260 MCGRAW-HILL’S SAT b – 1, is 6. The only positive integer pairs with a prod- uct of 6 are 2 × 3 and 1 × 6, so one possibility is that a = 2 and b = 4. This gives a(b – 1) = 2(4 – 1) = 2(3) = 6, and it satisfies the condition that a < b. Now check the statements. Statement I is true here because 4/2 = 2, which is an integer. Statement II is also true here because 4 is an even number. Statement III is also true because 2 × 4 = 8, which is 6 greater than 2. So the answer is (E) I, II and III, right? Wrong. Remember that the question asks what must be true, not just what can be true. We’ve only shown that the statements can be true. We can prove that statement I must be true by testing all the possi- ble cases. Since there is only one other possible solu- tion that satisfies the conditions: a = 1 and b = 7, and since 7/1 = 7 is an integer, we can say with confidence that statement I must be true. But statement II does- n’t have to be true because b can equal 7, which is not even. We have found a counterexample. Next, we can prove that statement III must be true by checking both cases: 2 × 4 is 6 greater than 2, and 1 × 7 is 6 greater than 1. (We can prove it algebraically too! If we add a to both sides of the original equation, we get ab = a + 6, which proves that ab is 6 greater than a.) Process of Elimination (POE) On multiple-choice questions (and especially “must be true” questions), it helps to cross off wrong answers right away. Sometimes POE simplifies the problem dramatically. What if, in the preceding question, the first solu- tion we found was a = 1 and b = 7. For this solution, statements I and III are true, but statement II is not. Therefore, we could eliminate those choices contain- ing II—(B), (D), and (E). Since the two remaining choices contain statement I, it must be true—we don’t even need to prove it! Lesson 7: Thinking Logically Numerical and Algebraic Proof Logical proofs aren’t just for geometry class. They apply to arithmetic and algebra, too. In arithmetic, you often need to apply the laws of arithmetic (such as odd × even = even, negative ÷ positive = negative—see Chapter 9, Lesson 3) to prove what you’re looking for. When you solve an algebraic equation, you use logical laws of equality (such as the addition law of equality) to prove the equation you want. “Must Be True” Questions Logic is especially useful in solving SAT “must be true” questions. You know them and hate them—they usually have those roman numer- als I, II, and III. To prove that a statement “must be true,” apply the laws of equality or the laws of arithmetic. To prove that a state- ment doesn’t have to be true, just find one counterexample, a valid example for which the statement is false. If a and b are positive integers such that a < b and ab – a = 6, which of the following must be true? I. is an integer. II. b is an even number. III. ab is 6 greater than a. (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III This requires both numerical and algebraic logic. First, let’s see how far we can get trying to solve the equation for a and b. ab – a = 6 Factor out the a: a(b – 1) = 6 Okay, we’ve got a problem. We have two un- knowns but only one equation, which means we can’t solve it uniquely. Fortunately, we know that a and b must be positive integers, so the equation basically says that the product of two positive integers, a and b a CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 261 Concept Review 7: Thinking Logically 1. What is a proof, and why is understanding proofs helpful on the SAT? 2. How can POE help on the SAT? 3. What is the difference between geometric, algebraic, and numerical proofs? 4. Name two geometric theorems that are useful on the SAT. 5. Name two algebraic theorems that are useful on the SAT. 6. Name two numerical theorems that are useful on the SAT. 262 MCGRAW-HILL’S SAT 4. If (m + 1)(n + 1) = 1, which of the following can be true? I. m and n are both positive. II. m and n are both negative. III. m is positive and n is negative. (A) II only (B) III only (C) I and II only (D) I and III only (E) II and III only 5. If p is a prime number greater than 5 and q is an odd number greater than 5, which of the fol- lowing must be true? I. p + q is not a prime number. II. pq has at least three positive integer factors greater than 1. III. is not an integer. (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III q p SAT Practice 7: Thinking Logically Use logical methods to solve each of the following SAT questions. 1. If a > b and b(b – a) > 0, which of the following must be true? I. b < 0 II. a < 0 III. ab < 0 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III A, B, C, and D are the consecutive vertices of a quadrilateral. ∠ABC and ∠BCD are right angles. 