Tài liệu Master the Gre 2010 - Part 24 pptx

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 see 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 see 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. 25. In a geometric sequence, each term is a constant multiple of the preceding one. If the third and fourth numbers in the sequence 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. 26. 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, Chapter 9: Math Review: Number Forms, Relationships, and Sets 213 ALERT! You can’t calculate the average of terms in a geometric sequence by averaging the first and last term i n the sequence: The progression is geometric, not arithmetic. You need to add up the terms, then divide by the number of terms. www.petersons.com 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 sequence, 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. 27. In a geometric sequence, each term is a constant multiple of the preceding one. If the first three terms in a geometric sequence are 22, x, and 28, which of the following could be the sixth term in the sequence? (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. 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. PART IV: Quantitative Reasoning214 www.petersons.com 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 permuta- tions is 6 3 4 5 24. 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, here’s how to determine 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. 28. 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): Chapter 9: Math Review: Number Forms, Relationships, and Sets 215 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. www.petersons.com 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 B, C, E B, D, E C, D, E The total number of combinations is 10. 29. 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 (0) and 1). Use whichever method is quickest for the question at hand. PROBABILITY Probability refers to the statistical chances of an event occurring or not occurring. By definition, probability ranges from zero (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 PART IV: Quantitative Reasoning216 www.petersons.com 30. 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 . 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.) On the GRE, a tougher probability question would involve this basic formula, but it would 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 these three types of probability questions, which we’ll examine next, 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. To do so, just plug what you know into the basic formula and solve for the missing number. 31. 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) 18 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), Chapter 9: Math Review: Number Forms, Relationships, and Sets 217 TIP The probability of an event not occurring is 1 minus the probability that it will occur. www.petersons.com 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 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 next.) On the GRE, 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 prob- abilities: (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 determine the probability that three events will all occur, just multiply the third event’s probability by the other two. To calculate the probability that two events will not both occur, subtract from 1 the probability of both events occurring. 32. From a group of seven students, one student is called on at random to answer a question. Then one of the same seven students is called on at random to answer another question. What is the probability that the same student will NOT be called on to answer BOTH questions? (A) 1 14 (B) 1 2 (C) 6 7 (D) 13 14 (E) 48 49 The correct answer is (E). 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 not picking the same student twice is 1 2 1 49 = 48 49 . PART IV: Quantitative Reasoning218 www.petersons.com 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: Cal- culate individual probabilities, then combine them. 33. 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 probability problems like Question 33 (above) as combination problems. For Question 33, here are all the combinations: • 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. Chapter 9: Math Review: Number Forms, Relationships, and Sets 219 ALERT! Strategies such as plugging in test numbers, working backward, and sizing up answer choices don’t work for most probability questions. www.petersons.com SUMMING IT UP • Although the types of questions reviewed in the early sections of this chapter are the most basic of the math problems you’ll encounter on the GRE Quantitative Reasoning 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 GRE. If you have time, memorize the standard conversions to save yourself time on the actual exam. • Percent change questions are typical on the GRE Quantitative Reasoning 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 sequence. You may be able to “shortcut” the addition instead of calculating the average of a long sequence of evenly spaced integers. • Memorizing common factorial combinations will save you time when you encounter permutation questions on the GRE. • 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. PART IV: Quantitative Reasoning220 www.petersons.com Math Review: Number Theory and Algebra OVERVIEW • Basic properties of numbers • Factors, multiples, and divisibility • Prime numbers and prime factorization • Exponents (powers) • Roots and radicals • Linear equations with one variable • Linear equations with two variables • Simple algebraic functions • Linear equations that cannot be solved • Factorable quadratic expressions with one variable • Nonlinear equations with two variables • Solving algebraic inequalities • Solving algebra “story” problems • Summing it up In this chapter, you’ll first broaden your arithmetical horizons by dealing with numbers in more abstract, theoretical settings. You’ll examine the following topics: • The concept of absolute value • Number signs and integers—and what happens to them when you apply the four basic operations • Factors, multiples, divisibility, prime numbers, and the “prime factor- ization” method • The rules for combining exponential numbers (base numbers and “powers”) using the four basic operations • The rules for combining radicals using the four basic operations • The rules for simplifying terms containing radical signs Then you’ll review the following basic algebra skills: • Solving a linear equation with one variable chapter 10 221 • Solving a system of two equations with two variables—by substitution and by addition-subtraction • Recognizing unsolvable linear equations when you see them • Applying simple algebraic functions • Handling algebraic inequalities Later you’ll learn how the GRE tests your algebra skills through the use of “story” problems, which involve real-world scenarios. BASIC PROPERTIES OF NUMBERS First let’s review the basics about integers, number signs (positive and negative), and prime numbers. Make sure you’re up to speed on the following definitions, which you’ll need to know for this chapter as well as for the test: • Absolute value (of a real number): The number’s distance from zero (the origin) on the real number line. The absolute value of x is indicated as |x|.By definition, a number’s absolute value cannot be negative—that is, less than zero (0). • Integer: Any non-fraction number on the number line: {. . . 23, 22, 21, 0, 1, 2, 3 . }.Exceptfor the numberzero (0), every integer iseither positive or negative and either even or odd. • Factor (of an integer n): Any integer that you can multiply by another integer for a product of n. • Prime number: Any positive integer greater than one that has exactly two positive factors: 1 and the number itself. In other words, a prime number is not divisible by (a multiple of) any positive integer other than itself and 1. Number Signs and the Four Basic Operations The four basic operations are addition, subtraction, multiplication, and division. Be sure you know the sign of a number that results from combining numbers using these operations. Here’s a table that includes all the possibilities (a “?” indicates that the sign depends on which number has the greater absolute value): Addition Subtraction Multiplication Division ~1!1~1!51 ~2! 1 ~2! 5 2 ~1! 1 ~2! 5 ? ~2! 1 ~1! 5 ? ~1! 2 ~2! 5 ~1! ~2!2~1!5~2! ~1!2~1!5? ~2! 2 ~2! 5 ? ~1! 3 ~1! 5 1 ~1!3~2!52 ~2! 3 ~1! 5 2 ~2!3~2!51 ~1!4~1!51 ~1!4~2!52 ~2!4~1!52 ~2!4~2!51 GRE problems involving combining numbers by addition or subtraction usually incor- porate the concept of absolute value, as well as the rule for subtracting negative numbers. PART IV: Quantitative Reasoning222 ALERT! The factors of any integer n include 1 as well as n itself. Zero (0) and 1 are not prime numbers; 2 is the first prime number. www.petersons.com . independent if neither event affects the probability that the other will occur. (You’ll look at dependent events next.) On the GRE, look for either of these two scenarios. by averaging the first and last term i n the sequence: The progression is geometric, not arithmetic. You need to add up the terms, then divide by the number