ࡱ> }~MOQSwf  bjbj ee T$$$"$e\#9999t>\D$Eddddddd${h-kdF4>@t>FFd99>e[[[F99d[Fd[[$]p]9nU"1Vh<]de0eD],%l[v%lp]%lp]0FF[FFFFFdd[FFFeFFFF%lFFFFFFFFF  : The Redesigned S A T Mathematics Sample Sets Information for users of assistive technology The document(s) that accompany these instructions are designed to be accessible to individuals who use screen readers, text readers, or other assistive technology. You may wish to consult the manual or help system of your application software to learn how best to take advantage of the following features implemented in this document. Headings Some questions include passages or other material that you may find it useful to return to or skip past. To assist in this kind of navigation, the test documents use headings as follows. Heading level 3: section titles Heading level 4: directions for a group of questions or references to material on which one or more questions will be based (for example, Question 3 is based on the following text:) Heading level 5: question numbers, which directly precede the associated questions Heading level 6: indications of skippable content (For example, you may prefer to skip some sections of this script, such as those that provide figure descriptions or possible answers in context for questions that involve revision. This content is identified at the beginning by the phrase Begin skippable content and at the end by the phrase End skippable content. These phrases are formatted as level6 headings.) Links This document includes hyperlinked material. There are two ways to follow a link. One is to move the flashing text cursor, or caret, into the hyperlinked text and press the Enter key; the other is to place the mouse cursor, or pointer, over the hyperlinked text and press Ctrl+leftclick (that is, press and release the left button on the mouse while holding down the Ctrl key on the keyboard). Some application software includes commands for listing links in a document. In JAWS, for example, pressing Insert+F5 provides a list of links. After following a link in Microsoft Word, you can return to your previous location by pressing Alt+leftarrow. Text attributes Boldfacing and underlining are used in this document for emphasis and in defined heading styles. Italic type is not used as an emphasis indicator in this document but is used in defined heading styles as well as where standard typographic conventions require them, such as book titles and mathematical variables. Adjust the settings of your screen reader or other application software if you wish to be notified of text attribute changes. Except where stated otherwise, this formatting is not critical to the meaning of the test material. Text and graphics size The styles used in this document result in text that is moderately enlarged. To enlarge text further in Microsoft Word, the following is recommended, in order of preference: 1.Adjust the styles to meet your needs. You can adjust both font size and typeface if desired. 2.Manually adjust the font size or typeface as desired. 3.Use Microsoft Words zoom function. This is the easiest way to enlarge any figures, but note that some screen readers will not read text that has moved off screen as a result of zooming. Pronunciation Some changes to the text have been made to improve the way screen readers pronounce the text where doing so would not inappropriately change test content. For example, we have inserted spaces between the letters of initialisms to ensure that the individual letters are spoken separately. However, please note that pronunciation errors may remain. If unsure of a word, use the spelling or characterbycharacter navigation function of your application software to resolve any uncertainties. Punctuation Where punctuation or symbols are critical to the meaning of test material, we either convert the symbol or punctuation mark to words (for example, it apostrophes or its apostrophe) or else include a statement advising you to take note of punctuation for a particular question or portion of a question. Tables Some questions may include tables. Use the tablenavigation features of your application software. Figures This document may include figures, which appear on screen. Following each figure on screen is text describing that figure. Readers using visual presentations of the figures may choose to skip parts of the text describing the figure that begin with Begin skippable figure description and end with End skippable figure description. Your application software may speak unhelpful information when you arrive at the figures, such as the figures size. If your application software offers a method of configuring speech for graphics, you may wish to use that to prevent it from speaking the unwanted information. Mathematical Equations and Expressions Some of these documents include mathematical equations and expressions. Some of the mathematical equations and expressions are presented as graphics. In cases where a mathematical equation or expression is presented as a graphic, a verbal presentation is also given and the verbal presentation comes directly after the graphic presentation. The verbal presentation is in green font to assist readers in telling the two presentation modes apart. Readers using audio alone can safely ignore the graphical presentations, and readers using visual presentations may ignore the verbal presentations. MathematicsNo Calculator The directions and question numbers below are representative of what students will encounter on test day. Some math sections allow the use of a calculator, while others do not, as indicated in the directions. Turn to Section 3 of your answer sheet to answer the questions in this section. For questions 1 through 15, solve each problem, choose the best answer from the choices provided, and fill in the corresponding circle on your answer sheet. For questions 16 through 20, solve the problem and enter your answer in the grid on the answer sheet. Please refer to the directions before question 16 on how to enter your answers in the grid. You may use any available space in your test booklet for scratch work. 1.The use of a calculator is not permitted. 2.All variables and expressions used represent real numbers unless otherwise indicated. 3.Figures provided in this test are drawn to scale unless otherwise indicated. 4.All figures lie in a plane unless otherwise indicated. 5.Unless otherwise indicated, the domain of a given function f is the set of all real numbers x for which  EMBED Equation.DSMT4  f of x is a real number.  