The Redesigned SAT®
Mathematics Sample Sets
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Heading level 3: section titles
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The Redesigned S A T®: Mathematics—No Calculator
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Tables
Some questions may include tables. Use the table-navigation features of your
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Mathematical Equations and Expressions
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Mathematics—No 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 f  x  f of x is a real number.
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Begin skippable figure descriptions.
The figure presents information for your reference in solving some of
the problems.
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Reference figure 1 is a circle with radius r. Two equations are presented below
reference figure 1.
A equals pi times the square of r.
C equals 2 pi r.
Reference figure 2 is a rectangle with length ℓ and width w. An equation is
presented below reference figure 2.
A equals ℓ w.
Reference figure 3 is a triangle with base b and height h. An equation is presented
below reference figure 3.
A equals one-half b h.
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 figure 4.
c squared equals a squared plus b squared.
Special Right Triangles
Reference figure 5 is a right triangle with a 30-degree angle and a 60-degree angle.
The side opposite the 30-degree angle is labeled x. The side opposite the 60-degree
angle is labeled x times the square root of 3. The side opposite the right angle is
labeled 2 x.
Reference figure 6 is a right triangle with two 45-degree angles. Two sides are
each labeled s. The side opposite the right angle is labeled s times the square root
of 2.
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Reference figure 7 is a rectangular solid whose base has length ℓ and width w and
whose height is h. An equation is presented below reference figure 7.
V equals ℓ w h.
Reference figure 8 is a right circular cylinder whose base has radius r and whose
height is h. An equation is presented below reference figure 8.
V equals pi times the square of r times h.
Reference figure 9 is a sphere with radius r. An equation is presented below
reference figure 9.
V equals four-thirds pi times the cube of r.
Reference figure 10 is a cone whose base has radius r and whose height is h.
An equation is presented below reference figure 10.
V equals one-third times pi times the square of r times h.
Reference figure 11 is an asymmetrical pyramid whose base has length ℓ and
width w and whose height is h. An equation is presented below reference
figure 11.
V equals one-third ℓ w h.
Additional Reference Information
The number of degrees of arc in a circle is 360.
The number of radians of arc in a circle is 2 pi.
The sum of the measures in degrees of the angles of a triangle is 180.
End skippable figure descriptions.
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For student-produced 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.
1
5. Mixed numbers such as 3 three and one-half must be recorded as
2
1
3.5 three point five or 7 / 2. seven slash two. (If 3 three and one-half is entered
2
into the grid as
, three, one, slash, two, it will be interpreted as
31
1
, thirty-one halves, not 3 . three and one-half).
2
2
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.
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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 seven-twelfths, it should be
recorded as follows. Enter seven in the first space, the fraction bar (a slash) 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 two point five, it could be
recorded as follows. Enter two in the second space, the decimal point in the third
space, and five in the fourth space. Only three spaces would be used in
this example.
End skippable figure description.
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Example 3
Beging skippable figure description.
Example 3: Acceptable ways to record two-thirds are: two slash three, point
six six six, and point six six seven.
End skippable figure description.
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Example 4
Note: You may start your answers in any column, space permitting. Columns you
don’t 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.
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Mathematics Sample Questions
Question 1 is based on the following figure, which presents the
graph of line ℓ in the x y-plane.
Begin skippable figure description.
The figure presents the graph of line ℓ in the x y-plane. The x- and y-axes have
tick marks. The fifth tick marks from the origin in the positive and negative
directions on both axes are labeled 5 and negative 5, respectively.
Six points are marked on line ℓ. The following are the coordinates of the six points
on line ℓ.
 2,10 parenthesis, negative 2 comma 10, close parenthesis
 0,7  parenthesis, zero comma 7, close parenthesis
 2,4  parenthesis, 2 comma 4, close parenthesis
 4,1 parenthesis, 4 comma 1, close parenthesis
 6, 2 parenthesis, 6 comma negative 2, close parenthesis
8, 5 parenthesis, 8 comma negative 5, close parenthesis
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From left to right, line ℓ extends from quadrant 2 downward to the right through
quadrant 1 and into quadrant 4. Line ℓ intersects the y-axis at value 7 and
intersects the x-axis between values 4 and 5.
End skippable figure description.
Question 1.
If line ℓ is translated up 5 units and right 7 units, then what is the slope of the
new line?
2
negative two-fifths
5
3
B.  negative three-halves
2
8
C.  negative eight-ninths
9
11
D.  negative eleven-fourteenths
14
A. 
Answer and Explanation. (Follow link to explanation of question 1.)
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Question 2.
The mean number of students per classroom, y, at Central High School can be
estimated using the equation y  0.863x  27.227, y equals 0.8636 x plus 27.227,
where x represents the number of years since 2004 and x  10. x is less than or
equal to 10. 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.
Answer and Explanation. (Follow link to explanation of question 2.)
Question 3.
If a 2  14a  51 a, squared plus 14 a, equals 51 and a  0, a, is greater than zero,
what is the value of a  7? a, plus 7?
Answer and Explanation. (Follow link to explanation of question 3.)
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Question 4.
If
2
4
 the fraction whose numerator is 2, and whose denominator is a, minus
a 1 y
1, equals, the fraction whose numerator is 4 and whose denominator is y, where
y  0 y does not equal zero and a  1, a does not equal 1, what is y in terms of a?
A. y  2a  2 y equals 2 a, minus 2.
B. y  2a  4 y equals 2 a, minus 4.
C. y  2a 
1
y equals 2 a, minus one-half.
2
1
D. y  a  1 y equals one-half a, plus 1.
2
Answer and Explanation. (Follow link to explanation of question 4.)
Question 5.
If y  x3  2 x  5 y equals x cubed, plus 2 x, plus 5 and z  x2  7 x  1, z equals
x squared, plus 7 x, plus 1, what is 2y  z 2 y plus z in terms of x?
A. 3x3  11x  11 3 x cubed, plus 11 x, plus 11.
B. 2 x3  x 2  9 x  6 2 x cubed, plus x squared, plus 9 x, plus 6.
C. 2 x3  x 2  11x  11 2 x cubed, plus x squared, plus 11 x, plus 11.
D. 2 x3  2 x 2  18 x  12 2 x cubed, plus 2 x squared, plus 18 x, plus 12.
Answer and Explanation. (Follow link to explanation of question 5.)
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Question 6.
1
If a 2  x, a, to the power of negative one-half, equals x, where a  0 a is greater

