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Section 1—Essay The following are guidelines for scoring the essay. 706-1045.qxd 5/1/08 3:24 PM Page 715 Score of 5 An essay in this category is effective, demonstrating reasonably consistent mastery, although it will have occa- sional errors or lapses in quality. A typical essay • effectively develops a point of view on the issue and demonstrates strong critical thinking, generally using appropriate examples, rea- sons, and other evidence to sup- port its position • is well organized and focused, demonstrating coherence and pro- gression of ideas • exhibits facility in the use of lan- guage, using appropriate vocabu- lary • demonstrates variety in sentence structure • is generally free of most errors in grammar, usage, and mechanics Score of 2 An essay in this category is seriously limited, demonstrating little mastery, and is flawed by ONE OR MORE of the following weaknesses: • develops a point of view on the issue that is vague or seriously lim- ited, demonstrating weak critical thinking, providing inappropriate or insufficient examples, reasons, or other evidence to support its position • is poorly organized and/or focused, or demonstrates serious problems with coherence or progression of ideas • displays very little facility in the use of language, using very limited vocabulary or incorrect word choice • demonstrates frequent problems in sentence structure • contains errors in grammar, usage, and mechanics so serious that meaning is somewhat obscured Score of 6 An essay in this category is outstanding, demonstrating clear and consistent mastery, although it may have a few minor errors. A typical essay • effectively and insightfully devel- ops a point of view on the issue and demonstrates outstanding criti- cal thinking, using clearly appropri- ate examples, reasons, and other evidence to support its position • is well organized and clearly focused, demonstrating clear coherence and smooth progression of ideas • exhibits skillful use of language, using a varied, accurate, and apt vocabulary • demonstrates meaningful variety in sentence structure • is free of most errors in grammar, usage, and mechanics Score of 3 An essay in this category is inade- quate, but demonstrates developing mastery, and is marked by ONE OR MORE of the following weaknesses: • develops a point of view on the issue, demonstrating some critical thinking, but may do so inconsis- tently or use inadequate examples, reasons, or other evidence to sup- port its position • is limited in its organization or focus, or may demonstrate some lapses in coherence or progression of ideas • displays developing facility in the use of language, but sometimes uses weak vocabulary or inappro- priate word choice • lacks variety or demonstrates problems in sentence structure • contains an accumulation of errors in grammar, usage, and mechanics Score of 4 An essay in this category is compe- tent, demonstrating adequate mastery, although it will have lapses in quality. A typical essay • develops a point of view on the issue and demonstrates competent critical thinking, using adequate examples, reasons, and other evi- dence