – 4x ⇒ 7x = 10 ⇒ x = 

and save backsolving for a harder problem. You have to determine which method is best for you.

 Replace Variables with Numbers

Mastery of TACTIC 2 is critical for anyone developing good test-taking skills. This tactic can be used whenever the five choices involve the variables in the question. There are three steps:

1.            Replace each letter with an easy-to-use number.

2.            Solve the problem using those numbers.

3.            Evaluate each of the five choices with the numbers you picked to see which choice is equal to the answer you obtained.

Examples 4 and 5 illustrate the proper use of TACTIC 2.

If a is equal to the sum of b and c, which of the following is equal to the difference of b and c?

(A) a – b – c     (B) a – b + c     (C) a – c     (D) a – 2c     (E) a – b – 2c

SOLUTION.

·        Pick three easy-to-use numbers which satisfy a = b + c: for example, a = 5, b = 3, c = 2.

·        Then, solve the problem with these numbers: the difference of b and c is 3 – 2 = 1.

·        Finally, check each of the five choices to see which one is equal to 1:

(A) Does a – b – c = 1?     NO. 5 – 3 – 2 = 0
(B) Does a – b + c = 1?     NO. 5 – 3 + 2 = 4
(C) Does a – c = 1?           NO. 5 – 2 = 3
(D) Does a – 2c = 1?         YES! 5 – 2(2) = 5 – 4 = 1
(E) Does a – b – 2c = 1?    NO. 5 – 3 – 2(2) = 2 – 4 = –2

·        The answer is D.

If the sum of five consecutive even integers is t, then, in terms of t, what is the greatest of these integers?

SOLUTION.

·        Pick five easy-to-use consecutive even integers: say, 2, 4, 6, 8, 10. Then t, their sum, is 30.

·        Solve the problem with these numbers: the greatest of these integers is 10.

·        When t = 30, the five choices are 

·        Only , Choice E, is equal to 10.

Of course, Examples 4 and 5 can be solved without using TACTIC 3 if your algebra skills are good. Here are the solutions.

SOLUTION 4. a = b + c ⇒ b = a – c ⇒ b – c = (a – c) – c = a – 2c.

SOLUTION 5. Let n, n + 2, n + 4, n + 6, and n + 8 be five consecutive even integers, and let t be their sum. Then,

t = n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 5n + 20

So, 

The important point is that if you can’t do the algebra, you can still use TACTIC 2 and always get the right answer. Of course, you should use TACTIC 2 even if you can do the algebra, if you think that by using this tactic you will solve the problem faster or will be less likely to make a mistake. This is a good example of what we mean when we say that with the proper use of these tactics, you can correctly answer many questions for which you may not know the correct mathematical solution.

Examples 6 and 7 are somewhat different. You are asked to reason through word problems involving only variables. Most students find problems like these mind-boggling. Here, the use of TACTIC 2 is essential. Without it, Example 6 is difficult and Example 7 is nearly impossible. This is not an easy tactic to master, but with practice you will catch on.

If a school cafeteria needs c cans of soup each week for each student, and if there are s students in the school, for how many weeks will x cans of soup last?

HELPFUL HINT

Replace the letters with numbers that are easy to use, not necessarily ones that make sense. It is perfectly OK to ignore reality. A school can have 5 students, apples can cost 10 dollars each, trains can go 5 miles per hour or 1000 miles per hour — it doesn’t matter.

SOLUTION.

·        Replace c, s, and x with three easy-to-use numbers. If a school cafeteria needs 2 cans of soup each week for each student, and if there are 5 students in the school, how many weeks will 20 cans of soup last?

·        Since the cafeteria needs 2 × 5 = 10 cans of soup per week, 20 cans will last 2 weeks.

·        Which of the choices equals 2 when c = 2, s = 5, and x = 20?

·        csx = 200; 

      The answer is  D.

NOTE: You do not need to get the exact value of each choice. As soon as you see that a choice does not equal the value you are looking for, stop—eliminate that choice and move on. For example, in the preceding problem, it is clear that csx is much greater than 2, so eliminate it immediately; you do not need to multiply it out to determine that the value is 200.

CAUTION

In this type of problem it is not a good idea to replace any of the variables by 1. Since multiplying and dividing by 1 give the same result, you would not be able to distinguish between  and , both of which are equal to 4 when c = 1, s = 5, and x = 20. It is also not a good idea to use the same number for different variables:  and  are each equal to x when c and s are equal.

