The sat math section 8

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5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 177 – THE SAT MATH SECTION – 33. The probability of selecting a green marble at ran- 36. If x + 2x + 3x + 4x = 1, then what is the value of x2? dom from a jar that contains only green, white, 1 and yellow marbles is 4 . The probability of select- 37. What is the least positive integer p for which ing a white marble at random form the same jar is 441p is the cube of an integer? 1 3 . If this jar contains 10 yellow marbles, what is the total number of marbles in the jar? 38. The average of 14 scores is 80. If the average of four of these scores is 75, what is the average of 34. If the operation is deﬁned by the equation the remaining 10 scores? x y = 2x + 3y, what is the value of a in the x 39. If 3x – 1 = 9 and 4y + 2 = 64, what is the value of y ? equation a 4 = 1 a? p+p+p 35. x, y, 22, 14, 10, . . . 40. If = 12 and p > 0, what is the value of p? p×p×p In the sequence above, each term after the ﬁrst term, x, is obtained by halving the term that comes before it and then adding 3 to that num- ber. What is the value of x – y? 177
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 178 – THE SAT MATH SECTION – G rid-In Answers If (3p + 2)2 = 64 and p > 0, the expression 11. 2. inside the parentheses is either 8 or –8. Since p > 0, let 3p + 2 = 8; then 3p = 6 and p 1. 158. If you are given two numbers, A and B, then = 2. A possible value of p is 2. B – A + 1 is the formula for ﬁnding the If (x – 1)(x – 3) = –1, then x2 – 4x + 3 = –1, 12. 2. quantity of items between the two numbers. so x2 – 4x + 4 = 0. Factoring this equation Therefore, 2,177 – 2,020 + 1 = 157 + 1. gives (x – 2)(x – 2) = 0; x = 2. Thus, a possi- 2. 24. If the value of x is increased by 3, then the value of y is increased by 8 × 3 = 24. Since ble value for x is 2. 13. 4. If 2y – 3 < 7, then 2y < 10, so y < 5. Since y + after x is increased by 3, 8(x + 4) = z, the 5 > 8 and 2y – 3 < 7, then y > 3 and at the value of z – y = 24. same time y < 5. Thus, the integer must be 4. 3. 3. Find the value of x by expressing each side 5 3 If 3 of x is 15, then 5 x = 15, so x = 5 (15) = 14. 8. of the equation as a power of the same base. 3 28 × 24 = 24x 9. Therefore, x decreased by 1 is 9 – 1 = 8. 212 = 24x 15. 46. If half the difference of two positive num- bers is 20, then the difference of the two 12 = 4x, so 3 = x 3 = 27 and (y – 1)3 = 27, then y – 1 = 3, positive numbers is 40. If the smaller of the 4. 25. Since 3 so y = 4. Thus, (y + 1)2 = (4 + 1)2 = 52 = 25. two numbers is 3, then the other positive number must be 43 since 43 –3 = 40. Thus, 6. 0. Let k equal 9, then 6k = 54. When 54 is the sum of the two numbers is 43 + 3 = 46. divided by 3 the remainder is 0. Since 80% of 35 = .80 × 35 = 28 and 25% of 28 16. 55.5. 7. 50. After three boys are dropped from the class, = .25 × 28 = 7, then 35 of the 63 boys and girls 25 students remain. Of those 25, 44% are have been club members for more than two boys, so 56% are girls. Since 56% of 25 = .56 35 × 25 = 14, 14 girls are enrolled in biology. years. Since 63 = .5555 . . . , 55.5% of the club have been members for more than two years. Hence, 14 of the 28 students in the original 17. 27. Since the lengths of the two pieces of string class were girls. Thus, the number of girls in 14 are in the ratio 3:8, let 3x and 8x represent the original class comprised 28 or 50%. 8. 4 . If 3x – 1 = 11, then 3x = 12, and x = 4. Since their lengths: 3 4 4y = 12, then y = 3. Therefore, x = 3 . You 8x – 3x = 45 y 4 5x = 45 can grid this in as 3 or 1.333. 2 x=9 9. 