to support its position • is generally organized and focused, demonstrating some coherence and progression of ideas • exhibits adequate but inconsistent facility in the use of language, using generally appropriate vocabulary • demonstrates some variety in sen- tence structure • has some errors in grammar, usage, and mechanics Score of 1 An essay in this category is funda- mentally lacking, demonstrating very little or no mastery, and is severely flawed by ONE OR MORE of the fol- lowing weaknesses: • develops no viable point of view on the issue, or provides little or no evidence to support its position • is disorganized or unfocused, resulting in a disjointed or incoher- ent essay • displays fundamental errors in vocabulary • demonstrates severe flaws in sen- tence structure • contains pervasive errors in gram- mar, usage, or mechanics that per- sistently interfere with meaning Essays not written on the essay assignment will receive a score of zero. The SAT Scoring Guide 706-1045.qxd 5/1/08 3:24 PM Page 716 717 Section 2: Math As you read these solutions, you are advised to do two things if you answered the Math question incorrectly: 1. When a specific Strategy is referred to in the solution, study that strategy, which you will find in “Using Critical Thinking Skills in Math Questions” (beginning on page 69). 2. When the solution directs you to the “Math Refresher” (beginning on page 191)—for example, Math Refresher Ե305—study the 305 Math principle to get a clear idea of the Math operation that was necessary for you to know in order to answer the question correctly. Explanatory Answers for Practice Test 2 (continued) 1. Choice D is correct. Given that 500w ϭ 3 ϫ 700 (Use Strategy 13: Find an unknown by divid- ing.) Divide by 500, giving ᎏ 5 5 0 0 0 0 w ᎏ ϭ ᎏ 3 ϫ 50 7 0 00 ᎏ (Use Strategy 19: Factor and reduce first. Then multiply.) w ϭ ᎏ 3 ϫ 5 ϫ 7 ϫ 10 1 0 00 ᎏ w ϭ ᎏ 2 5 1 ᎏ (Math Refresher 406) 2. Choice E is correct. Given: ᎏ 3 ϩ y y ᎏ ϭ 7 (Use Strategy 13: Find an unknown by multi- plying.) Multiply by y, to get y ᎏ 3 ϩ y y ᎏ ϭ (7)y 3 ϩ y ϭ 7y 3 ϭ 6y ᎏ 6 3 ᎏ ϭ y ᎏ 1 2 ᎏ ϭ y (Math Refresher 406) 3. Choice D is correct. (Use Strategy 2: Translate from words to algebra.) x is a multiple of 9, gives x ε {9, 18, 27, 36, 45, 54, . . . . . . } x is a multiple of 12, gives x ε {12, 24, 36, 48, 60, 72, . . . . . . } The smallest value that appears in both sets and is 36. (Logical Reasoning) 2 1 2 1 1 1 1 1 706-1045.qxd 5/1/08 3:24 PM Page 717 4. Choice E is correct. Method 1: Given: (r Ϫ s) (t Ϫ s) ϩ (s Ϫ r) (s Ϫ t) (Use Strategy 17: Use the given information effectively.) Recognizing that (s Ϫ r) ϭϪ1(r Ϫ s) (s Ϫ t) ϭϪ1(t Ϫ s) Substituting and into , we get (r Ϫ s)(t Ϫ s) ϩ [Ϫ1(r Ϫ s)] [Ϫ1(t Ϫ s)] ϭ (r Ϫ s)(t Ϫ s) ϩ (r Ϫ s)(t Ϫ s) ϭ 2(r Ϫ s)(t Ϫ s) Method 2: Given: (r Ϫ s)(t Ϫ s) ϩ (s Ϫ r)(s Ϫ t) Multiply both pairs of quantities from , giving rt Ϫ rs Ϫ st ϩ s 2 ϩ s 2 Ϫ st Ϫ rs ϩ rt ϭ 2rt Ϫ 2rs Ϫ 2st ϩ 2s 2 ϭ 2(rt Ϫ rs Ϫ st ϩ s 2 ) ϭ 2[r(t Ϫ s) Ϫ s(t Ϫ s)] ϭ 2(r Ϫ s)(t Ϫ s) (Math Refresher 409) 5. Choice C is correct. We want to find the area of the middle square, which is (CB) 2 . (Use Strategy 3: The whole equals the sum of