Begin skippable figure descriptions. The figure presents information for your reference in solving some of theproblems. Reference figure 1 is a circle with radius r. Two equations are presented below reference figure1. A equals pi times the square of r. C equals 2 pi r. Reference figure 2 is a rectangle with length ! and width w. An equation is presented below reference figure2. A equals !w. Reference figure 3 is a triangle with base b and height h. An equation is presented below reference figure3. A equals onehalfbh. Reference figure 4 is a right triangle. The two sides that form the right angle are labeled a and b, and the side opposite the right angle is labeled c. An equation is presented below reference figure4. c squared equals a squared plus b squared. Special Right Triangles Reference figure 5 is a right triangle with a 30degree angle and a 60degree angle. The side opposite the 30degree angle is labeled x. The side opposite the 60degree angle is labeledxtimes the squareroot of3. The side opposite the right angle is labeled2x. Reference figure 6 is a right triangle with two 45degree angles. Two sides are each labeleds. The side opposite the rightangle is labeled stimes the squareroot of2. Reference figure 7 is a rectangular solid whose base has length! and widthw and whose height ish. An equation is presented below reference figure7. V equals!wh. Reference figure 8 is a rightcircularcylinder whose base has radiusr and whose height ish. An equation is presented below reference figure8. V equalspitimes the square of rtimesh. Reference figure 9 is a sphere with radiusr. An equation is presented below reference figure9. V equalsfourthirds pi times the cube ofr. Reference figure 10 is a cone whose base has radiusr and whose height ish. Anequation is presented below reference figure10. V equals onethird times pi times the square of rtimesh. Reference figure 11 is an asymmetrical pyramid whose base has length! and widthw and whose height ish. An equation is presented below reference figure11. V equalsonethird!wh. Additional Reference Information The number of degrees of arc in a circle is360. The number of radians of arc in a circle is 2pi. The sum of the measures in degrees of the angles of a triangle is180. End skippable figure descriptions. For studentproduced response questions, students will also see the following directions: For questions 16 through 20, solve the problem and enter your answer in the grid, as described below, on the answer sheet. 1.Although not required, it is suggested that you write your answer in the boxes at the top of the columns to help you fill in the circles accurately. You will receive credit only if the circles are filled in correctly. 2.Mark no more than one circle in any column. 3.No question has a negative answer. 4.Some problems may have more than one correct answer. In such cases, grid only one answer. 5.Mixed numbers such as  EMBED Equation.DSMT4 three and onehalf must be recorded as  EMBED Equation.DSMT4 threepointfive or  EMBED Equation.DSMT4 sevenslashtwo. (If  EMBED Equation.DSMT4 three and onehalf is entered into the grid as , three,one,slash,two, it will be interpreted as  EMBED Equation.DSMT4 thirtyone halves, not  EMBED Equation.DSMT4 three and onehalf). 6.Decimal answers: If you obtain a decimal answer with more digits than the grid can accommodate, it may be either rounded or truncated, but it must fill the entire grid. The following are four examples of how to record your answer in the spaces provided. Keep in mind that there are four spaces provided to record each answer. Examples 1 and 2  Beging skippable figure description. Example 1: If your answer is a fraction such as seventwelfths, it should be recorded as follows. Enter seven in the first space, the fractionbar (aslash) in the second space, one in the third space, and two in the fourth space. All four spaces would be used in this example. Example 2: If your answer is a decimal value such as twopointfive, it could be recorded as follows. Enter two in the secondspace, the decimal point in the third space, and five in the fourth space. Only three spaces would be used in thisexample. End skippable figure description. Example 3  Beging skippable figure description. Example 3: Acceptable ways to record twothirds are: two slash three, point sixsixsix, and point sixsixseven. End skippable figure description. Example 4  Note: You may start your answers in any column, spacepermitting. Columns you dont need should be left blank. Beging skippable figure description. Example 4: It is not necessary to begin recording answers in the first space unless all four spaces are needed. For example, if your answer is 201, you may record two in the first space, zero in the second space, and one in the third space. Alternatively, you may record two in the second space, zero in the third space, and one in the fourth space. Spaces not needed should be left blank. End skippable figure description. Mathematics Sample Questions Question 1 is based on the following figure, which presents the graph of line! in the xyplane.  Begin skippable figure description. The figure presents the graph of line! in the xyplane. The x and yaxes have tick marks. The fifth tick marks from the origin in the positive and negative directions on both axes are labeled 5 and negative5, respectively. Six points are marked on line!. The following are the coordinates of the six points on line!.  EMBED Equation.DSMT4  parenthesis, negative2 comma 10, close parenthesis  EMBED Equation.DSMT4  parenthesis, zero comma 7, close parenthesis  EMBED Equation.DSMT4  parenthesis, 2 comma 4, close parenthesis  EMBED Equation.DSMT4  parenthesis, 4 comma 1, close parenthesis  EMBED Equation.DSMT4  parenthesis, 6 comma negative2, close parenthesis  EMBED Equation.DSMT4  parenthesis, 8 comma negative5, close parenthesis From left to right, line! extends from quadrant2 downward to the right through quadrant1 and into quadrant4. Line! intersects the yaxis at value 7 and intersects the xaxis between values 4 and5. End skippable figure description. Question 1. If line! is translated up 5 units and right 7 units, then what is the slope of the newline? A. EMBED Equation.DSMT4  negative twofifths B. EMBED Equation.DSMT4  negative threehalves C. EMBED Equation.DSMT4  negative eightninths D. EMBED Equation.DSMT4  negative elevenfourteenths  HYPERLINK \l "_Explanation_for_question" Answer and Explanation. (Follow link to explanation of question1.) Question 2. The mean number of students per classroom, y, at Central High School can be estimated using the equation EMBED Equation.DSMT4  yequals 0.8636x plus 27.227, where x represents the number of years since 2004 and  EMBED Equation.DSMT4  x is lessthan or equalto10. Which of the following statements is the best interpretation of the number 0.8636 in the context of this problem? A.The estimated mean number of students per classroom in 2004. B.The estimated mean number of students per classroom in 2014. C.The estimated yearly decrease in the mean number of students per classroom. D.The estimated yearly increase in the mean number of students per classroom.  HYPERLINK \l "_Explanation_for_question_1" Answer and Explanation. (Follow link to explanation of question2.) Question 3. If  EMBED Equation.DSMT4  a, squared plus 14a, equals51 and  EMBED Equation.DSMT4  a, is greater than zero, what is the value of  EMBED Equation.DSMT4  a, plus7?  HYPERLINK \l "_Explanation_for_question_2" Answer and Explanation. (Follow link to explanation of question3.) Question 4. If  EMBED Equation.DSMT4  the fraction whose numerator is 2, and whose denominator is a, minus 1, equals, the fraction whose numerator is 4 and whose denominator is y, where  EMBED Equation.DSMT4  y does not equal zero and  EMBED Equation.DSMT4  a does not equal 1, what is y in terms of a? A. EMBED Equation.DSMT4  y equals 2 a, minus 2. B. EMBED Equation.DSMT4  y equals 2 a, minus 4. C. EMBED Equation.DSMT4  y equals 2 a, minus onehalf. D. EMBED Equation.DSMT4  y equals onehalf a, plus 1.  