14 . If 1 – x – 2x – 4x – 7x = 7x – 1, then 1 – 7x = Since 3x = (3)9 = 27, the length of the 7x – 1, so 1 + 1 = 7x + 7x and 2 = 14x. 2 shorter piece of string is 27 inches. Hence, 14 = x. 10 10. 20. 4x2 + 20x + r = (2x + s)2 18. 32. If 10 pounds of fruit serve 36 people, then 36 10 pound serves one person. So, 48 × 36 = 4 × = (2x + s)(2x + s) 10 40 3 = 3 pounds. Since the fruit costs $2.40 a = (2x)(2x) + (2x)(s) + (s)(2x) + (s)(s) = 4x2 + 2xs + 2xs + s2 pound, the cost of the fruit needed to serve 40 48 people is 3 × $2.40 = 40 × .80 = $32.00. = 4x2 + 4xs + s2 19. 20. The measures of vertical angles are equal, so Since the coefﬁcients of x on each side of ∠EFC = 50. In right triangle CEF, the meas- the equation must be the same, 20 = 4s, so ures of the acute angles add up to 90, so s = 5. Comparing the last terms of the poly- ∠ECF + 50 = 90 or ∠ECF = 90 – 50 = 40. nomials on the two sides of the equation makes r = s2 = 52 = 25. Therefore, r – s = 25 – 5 = 20. 178
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 179 – THE SAT MATH SECTION – Since the measures of acute angles formed 26. 90. From 12:25 a .m. to 12:40 a .m. of the same by parallel lines are equal, y + ∠ECF = 50. day, the minute hand of the clock moves 15 Hence, y + 40 = 50, so y = 10. minutes since 40 – 25 = 15. There are 60 20. 35. In triangle ABC, ∠ACB = 180 – 35 – 25 = minutes in an hour, so 15 minutes repre- 15 120. Since angles ACB and DCE form a sents 60 of a complete rotation. Since there straight line, ∠DCE = 180 – 120 = 60. are 360° in a complete rotation, the minute hand moves 15 × 360 = 15 × 6 = 90. Angle BED is an exterior angle of triangle 60 27. 3 . ECD. Therefore, 3x = 60 + x Since the line that passes through points 4 2x = 60 3 (7,3k) and (0,k) has a slope of 14 , then: x = 30 3 3k – k 7 – 0 = 14 21. 1.5. Angle B measures 15 + 30 + 15 or 60°, so the 2k 3 7 = 14 sum of the measures of angles A and C is 28k = 21 120. Since AB = BC = 10, then ∠A = ∠C = 21 3 k = 28 = 4 60, so triangle ABC is equiangular. A trian- 3 5 28. 2 . Since the slope of the line l1 is 6 , then gle that is equiangular is also equilateral, so y1 – 0 15 5 AC = 10. Angles BDE and BED each meas- 5 or y1 = = = 6 3–0 6 2 ure 60 + 15 = 75°, since they are exterior 1 The slope of line l2 is 3 , so angles of triangles ADB and CEB. Therefore, y2 – 0 1 3 triangles ADB and CEB have the same shape or y2 = 3 = 1 = 3–0 3 and size, so AD = CE. Since you are given Since points A and B have the same that DE = 7, then AD + CE = 3, so AD = 1.5. x-coordinates, they lie on the same vertical 22. 49. If the lengths of two sides of an isosceles tri- 2 line, so the distance from A to B = y1 – y angle are 9 and 20, then the third side must 5 3 be 9 or 20. Since 20 is not less than 9 + 9, the = 2 – 1 or 2 3 third side cannot be 9. Therefore, the Grid as 2 . lengths of the three sides of the triangle 29. 60. You are given that all the dimensions of a must be 9, 20, and 20. The perimeter is 9 + rectangular box are integers greater than 1. 20 + 20 = 49. Since the area of one side of this box is 12, 23. 21. In the given triangle, 10 – 5 < x < 10 + 5 or, the dimensions of this side must be either 2 equivalently, 5 < x < 15. Since the smallest by 6 or 3 by 4. The area of another side of possible integer value of x is 6, the least possi- the box is given as 15, so the dimensions of ble perimeter of the triangle is 5 + 6 + 10 = 21. this side must be 3 and 5. Since the two 24. 15. Factor 120 as 4 × 5 × 6. Since each number of sides must have at least one dimension in the set 4, 5, and 6 is less than the sum, and common, the dimensions of the box are 3 by 4 by 5, so its volume is 3 × 4 × 5 = 60. greater than the difference of the other two, a possible perimeter of the triangle is 4 + 5 + 30. 96. A cube whose volume is 8 cubic inches has an edge length of 2 inches, since 2 × 2 × 2 = 8. 6 = 15. 1 25. 9 . Since the ﬁgure is a square, x = 4x – 1 Since a cube has six square faces of equal area, the surface area of this cube is 6 × 22 or 6 × 4 1 = 3x 1 3 =x or 24. The minimum length, or L, of 1 1 1 1 To ﬁnd the area of the square, square 3 = ( 3 )2 = 9 . 4 -inch-wide tape needed to completely cover the cube must have the same surface area of 179