its parts.) OA ϭ OC ϩ CB ϩ BA From the diagram, we get OA ϭ 21 AE ϭ 4 OD ϭ 10 Since each figure is a square, we get BA ϭ AE OC ϭ OD Substituting into , we get AE ϭ BA ϭ 4 7 35 6 5 4 3 2 1 1 1 132 3 2 1 Substituting into , we get OD ϭ OC ϭ 10 Substituting , , and into , we get 21 ϭ 10 ϩ CB ϩ 4 21 ϭ 14 ϩ CB 7 ϭ CB Area of square II ϭ (CB) 2 Area of square II ϭ 7 2 (From ) Area of square II ϭ 49 (Math Refresher 410 and 303) 6. Choice E is correct. Given: 1 cup ϭ 100 grams 1 cake ϭ 75 grams 1 pie ϭ 225 grams Using , we get 4 cups ϭ 4 (100 grams) 4 cups ϭ 400 grams (Using Strategy 8: When all choices must be tested, start with E and work backward.) 2 cakes and 1 pie is Choice E. Substituting and in , we get 2(75 grams) ϩ 225 grams ϭ 150 grams ϩ 225 grams ϭ 375 grams Since is less than , there is enough in 4 cups. So Choice E is correct. (Math Refresher 121 and 431) 7. Choice E is correct. I: Slope is defined as ᎏ x y 2 2 Ϫ Ϫ y x 1 1 ᎏ where (x 1 , y 1 ) and (x 2 , y 2 ) are points on the line. Thus here 0 ϭ x 1 , a ϭ y 1 , a ϭ x 2 , and 0 ϭ y 2 Thus ᎏ x y 2 2 Ϫ Ϫ y x 1 1 ᎏ ϭ ᎏ o a Ϫ Ϫ a o ᎏ ϭϪ1: I is therefore true. II: The triangle created above is an isosceles right triangle with sides a, a, a͙2 ෆ . Thus II is true. III: In an isosceles right triangle, the interior angles of the triangle are 90–45–45 degrees. Thus III is true. (Math Refresher 416, 411, 509) 4 6 6 532 5 4 1 3 2 1 9 9 1872 8 46 718 • SAT PRACTICE TEST 2 – SECTION 2 ANSWERS 706-1045.qxd 5/1/08 3:24 PM Page 718 8. Choice E is correct. Choice A is incorrect: On the number line b is to the left of Ϫ2 so this implies that b is less than Ϫ2 (written as b ϽϪ2). Since bϽϪ2, b is certainly less than Ϫ1 (written as b ϽϪ1). Thus Choice A is incorrect. Choice B is false because if b ϽϪ2, the absolute value of b (denoted as ͿbͿ) must be greater than 2. Choice C is false: c is positive (c Ͼϩ3 Ͼ 0) so c ϪͿcͿ, since ϪͿcͿ is negative. Choice D is false: Since a and b are negative num- bers and since a Ͻ b, ͿaͿϾ ͿbͿ, Choice E is correct and Choice D is incorrect. (Math Refresher ԵԵ 419, ԵԵ 615, ԵԵ 410, ԵԵ 129) 9. Choice B is correct. (Use Strategy 2: Translate from words to algebra.) We are told: A ϩ 8 ϩ A ϩ 1 ϩ A ϩ 2 ϭ A ϩ A ϩ 1 ϩ A ϩ 2 ϩ A ϩ 3 (Use Strategy 1: Cancel expressions that appear on both sides of an equation.) Each side contains an A, A ϩ 1, and A ϩ 2. Canceling each of these from each side, we get A ϩ 8 ϩ A ϩ 1 ϩ A ϩ 2 ϭ A ϩ A ϩ 1 ϩ A ϩ 2 ϩ A ϩ 3. Thus, 8 ϭ A ϩ 3 5 ϭ A (Math Refresher 406) 10. Choice E is correct. (Use Strategy 11: New defi- nitions lead to easy questions.) By the definition of a move, every 4 moves brings each hand back to 12. Thus, after 4, 8, 12, and 16 moves, respectively, each hand is at 12. Hand A, moving counterclockwise, moves to 9 on its 17th move. Hand B, moving clockwise, moves to 3 on its 17th move. (Logical Reasoning) 1 11. Choice E is correct. (Use Strategy 17: Use the given information effectively.) Given: w ϭ 7r ϩ 6r ϩ 5r ϩ 4r ϩ 3r Then, w ϭ 25r We are told we must add something to w so that the resulting sum will be divisible by 7 for every positive integer r. Check the choices. (Use Strategy 8: Start with Choice E.) Add 3r to 25r ϩ 