HYPERLINK \l "_Explanation_for_question_3" Answer and Explanation. (Follow link to explanation of question4.) Question 5. If  EMBED Equation.DSMT4  y equals x cubed, plus 2x, plus 5 and  EMBED Equation.DSMT4  z equals xsquared, plus 7x, plus 1, what is  EMBED Equation.DSMT4  2y plus z in terms of x? A. EMBED Equation.DSMT4  3x cubed, plus 11x, plus 11. B. EMBED Equation.DSMT4  2x cubed, plus x squared, plus 9x, plus 6. C. EMBED Equation.DSMT4  2x cubed, plus x squared, plus 11x, plus 11. D. EMBED Equation.DSMT4  2x cubed, plus 2x squared, plus 18x, plus12.  HYPERLINK \l "_Explanation_for_question_4" Answer and Explanation. (Follow link to explanation of question5.) Question 6. If  EMBED Equation.DSMT4  a, to the power of negative onehalf, equals x, where  EMBED Equation.DSMT4  a is greater than zero and  EMBED Equation.DSMT4  x is greater than zero, which of the following equations gives a in terms ofx? A. EMBED Equation.DSMT4  a, equals the fraction 1 over the square root of x B. EMBED Equation.DSMT4  a, equals the fraction 1 over x squared C. EMBED Equation.DSMT4  a, equals the square root of x D. EMBED Equation.DSMT4  a, equals negative x squared  HYPERLINK \l "_Explanation_for_question_6" Answer and Explanation. (Follow link to explanation of question6.) Question 7. The graph of  EMBED Equation.DSMT4  y equals, parenthesis 2x minus 4 close parenthesis, parenthesis xminus 4 close parenthesis is a parabola in the xyplane. In which of the following equivalent equations do the x- and ycoordinates of the vertex of the parabola appear as constants or coefficients? A. EMBED Equation.DSMT4  y equals 2xsquared minus 12x plus16. B. EMBED Equation.DSMT4  y equals 2x, parenthesis x minus 6 close parenthesis, plus16. C. EMBED Equation.DSMT4  y equals 2, parenthesis, x minus 3, close parenthesis, squared, plus, parenthesis, negative 2, close parenthesis. D. EMBED Equation.DSMT4  y equals parenthesis x minus 2 close parenthesis, parenthesis 2x minus 8 close parenthesis.  HYPERLINK \l "_Explanation_for_question_5" Answer and Explanation. (Follow link to explanation of question7.) Question 8. Which of the following is equal to EMBED Equation.DSMT4  parenthesis, 14 minus 2i, close parenthesis, parenthesis, 7 plus 12i, close parenthesis? (Note: EMBED Equation.DSMT4  iequals the square root of negative1.) A.74 B.122 C. EMBED Equation.DSMT4  74 plus 154i D. EMBED Equation.DSMT4  122 plus 154i  HYPERLINK \l "_Explanation_for_question_7" Answer and Explanation. (Follow link to explanation of question8.) Question 9.  EMBED Equation.DSMT4 the fraction whose numerator is 5, parenthesis, k plus2, close parenthesis, minus 7, and whose denominator is 6, equals the fraction whose numerator is 13 minus, parenthesis, 4 minus k, close parenthesis, and whose denominator is9. In the equation above, what is the value of k? A. EMBED Equation.DSMT4  9 over 17. B. EMBED Equation.DSMT4  9 over 13. C. EMBED Equation.DSMT4  33 over 17. D. EMBED Equation.DSMT4  33 over 13.  HYPERLINK \l "_Explanation_for_question_8" Answer and Explanation. (Follow link to explanation of question9.) Question 10.  EMBED Equation.DSMT4 4x minusy, equals 3y plus7.  EMBED Equation.DSMT4 x plus 8y equals4. Based on the system of equations above, what is the value of the product xy? A. EMBED Equation.DSMT4  negative threehalves B. EMBED Equation.DSMT4  onefourth C. EMBED Equation.DSMT4  onehalf D. EMBED Equation.DSMT4  elevenninths  HYPERLINK \l "_Explanation_for_question_9" Answer and Explanation. (Follow link to explanation of question10.) Question 11. If  EMBED Equation.DSMT4  onehalfx, plus onethirdy, equals4, what is the value of  EMBED Equation.DSMT4  3x plus 2y?  HYPERLINK \l "_Explanation_for_question_10" Answer and Explanation. (Follow link to explanation of question11.) Question 12. Which of the following is equal to  EMBED Equation.DSMT4  sine of the fraction pi over 5? A. EMBED Equation.DSMT4  negative cosine of the fraction pi over 5. B. EMBED Equation.DSMT4  negative sine of the fraction pi over 5. C. EMBED Equation.DSMT4  cosine of the fraction 3 pi over 10. D. EMBED Equation.DSMT4  sine of the fraction 7 pi over 10.  HYPERLINK \l "_Explanation_for_question_11" Answer and Explanation. (Follow link to explanation of question12.) Question 13.  EMBED Equation.DSMT4  onehalf x minus onefourth y, equals5.  EMBED Equation.DSMT4  ax minus 3y equals20. In the system of linear equations above, a is a constant. If the system has no solution, what is the value ofa? A. EMBED Equation.DSMT4  onehalf B.2 C.6 D.12  HYPERLINK \l "_Explanation_for_question_12" Answer and Explanation. (Follow link to explanation of question13.) Question 14.  EMBED Equation.DSMT4  the fraction one over x, plus the fraction 2 over x, equals onefifth. Anise needs to complete a printing job using both of the printers in her office. One of the printers is twice as fast as the other, and together the printers can complete the job in 5hours. The equation above represents the situation described. Which of the following describes what the expression  EMBED Equation.DSMT4  the fraction oneoverx represents in this equation? A.The time, in hours, that it takes the slower printer to complete the printingjob alone B.The portion of the job that the slower printer would complete in onehour C.The portion of the job that the faster printer would complete in twohours D.The time, in hours, that it takes the slower printer to complete  EMBED Equation.DSMT4  onefifth of the printingjob  HYPERLINK \l "_Explanation_for_question_13" Answer and Explanation. (Follow link to explanation of question14.) Question 15 is based on the following figure.  Begin skippable figure description. The figure presents a semicircle with a horizontal diameter labeled AB and a horizontal line segment labeled CD intersecting the semicircle at points C andD. End skippable figure description. Question 15. The semicircle above has a radius of rinches, and chord  EMBED Equation.DSMT4  line segment CD is parallel to the diameter  EMBED Equation.DSMT4  line segment AB. If the length of  EMBED Equation.DSMT4  line segment CD is  EMBED Equation.DSMT4  twothirds of the length of  EMBED Equation.DSMT4  line segment AB, what is the distance between the chord and the diameter in terms ofr? A. EMBED Equation.DSMT4  onethird pi times r. B. EMBED Equation.DSMT4  twothirds pi times r. C. EMBED Equation.DSMT4  the fraction the square root of 2 over 2, times r. D. EMBED Equation.DSMT4  the fraction the square root of 5 over 3, times r.  HYPERLINK \l "_Explanation_for_question_14" Answer and Explanation. (Follow link to explanation of question15.) Question 16. It is given that  EMBED Equation.DSMT4  sinex equalsa, where x is the radian measure of an angle and  EMBED Equation.DSMT4 the fraction piover2, is less thanx, and x is less thanpi. If  EMBED Equation.DSMT4  sinew equals negativea, which of the following could be the value ofw? A. EMBED Equation.DSMT4 pi minusx B. EMBED Equation.DSMT4 x minus pi C. EMBED Equation.DSMT4 2 pi plusx D. EMBED Equation.DSMT4 x minus 2pi  HYPERLINK \l "_Explanation_for_question_15" Answer and Explanation. (Follow link to explanation of question16.) Question 17.  