Tài liệu Master the Gre 2010 - Part 27 ppt

24. a + b =30 2b =602 2a Column A Column B ab (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (D). An unwary test taker might waste time trying to find the values of a and b, because the centered data appears at first glance to provide a system of two linear equations with two unknowns. But you can easily manipulate the second equation so that it is identical to the first: 2b =602 2a 2b =2~30 2 a! b =302 a a + b =30 As you can see, you’re really dealing with only one equation. Since you cannot solve one equation in two unknowns, you cannot make the comparison. Whenever you encounter a Quantitative Comparison question that calls for solving one or more linear equations, stop in your tracks before taking pencil to paper. Size up the equation to see whether it’s one of the two unsolvable kinds you learned about here. If so, unless you’re given more information, the correct answer will be choice (D). FACTORABLE QUADRATIC EXPRESSIONS WITH ONE VARIABLE A quadratic expression includes a “squared” variable, such as x 2 . An equation is quadratic if you can express it in this general form: ax 2 1 bx 1 c 5 0 Where: x is the variable a, b, and c are constants (numbers) a Þ 0 b can equal 0 c can equal 0 Here are four examples (notice that the b-term and c-term are not essential; in other words, either b or c, or both, can equal zero (0)): Chapter 10: Math Review: Number Theory and Algebra 243 ALERT! If the centered information in a Quantitative Comparison consists of one or more linear equations, never assume you can solve for the variable(s). www.petersons.com Quadratic Equation Same Equation, but in the Form: ax 2 1 bx 1 c 5 0 2w 2 5 16 x 2 5 3x 3y 5 4 2 y 2 7z 5 2z 2 2 15 2w 2 2 16 5 0(nob-term) x 2 2 3x 5 0(noc-term) y 2 1 3y 2 4 5 0 2z 2 2 7z 2 15 5 0 Every quadratic equation has exactly two solutions, called roots. (But the two roots might be the same.) On the GRE, you can often find the two roots by factoring.To solve any factorable quadratic equation, follow these three steps: Put the equation into the standard form: ax 2 1 bx 1 c 5 0. Factor the terms on the left side of the equation into two linear expressions (with no exponents). Set each linear expression (root) equal to zero (0) and solve for the variable in each one. Factoring Simple Quadratic Expressions Some quadratic expressions are easier to factor than others. If either of the two constants b or c is zero (0), factoring requires no sweat. In fact, in some cases, no factoring is needed at all: A Quadratic with No c-term A Quadratic with No b-term 2x 2 5 x 2x 2 2 x 5 0 x~2x 2 1!50 x 5 0, 2x 2 1 5 0 x 5 0, 1 2 2x 2 2 4 5 0 2~x 2 2 2!50 x 2 2 2 5 0 x 2 5 2 x 5 = 2, 2 = 2 Factoring Quadratic Trinomials A trinomial is simply an algebraic expression that contains three terms. If a quadratic expression contains all three terms of the standard form ax 2 1 bx 1 c, then factoring becomes a bit trickier. You need to apply the FOIL method, in which you add together these terms: (F) the product of the first terms of the two binomials (O) the product of the outer terms of the two binomials (I) the product of the inner terms of the two binomials (L) the product of the last (second) terms of the two binomials PART IV: Quantitative Reasoning244 ALERT! When dealing with a quadratic equation, your first step is usually to put it into the general form ax 2 1 bx 1 c 5 0. But keep in mind: The only essential term is ax 2 . www.petersons.com Note the following relationships: (F) is the first term (ax 2 ) of the quadratic expression (O 1 I) is the second term (bx) of the quadratic expression (L) is the third term (c) of the quadratic expression You’ll find that the two factors will be two binomials. The GRE might ask you to recognize one or both of these binomial factors. 25. Which of the following is a factor of x 2 2 x 2 6? (A) (x 1 6) (B) (x 2 3) (C) (x 1 1) (D) (x 2 2) (E) (x 1 3) The correct answer is (B). Notice that x 2 has no coefficient. This makes the process of factoring into two binomials easier. Set up two binomial shells: (x )(x ) The product of the two missing second terms (the “L” term under the FOIL method) is 26. The possible integral pairs that result in this product are (1,26), (21,6), (2,23,), and (22,3). Notice that the second term in the trinomial is 2x. This means that the sum of the two integers whose product is 26 must be 21. The pair (2,23) fits the bill. Thus, the trinomial is equivalent to the product of the two binomials (x 1 2) and (x 2 3). To check your work, multiply the two binomials using the FOIL method: ~x 1 2!