3r ϭ 28r ϭ 7(4r) will always be divisible by 7. Thus, Choice E is correct. (Math Refresher 431) 12. Choice D is correct. To obtain the maximum num- ber of members of S, choose the numbers as small as possible; hence 1 ϩ 3 ϩ 5 ϩ 7 ϩ 9 ϩ 11ϩ 13ϩ 15ϭ 64 Hence, the maximum is 8. (Math Refresher 801) 13. Choice E is correct. I, II, and III are correct. Examples: (2 3 ) 2 ϭ 2 6 ϭ 64, 2 3ϩ2 ϭ 2 3 2 2 ϭ 32, (2 ϫ 3) 2 ϭ 2 2 3 2 ϭ 36. (Math Refresher 429) 14. Choice C is correct. (Use Strategy 2: Translate from words to algebra.) The number of hours from 7:00 A.M. to 5:00 P.M. is 10. The number of hours from 1:00 P.M. to 7:00 P.M. is 6. He worked 10 hours for 3 days and 6 hours for 3 days. Thus, Total Hours ϭ 3(10) ϩ 3(6) ϭ 30 ϩ 18 Total hours ϭ 48 Total Earnings ϭ Hours worked ϫ Hourly rate Given: He earns $10 per hour Substituting and into , we get Total Earnings ϭ 48 ϫ $10 Total Earnings ϭ $480 (Math Refresher 200 and 406) 2 31 3 2 1 1 1 SAT PRACTICE TEST 2 – SECTION 2 ANSWERS • 719 706-1045.qxd 5/1/08 3:24 PM Page 719 15. Choice E is correct. (Use Strategy 3: The whole equals the sum of its parts.) The sum of the angles in a Δ ϭ 180. For the small triangle we have 120 ϩ a ϩ a ϭ 180 120 ϩ 2a ϭ 180 2a ϭ 60 a ϭ 30 For Δ RST, we have 100 ϩ mЄSRT ϩ mЄSTR ϭ 180 From the diagram, we get mЄSRT ϭ a ϩ b mЄSTR ϭ a ϩ b Substituting and into , we get 100 ϩ a ϩ b ϩ a ϩ b ϭ 180 100 ϩ 2a ϩ 2b ϭ 180 2a ϩ 2b ϭ 80 Substituting into , we get 2(30) ϩ 2b ϭ 80 60 ϩ 2b ϭ 80 2b ϭ 20 b ϭ 10 (Math Refr eshers 505 and 406) Question 16 51 5 243 4 3 2 1 16. Choice C is correct. In ascending order, the wages for the six days are: 35 35 40 45 60 75 The median is the middle number. But wait! There is no middle number. So we average the two middle numbers, 40 and 45, to get 42.5. The mode is the number appearing most fre- quently, that is, 35. So 42.5 Ϫ 35 ϭ 7.5. (Math Refresher 601a, 601b) 17. Choice E is correct. (Use Strategy 11: Use new definitions carefully.) (Use Strategy 8: When all choices must be tested, start with E and work backward.) Given: ϭ ᎏ a b Ϫ ϩ 1 1 ᎏ Choice E: ϭ ᎏ 5 3 ϩ Ϫ 1 1 ᎏ ϭ ᎏ 6 2 ᎏ ϭ 3 Choice E is the only choice with a > b. Therefore, it must be the largest. The remaining choices are shown below. Choice D: ϭ ᎏ 4 5 ϩ Ϫ 1 1 ᎏ ϭ ᎏ 5 4 ᎏ ϭ 1 ᎏ 1 4 ᎏ Choice C: ϭ ᎏ 3 5 ϩ Ϫ 1 1 ᎏ ϭ ᎏ 4 4 ᎏ ϭ 1 Choice B: ϭ ᎏ 3 3 ϩ Ϫ 1 1 ᎏ ϭ ᎏ 4 2 ᎏ ϭ 2 Choice A: ϭ ᎏ 2 3 ϩ Ϫ 1 1 ᎏ ϭ ᎏ 3 2 ᎏ ϭ 1 ᎏ 1 2 ᎏ (Math Refresher 431) 18. Choice E is correct. (Use Strategy 17: Use the given information effectively.) 720 • SAT PRACTICE TEST 2 – SECTION 2 ANSWERS a b 5 5 4 5 3 3 33 32 706-1045.qxd 5/1/08 3:24 PM Page 720 We know that Area of Δ ϭ ᎏ 1 2 ᎏ ϫ base ϫ height We are given that RS ϭ ST ϭ an integer Substituting into , we get Area ΔRST ϭ ᎏ 1 2 ᎏ ϫ (An integer) ϭ (same integer) Area ΔRST ϭ ᎏ 1 2 ᎏ ϫ (An integer) 2 Multiplying by 2, we have 2(Area ΔRST) ϭ (An integer) 2 (Use Strategy 8: When all choices must be tested, start with E and work backward.) Substituting Choice E, 20, into , we get 2(20) ϭ (An integer) 2 40 ϭ (An integer) 2 is not possible, since 40 isn’t the square of an integer. (Math Refresher 307, 406, and 431) 19. Choice E is correct. (Use Strategy 17: Use