than zero and x  0, x is greater than zero, which of the following equations gives
a in terms of x ?
1
a, equals the fraction 1 over the square root of x
x
1
B. a 
a, equals the fraction 1 over x squared
x2
A. a 
C. a  x a, equals the square root of x
D. a   x 2 a, equals negative x squared
Answer and Explanation. (Follow link to explanation of question 6.)
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Question 7.
The graph of y   2 x  4  x  4  y equals, parenthesis 2 x minus 4 close
parenthesis, parenthesis x minus 4 close parenthesis is a parabola in the x y-plane.
In which of the following equivalent equations do the x- and y-coordinates of the
vertex of the parabola appear as constants or coefficients?
A. y 5 2 x 2  12 x  16 y equals 2 x squared minus 12 x plus 16.
B. y 5 2 x  x  6   16 y equals 2 x, parenthesis x minus 6 close parenthesis,
plus 16.
C. y 5 2  x  32 1  2  y equals 2, parenthesis, x minus 3, close parenthesis,
squared, plus, parenthesis, negative 2, close parenthesis.
D. y 5  x  2  2 x  8 y equals parenthesis x minus 2 close parenthesis,
parenthesis 2 x minus 8 close parenthesis.
Answer and Explanation. (Follow link to explanation of question 7.)
Question 8.
Which of the following is equal to 14  2i  7  12i  ? parenthesis, 14 minus 2 i,
close parenthesis, parenthesis, 7 plus 12 i, close parenthesis? (Note: i  1)
i equals the square root of negative 1.)
A. 74
B. 122
C. 78  154i 74 plus 154 i
D. 122  154i 122 plus 154 i
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Answer and Explanation. (Follow link to explanation of question 8.)
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Question 9.
5  k + 2  - 7 13 -  4 - k 
the fraction whose numerator is 5, parenthesis, k
=
6
9
plus 2, 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 is 9.
In the equation above, what is the value of k ?
A.
9
9 over 17.
17
B.
9
9 over 13.
13
C.
33
33 over 17.
17
D.
33
33 over 13.
13
Answer and Explanation. (Follow link to explanation of question 9.)
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Question 10.
4 x - y = 3 y + 7 4 x minus y, equals 3 y plus 7.
x + 8 y = 4 x plus 8 y equals 4.
Based on the system of equations above, what is the value of the product x y ?
A. -
3
negative three-halves
2
1
one-fourth
4
1
C.
one-half
2
11
D.
eleven-ninths
9
B.
Answer and Explanation. (Follow link to explanation of
question 10.)
Question 11.
1
1
x  y  4, one-half x, plus one-third y, equals 4, what is the value of
2
3
3x  2 y ? 3 x plus 2 y?
If
Answer and Explanation. (Follow link to explanation of
question 11.)
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Question 12.
 