~x 2 3!5x 2 2 3x 1 2x 2 6 5 x 2 2 x 1 6 If the preceding question had asked you to determine the roots of the equation x 2 2 x 2 6 5 0, you’d simply set each of the binomial factors equal to zero (0), then solve for x in each one. The solution set (the two possible values of x) includes the roots 22 and 3. 26. How many different values of x does the solution set for the equation 4x 2 5 4x 2 1 contain? (A) none (B) one (C) two (D) four (E) infinitely many The correct answer is (B). First, express the equation in standard form: 4x 2 2 4x 1 1 5 0. Notice that the c-term is 1. The only two integral pairs that result in Chapter 10: Math Review: Number Theory and Algebra 245 www.petersons.com this product are (1,1) and (21,21). Since the b-term (24x) is negative, the integral pair whose product is 1 must be (21,21). Set up a binomial shell: (? 2 1)(? 2 1) Notice that the a-term contains the coefficient 4. The possible integral pairs that result in this product are (1,4), (2,2), (21,24), and (22,22). A bit of trial-and- error reveals that only the pair (2,2) works. Thus, in factored form, the equation becomes (2x 2 1)(2x 2 1) 5 0. To check your work, multiply the two binomials using the FOIL method: ~2x 2 1!~2x 2 1!54x 2 2 2x 2 2x 1 1 5 4x 2 2 4x 1 1 Since the two binomial factors are the same, the two roots of the equation are the same. In other words, x has only one possible value. (Although you don’t need to find the value of x in order to answer the question, solving for x in the equation 2x21 5 0 yields x 5 1 2 . Stealth Quadratic Equations Some equations that appear linear (variables include no exponents) may actually be quadratic. On the GRE, be on the lookout for either of two situations: The same variable inside a radical also appears outside: = x 5 5x ~ = x! 2 5~5x! 2 x 5 25x 2 25x 2 2 x 5 0 The same variable that appears in the denominator of a fraction also appears elsewhere in the equation: 2 x 5 3 2 x 2 5 x~3 2 x! 2 5 3x 2 x 2 x 2 2 3x 1 2 5 0 In both scenarios, you’re dealing with a quadratic (nonlinear) equation with one variable. So, in either equation, there are two roots. (Both equations are factorable, so go ahead and find their roots.) The test makers often use the Quantitative Comparison format to cover this concept. PART IV: Quantitative Reasoning246 www.petersons.com 27. 63xx= Column A Column B x 1 12 (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (D). An unwary test taker might assume that the equation is linear, because x is not squared. Substituting 1 12 for x satisfies the centered equation. But the two quantities are not necessarily equal. If you clear the radical by squaring both sides of the equation, then isolate the x-terms on one side of the equation, you’ll see that the equation is quadratic: 36 3 2 xx= 36 3 0 2 xx−= To find the two roots, factor out 3x, then solve for each root: 3x~12x 2 1! =0 x =0,12x 2 1=0 x =0, 1 12 Since 1 12 is only one of two possible values for x, you cannot make a definitive comparison. NONLINEAR EQUATIONS WITH TWO VARIABLES In the world of math, solving nonlinear equations with two or more variables can be very complicated, even for bona-fide mathematicians. But on the GRE, all you need to remember are these three general forms: Sum of two variables, squared: (x 1 y) 2 5 x 2 1 2xy 1 y 2 Difference of two variables, squared: (x 2 y) 2 5 x 2 2 2xy 1 y 2 Difference of two squares: x 2 2 y 2 5 (x 1 y)(x 2 y) Chapter 10: Math Review: Number Theory and Algebra 247 www.petersons.com You can verify these equations using the FOIL method: (x 1 y) 2 5 (x 1 y)(x 1 y) 5 x 2 1 xy 1 xy 1 y 2 5 x 2 1 2xy 1 y 2 (x 2 y) 2 5 (x 2 y)(x 2 y) 5 x 2 2 xy 2 xy 1 y 2 5 x 2 2 2xy 1 y 2 (x 1 y)(x 2 y) 5 x 2 1 xy 2 xy 2 y 2 5 x 2 2 y 2 For the GRE, memorize the three equations listed here. When you see one form on the exam, it’s a sure bet that your task is to rewrite it as the other form. 28. If x 2 2 y 2 5 100, and if x 1 y 5 2, then x 2 y 5 (A) 22 (B) 10 (C) 20 (D) 50 (E) 200 The correct answer is (D). If you’re on the lookout for the difference of two squares, you can handle this question with no sweat. Use the third equation you just learned, substituting 2 for (x 1 y), then solving for (x 2 y): x 2 2 y 2 5~x 1 y!~x 2 y! 100 5~x 1 y!~x 2 y! 100 5~2!~x 2 y! 50 5~x 2 y! SOLVING ALGEBRAIC INEQUALITIES You can solve algebraic inequalities in the same manner as equations. Isolate the variable on one side of the inequality symbol, factoring and eliminating terms wherever possible. However, one important rule distinguishes inequalities from equa- tions: Whenever you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. Simply put: If a . b, then 2a ,2b. 12 2 4x , 8 Original inequality 24x ,24 12 subtracted from each side; inequality unchanged x . 