the given information effectively.) Volume of rectangler solid ϭ l ϫ w ϫ h Substituting the given dimensions into , we get Volume of solid ϭ 2 feet ϫ 2 feet ϫ 1 foot Volume of solid ϭ 4 cubic feet Volume of cube ϭ (edge) 3 Substituting edge ϭ .1 foot into , we get Volume of cube ϭ (.1 foot) 3 Volume of cube ϭ .001 cubic feet (Use Strategy 3: The whole equals the sum of its parts.) Since the volume of the rectangular solid must equal the sum of the small cubes, we need to know ϭ Number of cubes Substituting and into , we get ϭ Number of cubes Volume of rectangular solid ᎏᎏᎏᎏ Volume of cube 5 42 5 Volume of rectangular solid ᎏᎏᎏᎏ Volume of cube 4 3 3 2 1 1 5 5 4 4 3 3 12 2 1 ᎏ .0 4 01 cu c b u i b c ic fe f e e t et ᎏ ϭ Number of cubes ᎏ .0 4 01 ᎏϭNumber of cubes Multiplying numerator and denominator by 1,000, we get ᎏ .0 4 01 ᎏϫ ᎏ 1 1 , , 0 0 0 0 0 0 ᎏ ϭ Number of cubes ᎏ 4,0 1 00 ᎏ ϭ Number of cubes 4,000 ϭ Number of cubes (Math Refresher 312 and 313) 20. Choice D is correct. (Use Strategy 2: Translate from words to algebra.) (Use Strategy 17: Use the given information effectively.) Given the perimeter of the square ϭ 40 Thus, 4(side) ϭ 40 side ϭ 10 A side of the square ϭ length of diameter of circle. Thus, diameter ϭ 10 from Since diameter ϭ 2 (radius) 10 ϭ 2 (radius) 5 ϭ radius Area of a circle ϭ p r 2 Substituting into , we have Area of circle ϭ p r 2 Area of circle ϭ 25p (Math Refresher 303 and 310) 3 2 3 2 1 1 SAT PRACTICE TEST 2 – SECTION 2 ANSWERS • 721 706-1045.qxd 5/1/08 3:24 PM Page 721 722 1. Choice D is correct. (Use Strategy 2: Translate from words to algebra.) Let n ϭ the number. Then ᎏ n ϩ 4 3 ᎏ ϭ 6 Multiplying both sides by 4, we have 4 ᎏ n ϩ 4 3 ᎏ ϭ (6)4 n ϩ 3 ϭ 24 n ϭ 21 (Math Refresher 200) 2. Choice C is correct. (Use Strategy 17: Use the given information effectively.) Given: ᎏ 4 3 ᎏ Ͻ x Ͻ ᎏ 4 5 ᎏ Change both fractions to fractions with the same denominator. Thus, ᎏ 3 4 ᎏ Ͻ x Ͻ ᎏ 4 5 ᎏ becomes ᎏ 1 2 5 0 ᎏ Ͻ x Ͻ ᎏ 1 2 6 0 ᎏ (Use Strategy 15: Certain choices may be eas- ily eliminated.) Choice B ϭ ᎏ 1 2 3 0 ᎏ can be instantly eliminated. Choice D ϭ ᎏ 1 2 6 0 ᎏ can be instantly eliminated. Change both fractions to 40 ths to compare Choice C. Thus, ᎏ 3 4 0 0 ᎏ Ͻ x Ͻ ᎏ 3 4 2 0 ᎏ Choice C ϭ ᎏ 3 4 1 0 ᎏ is a possible value of x. (Math Refresher 108 and 419) Explanatory Answers for Practice Test 2 (continued) As you read these solutions, you are advised to do two things if you answered the Math question incorrectly: 1. When a specific Strategy is referred to in the solution, study that strategy, which you will find in “Using Critical Thinking Skills in Math Questions” (beginning on page 69). 2. When the solution directs you to the “Math Refresher” (beginning on page 191)—for example, Math Refresher Ե305—study the 305 Math principle to get a clear idea of the Math operation that was necessary for you to know in order to answer the ques- tion correctly. Section 3: Math 706-1045.qxd 5/1/08 3:24 PM Page 722 3. Choice B is correct. (Use Strategy 2: Translate from words to algebra.) Perimeter of a square ϭ 4 ϫ side. We are given that Perimeter ϭ 20 meters Substituting into , we get 20 meters ϭ 4 ϫ side 5 meters ϭ side