make the comparison without precise calculations—by remembering to follow this general rule: The greater the data are spread away from the mean, the greater the standard deviation. For example, consider these two distributions: Distribution A: {1, 2.5, 4, 5.5, 7} Distribution B: {1, 3, 4, 5, 7} In both sets, the mean and median is 4, and the range is 6. But the standard deviation of A is greater than that of B, because 2.5 and 5.5 are further away than 3 and 5 from the mean. 18. Which of the following distributions has the greatest standard deviation? (A) {21, 1, 3} (B) {1,2,5} (C) {0,4,5} (D) {23, 21, 2} (E) {2,3,6} The correct answer is (C). Notice that in each of the choices (A), (B), and (E), the distribution’s range is 4. But in choice (C) and choice (D), the range is 5. So the correct answer is probably either (C) or (D). Focusing on these two choices, notice that the middle term in choice (C), 4, is skewed further away from the mean than the middle term in choice (D). That’s a good indication that (C) provides the distribution having the greatest standard deviation. GEOMETRIC SEQUENCES In a geometric sequence of numbers, each term is a constant multiple of the preceding one; in other words, the ratio between any term and the next one is constant. The multiple (or ratio) might be obvious by examining the sequence—for example: In the geometric sequence 2, 4, 8, 16, ,youcaneasily determine that the constant multiple is 2 (and the ratio of each term to the next is 1:2). In the geometric sequence 1, 23, 9, 227, ,youcaneasily determine that the constant multiple is 23 (and the ratio of each term to the next is 1:23). Once you know the multiple (or ratio), you can answer any question asking for an unknown term—or for either the sum or the average of certain terms. 19. In a geometric sequence, each term is a constant multiple of the preceding one. If the third and fourth numbers in the series are 8 and 216, respectively, what is the first term in the sequence? (A) 232 (B) 24 (C) 2 (D) 4 (E) 64 The correct answer is (C). The constant multiple is 22. But since you need to work backward from the third term (8), apply the reciprocal of that multiple twice. The second term is ~8! S 2 1 2 D 524. The first term is ~24! S 2 1 2 D 5 2. Chapter 9: Math Review: Number Forms, Relationships, and Sets 223 www.petersons.com 20. In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6? (A) 16 (B) 19 (C) 22 1 2 (D) 21 2 3 (E) 20 The correct answer is (D). The constant multiple is 3 2 . In other words, the ratio of each term to the next is 2:3. Since the second term is 4, the first term is 4 3 2 3 5 8 3 . Since the third term is 6, the fourth term is 6 3 3 2 5 18 2 , or 9. The sum of the four terms 5 8 3 1 4 1 6 1 9 5 21 2 3 . You can also solve geometric sequence problems by applying a special formula. But you’ll need to memorize it because the test won’t provide it. In the following formula, r 5 the constant multiple (or the ratio between each term and the preceding one), a 5 the first term in the sequence, n 5 the position number for any particular term in the series, and T 5 the particular term itself: ar (n 2 1) 5 T You can solve for any of the formula’s variables, as long as you know the values for the other three. Following are two examples: If a 5 3 and r 5 2, then the third term 5 (3)(2) 2 5 12, and the sixth term 5 (3)(2) 5 5 (3)(32) 5 96. If the sixth term is 2 1 16 and the constant ratio is 1 2 , then the first term (a) 522: a S 1 2 D 5 5 2 1 16 a S 1 32 D 5 2 1 16 a 5 S 2 1 16 D ~32! 522 The algebra is simple enough—but you need to know the formula, of course. 224 PART IV: GMAT Quantitative Section ALERT! You can’t calculate the average of terms in a geometric series by averaging the first and last term in the series: The progression is geometric, not arithmetic. You need to add up the terms, then divide by the number of terms. www.petersons.com 21. In a geometric series, each term is a constant multiple of the preceding one. If the first three terms in a geometric series are 22, x, and 28, which of the following could be the sixth term in the series? (A) 232 (B) 216 (C) 16 (D) 32 (E) 64 The correct answer is (E). Since all pairs of successive terms must have the same ratio, 22 x 5 x 28 . By the cross-product method, x 2 5 16, and hence x 564. For x = 4, the ratio is 4 22 522. Applying the formula you just learned, the sixth term would be (22)(22) 5 5 64. For x 524, the ratio is 24 22 5 2. The sixth term would be (22)(2) 5 5264. ARITHMETIC SEQUENCES In an arithmetic sequence of numbers, there is a constant (unchanging) difference between successive numbers in the sequence. In other words, all numbers in an arithmetic sequence are evenly spaced. All of the following are examples of an arithmetic sequence: • Successive integers • Successive even integers • Successive odd integers • Successive multiples of the same number • Successive integers ending in the same digit On the GMAT, questions involving an arithmetic sequence might ask for the average or the sum of a sequence. When the numbers to be averaged form an arithmetic (evenly spaced) sequence, the average is simply the median (the middle number or the average of the two middle numbers if the number of terms is even). In other words, the mean and median of the set of numbers are the same. Faced with calculating the average of a long sequence of evenly-spaced integers, you can shortcut the addition. 