A vendor sells h hot dogs and s sodas. If a hot dog costs twice as much as a soda, and if the vendor takes in a total of d dollars, how many cents does a soda cost?

SOLUTION.

·        Replace h, s, and d with three easy-to-use numbers. Suppose a soda costs 50¢ and a hot dog $1.00. Then, if he sold 2 sodas and 3 hot dogs, he took in 4 dollars.

·        Which of the choices equals 50 when s = 2, h = 3, and d = 4?

·        Only 

  Now, practice TACTIC 3 on the following problems.

Yann will be x years old y years from now. How old was he z years ago?

(A) x + y + z    (B) x + y – z    (C) x – y – z    (D) y – x – z    (E) z – y – x

Stan drove for h hours at a constant rate of r miles per hour. How many miles did he go during the final 20 minutes of his drive?

SOLUTION. 8. Assume that Yann will be 10 in 2 years. How old was he 3 years ago? If he will be 10 in 2 years, he is 8 now and 3 years ago he was 5. Which of the choices equals 5 when x = 10, y = 2, and z = 3? Only x – y – z, C.

SOLUTION 9. If Stan drove at 60 miles per hour for 2 hours, how far did he go in the last 20 minutes? Since 20 minutes is  of an hour, he went 20 ( of 60) miles.

Only  is 20 when r = 60 and h = 2. Notice that h is irrelevant. Whether he had been driving for 2 hours or 20 hours, the distance he covered in the last 20 minutes would be the same.

 Choose an Appropriate Number

TACTIC 3 is similar to TACTIC 2, in that we pick convenient numbers. However, here no variable is given in the problem. TACTIC 3 is especially useful in problems involving fractions, ratios, and percents.

At Madison High School each student studies exactly one foreign language. Three-fifths of the students take Spanish, and one-fourth of the remaining students take German. If all of the others take French, what percent of the students take French?

(A) 10    (B) 15    (C) 20    (D) 25 (E)    30

HELPFUL HINT

In problems involving fractions, the best number to use is the least common denominator of all the fractions. In problems involving percents, the easiest number to use is 100. (See Sections 14-B and 14-C.)

SOLUTION. The least common denominator of  and  is 20, so assume that there are 20 students at Madison High. (Remember the numbers don’t have to be realistic.) The number of students taking Spanish is 12 ( of 20). Of the remaining 8 students, 2 of them ( of 8) take German. The other 6 take French. Finally, 6 is 30% of 20. The answer is E.

From 1994 to 1995 the sales of a book decreased by 80%. If the sales in 1996 were the same as in 1994, by what percent did they increase from 1995 to 1996?

(A) 80%    (B) 100%    (C) 120%    (D) 400%    (E) 500%

SOLUTION. Since this problem involves percents, assume that 100 copies of the book were sold in 1994 (and 1996). Sales dropped by 80 (80% of 100) to 20 in 1995 and then increased by 80, from 20 back to 100, in 1996. The percent increase was

 Eliminate Absurd Choices and Guess

When you have no idea how to solve a problem, you are forced to guess since you can’t get to the next question until you answer and confirm the one on the screen; but first eliminate all the absurd choices. Then guess from among the remaining ones.

During the course of a GRE, you will probably find at least a few multiple-choice questions that you have no idea how to solve. Since you can’t omit them, you have to guess. But take a moment to look at the answer choices. Often two or three of them are absurd. Eliminate those and then guess one of the others. Occasionally, four of the choices are absurd. When this occurs, your answer is no longer a guess.

What makes a choice absurd? Lots of things. Here are a few. Even if you don’t know how to solve a problem you may realize that

·        the answer must be positive, but some of the choices are negative;

·        the answer must be even, but some of the choices are odd;

·        the answer must be less than 100, but some choices exceed 100;

·        a ratio must be less than 1, but some choices are greater than 1.

Let’s look at several examples. In a few of them the information given is intentionally insufficient to solve the problem; but you will still be able to determine that some of the answers are absurd. In each case the “solution” will indicate which choices you should have eliminated. At that point you would simply guess. Remember, on the GRE when you have to guess, don’t agonize. Just guess and move on.

A region inside a semicircle of radius r is shaded and you are asked for its area.