Which of the following is equal to sin   ? sine of the fraction pi over 5?
5
 
A.  cos   negative cosine of the fraction pi over 5.
5
 
B.  sin   negative sine of the fraction pi over 5.
5
 3 
C. cos   cosine of the fraction 3 pi over 10.
 10 
 7 
D. sin 
 sine of the fraction 7 pi over 10.
 10 
Answer and Explanation. (Follow link to explanation of
question 12.)
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Question 13.
1
1
x  y  5 one-half x minus one-fourth y, equals 5.
2
4
ax  3 y  20 a x minus 3 y equals 20.
In the system of linear equations above, a is a constant. If the system has no
solution, what is the value of a?
1
one-half
2
B. 2
C. 6
D. 12
A.
Answer and Explanation. (Follow link to explanation of
question 13.)
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Question 14.
1 2 1
  the fraction one over x, plus the fraction 2 over x, equals one-fifth.
x x 5
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 5 hours. The equation above represents the situation described. Which of
1
the following describes what the expression the fraction one over x represents in
x
this equation?
A. The time, in hours, that it takes the slower printer to complete the printing job
alone
B. The portion of the job that the slower printer would complete in one hour
C. The portion of the job that the faster printer would complete in two hours
1
D. The time, in hours, that it takes the slower printer to complete one-fifth of
5
the printing job
Answer and Explanation. (Follow link to explanation of
question 14.)
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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 and D.
End skippable figure description.
Question 15.
The semicircle above has a radius of r inches, and chord CD line segment CD is
parallel to the diameter AB. line segment AB. If the length of CD line segment
2
CD is
two-thirds of the length of AB, line segment AB, what is the distance
3
between the chord and the diameter in terms of r?
1
 r one-third pi times r.
3
2
B.  r two-thirds pi times r.
3
A.
C.
2
r the fraction the square root of 2 over 2, times r.
2
D.
5
r the fraction the square root of 5 over 3, times r.
3
Answer and Explanation. (Follow link to explanation of
question 15.)
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Question 16.
It is given that sin x  a, sine x equals a, where x is the radian measure of an angle
and

2
 x   . the fraction pi over 2, is less than x, and x is less than pi.
If sin w  a, sine w equals negative a, which of the following could be the value
of w?
A.
B.
C.
D.
  x pi minus x
x   x minus pi
2  x 2 pi plus x
x  2 x minus 2 pi
Answer and Explanation. (Follow link to explanation of
question 16.)
Question 17.
x 2  y 2  6 x  8 y  144 x squared plus y squared, minus 6 x plus 8 y, equals 144.
The equation of a circle in the x y-plane is shown above. What is the diameter of
the circle?
Answer and Explanation. (Follow link to explanation of
question 17.)
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Question 18.
24
12