22. What is the average of the first 20 positive integers? (A) 7 1 2 (B) 10 (C) 10 1 2 (D) 15 (E) 20 Chapter 9: Math Review: Number Forms, Relationships, and Sets 225 www.petersons.com The correct answer is (C). Since the terms are evenly spaced, the average is halfway between the 10th and 11th terms—which happen to be the integers 10 and 11. So the average is 10 1 2 . (This number is also the median.) If you take the average of the first term (1) and the last term (20), you get the same result: 1 1 20 2 5 21 2 ,or10 1 2 Finding the sum (rather than the average) of an arithmetic (evenly spaced) sequence of numbers requires only one additional step: multiplying the average (which is also the median) by the number of terms in the sequence. The trickiest aspect of this type of question is determining the number of terms in the sequence. 23. What is the sum of all odd integers between 10 and 40? (A) 250 (B) 325 (C) 375 (D) 400 (E) 450 The correct answer is (C). The average of the described numbers is 25—halfway between 10 and 40 (in other words, half the sum of 10 and 40). The number of terms in the sequence is 15. (The first term is 11, and the last term is 39.) The sum of the described series of integers 5 25 3 15 5 375. When calculating the average or sum of a sequence of evenly spaced numbers, be careful counting the number of terms in the series. For instance, the number of positive odd integers less than 50 is 25, but the number of positive even integers less than 50 is only 24. PERMUTATIONS A permutation is an arrangement of objects in which the order (sequence) is important. Each arrangement of the letters A, B, C, and D, for example, is a different permutation of the four letters. There are two different ways to determine the number of permutations for a group of distinct objects. List all the permutations, using a methodical process to make sure you don’t overlook any. For the letters A, B, C, and D, start with A in the first position, then list all possibilities for the second position, along with all possibilities for the third and fourth positions (you’ll discover six permutations): A B CD AC BD AD BC A B DC AC DB AD CB Placing B in the first position would also result in 6 permutations. The same applies to either C or D in the first position. So the total number of permutations is 6x45 24. 226 PART IV: GMAT Quantitative Section www.petersons.com Use the following formula (let n 5 the number of objects) and limit the number of terms to the counting numbers, or positive integers: Number of permutations 5 n(n 2 1)(n 2 2)(n 2 3) (1) The number of permutations can be expressed as n!(“n” factorial). Using the factorial is much easier than compiling a list of permutations. For example, the number of arrangements (permutations) of the four letters A, B, C, and D: 4! 5 4(4 2 1)(4 2 2)(4 2 3) 5 4 3 3 3 2 3 1 5 24 24. Five tokens—one red, one blue, one green, and two white—are arranged in a row, one next to another. If the two white tokens are next to each other, how many arrangements according to color are possible? (A) 12 (B) 16 (C) 20 (D) 24 (E) 30 The correct answer is (D). The two white tokens might be in positions 1 and 2, 2 and 3, 3 and 4, or 4 and 5. For each of these four possibilities, there are 6 possible color arrangements (3!) for the other three tokens (which all differ in color). Thus, the total number of possible arrangements is 4 3 6, or 24. COMBINATIONS A combination is a group of certain objects selected from a larger set. The order of objects in the group is not important. You can determine the total number of possible combinations by listing the possible groups in a methodical manner. For instance, to determine the number of possible three-letter groups among the letters A, B, C, D, and E, work methodically, starting with A as a group member paired with B, then C, then D, then E. Be sure not to repeat combinations (repetitions are indicated in parentheses here). A, B, C (A, C, B) (A, D, B) (A, E, B) A, B, D A, C, D (A, D, C) (A, E, C) A, B, E A, C, E A, D, E (A, E, D) Perform the same task assuming B is in the group, then assuming C is in the group (all combinations not listed here repeat what’s already listed). B, C, D C, D, E B, C, E B, D, E The total number of combinations is 10. TIP You can shortcut common factorial calculations by memorizing them: 3! 5 6, 4! 5 24, and 5! 