1 Both sides divided by 24; inequality reversed Here are five general rules for dealing with algebraic inequalities. Study them until they’re second nature to you because you’ll put them to good use on the GRE. Adding or subtracting unequal quantities to (or from) equal quantities: If a . b, then c 1 a . c 1 b If a . b, then c 2 a , c 2 b Adding unequal quantities to unequal quantities: If a . b, and if c . d, then a 1 c . b 1 d PART IV: Quantitative Reasoning248 TIP You usually can’t solve quadratics using a shortcut. Always look for one of the three common quadratic forms. If you see it, rewrite it as its equivalent form to answer the question as quickly and easily as possible. ALERT! Be careful when handling inequality problems: The wrong answers might look right, depending on the values you use for the different variables. www.petersons.com Comparing three unequal quantities: If a . b, and if b . c, then a . c Combining the same positive quantity with unequal quantities by multiplication or division: If a . b, and if x . 0, then xa . xb Combining the same negative quantity with unequal quantities by multiplication or division: If a . b, and if x , 0, then xa , xb 29. If a . b, and if c . d, then which of the following must be true? (A) a 2 b . c 2 d (B) a 2 c . b 2 d (C) c 1 d , a 2 b (D) b 1 d , a 1 c (E) a 2 c , b 1 d The correct answer is (D). Inequality questions can be a bit confusing, can’t they? In this problem, you need to remember that if unequal quantities (c and d) are added to unequal quantities of the same order (a and b), the result is an inequality in the same order. This rule is essentially what answer choice (D) says. SOLVING ALGEBRA “STORY” PROBLEMS The remainder of this chapter is devoted exclusively to algebra “story” problems, in which you translate real-world scenarios into algebraic expressions and equations. Here are the types of story problems you’ll review in the pages ahead (some call for the application of specific formulas): • weighted-average problems • currency problems • mixture problems • investment problems • problems involving rate of production and rate of travel • problems involving overlapping sets To illustrate each type, we’ll show you one or two GRE-style questions. Read the analysis of each question to find out what formulas to apply and what shortcuts (if any) and strategies you can use to solve the problem. Weighted-Average Problems You solve weighted-average problems using the arithmetic mean (simple average) formula, except you give the set’s terms different weights. For example, if a final exam score of 90 receives twice the weight of each of two midterm exam scores 75 and 85, Chapter 10: Math Review: Number Theory and Algebra 249 NOTE For algebra story problems, the GRE test makers generally use the Problem Solving format (the same format we use here). Don’t be surprised, though, if your exam includes one or two Quantitative Comparisons that are actually story problems in disguise. www.petersons.com think of the final exam score as two scores of 90—and the total number of scores as 4 rather than 3: WA 5 75 1 85 1~2!~90! 4 5 340 4 5 85 Similarly, when some numbers among terms appear more often than others, you must give them the appropriate “weight” before computing an average. 30. During an 8-hour trip, Brigitte drove 3 hours at 55 miles per hour and 5 hours at 65 miles per hour. What was her average rate, in miles per hour, for the entire trip? (A) 58.5 (B) 60 (C) 61.25 (D) 62.5 (E) 66.25 The correct answer is (C). Determine the total miles driven: (3)(55) 1 (5)(65) 5 490. To determine the average over the entire trip, divide this total by 8, which is the number of total hours: 490 4 8 5 61.25. A tougher weighted-average problem might provide the weighted average and ask for one of the terms, or require conversions from one unit of measurement to another—or both. 31. A certain olive orchard produces 315 gallons of oil annually, on average, during four consecutive years. How many gallons of oil must the orchard produce annually, on average, during the next six years, if oil production for the entire 10-year period is to meet a goal of 378 gallons per year? (A) 240 (B) 285 (C) 396 (D) 420 (E) 468 The correct answer is (D). In the weighted-average formula, 315 annual gallons receives a weight of 4, while the average annual number of gallons for the next six years (x) receives a weight of 6: 378 5 1,260 1 6x 10 