 1? the fraction whose
x 1 x 1
numerator is 24 and whose denominator is x plus one, minus, the fraction whose
numerator is 12 and whose denominator is x minus one, equals one?
What is one possible solution to the equation
Answer and Explanation. (Follow link to explanation of
question 18.)
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Mathematics Sample Question Answers
and Explanations
The following are explanations of answers to sample questions 1 through 18. 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+left-click (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+left arrow.
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Explanation for question 1. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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 y-coordinates divided by the difference in the x-coordinates for any two
3
points on the line. Using the points indicated, the slope is  .
2
negative three-halves. 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
3
slope of the image is  . negative three-halves.
2
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
2 31 5 . the fraction whose numerator is
2 2 1 7 
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.
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Choice D is not the correct answer. This answer may result from adding
5
7
five-sevenths to the slope of the line.
Link back to question 1.
Link back to question 2.
Explanation for question 2. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 10
Subscore: Heart of algebra
Calculator Usage: No
Objective: Students must interpret the slope of an equation in relation to the
real-world 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 y  mx  b, y equals
m x plus b, the coefficient of the x-term (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 by y) changes per year (represented by x).
Choice A is not the correct answer. This answer may result from a
misunderstanding of slope and y-intercept. The y-intercept of the equation
represents the estimated mean number of students per classroom in 2004.
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Choice B is not the correct answer. This answer may result from a
misunderstanding of the limitations of the model. Students may see that x  x is
less than or equal to 10 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 than 1 represents a decreasing function.
Link back to question 2.
Link back to question 3.
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Sample Items
Explanation for question 3. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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 a 2  14a  51 a, squared plus 14 a, equals 51
as a 2  14a  51  0 a, squared plus 14 a, minus 51 equals zero and then factoring.
Since the coefficient of a is 14 and the constant term is 51, negative 51, factoring
a 2  14a  51  0 a, squared plus 14 a, minus 51 equals zero requires writing 51as
the product of two numbers that differ by 14. This is 51  (3)(17), 51 equals
parenthesis 3 close parenthesis, parenthesis 17 close parenthesis, which gives the
factorization a 2  14a  51  (a  17)(a  3)  0. a, squared plus 14 a, minus 51
equals parenthesis, a plus 17, close parenthesis, parenthesis, a minus 3, close
parenthesis, equals zero. The possible values of a are a  17 a, equals
negative 17 and a  3. a, equals 3. Since it is given that a  0, a, is greater than
zero, it must be true that a  3. a, equals 3. Thus, the value of a  7 a, plus 7 is
3  7  10. 3 plus 7 equals 10.
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Sample Items
A student could also use the quadratic formula to find the possible values of
14  142  4(1)(51) 14  196  (204) 14  400 14  20
a: a 



.
2(1)
2
2
2
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 is 2.
14  20
 17 a, equals the fraction whose
2
numerator is negative 14 minus 20, and whose denominator is 2, equals
14  20
 3. a, equals the fraction whose numerator is
negative 17 and a 
2
negative 14 plus 20, and whose denominator is 2, equals 3. Again, since it is given
that a  0, a, is greater than zero, it must be true that a  3. a, equals 3. Thus, the
The possible values of a are a 
value of a  7 a, plus 7 is 3  7  10. 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
a 2  14a  49  51  49, a, squared plus 14 a plus 49 equals 51 plus 49, or rather
(a  7)2  100, parenthesis a, plus 7 close parenthesis, squared equals 100, which
has a perfect square on each side. Since a  0, a, is greater than zero, the solution
a  7  10 a, plus 7 equals 10 is evident.
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Sample Items
Link back to question 3.
Link back to question 4.
Explanation for question 4. (Follow link back to original question.)
Program: S A T
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 2 y  4a  4. 2 y equals 4 a minus 4.
The student should then divide both sides by 2 to isolate the y variable, yielding
the equation y  2a  2. y equals 2 a 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 2 y  4a  4.
2 y equals 4 a, minus 4.
Choice C is not the correct answer. This equation may result from the student not
distributing the 4 when multiplying 4 and  a  1 . parenthesis, a minus 1,
close parenthesis.
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Sample Items
Choice D is not the correct answer. This equation may result from solving
1
2 y  4a  4 2 y equals 4 a minus 4 for a, yielding a  y  1. a equals one-half y
2
plus 1. A student who does not understand the meaning of the variables may then
switch them to match the answer choice.
Link back to question 4.
Link back to question 5.
Explanation for question 5. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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