2. If the two statements above are true, then which of the following also must be true? (A) ABCD is a rectangle. (B) is parallel to . (C) is parallel to . (D) Triangle ACD is a right triangle. (E) Triangle ABD is a right triangle. 3. The statement a ⇔ b is defined to be true if and only if . Which of the following is true? (A) 3 ⇔ 5 (B) 5 ⇔ 3 (C) 4 ⇔ 2 (D) 6 ⇔ 4 (E) 7 ⇔ 5 ab 53 > AD BC DC AB CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 263 SAT Practice 7 1. A Since a is greater than b, b – a must be a negative number. Since b(b – a) must be positive, but b – a is negative, b also must be negative because negative × negative = positive, but positive × negative = negative. This proves that statement I must be true. However, statement II does not have to be true because a coun- terexample is a = 1 and b = −1. Notice that this satis- fies the conditions that a > b and that b(b – a) > 0. Statement III also isn’t necessarily true because a counterexample is a = −1 and b = −2. Notice that this also satisfies the conditions that a > b and b(b – a) > 0 but contradicts the statement that ab < 0. 2. B First draw a diagram that illustrates the given conditions, such as the one above. This diagram shows that the only true statement among the choices is (B). This fact follows from the fact that “if a line (BC), crosses two other lines (AB and DC) in a plane so that same-side interior angles are supplementary, then the two lines are parallel.” 3. C First, translate each choice according to the definition of the bizarre new symbol. This gives us (A) 3/5 > 5/3, (B) 5/5 > 3/3, (C) 4/5 > 2/3, (D) 6/5 > 4/3, and (E) 7/5 > 5/3. The only true statement among these is (C). 4. E The question asks whether the statements can be true, not whether they must be true. The equa- tion says that two numbers have a product of 1. You might remember that such numbers are reci- procals, so we want to find values such that m + 1 and n + 1 are reciprocals of each other. One pair of reciprocals is 2 and 1 / 2 , which we can get if m = 1 and n = − 1 / 2 . Therefore, statement III can be true, and we can eliminate choices (A) and (C). Next, think of negative reciprocals, such as −2 and − 1 / 2 , which we can get if m = −3 and n = − 1 / 2 . Therefore, statement II can be true, and we can eliminate choices (B) and (D), leaving only (E), the correct answer. Statement I can’t be true because if m and n are both positive, then both m + 1 and n + 1 are greater than 1. But, if a number is greater than 1, its reciprocal must be less than 1. 5. A You might start by just choosing values for p and q that satisfy the conditions, such as p = 7 and q = 9. When you plug these values in, all three statements are true. Bummer, because this nei- ther proves any statement true nor proves any statement false. Are there any interesting possible values for p and q that might disprove one or more of the statements? Notice that nothing says that p and q must be different, so choose p = 7 and q = 7. Now pq = 49, which only has 1, 7, and 49 as fac- tors. Therefore, it does not have at least three pos- itive integer factors greater than 1, and statement II is not necessarily true. Also, q/p = 1, which is an integer, so statement III is not necessarily true. So we can eliminate any choices with II or III, leav- ing only choice (A). Concept Review 7 1. A proof is a sequence of logical statements that begins with a set of assumptions and proceeds to a desired conclusion. You construct a logical proof every time you solve an equation or determine a geometric or arithmetic fact. 2. The process of elimination (POE) is the process of eliminating wrong answers. Sometimes it is easier to show that one choice is wrong than it is to show that another is right, so POE may provide a quicker path to the right answer. 3. Geometric proofs depend on geometric facts such as “angles in a triangle have a sum of 180°,” algebraic proofs use laws of equality such as “any number can be added to both sides of an equation,” and numeri- cal proofs use facts such as “an odd number plus an odd number always equals an even number.” 4. The most important geometric theorems for the SAT are given in Chapter 10. They include parallel lines theorems such as “if two parallel lines are cut by a transversal, then alternate interior angles are congruent” and triangle theorems such as “if two sides of a triangle are congruent, then the angles opposite those sides are also congruent.” 5. The most important algebraic theorems are the laws of equality, such as “you can subtract any number from both sides of an equation.” 