EMBED Equation.DSMT4  xsquared plus ysquared, minus 6x plus 8y, equals144. The equation of a circle in the xyplane is shown above. What is the diameter of thecircle?  HYPERLINK \l "_Explanation_for_question_16" Answer and Explanation. (Follow link to explanation of question17.) Question 18. What is one possible solution to the equation  EMBED Equation.DSMT4  the fraction whose numerator is 24 and whose denominator is xplusone, minus, the fraction whose numerator is 12 and whose denominator is xminusone, equalsone?  HYPERLINK \l "_Explanation_for_question_17" Answer and Explanation. (Follow link to explanation of question18.) Mathematics Sample Question Answers andExplanations The following are explanations of answers to sample questions 1 through18. The heading of each explanation is hyperlinked to the actual question. In addition, each explanation is followed by two hyperlinks: one to the question explained and one to the next question. There are two ways to follow a link. One is to move the flashing text cursor, or caret, into the hyperlinked text and press the Enter key; the other is to place the mouse cursor, or pointer, over the hyperlinked text and press Ctrl+leftclick (that is, press and release the left button on the mouse while holding down the Ctrl key on the keyboard). After following a link in Microsoft Word, you can return to your previous location (for example, the answer explanation) by pressing Alt+leftarrow. Explanation for  HYPERLINK \l "_Question_1." question 1. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: Students must make a connection between the graphical form of a relationship and a numerical description of a key feature. Difficulty: Easy Key: B Choice B is correct. The slope of a line can be determined by finding the difference in the ycoordinates divided by the difference in the xcoordinates for any two points on the line. Using the points indicated, the slope is  EMBED Equation.DSMT4  negativethreehalves. Translating the line moves all the points on the line the same distance in the same direction, and the image will be a parallel line. Therefore, the slope of the image is  EMBED Equation.DSMT4  negativethreehalves. Choice A is not the correct answer. This value may result from a combination of errors. The student may misunderstand how the negative sign affects the fraction and apply the transformation as  EMBED Equation.DSMT4  the fraction whose numerator is parenthesis, negative 3 plus 5, close parenthesis, and whose denominator is parenthesis, negative 2 plus 7, close parenthesis. Choice C is not the correct answer. This value may result from finding the slope of the line and then subtracting 5 from the numerator and 7 from the denominator. Choice D is not the correct answer. This answer may result from adding  EMBED Equation.DSMT4  fivesevenths to the slope of the line. Link back to  HYPERLINK \l "_Question_1." question 1. Link back to  HYPERLINK \l "_Question_2." question 2. Explanation for  HYPERLINK \l "_Question_2." question 2. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: Students must interpret the slope of an equation in relation to the realworld situation it models. Furthermore, when the models are created from data, students must recognize that these models only estimate the independent variable, y, for a given value of x. Difficulty: Easy Key: D Choice D is correct. When an equation is written in the form  EMBED Equation.DSMT4  y equals mx plusb, the coefficient of thexterm (in this case 0.8636) is the slope. The slope of a linear equation gives the amount that the mean number of students per classroom (represented byy) changes per year (represented byx). Choice A is not the correct answer. This answer may result from a misunderstanding of slope and yintercept. The yintercept of the equation represents the estimated mean number of students per classroom in2004. Choice B is not the correct answer. This answer may result from a misunderstanding of the limitations of the model. Students may see that  EMBED Equation.DSMT4  x is less than or equal to10 and erroneously use this statement to determine that the model finds the mean number of students in 2014. Choice C is not the correct answer. This answer may result from a misunderstanding of slope. The student recognizes that slope models the rate of change, but may think that a slope of less than1 represents a decreasing function. Link back to  HYPERLINK \l "_Question_2." question 2. Link back to  HYPERLINK \l "_Question_3." question 3. Explanation for  HYPERLINK \l "_Question_3." question 3. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Passport to advanced math Calculator Usage: No Objective: Students must solve a quadratic equation. Difficulty: Easy Key: 10 There is more than one way to solve this problem. A student can apply standard techniques by rewriting the equation  EMBED Equation.DSMT4  a, squared plus 14a, equals51 as  EMBED Equation.DSMT4  a, squared plus 14a, minus 51 equals zero and then factoring. Since the coefficient of a is 14 and the constant term is  EMBED Equation.DSMT4 negative51, factoring  EMBED Equation.DSMT4  a, squared plus 14a, minus 51 equals zero requires writing 51as the product of two numbers that differ by 14. This is  EMBED Equation.DSMT4 51 equals parenthesis 3 close parenthesis, parenthesis 17 close parenthesis, which gives the factorization  EMBED Equation.DSMT4  a, squared plus 14a, minus51 equals parenthesis, a plus 17, close parenthesis, parenthesis, a minus 3, close parenthesis, equals zero. The possible values of a are  EMBED Equation.DSMT4  a, equals negative17 and  EMBED Equation.DSMT4  a, equals 3. Since it is given that  EMBED Equation.DSMT4  a, is greater than zero, it must be true that  EMBED Equation.DSMT4  a, equals 3. Thus, the value of  EMBED Equation.DSMT4  a, plus 7 is  EMBED Equation.DSMT4 3 plus7 equals10. A student could also use the quadratic formula to find the possible values of a: EMBED Equation.DSMT4  a,equals the fraction whose numerator is negative 14, plus or minus the square root of the quantity, 14 squared minus 4, parenthesis one close parenthesis, parenthesis negative 51 close parenthesis, and whose denominator is 2, parenthesis one close parenthesis, equals the fraction whose numerator is negative 14, plus or minus the square root of the quantity 196 minus parenthesis negative 204 close parenthesis, and whose denominator is 2, equals the fraction whose numerator is negative 14, plus or minus the square root of 400, and whose denominator is 2, equals the fraction whose numerator is negative 14, plus or minus 20 and whose denominator is2. The possible values of a are  EMBED Equation.DSMT4 a, equals the fraction whose numerator is negative 14 minus 20, and whose denominator is 2, equals negative17 and  EMBED Equation.DSMT4  a, equals the fraction whose numerator is negative14 plus20, and whose denominator is 2, equals3. Again, since it is given that  EMBED Equation.DSMT4  a, is greater than zero, it must be true that  EMBED Equation.DSMT4  a, equals 3. Thus, the value of  EMBED