Area of square ϭ (side) 2 Substituting into , we get Area of square ϭ (5 meter) 2 Area of square ϭ 25 square meters (Math Refresher 303) 4. Choice E is correct. (Use Strategy 17: Use the given information effectively.) Given: 80 ϩ a ϭϪ32 ϩ b Subtract a from both sides, getting 80 ϩ a ϭϪ32 ϩ b Ϫ a Ϫ a 80 ϭϪ32 ϩ b Ϫ a Add 32 to both sides, giving 80 ϭϪ32 ϩ b Ϫ a ϩ 32 ϩ32 112 ϭ b Ϫ a (Math Refresher 406) 5. Choice E is correct. (Use Strategy 8: When all choices must be tested, start with E and work backward.) Choice E is x 2 ϩ x ϩ 2 (Use Strategy 7: Use specific number examples.) Let x ϭ 3 (an odd positive integer) Then x 2 ϩ x ϩ 2 ϭ 3 2 ϩ 3 ϩ 2 ϭ 9 ϩ 3 ϩ 2 ϭ 14 ϭ (an even result) Now let x ϭ 2 (an even positive integer) Then x 2 ϩ x ϩ 2 ϭ 2 2 ϩ 2 ϩ 2 ϭ 4 ϩ 2 ϩ 2 ϭ 8 ϭ (an even result) Whether x is odd or even, Choice E is even. (Math Refresher 431) 4 3 4 3 12 2 1 The more sophisticated way of doing this is to use Strategy 8: When all choices must be tested, start with Choice E and work backward. Choice E is x 2 + x + 2. Now factor: x 2 + x + 2 = x(x + 1) + 2. Note that since x is an integer, x(x + 1) is always the product of an even integer multiplied by an odd integer. So x(x + 1) is 2 times an integer and there- fore even. + 2 is even so x(x + 1)+ 2 is even. And since x(x + 1) + 2 = x 2 + x + 2, x 2 + x + 2 is even. (Math Refresher #409) 6. Choice A is correct. (Use Strategy 17: Use the given information effectively.) Given: ax ϭ r by ϭ r Ϫ 1 The quick method is to substitute into , giving by ϭ ax Ϫ 1 by ϩ 1 ϭ ax ᎏ by a ϩ 1 ᎏ ϭ x (Math Refresher 431 and 406) 7. Choice B is correct. Let the capacity of container B be x. Then the capacity of container A will be 2x, and that of container C will be 3x. The amount poured into container C is equal to half of 2x plus one-third of x, or ᎏ 2 2 x ᎏ ϩ ᎏ 3 x ᎏ ϭ x ϩ ᎏ 3 x ᎏ ϭ ᎏ 4 3 x ᎏ . Dividing this amount by the total capacity of con- tainer C, we find the fraction that was filled: ϭ ᎏ 4 9 ᎏ . (Math Refresher 406) 8. Choice D is correct. In 12 seconds, the wheel trav- els through 2 revolutions (since 12 seconds is 1 / 5 of the minute it would take for ten revolutions). Since this distance is equal to 16 feet, the wheel travels 8 feet per revolution; thus, 8 feet must be the circum- ference of the wheel. To find the diameter, we divide this figure by p (because the circum- ference of a circle is p times its diameter). Thus, the diameter is ᎏ p 8 ᎏ feet. (Math Refresher 310) ᎏ 4 3 x ᎏ ᎏ 3x 2 1 2 1 SAT PRACTICE TEST 2 – SECTION 3 ANSWERS • 723 706-1045.qxd 5/1/08 3:24 PM Page 723 13. 125 (Use Strategy 18: Remember the isos- celes triangle.) Since AB ϭ BC in Δ ABC, it is isosceles, and the opposite angles are equal. So m ѯ A ϭ Єy (Use Strategy 3: The whole equals the sum of its parts.) The sum of the angles in a triangle is 180Њ, so m ѯ A ϩ y ϩ 90 ϭ 180Њ Subtracting 90 from both sides gives m ѯ A ϩ y ϭ 90 From , the angles are ϭ, so substituting y for Є A in gives y ϩ y ϭ 90 ᎏ 2 2 y ᎏ ϭ ᎏ 9 2 0 ᎏ y ϭ 45 x° is a central angle, so it is measured by the intercepted arc AD. <DCA = 40° is an inscribed angle and measured by 1 ⁄ 2 its intercepted arc AD. Therefore, the intercepted arc AD = 80°. So x = 80, thus x + y = 80 + 45 = 125. (Math Refresher #507, #406, and #527) 14. 