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 180 – THE SAT MATH SECTION – 1 the cube. Therefore, L × 4 = 24 and L = 36. 0.01. If x + 2x + 3x + 4x = 1, then 10x = 1, so x = 1 12 1 1 24 × 4 = 96 inches. 2 10 and x = ( 10 ) = 100 . Since 100 does not 31. 22. Let x, x + 2, x + 4, and x + 6 represent four ﬁt in the grid, grid in .01 instead. 37. 21. Since 441p = 9 × 49 × p = 32 × 72 × p, let p = consecutive even integers. If their average is 3 × 7, which makes 441p = 33 × 73 = (3 × 7)3 19, then = 213. x + (x + 2) + (x + 4) + (x + 6) = 19 4 38. 82. If the average of 14 scores is 80, the sum of or the 14 scores is 14 × 80 or 1,120. If the aver- 4x + 12 = 4 × 19 = 76 age of four of these scores is 75, the sum of Then, these four scores is 300, so the sum of the 4x = 76 – 12 = 64 remaining 10 scores is 11,200 – 300 or 820. 64 x = 4 = 16 820 The average of these 10 scores is 10 or 82. Therefore, x + 6, the greatest of the four 39. 3. To ﬁnd the value of x , given that 3x – 1 = 9 y consecutive integers is 16 + 6 or 22. and 4y + 2 = 64, use the two equations to 32. 36. Since the average of x, y, and z is 12, then x ﬁnd the values of x and y. + y + z = 3 × 12 = 36. Thus, 3x + 3y + 3z = Since, 3x – 1 = 9 = 32, then x – 1 = 2, so x = 3. 3(36) or 108. And 4y + 2 = 64 = 43, then y + 2 = 3, so y = 1. The average of 3x, 3y, and 3z is their sum, x 3 Therefore, y = 1 = 3. 108, divided by 3 since three values are being 108 1 3 p+p+p 3p added: 3 or 36. 40. 2 . Since p × p × p = p × p = p2 = 12 1 1 7 p2 1 3 1 1 33. 24. Since 1 – ( 4 + 3 ) = 1 – 12 , the probability of then 3 = 12 , so p2 = 12 = 4 . Therefore, p = 2 , 5 selecting a yellow marble is 12 . If 10 of the x 11 1 1 since 4 × 4 = 2 . Grid in as 2 . 5 10 marbles in the jar are yellow, then 12 = x . Since 10 is two times 5, x must be two times 12 or 24. F inally 34. 10. Since x y = 2x + 3y, evaluate a 4 by let- ting x = a and y = 4: Don’t forget to keep track of your time during the Math a 4 = 2a + 3(4) = 2a + 12 section. Although most questions will take you about a Evaluate 1 a by letting x = 1 and y = 4: minute or so—the amount of time it takes to answer a 1 a = 2(1) + 3a = 2 + 3a particular question can vary according to difﬁculty. Since a 4 = 1 a, then 2a + 12 = 2 + 3a, Don’t hold yourself to a strict schedule, but learn to be or 12 – 2 = 3a – 2a, so 10 = a. aware of the time you are taking. Never spend too much 35. 32. In the sequence x, y, 22, 14, 10, . . . each term time on any one question. Feel free to skip around and after the ﬁrst term, x, is obtained by halving answer any questions that are easier for you, but be sure the term that comes before it and then to keep track of which questions you have skipped. adding 3 to that number. Hence, to obtain y, Remember that, in general, each set of questions begins do the opposite to 22: Subtract 3 and then with easy problems and becomes increasingly harder. double the result, getting 38. To obtain x, Finally, if you can eliminate one or more answers on a subtract 3 from 38 and then double the tough question, go ahead and make a guess, and if you result, getting 70. Thus, x – y = 70 – 38 = 32. have time at the end of the section, go back and check your answers. Good luck! 180