5 120. ALERT! Notice that each parenthetical combination backtracks to an earlier letter. Be sure you don’t repeat any combination and make sure you don’t backtrack to an earlier object. Chapter 9: Math Review: Number Forms, Relationships, and Sets 227 www.petersons.com 25. How many two-digit numbers can be formed from the digits 1 through 9, if no digit appears twice in a number? (A) 36 (B) 72 (C) 81 (D) 144 (E) 162 The correct answer is (B). Each digit can be paired with any of the other 8 digits. To avoid double counting, account for the possible pairs as follows: 1 and 2–9 (8 pairs), 2 and 3–9 (7 pairs), 3 and 4–9 (6 pairs), and so forth. The total number of distinct pairs is 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 5 36. Since the digits in each pair can appear in either order, the total number of possible two-digit numbers is 2 3 36, or 72. Here’s something to consider: You can approach combination problems as probability problems as well. Think of the “probability” of any single combination as “one divided by” the total number of combinations (a fraction between zero and 1). Use whichever method is quickest for the question at hand. We’ll review probability next. PROBABILITY Probability refers to the statistical chances of an event occurring (or not occurring). By definition, probability ranges from 0 to 1. (Probability is never negative, and it’s never greater than 1.) Here’s the basic formula for determining probability: Probability 5 number of ways the event can occur total number of possible occurrences 26. If you randomly select one candy from a jar containing two cherry candies, two licorice candies, and one peppermint candy, what is the probability of selecting a cherry candy? (A) 1 6 (B) 1 5 (C) 1 3 (D) 2 5 (E) 3 5 The correct answer is (D). There are two ways among five possible occurrences that a cherry candy will be selected. Thus, the probability of selecting a cherry candy is 2 5 . 228 PART IV: GMAT Quantitative Section www.petersons.com To calculate the probability of an event not occurring, just subtract the probability of the event occurring from 1. So, referring to the preceding question, the probability of not selecting a cherry candy is 3 5 . (Subtract 2 5 from 1.) Here’s another example of probability, but a bit tougher this time. A standard deck of 52 playing cards contains 12 face cards. The probability of selecting a face card from a standard deck is 12 52 ,or 3 13 . On the GMAT, a tougher probability question will involve this basic formula, but it will also add a complication of some kind. It might require you to determine any of the following: • Certain missing facts needed for a given probability • Probabilities involving two (or more) independent events • Probabilities involving an event that is dependent on another event For the next three types of probability questions, don’t try to “intuit” the answer. Probabilities involving complex scenarios such as these are often greater or less than you might expect. Missing Facts Needed for a Given Probability In this question type, instead of calculating probability, you determine what missing number is needed for a given probability. Don’t panic; just plug what you know into the basic formula and solve for the missing number. 27. A piggy-bank contains a certain number of coins, of which 53 are dimes and 19 are nickels. The remainder of the coins in the bank are quarters. If the probability of selecting a quarter from this bank is 1 4 , how many quarters does the bank contain? (A) 30 (B) 27 (C) 24 (D) 21 (E) 16 The correct answer is (C). On its face, this question looks complicated, but it’s really not. Just plug what you know into the probability formula. Let x 5 the number of quarters in the bank (this is the numerator of the formula’s fraction), and let x 1 72 5 the total number of coins (the fraction’s denominator). Then solve for x (use the cross-product method to clear fractions): 1 4 5 x x 1 72 x 1 72 5 4x 72 5 3x 24 5 x Chapter 9: Math Review: Number Forms, Relationships, and Sets 229 www.petersons.com Probability Involving Two (or More) Independent Events Two events are independent if neither event affects the probability that the other will occur. (You’ll look at dependent events on the next page.) On the GMAT, look for either of these two scenarios involving independent events: The random selection of one object from each of two or more groups The random selection of one object from a group, then replacing it and selecting again (as in a “second round” or “another turn” of a game) In either scenario, the simplest calculation involves finding the probability of two events both occurring. All you need to do is multiply together their individual probabilities: (probability of event 1 occurring) 3 (probability of event 2 occurring) 5 (probability of both events occurring). For example, assume that you randomly select one letter from each of two sets: {A,B} and {C,D,E}. The probability of selecting A and C 5 1 2 3 1 3 ,or 1 6 . To calculate the probability that two events will not both occur, subtract from 1 the probability of