3,780 5 1,260 1 6x 3,780 2 1,260 5 6x 2,520 5 6x 420 5 x This solution (420) is the average number of gallons needed per year, on average, during the next 6 years. PART IV: Quantitative Reasoning250 www.petersons.com To guard against calculation errors, check your answer by sizing up the question. Generally, how great a number are you looking for? Notice that the stated goal is a bit greater than the annual average production over the first four years. So you’re looking for an answer that is greater than the goal—a number somewhat greater than 378 gallons per year. You can eliminate choices (A) and (B) out of hand. The number 420 fits the bill. Currency Problems Currency problems are similar to weighted-average problems in that each item (bill or coin) is weighted according to its monetary value. Unlike weighted-average problems, however, the “average” value of all the bills or coins is not at issue. In solving currency problems, remember the following: • You must formulate algebraic expressions involving both number of items (bills or coins) and value of items. • You should convert the value of all moneys to a common currency unit before formulating an equation. If converting to cents, for example, you must multiply the number of nickels by 5, dimes by 10, and so forth. 32. Jim has $2.05 in dimes and quarters. If he has four fewer dimes than quarters, how much money does he have in dimes? (A) 20 cents (B) 30 cents (C) 40 cents (D) 50 cents (E) 60 cents The correct answer is (B). Letting x equal the number of dimes, x 1 4 represents the number of quarters. The total value of the dimes (in cents) is 10x, and the total value of the quarters (in cents) is 25(x 1 4) or 25x 1 100. Given that Jim has $2.05, the following equation emerges: 10x 1 25x 1 100 5 205 35x 5 105 x 5 3 Jim has three dimes, so he has 30 cents in dimes. You could also solve this problem without formal algebra, by plugging in each answer choice in turn. Let’s try this strategy for choices (A) and (B): (A) 20 cents is 2 dimes, so Jim has 6 quarters. 20 cents plus $1.50 adds up to $1.70. Wrong answer! (B) 30 cents is 3 dimes, so Jim has 7 quarters. 30 cents plus $1.75 adds up to $2.05. Correct answer! Chapter 10: Math Review: Number Theory and Algebra 251 TIP You can solve most GRE currency problems by working backward from the answer choices. www.petersons.com Mixture Problems In GRE mixture problems, you combine substances with different characteristics, resulting in a particular mixture or proportion, usually expressed as percentages. Substances are measured and mixed by either volume or weight—rather than by number (quantity). 33. How many quarts of pure alcohol must you add to 15 quarts of a solution that is 40% alcohol to strengthen it to a solution that is 50% alcohol? (A) 4.0 (B) 3.5 (C) 3.25 (D) 3.0 (E) 2.5 The correct answer is (D). You can solve this problem by working backward from the answer choices—trying out each one in turn. Or, you can solve the problem algebraically. The original amount of alcohol is 40% of 15. Letting x equal the number of quarts of alcohol that you must add to achieve a 50% alcohol solution, .4(15) 1 x equals the amount of alcohol in the solution after adding more alcohol. You can express this amount as 50% of (15 1 x). Thus, you can express the mixture algebraically as follows: ~0.4!~15!1x 5~0.5!~15 1 x! 6 1 x 5 7.5 1 0.5x 0.5x 5 1.5 x 5 3 You must add 3 quarts of alcohol to obtain a 50% alcohol solution. Investment Problems GRE investment problems involve interest earned (at a certain percentage rate) on money over a certain time period (usually a year). To calculate interest earned, multiply the original amount of money by the interest rate: amount of money 3 interest rate 5 amount of interest on money For example, if you deposit $1,000 in a savings account that earns 5% interest annually, the total amount in the account after one year will be $1,000 1 .05($1,000) 5 $1,000 1 $50 5 $1,050. GRE investment questions usually involve more than simply calculating interest earned on a given principal amount at a given rate. They usually call for you to set up and solve an algebraic equation. When handling these problems, it’s best to eliminate percent signs. PART IV: Quantitative Reasoning252 www.petersons.com . product of the first terms of the two binomials (O) the product of the outer terms of the two binomials (I) the product of the inner terms of the two binomials (L). equal. If you clear the radical by squaring both sides of the equation, then isolate the x-terms on one side of the equation, you’ll see that the equation is