SOLUTION. You may have no idea how to find the area of the shaded region, but you should know that since the area of a circle is πr², the area of a semicircle is πr². Therefore, the area of the shaded region must be less than πr², so eliminate C, D, and E. On an actual GRE problem, you may be able to make an educated guess between A and B. If so, terrific; if not, just choose one or the other.

The average (arithmetic mean) of 5, 10, 15, and z is 20. What is z?

(A) 0    (B) 20    (C) 25    (D) 45    (E) 50

SOLUTION. If the average of four numbers is 20, and three of them are less than 20, the other one must be greater than 20. Eliminate A and B and guess. If you further realize that since 5 and 10 are a lot less than 20, z will probably be a lot more than 20; eliminate C, as well.

If 25% of 260 equals 6.5% of a, what is a?

(A) 10    (B) 65    (C) 100    (D) 130    (E) 1000

SOLUTION. Since 6.5% of a equals 25% of 260, which is surely greater than 6.5% of 260, a must be greater than 260. Eliminate A, B, C, and D. The answer must be E!

Example 14 illustrates an important point. Even if you know how to solve a problem, if you immediately see that four of the five choices are absurd, just pick the fifth choice and move on.

A jackpot of $39,000 is to be divided in some ratio among three people. What is the value of the largest share?

(A) $23,400    (B) $19,500    (C) $11,700    (D) $7800    (E) $3900

SOLUTION. If the prize were divided equally, each of the three shares would be worth $13,000. If it is divided unequally, the largest share is surely worth more than $13,000. Eliminate C, D, and E. In an actual question, you would be told what the ratio is, and that might enable you to eliminate A or B. If not, you just guess.

In a certain club, the ratio of the number of boys to girls is 5:3. What percent of the members of the club are girls?

(A) 37.5%    (B) 50%    (C) 60%    (D) 62.5%    (E) 80%

SOLUTION. Since there are 5 boys for every 3 girls, there are fewer girls than boys. Therefore, fewer than half (50%) of the members are girls. Eliminate B, C, D, and E. The answer is A.

In the figure below, four semicircles are drawn, each one centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?

(A) 8π    (B) 32 – 8π    (C) 16 – 8π    (D) 8π  – 32    (E) 8π  – 16

SOLUTION. Since the diagram is drawn to scale, you may trust it in making your estimate (TACTIC 2, Chapter 10).

·        Since the shaded area appears to take up a little more than half of the square, it does.

·        The area of the square is 16, and so the area of the shaded region must be about 9.

·        Check each choice. Since π is slightly more than 3 (π ≈ 3.14), 8π is somewhat greater than 24, approximately 25.

·        (A) 8π ≈ 25. More than the area of the whole square: way too big.

·        (B) 32 – 8π ≈ 32 – 25 = 7. Too small (but close enough to consider if nothing is closer).

·        (C) 16 – 8π is negative. Clearly impossible!

·        (D) 8π – 32 is also negative.

·        (E) 8π – 16 ≈ 25 – 16 = 9. Finally! The answer is E.

NOTE: Three of the choices are absurd: A is more than the area of the entire square and C and D are negative; they can be eliminated immediately. No matter what your estimate was, at worst you had to guess between two choices.

Now use TACTIC 5 on each of the following problems. Even if you know how to solve them, don’t. Practice this technique and see how many choices you can eliminate without actually solving.

In the figure at the right, diagonal EG of square EFGH is  of diagonal AC of the square ABCD. What is the ratio of the area of the shaded region to the area of ABCD?

(A) :1    (B) 3:4    (C) :2    (D) 1:2    (E) 1:2

Shari receives a commission of 25¢ for every $20.00 worth of merchandise she sells. What percent is her commission?

(A) %    (B) %    (C) 5%    (D) 25%    (E) 125%

From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 1980?

(A) 1.75k    (B) 1.25k    (C) 1.20k    (D) .80k    (E) .75k

The average of 10 numbers is –10. If the sum of 6 of them is 100, what is the average of the other 4?

(A) –100    (B) –50    (C) 0    (D) 50    (E) 100

SOLUTION 18. Obviously, the shaded region is smaller than square ABCD, so the ratio must be less than 1. Eliminate A. Also, from the diagram, it is clear that the shaded region is more than half of square ABCD, so the ratio is greater than 0.5. Eliminate D and E. Since 3:4 = .75 and :2 ≈ .71, B and C are too close to tell which is correct just by looking; so guess. The answer is B.