6. The most important numerical theorems are dis- cussed in Chapter 9, Lesson 3, and Chapter 7, Les- son 7. They include “odd × odd = odd” and “positive × negative = negative.” Answer Key 7: Thinking Logically A B C D 264 MCGRAW-HILL’S SAT Check the Question Always quickly reread the question before marking your answer to make sure that you’ve answered the right question and to make sure that your solution makes sense in the context of the question. A bin contains 20 basketballs and soccer balls. If there are 8 more basketballs than soccer balls in the bin, how many soccer balls are in the bin? (A) 4 (B) 6 (C) 8 (D) 10 (E) 12 Many students think that since there are 8 more basketballs than soccer balls, they should just subtract 8 from the total to get the number of soccer balls, get- ting 20 – 8 = 12 soccer balls. The answer is (E), right? Wrong. If there are 12 soccer balls, then there must be 8 basketballs, for a total of 20. But the ques- tion says that there are 8 more basketballs than soc- cer balls, and 8 sure isn’t more than 12! So now what? Eliminate choice (E) first of all. Since there are fewer basketballs than soccer balls, soccer balls must make up fewer than half the balls, so there must be fewer than 10 soccer balls, eliminating choice (D). Checking the remaining choices shows (B) 6 works because if there are 6 soccer balls, there are 14 basketballs, and 14 is 8 greater than 6! The “direct” method for solving is to subtract 8 from 20 and then divide the result by 2 to get the num- ber of soccer balls. Check Your Algebra When solving an equation, check two things: first your answer, and then your steps. If the answer works when you plug it back into the equation, there’s no need to check the steps. If it doesn’t, then check your steps. When solving equations, write out every step, and make sure that each one is logical. You’re likely to make mistakes if you skip steps or do them too quickly in your head. If , then what is the value of x? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 You might notice that since both fractions have the same numerator, the equation can be simplified and solved without much trouble. Multiply by x – 2: 2x = x 2 Divide by x:2 = x Piece of cake. The answer is (C), right? Wrong. Notice that substituting x = 2 into the original equation gives you 4/0 = 4/0. Although this seems true at first, it’s not, be- cause 4/0 isn’t a number! (Just ask your calculator.) What now? The check suggests a solution: you can just test the choices. Plugging in the other choices for x shows that only (A) 0 produces a true equation: 0/–2 = 0/–2. What went wrong the first time? Check the steps. Our second step was to divide by x. We’re allowed to divide both sides of an equation by any number except 0 because division by 0 is undefined. That’s where we went wrong: We didn’t check division by 0. Notice that division by 0 also explains why x = 2 doesn’t work in the original equation. Check by Estimating Estimation is one of the simplest and most ef- fective checking strategies you can use on the SAT. Getting an approximate answer can help you to narrow down the choices quickly. If Carla drives at 40 miles per hour for n miles and then drives at 60 miles per hour for another n miles, what is her average speed, in miles per hour, for the entire trip? (A) 42 (B) 48 (C) 50 (D) 52 (E) 54 Many students average 40 and 60 and get 50. But this is wrong because Carla is not spending equal times at 40 and 60 miles an hour. Since 40 mph is slower than 60 mph, she spends more time at that speed. So her av- erage speed is closer to 40 than 60. This eliminates choices (C), (D), and (E). The correct answer is (B). (For more on rate problems, see Chapter 9, Lesson 4.) 2 22 2 x x x x− = − 2 22 2 x x x x− = − Lesson 8: Checking Your Work CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 265 Concept Review 8: Checking Your Work 1. What are the best strategies for avoiding mistakes when solving an equation? 2. When should you estimate on an SAT math question? 3. Why should you estimate on certain SAT math questions? 4. What is the last thing to check before marking an answer to an SAT math question? 5. What steps “aren’t allowed” when solving equations? 266 MCGRAW-HILL’S SAT 4. If the 30 students in Ms. Harkin’s class scored an average of 80% on their final exams, and if the 20 students in Ms. Johnson’s class scored an aver- age of 70% on their final exams, what was the av- erage score for the two classes combined? (A) 74% (B) 75% (C) 76% (D) 77% (E) 78% 5. What is the least positive integer m such that 168m is the square of an integer? SAT Practice 8: Checking Your Work Check your work carefully before choosing your an- swer to the following questions. 