Equation.DSMT4  a, plus 7 is  EMBED Equation.DSMT4 3 plus 7 equals 10. There is another way to solve this problem that will reward the student who recognizes that adding 49 to both sides of the equation yields  EMBED Equation.DSMT4  a, squared plus 14 a plus 49 equals 51 plus 49, or rather  EMBED Equation.DSMT4 parenthesis a, plus 7 close parenthesis, squared equals 100, which has a perfect square on each side. Since  EMBED Equation.DSMT4  a, is greater than zero, the solution  EMBED Equation.DSMT4  a, plus 7 equals 10 is evident. Link back to  HYPERLINK \l "_Question_3." question 3. Link back to  HYPERLINK \l "_Question_4." question 4. Explanation for  HYPERLINK \l "_Question_4." question 4. (Follow link back to original question.) Program: SAT Subscore: Passport to advanced math Calculator Usage: No Objective: Students must complete operations with multiple terms and manipulate an equation to isolate the variable of interest. Difficulty: Medium Key: A Choice A is correct. Multiplying both sides of the equation by the denominators of the rational expressions in the equation gives  EMBED Equation.DSMT4  2y equals 4a minus 4. The student should then divide both sides by 2 to isolate the y variable, yielding the equation  EMBED Equation.DSMT4  y equals 2a minus 2. Choice B is not the correct answer. This equation may result from a student who does not divide both terms by 2 when isolating y in the equation  EMBED Equation.DSMT4  2yequals 4a, minus4. Choice C is not the correct answer. This equation may result from the student not distributing the 4 when multiplying 4 and  EMBED Equation.DSMT4  parenthesis, a minus1, closeparenthesis. Choice D is not the correct answer. This equation may result from solving  EMBED Equation.DSMT4  2y equals 4a minus 4 for a, yielding  EMBED Equation.DSMT4  aequals onehalfy plus1. A student who does not understand the meaning of the variables may then switch them to match the answer choice. Link back to  HYPERLINK \l "_Question_4." question 4. Link back to  HYPERLINK \l "_Question_5." question 5. Explanation for  HYPERLINK \l "_Question_5." question 5. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Passport to advanced math Calculator Usage: No Objective: Students must substitute polynomials into an expression and then simplify the resulting expression by combining like terms. Difficulty: Medium Key: C Choice C is correct. Substituting the expressions equivalent to y and z into  EMBED Equation.DSMT4  2y plus z results in the expression  EMBED Equation.DSMT4  2 parenthesis, xcubed plus 2x plus 5, close parenthesis, plus x squared, plus 7x, plus1. The student must apply the distributive property to multiply  EMBED Equation.DSMT4  xcubed plus 2x plus5 by 2and then combine the like terms in the expression. Choice A is not the correct answer. This answer may result if a student correctly finds 2y in terms of x but does not pay careful attention to exponents when adding to  EMBED Equation.DSMT4  x squared plus 7x plus 1 and then combines the  EMBED Equation.DSMT4  xcubed and  EMBED Equation.DSMT4  xsquaredterms. Choice B is not the correct answer. This answer may result if a student fails to distribute the 2 when multiplying  EMBED Equation.DSMT4  2, parenthesis, x cubed plus 2x plus 5, closeparenthesis. Choice D is not the correct answer. This answer may result from a student finding  EMBED Equation.DSMT4  2, parenthesis, y plus z, close parenthesis, instead of  EMBED Equation.DSMT4  2y plusz. Link back to  HYPERLINK \l "_Question_5." question 5. Link back to  HYPERLINK \l "_Question_6." question 6. Explanation for  HYPERLINK \l "_Question_6." question 6. (Follow link back to original question.) Program: SAT Subscore: Passport to advanced math Calculator Usage: No Objective: Students must demonstrate fluency with the properties of exponents. They must be able to relate fractional exponents to radicals as well as demonstrate understanding of negative exponents. Difficulty: Medium Key: B Choice B is correct. There are multiple ways to approach this problem, but all require an understanding of the properties of exponents. The student may rewrite the equation as  EMBED Equation.DSMT4  the fraction 1 over the square root of a, equalsx and then proceed to solve for a, first by squaring both sides, which gives  EMBED Equation.DSMT4  the fraction 1 over a, equals, xsquared, and then multiplying both sides by a to find  EMBED Equation.DSMT4  1 equals axsquared. Finally, dividing both sides by  EMBED Equation.DSMT4  xsquared isolates the desired variable. Choice A is not the correct answer. This answer may result from a misunderstanding of the properties of exponents. The student may understand that a negative exponent can be translated to a fraction but misapply the fractional exponent. Choice C is not the correct answer. This answer may result from a misunderstanding of the properties of exponents. A student may recognize that an exponent of  EMBED Equation.DSMT4  onehalf is the same as the square root but misapply this information. Choice D is not the correct answer. This answer may result from a misunderstanding of the properties of exponents. The student may recognize that the fractional exponent on a is the same as the square root, and that therefore a can be isolated by squaring both sides. However, the student may not understand how the negative exponent affects the base. Link back to  HYPERLINK \l "_Question_6." question 6. Link back to  HYPERLINK \l "_Question_7." question 7. Explanation for  HYPERLINK \l "_Question_7." question 7. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Passport to advanced math Calculator Usage: No Objective: Students must be able to see structure in expressions and equations and create a new form of an expression which reveals a specific property. Difficulty: Medium Key: C Choice C is correct. The equation  EMBED Equation.DSMT4  y equals, parenthesis, 2x minus 4, close parenthesis, parenthesis, x minus 4, close parenthesis, can be written in vertex form,  EMBED Equation.DSMT4  y equals a, parenthesis, x minus h, close parenthesis, squared, plus k, to reveal the vertex,  EMBED Equation.DSMT4  parenthesis, h comma k, close parenthesis, of the parabola. To put the equation in vertex form, first multiply:  EMBED Equation.DSMT4  parenthesis 2x minus 4 close parenthesis, parenthesis x minus 4 close parenthesis, equals 2x squared minus 8x, minus 4x plus16. Then add like terms,  EMBED Equation.DSMT4 2xsquared, minus 8x, minus 4x, plus 16, equals2xsquared, minus 12x, plus16. The next step is completing the square.  EMBED Equation.DSMT4  y equals 2x squared, minus 12x, plus 16.  EMBED Equation.DSMT4  y equals 2, parenthesis, x squared minus 6x, close parenthesis, plus 16. Isolate the  EMBED Equation.DSMT4 x squared term by factoring.  EMBED Equation.DSMT4  y equals 2, parenthesis, x squared minus 6x, plus 9, minus 9, close parenthesis, plus 16. Make a perfect square in the parentheses.  EMBED Equation.DSMT4  y equals 2, parenthesis, x squared minus 6x, plus 9, close parenthesis, minus 18, plus 16. Move the extra term out of the parentheses.  EMBED Equation.DSMT4  y equals 2, parenthesis, x minus 3, close parenthesis, squared, minus 18, plus 16. Factor inside the parentheses.  