SOLUTION 19. Clearly, a commission of 25¢ on $20 is quite small. Eliminate D and E and guess one of the small percents. If you realize that 1% of $20 is 20¢, then you know the answer is a little more than 1%, and you should guess A (maybe B, but definitely not C). The answer is A.

SOLUTION 20. Since Lior’s weight increased, his weight in 1980 was less than k. Eliminate A, B, and C and guess. The answer is D.

SOLUTION 21. Since the average of all 10 numbers is negative, so is their sum. But the sum of the first 6 is positive, so the sum (and the average) of the others must be negative. Eliminate C, D, and E. B is correct.

Practice Exercises

Discrete Quantitative Questions

1.    Evan has 4 times as many books as David and 5 times as many as Jason. If Jason has more than 40 books, what is the least number of books that Evan could have?
(A) 200    (B) 205    (C) 210    (D) 220    (E) 240

2.    Judy plans to visit the National Gallery once each month in 2001 except in July and August when she plans to go three times each. A single admission costs $3.50, a pass valid for unlimited visits in any 3-month period can be purchased for $18, and an annual pass costs $60.00. What is the least amount, in dollars, that Judy can spend for her intended number of visits?
(A) 72    (B) 60    (C) 56    (D) 49.5   (E) 48

3.    Alison is now three times as old as Jeremy, but 5 years ago, she was 5 times as old as he was. How old is Alison now?
(A) 10    (B) 12    (C) 24    (D) 30    (E) 36

4.    What is the largest prime factor of 255?
(A) 5    (B) 15    (C) 17    (D) 51    (E) 255

5.    If c is the product of a and b, which of the following is the quotient of a and b?

6.    If w widgets cost c cents, how many widgets can you get for d dollars?

7.    If 120% of a is equal to 80% of b, which of the following is equal to a + b?
(A) 1.5a    (B) 2a    (C) 2.5a    (D) 3a    (E) 5a

8.    In the figure below, WXYZ is a square whose sides are 12. AB, CD, EF, and GH are each 8, and are the diameters of the four semicircles. What is the area of the shaded region?

(A 144 – 128π
(B) 144 – 64π
(C) 144 – 32π
(D) 144 – 16π
(E) 16π

9.    If x and y are integers such that x³ = y², which of the following could not be the value of y?
(A) –1    (B) 1    (C) 8    (D) 16    (E) 27

10.What is a divided by a% of a?

11.If an object is moving at a speed of 36 kilometers per hour, how many meters does it travel in one second?
(A) 10    (B) 36    (C) 100    (D) 360    (E) 1000

12.On a certain French-American committee,  of the members are men, and  of the men are Americans. If  of the committee members are French, what fraction of the members are American women?

13.For what value of x is 8^2x–4 = 16^x?
(A) 2    (B) 3    (C) 4    (D) 6    (E) 8

14.If 12a + 3b = 1 and 7b – 2a = 9, what is the average (arithmetic mean) of a and b?
(A) 0.1    (B) 0.5    (C) 1    (D) 2.5    (E) 5

15.If x% of y is 10, what is y?

ANSWER KEY

1. D	4. C	7. C	10. B	13. D
2. D	5. B	8. C	11. A	14. B
3. D	6. A	9. D	12. A	15. C

Answer Explanations

Two asterisks (**) indicate an alternative method of solving.

1.    (D) Test the answer choices starting with the smallest value. If Evan had 200 books, Jason would have 40. But Jason has more than 40, so 200 is too small. Trying 205 and 210, we see that neither is a multiple of 4, so David wouldn’t have a whole number of books. Finally, 220 works. (So does 240, but we shouldn’t even test it since we want the least value.)
   **Since Jason has at least 41 books, Evan has at least 41 × 5 = 205. But Evan’s total must be a multiple of 4 and 5, hence of 20. The smallest multiple of 20 greater than 205 is 220.

2.    (D) Judy intends to go to the Gallery 16 times during the year. Buying a single admission each time would cost 16 × $3.50 = $56, which is less than the annual pass. If she bought a 3-month pass for June, July, and August, she would pay $18 plus $31.50 for 9 single admissions (9 × $3.50), for a total expense of $49.50, which is the least expensive option.