5658 SAT2006[05](fin).qx 11/21/05 6:45 PM Page 181 CHAPTER 5 The SAT Writing Section W hat to Expect in the Writing Section In March 2005, the SAT® was revamped to include a Writing section that consists of 49 multiple-choice gram- mar and usage questions and an essay. The essay has essentially the same structure and content as the one on the old SAT II™ Writing Test, which means that you will be able to easily prepare for it. In the multiple-choice part of the Writing section, you will have 35 minutes, split into one 25-minute sec- tion and one 10-minute section. The multiple-choice questions, too, are essentially the same as the multiple-choice questions on the old SAT II Writing Test. They will ask you to identify errors in grammar and usage and/or select the most effective way to revise a sentence or passage. They are designed to measure your knowledge of basic gram- mar and usage rules as well as general writing and revising strategies. There are three types of multiple-choice questions: identifying sentence errors, improving sentences, and improving paragraphs. None of the multiple-choice questions ask you to formally name grammatical terms, or test you on spelling. 181
5658 SAT2006[05](fin).qx 11/21/05 6:45 PM Page 182 SAT Writing Section at a Glance There are four question types on the Writing section: ■ Identifying Sentence Errors—items require you to read a sentence and identify the error (if any) in gram- mar or usage ■ Improving Sentences—items require you to determine the best way to correct a sentence ■ Improving Paragraphs—items ask you how a draft essay could best be improved ■ Essay—requires you to write a coherent, well-constructed essay in response to a prompt I dentifying Sentence Errors Responding to Quotes. You will be given one or ■ Each sentence will have four underlined words or two quotes and asked to evaluate or compare phrases. You need to determine which underlined por- them by writing an essay. tion, if any, contains an error in grammar or usage. If Completing a Statement or Idea. You will be ■ none of the four underlined portions contain an error, given an incomplete statement and asked to ﬁll in you will need to select choice e, which is “No error.” the blank; then you will use the completed state- Approximately 18 of the 49 multiple-choice questions ment as the basis for an essay. in the Writing section will be this type. For both types of prompts, you will be asked to Improving Sentences develop a point of view and to back up your opinion With these question types, you will need to determine with examples from your own experience or from sub- which of ﬁve versions of a sentence is the most clear and jects you have studied. correct. Approximately 25 of the 49 questions in this section will be this type. W hy Write an Essay? Improving Paragraphs With these question types, you will be asked about Anyone who has gone to college can tell you that writ- ways in which a draft version of a short essay can be ing is a big part of the experience. Students have to take improved. These questions can cover everything from accurate notes in all classes, write essays and papers for grammar issues to matters of organization and devel- different subjects, and often have to respond to essay opment of ideas. Approximately 6 of the 49 questions questions on exams. Students need to be able to think will be this type. logically in order to do this, and be able to take a stance on an issue and defend their position in writing. Essay For the essay portion of the Writing section, you will have 25 minutes to respond to a prompt. This prompt will be one of two types: 182