EMBED Equation.DSMT4  y equals 2, parenthesis, x minus 3, close parenthesis, squared, minus 2. Simplify the remaining terms. Therefore, the coordinates of the vertex are  EMBED Equation.DSMT4  parenthesis, 3 comma negative2, close parenthesis, which are revealed in (C). Since students are told that all of the equations are equivalent, simply knowing the form that displays the coordinates of the vertex will save them all of these stepsthis is known as seeing structure in the expression or equation. Choice A is not the correct answer. This answer displays the location of the yvalue of the yintercept of the graph  EMBED Equation.DSMT4  parenthesis, zero comma 16, closeparenthesis, as a constant. Choice B is not the correct answer. This answer displays the location of the yvalue of the yintercept of the graph  EMBED Equation.DSMT4  parenthesis, zero comma 16, closeparenthesis, as a constant. Choice D is not the correct answer. This answer displays the location of the xvalue of one of the xintercepts of the graph  EMBED Equation.DSMT4  parenthesis, 2 comma zero, closeparenthesis, as a constant. Link back to  HYPERLINK \l "_Question_7." question 7. Link back to  HYPERLINK \l "_Question_8." question 8. Explanation for  HYPERLINK \l "_Question_8." question 8. (Follow link back to original question.) Program: SAT Subscore: No subscore (additional topics in math) Calculator Usage: No Objective: Students must apply the distributive property on two complex binomials and then simplify the result. Difficulty: Medium Key: D Choice D is correct. Applying the distributive property to multiply the binomials yields the expression  EMBED Equation.DSMT4  98 plus 168i, minus 14i, minus 24isquared. The note in the stem of the question reminds students that  EMBED Equation.DSMT4  iequals the square root of negative 1, therefore  EMBED Equation.DSMT4  i squared equals negative1. Substituting this value into the expression gives the student  EMBED Equation.DSMT4  98 plus 168i, minus 14i, minus, parenthesis, negative24,close parenthesis, and combining like terms results in  EMBED Equation.DSMT4  122plus154i. Choice A is not the correct answer. This answer may result from a combination of errors. The student may not correctly distribute when multiplying the binomials, multiplying only the first terms together and the second terms together. The student may also think that  EMBED Equation.DSMT4  i squared equals1. Choice B is not the correct answer. This answer may result from a misunderstanding of how to multiply binomials. The distributive property is not applied correctly and the student may only multiply the first two terms together and the last two terms together. Choice C is not the correct answer. This answer results from misapplying the statement  EMBED Equation.DSMT4  i equals the square root of negative1. Link back to  HYPERLINK \l "_Question_8." question 8. Link back to  HYPERLINK \l "_Question_9." question 9. Explanation for  HYPERLINK \l "_Question_9." question 9. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: Students must solve an algebraic equation in one variable. Difficulty: Medium Key: B Choice B is correct. Simplifying the numerators yields  EMBED Equation.DSMT4  the fraction whose numerator is 5k plus 3, and whose denominator is 6, equals, the fraction whose numerator is 9 plusk, and whose denominator is 9, and crossmultiplication gives  EMBED Equation.DSMT4  45k plus 27, equals 54 plus 6k. Solving for k yields  EMBED Equation.DSMT4  kequals 9 over13. Choice A is not the correct answer. This value may result from not correctly applying the distributive property on the righthand side, resulting in the expression  EMBED Equation.DSMT4  13 minus 4 minus k, in the numerator. Correctly applying the distributive property yields  EMBED Equation.DSMT4  13 minus, parenthesis, 4 minus k, close parenthesis, equals, 13 minus 4 plus k, in the numerator. Choice C is not the correct answer. This value may result from not correctly applying the distributive property on the lefthand side, resulting in the expression  EMBED Equation.DSMT4 5k plus 2 minus 7. Correctly applying the distributive property yields  EMBED Equation.DSMT4  5 parenthesis, k plus 2, close parenthesis, minus 7, equals, 5k plus3 in the numerator. Choice D is not the correct answer. This value may result from not using the appropriate order of operations when simplifying either numerator. Link back to  HYPERLINK \l "_Question_9." question 9. Link back to  HYPERLINK \l "_Question_10." question 10. Explanation for  HYPERLINK \l "_Question_10." question 10. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: Students must solve a system of linear equations. Difficulty: Medium Key: C Choice C is correct. There are several solution methods possible, but all involve persevering in solving for the two variables and calculating the product. For example, combining like terms in the first equation yields  EMBED Equation.DSMT4  4x minus 4y equals 7 and then multiplying that by 2 gives  EMBED Equation.DSMT4  8x minus 8y equals14. When this transformed equation is added to the second given equation, the yterms are eliminated, leaving an equation in just one variable:  EMBED Equation.DSMT4  9x equals18, or  EMBED Equation.DSMT4  xequals2. Substituting 2 for x in the second equation (one could use either to solve) yields  EMBED Equation.DSMT4  2 plus 8y equals4, which gives  EMBED Equation.DSMT4  yequals onefourth. Finally, the product xy is  EMBED Equation.DSMT4  2times onefourth equalsonehalf. Choice A is not the correct answer. Students who select this option have most likely made a calculation error in transforming the second equation (using  EMBED Equation.DSMT4  negative 4x minus 8y, equals negative16 instead of  EMBED Equation.DSMT4  negative4x minus 32y, equals negative16) and used it to eliminate the xterms. Choice B is not the correct answer. This is the value of y for the solution of the system, but it has not been put back into the system to solve for x to determine the productxy. Choice D is not the correct answer. Not understanding how to eliminate a variable when solving a system, a student may have added the equations  EMBED Equation.DSMT4  4x minus 4y equals7 and  EMBED Equation.DSMT4  xplus 8y equals 4 to yield  EMBED Equation.DSMT4  5x plus 4y equals11. From here, a student may mistakenly simplify the lefthand side of this resulting equation to yield  EMBED Equation.DSMT4  9xy equals11 and then proceed to use division by 9 on both sides in order to solve for xy. Link back to  HYPERLINK \l "_Question_10." question 10. Link back to  HYPERLINK \l "_Question_11." question 11. Explanation for  HYPERLINK \l "_Question_11." question 11. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: Students must use algebraic operations to determine the value of a new algebraic expression. Difficulty: Medium Key: 24 A student may find the solution to this problem by noticing the structure of the given equation and seeing that multiplying both sides of the equation  EMBED Equation.DSMT4  onehalfx, plus onethirdy, equals4, by 6 to clear fractions from the equation yields  EMBED Equation.DSMT4  3x plus 2y equals24. Link back to  HYPERLINK \l "_Question_11." question 11. Link back to  HYPERLINK \l "_Question_12." question 12. Explanation for  HYPERLINK \l "_Question_12." question 12. (Follow link back to original question.) Program: SAT Subscore: No subscore (additional topics in math) Calculator Usage: No Objective: Students must understand radian measure and have a conceptual understanding of trigonometric relationships. Difficulty: Hard Key: C Choice C is correct. Sine and cosine are related by the equation:  EMBED Equation.DSMT4  sineofx equals cosine of the quantity, the fraction pi over 2, minusx. Therefore,  EMBED Equation.DSMT4  sine of the fraction pi over 5, equals, cosine of the quantity, the fraction pi over 2, minus, the fraction pi over 5, which reduces to  EMBED Equation.DSMT4  cosineof the fraction 3 pi over10. Choice A is not the correct answer. This answer may result from a misunderstanding about trigonometric relationships. A student may think that cosine is the opposite function of sine, and therefore think that the negative of the cosine of an angle is equivalent to the sine of thatangle. Choice B is not the correct answer. This answer may result from a misunderstanding of the unit circle and how it relates to trigonometric expressions. A student may think that, on a coordinate grid, the negative sign only changes the orientation of the triangle formed, not the value of the trigonometric expression. Choice D is not the correct answer. The student mistakenly remembers the relationship between sine and cosine and adds  EMBED Equation.DSMT4  the fraction pi over 2 to the angle measure instead of subtracting the angle measure from  EMBED Equation.DSMT4  the fraction pi over 2. Link back to  HYPERLINK \l "_Question_12." question 12. Link back to  HYPERLINK \l "_Question_13." question 13. Explanation for  HYPERLINK \l "_Question_13." question 13. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: Heart of algebra Calculator Usage: No Objective: In addition to solving systems of linear equations that have a solution, students must be familiar with systems that have no solution or an infinite number of solutions. Difficulty: Hard Key: C Choice C is correct. If the system of equations has no solution, the graphs of the equations in the xyplane are parallel lines. To be parallel, the lines must have the same slope, and this will be true if the expression  EMBED Equation.DSMT4  a, x minus 3y is a multiple of the expression  EMBED Equation.DSMT4  onehalfx minus onefourthy. Since  EMBED Equation.DSMT4  negative3y equals 12, parenthesis, negative onefourthy, close parenthesis, the expression  EMBED Equation.DSMT4  a,xminus3y would have to be 12times the expression  EMBED Equation.DSMT4  onehalfx minus onefourthy. This means  EMBED Equation.DSMT4  a,x equals12, parenthesis, onehalfx, close parenthesis, so  EMBED Equation.DSMT4  a, equals6. The resulting system is  EMBED Equation.DSMT4  onehalfx minus onefourthy, equals5 and  EMBED Equation.DSMT4  6x minus 3y equals20, which is equivalent to  EMBED Equation.DSMT4  6x minus 3y equals60 and  EMBED Equation.DSMT4  6x minus 3y equals20, which has no solution. Choice A is not the correct answer. This may result from the misconception that ifeach equation in a system has the same xcoefficient, the system cannot have a solution. But if  EMBED Equation.DSMT4  a, equals onehalf, subtracting the two equations eliminatesx and produces a solution to the system. Choice B is not the correct answer. This may result from trying to make the second equation in the system a multiple of the first by looking at the ratio of the constants on the right sides,  EMBED Equation.DSMT4  the fraction 20 over 5, and wrongly concluding that the second equation must be 4 times the first, which would give  EMBED Equation.DSMT4  a, equals4, parenthesis, onehalf, close parenthesis, or EMBED Equation.DSMT4  a, equals 2. But the two equations in a system are multiples only if the system has infinitely many solutions, not if the system has no solution. Choice D is not the correct answer. The student may have found the factor, 12, that multiplies the left side of the first equation to yield the left side of the second, but then neglected to find  EMBED Equation.DSMT4  a, equals 12, parenthesis, onehalf, closeparenthesis, or  EMBED Equation.DSMT4  a, equals6. Link back to  HYPERLINK \l "_Question_13." question 13. Link back to  HYPERLINK \l "_Question_14." question 14. Explanation for  HYPERLINK \l "_Question_14." question 14. (Follow link back to original question.) Program: SAT Subscore: Passport to advanced math Calculator Usage: No Objective: Students must interpret an equation that models a realworld situation and be able to interpret the whole expression (or specific parts) in terms of its context. Difficulty: Hard Key: B Choice B is correct. From the description given,  EMBED Equation.DSMT4  onefifth is the portion of the job that the two printers, working together, can complete in one hour, and each term in the sum on the left side is the part of this  EMBED Equation.DSMT4  onefifth of the job that one of the printers contributes. Since one of the printers is twice as fast as the other,  EMBED Equation.DSMT4  thefraction 2overx describes the portion of the job that the faster printer is able to complete in one hour and  EMBED Equation.DSMT4  the fraction one over x describes the portion of the job that the slower printer is able to complete in one hour. Choice A is not the correct answer. The student may have not seen that in this context, the rates (that is, the work completed in a fixed time) of the printers can be added to get the combined rate, but the times it takes each printer to complete the job cannot be added to get the time for both printers working together, since the time for printers working together is less than, not greater than, the times for each printer alone. Hence the terms in the sum cannot refer to hours worked. In fact, the time it would take the slower printer to complete the whole job is xhours. Choice C is not the correct answer. The student may have seen that  EMBED Equation.DSMT4  the fraction oneoverx is the smaller term in the sum, wrongly concluded that the smaller term must apply to the faster printer, and then assumed the 2 in the numerator of the second term implies the equation describes work completed in 2hours. In fact, the portion of the job that the faster printer could complete in 2hours is  EMBED Equation.DSMT4  parenthesis, 2, close parenthesis, parenthesis, the fraction 2overx, close parenthesis, equals, the fraction 4 over x. Choice D is not the correct answer. The student may have correctly seen that the value  EMBED Equation.DSMT4  onefifth on the right side refers to the portion of the job completed, but not seen that in this context, the rates (that is, the work completed in a fixed time) of the printers can be added to get the combined rate, but the times it takes each printer to complete the job cannot be added to get the time for both printers working together. Hence the terms in the sum cannot refer to hours worked. In fact, the time it takes the slower printer to complete  EMBED Equation.DSMT4  onefifth of the job