both events occurring. To determine the probability that three events will all occur, just multiply the third event’s probability by the other two. 28. If one student is chosen randomly out of a group of seven students, then one student is again chosen randomly from the same group of seven, what is the probability that two different students will be chosen? (A) 36 49 (B) 6 7 (C) 19 21 (D) 13 14 (E) 48 49 The correct answer is (B). You must first calculate the chances of picking a particular student twice by multiplying together the two individual probabilities for the student: 1 7 3 1 7 5 1 49 . The probability of picking any one of the seven students twice is then 7 3 1 49 5 7 49 . The probability of picking the same student twice, added to the probability of not picking the same student twice, equals 1. So to answer the question, subtract 7 49 from 1. Beware: In one selection, the probability of not selecting a certain student from the group of seven is 6 7 (the probability of selecting the student, subtracted from 1). But does this mean 230 PART IV: GMAT Quantitative Section www.petersons.com that the probability of not selecting the same student twice 5 6 7 3 6 7 5 36 49 ? No, it doesn’t. Make sure you understand the difference. Probability Involving a Dependent Event Two distinct events might be related in that one event affects the probability of the other one occurring—for example, randomly selecting one object from a group, then selecting a second object from the same group without replacing the first selection. Removing one object from the group increases the odds of selecting any particular object from those that remain. You handle this type of problem as you would any other probability problem: Calculate individual probabilities, then combine them. 29. In a random selection of two people from a group of five—A, B, C, D, and E—what is the probability of selecting A and B? (A) 2 5 (B) 1 5 (C) 1 10 (D) 1 15 (E) 1 20 The correct answer is (C). You need to consider each of the two selections separately. In the first selection, the probability of selecting either A or B is 2 5 . But the probability of selecting the second of the two is 1 4 , because after the first selection only four people remain from whom to select. Since the question asks for the probability of selecting both A and B (as opposed to either one), multiply the two individual probabilities: 2 5 3 1 4 5 2 20 5 1 10 . You can also approach a question such as this one as a combination problem. For this question, here are all the possibilities: A and either B, C, D, or E (4 combinations) B and either C, D, or E (3 combinations) C and either D or E (2 combinations) D and E (1 combination) There are 10 possible combinations, so the probability of selecting A and B is 1 in 10. ALERT! Strategies such as plugging in test numbers, working backward, and sizing up answer choices don’t work for most probability questions. Chapter 9: Math Review: Number Forms, Relationships, and Sets 231 www.petersons.com SUMMING IT UP • Although the types of questions reviewed in this chapter are the most basic of the math problems you’ll encounter on the GMAT Quantitative section, don’t underestimate how useful they’ll be as building blocks for solving more complex problems. • Certain fraction-decimal-percent equivalents show up more frequently than others on the GMAT. If you have time, memorize the standard conversions to save yourself time on the actual exam. • Percent change questions are typical on the GMAT Quantitative section, so be ready for them. • As with fractions, you can simplify ratios by dividing common factors. • Review the definitions of arithmetic mean, median, mode, and range, so you’re better equipped to solve such problems on the exam. • Many arithmetic sequence questions ask for the average or sum of a series. You may be able to “shortcut” the addition instead of calculating the average of a long series of evenly spaced integers. • Memorizing common factorial combinations will save you time when you encounter permutation questions on the GMAT. • Work methodically on combination questions to avoid backtracking to an earlier object. • It’s wise not to try “intuiting” the answers to probability questions. Many of these problems are too complex to arrive at an accurate answer this way. 232 PART IV: GMAT Quantitative Section www.petersons.com . sum of all odd integers between 10 and 40? (A) 250 (B) 325 (C) 375 (D) 400 (E) 450 The correct answer is (C). The average of the described numbers is 25 halfway between 10 and 40 (in other words,. 5 S 2 1 16 D ~32! 522 The algebra is simple enough—but you need to know the formula, of course. 224 PART IV: GMAT Quantitative Section ALERT! You can’t calculate the average of terms in a geometric series. one), a 5 the first term in the sequence, n 5 the position number for any particular term in the series, and T 5 the particular term itself: ar (n 2 1) 5 T You can solve for any of the formula’s