3.    (D) Use TACTIC 1: backsolve starting with C. If Alison is now 24, Jeremy is 8, and 5 years ago, they would have been 19 and 3, which is more than 5 times as much. Eliminate A, B, and C, and try a bigger value. If Alison is now 30, Jeremy is 10, and 5 years ago, they would have been 25 and 5. That’s it; 25 is 5 times 5.
   **If Jeremy is now x, Alison is 3x, and 5 years ago they were x – 5 and 3x – 5, respectively. Now, solve:
3x – 5 = 5(x – 5) ⇒ 3x – 5 = 5x – 25 ⇒ 
2x = 20 ⇒ x = 10 ⇒ 3x = 30.

4.    (C) Test the choices starting with C: 255 is divisible by 17 (255 = 17 × 15), so this is a possible answer. Does 255 have a larger prime factor? Neither Choice D nor E is prime, so the answer must be Choice C.

5.    (B) Use TACTIC 2. Pick simple values for a, b, and c. Let a = 3, b = 2, and c = 6. Then a ÷ b = 3/2. Without these values of a, b, and c, only B is equal to 3/2.

6.    (A) Use TACTIC 2. If 2 widgets cost 10 cents, then widgets cost 5 cents each, and for 3 dollars, you can get 60. Which of the choices equals 60 when w = 2, c = 10, and d = 3? Only A.

7.    (C) Since 120% of 80 = 80% of 120, let a = 80 and b = 120. Then a + b = 200, and 200 ÷ 80 = 2.5.

8.    (C) If you don’t know how to solve this, you must use TACTIC 4 and guess after eliminating the absurd choices. Which choices are absurd? Certainly, A and B, both of which are negative. Also, since Choice D is about 94, which is much more than half the area of the square, it is much too big. Guess between Choice C (about 43) and Choice E (about 50). If you remember that the way to find shaded areas is to subtract, guess C.
   **The area of the square is 12² = 144. The area of each semicircle is 8π, one-half the area of a circle of radius 4. So together the areas of the semicircles is 32π.

9.    (D) Test each choice until you find the correct answer. Could y = –1? Is there an integer x such that x³ = (–1)² = 1? Yes, x = 1. Similarly, if y = 1, x = 1. Could y = 8? Is there an integer x such that x³ = (8)² = 64? Yes, x = 4. Could y = 16? Is there an integer x such that x³ = 16² = 256? No, 6³ = 216, which is too small; and 7³ = 343, which is too big. The answer is D.

10.(B) a ÷ (a% of a) = a ÷ 
   **Use TACTICS 2 and 3: replace a by a number, and use 100 since the problem involves percents. 100 ÷ (100% of 100) = 100 ÷ 100 = 1. Test each choice; which one equals 1 when a = 100. Both A and B:  = 1. Eliminate Choices C, D, and E, and test A and B with another value for a. 50 ÷ (50% of 50) = 50 ÷ (25) = 2. Now, only B works .

11.(A) Set up a ratio:

   **Use TACTIC 1: Test choices starting with C:
100 meters/second = 6000 meters/minute =
360,000 meters/hour = 360 kilometers/hour.
   Not only is that too big, it is too big by a factor of 10. The answer is 10.

12.(A) Use TACTIC 3. The LCM of all the denominators is 120, so assume that the committee has 120 members. Then there are  × 120 = 80 men and 40 women. Of the 80 men 30  are American. Since there are 72  French members, there are 120 – 72 = 48 Americans, of whom 30 are men, so the other 18 are women. Finally, the fraction of American women is .
This is illustrated in the Venn diagram below.

13.(D) Use the laws of exponents to simplify the equation, and then solve it: 8^2x–4 = 16^x ⇒ (2³)^2x–4 = (2⁴)^x ⇒ 3(2x – 4) = 4x ⇒ 6x – 12 = 4x ⇒ 2x = 12 ⇒ x = 6.

14.(B) Add the two equations:
10a + 10b = 10 ⇒ a + b = 1 ⇒ .
Do not waste time solving for a and b.

15.(C) Pick easy-to-use numbers. Since 100% of 10 is 10, let x = 100 and y = 10. When x = 100, Choices C and E are each 10. Eliminate Choices A, B, and D, and try some other numbers: 50% of 20 is 10. Of Choices C and E, only C = 20 when x = 50.