25 Use Strategy 5: Average ϭ Average age of students in a class ϭ Thus, Average age of all 80 students ϭ Using , we know that 20 ϭ and 40 ϭ Thus, sum of the ages of the 60 students ϭ (60)(20) ϭ 1200 and the sum of the ages of the 20 students ϭ (40)(20) ϭ 800 sum of the ages of the 20 students ᎏᎏᎏᎏ 20 sum of the ages of the 60 students ᎏᎏᎏᎏ 60 1 2 sum of the ages of the 80 students ᎏᎏᎏᎏ 80 1 sum of the ages of students in the class ᎏᎏᎏᎏᎏ number of students in the class sum of values ᎏᎏᎏ total number of values 3 2 1 2 1 724 • SAT PRACTICE TEST 2 – SECTION 3 ANSWERS 9. 24 (Use Strategy 2: Translate from words to algebra.) Let n ϭ the number We are given: ᎏ 5 8 ᎏ n ϭ ᎏ 3 4 ᎏ n Ϫ 3 (Use Strategy 13: Find unknowns by multipli- cation.) Multiply by 8. We get 8 ᎏ 5 8 ᎏ n ϭ 8 ᎏ 3 4 ᎏ n Ϫ 3 5n ϭ ᎏ 2 4 4 ᎏ n Ϫ 24 5n ϭ 6n Ϫ 24 24 ϭ n (Answer) (Math Refresher 200 and 406) 10. 10 (Use Strategy 11: Use new definitions care- fully.) By definition (Math Refresher 603 and 607) 11. 2300 (Use Strategy 12: Try not to make tedious calculations.) 23m ϩ 23n ϭ 23(m ϩ n) ϭ 23(94 ϩ 6) ϭ 23(100) ϭ 2300 Multiplying 23(94) and 23(6) and adding would be time consuming and therefore tedious. (Math Refresher 431) 12. 11 Since lines are drawn every 10 yards after the first one, ᎏ 1 1 0 0 0 ᎏ lines or 10 additional lines are drawn. (Use Strategy 2: Translate from words to alge- bra.) The total number of lines on the field ϭ the original line ϩ the number of additional lines ϭ 1 ϩ 10 ϭ 11 (Math Refresher 200 and Logical Reasoning) 1 1 706-1045.qxd 5/1/08 3:24 PM Page 724 [...]... 3, and 4 are parallel 5, 6, 7, and 8 are parallel 9, 10, 11, and 12 are parallel 706-1045.qxd 5/1/ 08 3:24 PM Page 734 734 • SAT PRACTICE TEST 2 – SECTION 6 ANSWERS For each group, there are 6 different pairs of edges Thus, 3 ϫ 6 ϭ 18 different pairs of edges in all Below is a listing of all the pairs: 1–2 1–3 1–4 2–3 2–4 3–4 5–6 5–7 5 8 9–10 9–11 9–12 6–7 6 8 7 8 10–11 10–12 11–12 (Math Refresher... Ե312) 14 1.999, 1.9 98 001, or any number 1 1 0 Ͻ r Ͻ 2, like ᎏᎏ, ᎏᎏ, etc (Use Strategy 2: 2 4 Translate from words to algebra.) 2r ϩ 2r ϩ 3 Ͻ 11 4r ϩ 3 Ͻ 11 4r Ͻ 8 rϽ 2 (Math Refresher mЄACB ϩ mЄBCD ϭ mЄACD 1 We are given that AD is a straight line segment We know that 2 mЄACD ϭ 180 3 Given: mЄACD ϭ y 4 mЄBCD ϭ 80 ϩ x We substitute 2 , 3 , and 4 into 1 Thus, y ϩ 80 ϩ x ϭ 180 Subtract 80 : y ϩ x ϭ 100... y-axis (x ϭ 0) at y ϭ 4 (Math Refresher 8 Choice B is correct f(2x) ϭ Ϳ2xͿ Ϫ 2x ϭ 2ͿxͿ Ϫ 2x ϭ 2f(x) (Math Refresher 9 Ե416 and Ե414) Ե616 and Ե615) 1/1, 1, 1.0, etc This means that when x ϭ 5, y ϭ 6 and when x ϭ 7, y ϭ 8 706-1045.qxd 5/1/ 08 3:24 PM Page 733 SAT PRACTICE TEST 2 – SECTION 6 ANSWERS • 733 The slope is ( y2 Ϫ y1)/(x 2 Ϫ x1 ) Thus ( y2 Ϫ y1)/(x 2 Ϫ x1 )ϭ (8 Ϫ 6)/(7 Ϫ 5) ϭ 1/1 See diagram below:...706-1045.qxd 5/1/ 08 3:24 PM Page 725 SAT PRACTICE TEST 2 – SECTION 3 ANSWERS • 725 Hence, the sum of the ages of the 80 students ϭ sum of the ages of the 60 students ϩ sum of the ages of the 20 students ϭ 1,200 ϩ 80 0 ϭ 2,000 radius ϭ distance from (7,5) to (4,5) ϭ ϭ7Ϫ4 radius ϭ 3 Ե410 and Ե524) 17 36.2 (Use Strategy 2: Translate from words to Substituting 3 into 2 , we get algebra.) 