2y  z 2 y plus z results in the expression 2 x3  2 x  5  x 2  7 x  1. 2
parenthesis, x cubed plus 2 x plus 5, close parenthesis, plus x squared, plus 7 x,
plus 1. The student must apply the distributive property to multiply x3  2 x  5
x cubed plus 2 x plus 5 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 2 y in terms of x but does not pay careful attention to exponents when adding
to x 2  7 x  1 x squared plus 7 x plus 1 and then combines the x 3 x cubed and x 2
x squared terms.
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Sample Items
Choice B is not the correct answer. This answer may result if a student fails to


distribute the 2 when multiplying 2 x3  2 x  5 . 2, parenthesis, x cubed plus 2 x
plus 5, close parenthesis.
Choice D is not the correct answer. This answer may result from a student finding
2  y  z  2, parenthesis, y plus z, close parenthesis, instead of 2 y  z. 2 y plus z.
Link back to question 5.
Link back to question 6.
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Sample Items
Explanation for question 6. (Follow link back to original question.)
Program: S A T
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
1
the equation as
 x the fraction 1 over the square root of a, equals x and then
a
1
proceed to solve for a, first by squaring both sides, which gives  x 2 , the
a
fraction 1 over a, equals, x squared, and then multiplying both sides by a to find
1  ax 2. 1 equals a x squared. Finally, dividing both sides by x 2 x squared 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
1
exponent of
one-half is the same as the square root but misapply this
2
information.
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Sample Items
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 question 6.
Link back to question 7.
Explanation for question 7. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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 y 5  2 x  4  x  4  y equals, parenthesis, 2 x
minus 4, close parenthesis, parenthesis, x minus 4, close parenthesis, can be written
in vertex form, y  a x  h 2  k , y equals a, parenthesis, x minus h, close


parenthesis, squared, plus k, to reveal the vertex,  h,k  , parenthesis, h comma k,
close parenthesis, of the parabola. To put the equation in vertex form, first
multiply: 2 x  4 x  4  2 x2  8x  4 x  16. parenthesis 2 x minus 4 close



parenthesis, parenthesis x minus 4 close parenthesis, equals 2 x squared minus 8 x,
minus 4 x plus 16. Then add like terms,
2 x 2  8 x  4 x  16  2 x 2  12 x  16. 2 x squared, minus 8 x, minus 4 x, plus 16,
equals 2 x squared, minus 12 x, plus 16. The next step is completing the square.
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Sample Items
y  2 x 2  12 x  16 y equals 2 x squared, minus 12 x, plus 16.


y  2 x 2  6 x  16 y equals 2, parenthesis, x squared minus 6 x, close
parenthesis, plus 16. Isolate the x 2 x squared term by factoring.


y  2 x 2  6 x  9  9  16 y equals 2, parenthesis, x squared minus 6 x, plus 9,
minus 9, close parenthesis, plus 16. Make a perfect square in the parentheses.