1. If 3x – 5 = 20, what is the value of 3x + 5? 2. If s 2 – 1 = 2s – 1 , which of the following gives all possible values of s 2 ? (A) 2 only (B) 4 only (C) 0 and 2 only (D) 0 and 4 only (E) All even integers 3. Last year, Tom was twice as old as Julio. This year, the sum of their ages is 65. How old is Tom now? (A) 22 (B) 32 (C) 41 (D) 42 (E) 43 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 267 Concept Review 8 1. Check your answer by plugging it back into the original equation and checking your steps. Write out each step, one beneath the other, so that checking your logic and arithmetic is easier. 2. Estimate only when it is easy to do. If the answer choices are numerically far apart, estimation can help you to eliminate obviously wrong answers. 3. If you can quickly “ballpark” a numerical answer and rule out those choices that are “out of the ball- park,” you can often avoid doing complicated cal- culations or algebra. 4. Reread the question and make sure that you’ve answered the right question, and make sure that your answer makes sense in the context of the question. 5. Dividing by 0 and taking the square root of an ex- pression that can be negative are not allowed be- cause they are undefined. Answer Key 8: Checking Your Work SAT Review 8 1. 30 The simplest way to solve this problem is to add 10 to both sides of the equation, which gives 3x + 5 = 30. However, many students do the “knee-jerk” of solving for x and become prone to silly arithmetic mistakes. If you did solve for x, you should have checked your answer by plug- ging it back in to the original equation. 2. D Did you say (C)? Then you misread the question. Always reread before marking your answer. It asks for the value of s 2 , not s. Although s can be 0 or 2, s 2 is either 0 or 4. Did you say (A) or (B)? Then you may have made this common mistake: s 2 – 1 = 2s – 1 Add 1: s 2 = 2s Divide by s: s = 2 What went wrong? In the second step, we divided by s without checking whether it could equal 0. (Remember that division by 0 is undefined and usually causes trouble.) Indeed, plugging in shows that s = 0 is in fact a solution. The correct method is s 2 – 1 = 2s – 1 Add 1: s 2 = 2s Subtract 2s: s 2 – 2s = 0 Factor: s(s – 2) = 0 Use 0 product property: s = 0 or 2 3. E You might start by approximating. Since the sum of their ages is about 60, and since Tom is about twice as old as Julio, Tom is about 40 and Julio is about 20. This rules out (A) and (B). From here, you may just want to “test” the remaining choices until you find what works. If you prefer algebra, you may want to let t equal Tom’s age now and j equal Julio’s age now. You are told that t + j = 65 and that 2(j – 1) = t – 1. Since you only need the value of t, solve the first equation for j, getting j = 65 – t, and sub- stitute this into the second equation. This gives 2(65 – t – 1) = t – 1 Simplify: 2(64 – t) = t – 1 Distribute: 128 – 2t = t – 1 Add 2t and 1: 129 = 3t Divide by 3: 43 = t Therefore, Tom is now 43 and Julio is now 65 – 43 = 22. Notice that last year they were 42 and 21, respec- tively, and 42 is twice as old as 21. 4. E Since more students averaged 80% than 70%, the overall average should be closer to 80%. This rules out (A) and (B). To get the precise answer, let x be the overall average. There are two ways to calculate the 268 MCGRAW-HILL’S SAT sum of all the scores: (30)(80) + (20)(70) and (50)(x). Since these must be equal, 2,400 + 1,400 = 50x Simplify: 3,800 = 50x Divide by 50: 76 = x 5. 42 Do the prime factorization of 168: 2 3 × 3 × 7. Since, in a perfect square, all prime factors “pair up,” we need to multiply at least one more factor of 2, one more factor of 3, and one more factor of 7 to make a perfect square: 2 4 × 3 2 × 7 2 = 7,056 = 84 2 . (Notice that now every factor appears an even number of times.) Therefore, k = 2 × 3 × 7 = 42. 269 ESSENTIAL PRE-ALGEBRA SKILLS CHAPTER 7 ✓ 1. Numbers and Operations 2. Laws of Arithmetic 3. Fractions 4. Ratios and Proportions 5. Percents 6. Negatives 7. Divisibility Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use. . theorems that are useful on the SAT. 5. Name two algebraic theorems that are useful on the SAT. 6. Name two numerical theorems that are useful on the SAT. 262 MCGRAW-HILL’S SAT 4. If (m + 1)(n + 1). dis- cussed in Chapter 9, Lesson 3, and Chapter 7, Les- son 7. They include “odd × odd = odd” and “positive × negative = negative.” Answer Key 7: Thinking Logically A B C D 264 MCGRAW-HILL’S SAT Check. certain SAT math questions? 4. What is the last thing to check before marking an answer to an SAT math question? 5. What steps “aren’t allowed” when solving equations? 266 MCGRAW-HILL’S SAT 4.