2,000 ᎏᎏ ϭ 25 80 Average... Refresher Ե431) Ե603, Ե604, Ե605, Ե702) 3 Choice A is correct Before the rotation, we have and 731 706-1045.qxd 5/1/ 08 3:24 PM Page 732 732 • SAT PRACTICE TEST 2 – SECTION 6 ANSWERS After the rotation, we have 5 Choice B is correct From the diagram, we get 1 2 3 a ϩ d ϭ 180 b ϩ e ϭ 180 c ϩ f ϭ 180 (Use Strategy 13: Find unknowns by adding equations.) Adding 1 ϩ 2 ϩ 3 , we get a ϩ b ϩ c ϩ d ϩ e ϩ f ϭ 540... 706-1045.qxd 5/1/ 08 3:24 PM Page 7 48 What You Must Do Now to Raise Your SAT Score 1 2 3) Study “Word Building with Roots, Prefixes, and Suffixes,” beginning on page 352 a) Follow the directions on (p 707) to determine your scaled score for the SAT Test you’ve just taken These results will give you a good idea about whether or not you ought to study hard in order to achieve a certain score on the actual SAT b) Using... Take the 100 SAT- type “tough word” Vocabulary Tests beginning on page 415 7 48 706-1045.qxd 5/1/ 08 3:24 PM Page 749 SAT PRACTICE TEST 2 • 749 For Both the Math and Critical Reading Parts Giant Step 5 You may want to take the Strategy Diagnostic Test on page 1 to assess whether you’re using the best strategies for the questions For the Writing Part Giant Step 6 Take a look at Part 9—The SAT Writing test... with the growing understanding of their problems and responsibilities”—the best meaning of “pragmatic” would be “practical.” See also Reading Comprehension Strategy 5 706-1045.qxd 5/1/ 08 3:24 PM Page 7 38 7 38 • SAT PRACTICE TEST 2 – SECTION 7 ANSWERS 21 Choice D is correct See the next-to-last paragraph: 23 Choice E is correct Given the context of the sen- “ so much difficulty in resolving our problems... since if x ϭ 4 and y ϭ 8, 8 0 ᎏᎏ ϭ 2 4Ϫ0 All the other choices give a different ratio from 2 (Math Refresher Ե416) 10 Choice A is correct (Use Strategy 3: The whole equals the sum of its parts.) The whole straight angle ABC is equal to the sum of the individual angles Thus, m ЄABC ϭ a ϩ a ϩ a ϩ b ϩ b ϩ bϩaϩaϩa m ЄABC ϭ 6a ϩ 3b We know m ЄABC ϭ 180 Њ Substituting 2 into 1 , we get 180 Њ ϭ 6a ϩ 3b (Use Strategy... also Reading Comprehension Strategy 5 17 Choice C is correct In lines 8 9 the author links “don” with “city disguise, cocktail parties, dinners.” It is logical to assume that “don” relates to “clothing.” See also Reading Comprehension Strategy 5 706-1045.qxd 5/1/ 08 3:24 PM Page 745 SAT PRACTICE TEST 2 – SECTION 9 ANSWERS • 745 18 Choice C is correct See lines 31–34 Choice A is incorrect: Although . Refresher 416, 411, 509) 4 6 6 532 5 4 1 3 2 1 9 9 187 2 8 46 7 18 • SAT PRACTICE TEST 2 – SECTION 2 ANSWERS 706-1045.qxd 5/1/ 08 3:24 PM Page 7 18 8. Choice E is correct. Choice A is incorrect: On. b Substituting and into , we get 100 ϩ a ϩ b ϩ a ϩ b ϭ 180 100 ϩ 2a ϩ 2b ϭ 180 2a ϩ 2b ϭ 80 Substituting into , we get 2(30) ϩ 2b ϭ 80 60 ϩ 2b ϭ 80 2b ϭ 20 b ϭ 10 (Math Refr eshers 505 and 406) Question. 18 different pairs of edges in all. Below is a listing of all the pairs: 1–2 2–3 5–6 6–7 1–3 2–4 5–7 6 8 1–4 3–4 5 8 7 8 9–10 10–11 9–11 10–12 9–12 11–12 (Math Refresher 312) 14. 1.999, 1.998