y  2 x 2  6 x  9  18  16 y equals 2, parenthesis, x squared minus 6 x, plus 9,
close parenthesis, minus 18, plus 16. Move the extra term out of the parentheses.
y  2  x  32  18  16 y equals 2, parenthesis, x minus 3, close parenthesis,
squared, minus 18, plus 16. Factor inside the parentheses.
y  2  x  32  2 y equals 2, parenthesis, x minus 3, close parenthesis, squared,
minus 2. Simplify the remaining terms.
Therefore, the coordinates of the vertex are  3,   , parenthesis, 3 comma
negative 2, 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 steps—this is known as
“seeing structure in the expression or equation.”
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Sample Items
Choice A is not the correct answer. This answer displays the location of the y-value
of the y-intercept of the graph  0,16
  parenthesis, zero comma 16,
close parenthesis, as a constant.
Choice B is not the correct answer. This answer displays the location of the y-value
  parenthesis, zero comma 16,
of the y-intercept of the graph  0,16
close parenthesis, as a constant.
Choice D is not the correct answer. This answer displays the location of the x-value
of one of the x-intercepts of the graph  2,  parenthesis, 2 comma zero,
close parenthesis, as a constant.
Link back to question 7.
Link back to question 8.
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Sample Items
Explanation for question 8. (Follow link back to original question.)
Program: S A T
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 98  168i  14i  24i 2. 98 plus 168 i, minus 14 i, minus
24 i squared. The note in the stem of the question reminds students that i  1,
i equals the square root of negative 1, therefore i 2 5 2 1. i squared equals
negative 1. Substituting this value into the expression gives the student
98  168i  14i   24  , 98 plus 168 i, minus 14 i, minus, parenthesis,
negative 24, close parenthesis, and combining like terms results in 122  154i.
122 plus 154 i.
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 i 2  1. i squared equals 1.
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.
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Sample Items
Choice C is not the correct answer. This answer results from misapplying the
statement i  1. i equals the square root of negative 1.
Link back to question 8.
Link back to question 9.
Explanation for question 9. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 10
Subscore: Heart of algebra
Calculator Usage: No
Objective: Students must solve an algebraic equation in one variable.
Difficulty: Medium
Key: B
5k + 3 9 + k
=
, the
6
9
fraction whose numerator is 5 k plus 3, and whose denominator is 6, equals, the
fraction whose numerator is 9 plus k, and whose denominator is 9, and
cross-multiplication gives 45 k + 27 = 54 + 6 k. 45 k plus 27, equals 54 plus 6 k.
Choice B is correct. Simplifying the numerators yields
Solving for k yields k =
9
. k equals 9 over 13.
13
Choice A is not the correct answer. This value may result from not correctly
applying the distributive property on the right-hand side, resulting in the
expression 13 - 4 - k 13 minus 4 minus k, in the numerator. Correctly applying
the distributive property yields 13 -  4 - k  = 13 - 4 + k 13 minus, parenthesis, 4
minus k, close parenthesis, equals, 13 minus 4 plus k, in the numerator.
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Sample Items
Choice C is not the correct answer. This value may result from not correctly
applying the distributive property on the left-hand side, resulting in the expression
5 k + 2 - 7. 5 k plus 2 minus 7. Correctly applying the distributive property yields
5  k + 2  - 7 = 5 k + 3 5 parenthesis, k plus 2, close parenthesis, minus 7, equals,
5 k plus 3 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 question 9.
Link back to question 10.
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Sample Items
Explanation for question 10. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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 4 x - 4 y = 7 4 x minus
4 y equals 7 and then multiplying that by 2 gives 8 x - 8 y = 14. 8 x minus 8 y
equals 14. When this transformed equation is added to the second given equation,
the y-terms are eliminated, leaving an equation in just one variable: 9x = 18, 9 x
equals 18, or x = 2. x equals 2. Substituting 2 for x in the second equation (one
could use either to solve) yields 2 + 8 y = 4, 2 plus 8 y equals 4, which gives
1
1 1
. y equals one-fourth. Finally, the product x y is 2  = . 2 times
4
4 2
one-fourth equals one-half.
y=
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
- 4 x - 8 y = - 16 negative 4 x minus 8 y, equals negative 16 instead of
- 4 x - 32 y = - 16) negative 4 x minus 32 y, equals negative 16) and used it to
eliminate the x-terms.
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
product x y.
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Sample Items
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 4 x - 4 y = 7 4 x
minus 4 y equals 7 and x + 8 y = 4 x plus 8 y equals 4 to yield 5x + 4 y = 11. 5 x
plus 4 y equals 11. From here, a student may mistakenly simplify the left-hand side
of this resulting equation to yield 9 xy = 11 9 x y equals 11 and then proceed to use
division by 9 on both sides in order to solve for x y.
Link back to question 10.
Link back to question 11.
Explanation for question 11. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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
1
1
given equation and seeing that multiplying both sides of the equation x  y  4
2
3
one-half x, plus one-third y, equals 4, by 6 to clear fractions from the equation
yields 3x  2 y  24. 3 x plus 2 y equals 24.
Link back to question 11.
Link back to question 12.
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Sample Items
Explanation for question 12. (Follow link back to original question.)
Program: S A T
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:


sin  x  5 cos   x  . sine of x equals cosine of the quantity, the fraction pi over 2,
2