is  EMBED Equation.DSMT4  thefractionxover5 hours. Link back to  HYPERLINK \l "_Question_14." question 14. Link back to  HYPERLINK \l "_Question_15." question 15. Explanation for  HYPERLINK \l "_Question_15." question 15. (Follow link back to original question.) Program: SAT Subscore: No subscore (additional topics in math) Calculator Usage: No Objective: Students must make use of properties of circles and parallel lines in an abstract setting. Difficulty: Hard Key: D Choice D is correct. This represents the length of the distance between the chord and the diameter, using a radius of the circle to create a triangle, and then the Pythagorean theorem to solve correctly:  EMBED Equation.DSMT4  rsquared equals xsquared plus, parenthesis, twothirdsr, close parenthesis, squared, where r represents the radius of the circle and x represents the distance between the chord and thediameter. Choice A is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have tried to use the circumference formula to determine the distance rather than making use of the radius of the circle to create a triangle. Choice B is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have tried to use the circumference formula to determine the distance rather than making use of the radius of the circle to create a triangle. Choice C is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have made a triangle within the circle, using a radius to connect the chord and the diameter, but then may have mistaken the triangle for a 454590 triangle and tried to use this relationship to determine the distance. Link back to  HYPERLINK \l "_Question_15." question 15. Link back to  HYPERLINK \l "_Question_16." question 16. Explanation for  HYPERLINK \l "_Question_16." question 16. (Follow link back to original question.) Program: SAT Subscore: No subscore (additional topics in math) Calculator Usage: No Objective: Students must be fluent with radian measure and have a conceptual understanding of trigonometric relationships. Difficulty: Hard Key: B Choice B is correct. If an angle with radian measure x such that  EMBED Equation.DSMT4  the fraction piover2 is less thanx, and x is less than pi is placed in standard position, its terminal side will fall in Quadrant2, and  EMBED Equation.DSMT4  sinex equals,a, will be the ycoordinate of the pointP where its terminal side intersects the unit circle. If  EMBED Equation.DSMT4  sinew equals negativea, then when the angle with radian measure w is placed in standard position, its terminal side will intersect the unit circle at a point with ycoordinate equal to  EMBED Equation.DSMT4  negativea. There are two such points on the unit circle: the reflection ofP across the xaxis, which would correspond to an angle with radian measure  EMBED Equation.DSMT4  negativex (and also with radian measures...  EMBED Equation.DSMT4  negative6pi minusx, negative4pi minusx, negative2pi minusx, 2pi minusx, 4pi minusx, 6pi minusx...); and the reflection of P through the origin, which would correspond to an angle with radian measure  EMBED Equation.DSMT4  x minus pi (and also with radian measures...  EMBED Equation.DSMT4  x minus 5pi, x minus 3pi, x pluspi, xplus 3pi, x plus5pi...). Thus, of the choices given, only  EMBED Equation.DSMT4  xminuspi could be the value ofw. Choice A is not the correct answer. In general,  EMBED Equation.DSMT4  sine of negativex, equals negativesinex and  EMBED Equation.DSMT4  sine of the quantity xpluspi, equals negative sinex, so  EMBED Equation.DSMT4  sine of the quantity piminusx, equals negativesine of negativex, equals negative, parenthesis, negativesinex, close parenthesis, equals sinex. Therefore,  EMBED Equation.DSMT4  sine of the quantity piminusx, equalsa, not  EMBED Equation.DSMT4  negativea. Choice C is not the correct answer. In general,  EMBED Equation.DSMT4  sine of the quantity 2pi plusx, equals sinex, so  EMBED Equation.DSMT4  sine of the quantity 2pi plusx, equalsa, not  EMBED Equation.DSMT4  negativea. Choice D is not the correct answer. In general,  EMBED Equation.DSMT4  sine of the quantity xminus 2pi, equals sinex, so  EMBED Equation.DSMT4  sine of the quantity xminus 2pi, equalsa, not  EMBED Equation.DSMT4  negativea. Link back to  HYPERLINK \l "_Question_16." question 16. Link back to  HYPERLINK \l "_Question_17." question 17. Explanation for  HYPERLINK \l "_Question_17." question 17. (Follow link back to original question.) Program: SAT, PSAT/NMSQT, PSAT 10 Subscore: No subscore (additional topics in math) Calculator Usage: No Objective: Students must determine a circle property given the equation of the circle. Difficulty: Hard Key: 26 Completing the square yields the equation  EMBED Equation.DSMT4  parenthesis, xminus3, close parenthesis, squared, plus, parenthesis, yplus4, close parenthesis, squared, equals 169, -. ,4D#&0d*+UXh- h8hF h-hFh}hF0JOJQJ h}hFhYFhhEHhFPJnHo(tHhFPJnHtH hF5 hhFh63hFEHh-hFCJaJhF7.\ p H ? A G %^+"  ^ `gd#gdaEgdaEgdogdogdogdogdo"X_0W e 6":"<"@"B""8#$F$ x^gdaEgdogdogdogdo3./0 !ԽԲԲԲԲԪԛԂxh>hdq6@h Eh E@<B*aJph# h>hdq6@2h>Qh E@(B*aJph# h#hdq5h4y h4yh4yhuh-hPhYF hKhdqhdqhFPJnHo(tHhFPJnHtHh5whF5hF h%5khF/! ! ! ! !!!"!#!"""" 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Understanding this form results in the equation  EMBED Equation.DSMT4  rsquared equals169, which when solved for r gives the value of the radius as 13. Diameter is twice the value of the radius; therefore, the diameter is26. Link back to  HYPERLINK \l "_Question_17." question 17. Link back to  HYPERLINK \l "_Question_18." question 18. Explanation for  HYPERLINK \l "_Question_18." question 18. (Follow link back to original question.) Program: SAT Subscore: Passport to advanced math Calculator Usage: No Objective: Students must solve a rational equation in one variable. Difficulty: Hard Key: 5 or 7 Students should look for the best solution methods for solving rational equations before they begin. Looking for structure and common denominators will prove very useful at the onset, and will help prevent complex computations that do not lead to a solution. In this problem, multiplying both sides of the equation by the common denominator  EMBED Equation.DSMT4  parenthesis, xplusone, close parenthesis, parenthesis, xminusone, close parenthesis, yields  EMBED Equation.DSMT4  24, parenthesis, xminusone, closeparenthesis, minus 12, parenthesis, xplusone, closeparenthesis, equals, parenthesis, xplusone, close parenthesis, parenthesis, xminusone, close parenthesis. Multiplication and simplification then yields  EMBED Equation.DSMT4  12xminus36 equals, xsquared minusone, or  EMBED Equation.DSMT4  xsquared minus 12x plus35 equalszero. Factoring the quadratic gives  EMBED Equation.DSMT4  parenthesis, xminus 5, close parenthesis, parenthesis, xminus7, close parenthesis, equals zero, so the solutions occur at  EMBED Equation.DSMT4  xequals 5 and  EMBED Equation.DSMT4 xequals7, both of which should be checked in the original equation to ensure that they are not extraneous. In this case, both values are solutions. Link back to  HYPERLINK \l "_Question_18." question 18. This is the end of Mathematics Sample Question Answers and Explanations.     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