 
  
minus x. Therefore, sin    cos    , sine of the fraction pi over 5, equals,
5
2 5
cosine of the quantity, the fraction pi over 2, minus, the fraction pi over 5, which
 3 
reduces to cos   . cosine of the fraction 3 pi over 10.
 10 
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 that angle.
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.
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Sample Items
Choice D is not the correct answer. The student mistakenly remembers the

the fraction pi over 2 to the angle
2

measure instead of subtracting the angle measure from . the fraction pi over 2.
2
relationship between sine and cosine and adds
Link back to question 12.
Link back to question 13.
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Explanation for question 13. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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 x y-plane are parallel lines. To be parallel, the lines must have the
same slope, and this will be true if the expression ax  3 y a, x minus 3 y is a
multiple of the expression
1
1
x  y. one-half x minus one-fourth y. Since
2
4
 1 
3 y  12   y  , negative 3 y equals 12, parenthesis, negative one-fourth y, close
 4 
parenthesis, the expression ax  3 y a, x minus 3 y would have to be 12 times the
expression
1
1
1 
x  y. one-half x minus one-fourth y. This means ax  12  x  ,
2
4
2 
a, x equals 12, parenthesis, one-half x, close parenthesis, so a  6. a, equals 6. The
1
1
resulting system is x  y  5 one-half x minus one-fourth y, equals 5 and
2
4
6 x  3 y  20, 6 x minus 3 y equals 20, which is equivalent to 6 x  3 y  60 6 x
minus 3 y equals 60 and 6 x  3 y  20, 6 x minus 3 y equals 20, which has no
solution.
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Choice A is not the correct answer. This may result from the misconception that
if each equation in a system has the same x-coefficient, the system cannot have a
1
solution. But if a  , a, equals one-half, subtracting the two equations
2
eliminates x 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
20
on the right sides,
, the fraction 20 over 5, and wrongly concluding that the
5
1
second equation must be 4 times the first, which would give a  4   , a, equals 4,
2
parenthesis, one-half, close parenthesis, or a  2. 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
1
then neglected to find a  12   , a, equals 12, parenthesis, one-half,
2
close parenthesis, or a  6. a, equals 6.
Link back to question 13.
Link back to question 14.
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Explanation for question 14. (Follow link back to original question.)
Program: S A T
Subscore: Passport to advanced math
Calculator Usage: No
Objective: Students must interpret an equation that models a real-world situation
and be able to interpret the whole expression (or specific parts) in terms of its
context.
Difficulty: Hard
Key: B
1
one-fifth is the portion of the
5
job that the two printers, working together, can complete in one hour, and each
1
term in the sum on the left side is the part of this one-fifth of the job that one of
5
2
the printers contributes. Since one of the printers is twice as fast as the other,
x
the fraction 2 over x describes the portion of the job that the faster printer is able to
1
complete in one hour and
the fraction one over x describes the portion of the
x
job that the slower printer is able to complete in one hour.
Choice B is correct. From the description given,
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 x hours.
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Sample Items
1
the fraction
x
one over x 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 2 hours. In fact, the
Choice C is not the correct answer. The student may have seen that
2 4
portion of the job that the faster printer could complete in 2 hours is (2)    .
 x x
parenthesis, 2, close parenthesis, parenthesis, the fraction 2 over x, close
parenthesis, equals, the fraction 4 over x.
Choice D is not the correct answer. The student may have correctly seen that the
1
value one-fifth on the right side refers to the portion of the job completed, but
5
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,
1
x
the time it takes the slower printer to complete
one-fifth of the job is
5
5
the fraction x over 5 hours.
Link back to question 14.
Link back to question 15.
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Sample Items
Explanation for question 15. (Follow link back to original question.)
Program: S A T
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
2
2 
2
2
Pythagorean theorem to solve correctly: r  x   r  , r squared equals
3 
x squared plus, parenthesis, two-thirds r, close parenthesis, squared, where r
represents the radius of the circle and x represents the distance between the chord
and the diameter.
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.
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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 45-45-90 triangle and tried
to use this relationship to determine the distance.
Link back to question 15.
Link back to question 16.
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Explanation for question 16. (Follow link back to original question.)
Program: S A T
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

 x   the
2
fraction pi over 2 is less than x, and x is less than pi is placed in standard position,
its terminal side will fall in Quadrant 2, and sin x  a sine x equals, a, will be the
y-coordinate of the point P where its terminal side intersects the unit circle. If
sin w  a, sine w equals negative a, then when the angle with radian measure w
Choice B is correct. If an angle with radian measure x such that
is placed in standard position, its terminal side will intersect the unit circle at a
point with y-coordinate equal to a. negative a. There are two such points on the
unit circle: the reflection of P across the x-axis, which would correspond to an
angle with radian measure  x negative x (and also with radian measures . . .
6  x,  4  x,  2  x, 2  x, 4  x,6  x ...); negative 6 pi minus x,
negative 4 pi minus x, negative 2 pi minus x, 2 pi minus x, 4 pi minus x, 6 pi
minus x . . .); and the reflection of P through the origin, which would correspond
to an angle with radian measure x   x minus pi (and also with radian
measures . . . x  5 , x  3 , x   , x  3 , x  5 ...). x minus 5 pi, x minus 3 pi, x
plus pi, x plus 3 pi, x plus 5 pi . . .). Thus, of the choices given, only x  
x minus pi could be the value of w.
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Choice A is not the correct answer. In general, sin( x)   sin x sine of negative x,
equals negative sine x and sin( x   )   sin x, sine of the quantity x plus pi,
equals negative sine x, so sin(  x)   sin( x)  ( sin x)  sin x. sine of the
quantity pi minus x, equals negative sine of negative x, equals negative,
parenthesis, negative sine x, close parenthesis, equals sine x. Therefore,
sin(  x)  a, sine of the quantity pi minus x, equals a, not a. negative a.
Choice C is not the correct answer. In general, sin(2  x)  sin x, sine of the
quantity 2 pi plus x, equals sine x, so sin(2  x)  a, sine of the quantity 2 pi
plus x, equals a, not a. negative a.
Choice D is not the correct answer. In general, sin( x  2 )  sin x, sine of the
quantity x minus 2 pi, equals sine x, so sin( x  2 )  a, sine of the quantity
x minus 2 pi, equals a, not a negative a.
Link back to question 16.
Link back to question 17.
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Explanation for question 17. (Follow link back to original question.)
Program: S A T, P S A T/N M S Q T, P S A T 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  x  32   y  4 2  169, parenthesis,
x minus 3, close parenthesis, squared, plus, parenthesis, y plus 4, close parenthesis,
squared, equals 169, the standard form of an equation of the circle. Understanding
this form results in the equation r 2  169, r squared equals 169, which when
solved for r gives the value of the radius as 13. Diameter is twice the value of the
radius; therefore, the diameter is 26.
Link back to question 17.
Link back to question 18.
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Explanation for question 18. (Follow link back to original question.)
Program: S A T
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 ( x  1)( x  1) parenthesis, x plus one, close parenthesis, parenthesis,
x minus one, close parenthesis, yields 24( x  1)  12( x  1)  ( x  1)( x  1). 24,
parenthesis, x minus one, close parenthesis, minus 12, parenthesis, x plus one,
close parenthesis, equals, parenthesis, x plus one, close parenthesis, parenthesis,
x minus one, close parenthesis. Multiplication and simplification then yields
12 x  36  x2  1, 12 x minus 36 equals, x squared minus one, or x 2  12 x  35  0.
x squared minus 12 x plus 35 equals zero. Factoring the quadratic gives
( x  5)( x  7)  0, parenthesis, x minus 5, close parenthesis, parenthesis, x minus 7,
close parenthesis, equals zero, so the solutions occur at x  5 x equals 5 and
x  7, x equals 7, 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 question 18.
This is the end of Mathematics Sample Question Answers and Explanations.
The Redesigned S A T®: Mathematics—No Calculator
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