Download - SAT Practice Test 2 Math Test: No Calculator, for Assistive Technology

Math Test—No Calculator20 Questions

Turn to Section 3 of your answer sheet to answer the questions in this section.

Directions

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 scratch paper for scratch work.

Notes

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 of x is a real number.

The S A T® Copyright 2015 by the College Board W F-5 L S A 0 7

Reference

Begin skippable figure descriptions.

The figure presents information for your reference in solving some of the problems.


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.


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.

End skippable figure descriptions.

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.


Question 1.If 5 x plus 6 equals 10, what is the value of 10 x plus 3 ?

A. 4

B. 9

C. 11

D. 20

Explanation for question 1.

Question 2 is based on the following set of equations. x plus y equals 0

3 x minus 2 y equals 10

Question 2.

Which of the following ordered pairs parenthesis, x comma y, close parenthesis, satisfies the preceding system of equations?

A. parenthesis, 3 comma negative 2, close parenthesis

B. parenthesis, 2 comma negative 2, close parenthesis

C. parenthesis, negative 2 comma 2, close parenthesis

D. parenthesis, negative 2 comma negative 2, close parenthesis



Question 3.

A landscaping company estimates the price of a job, in dollars, using the expression 60 plus 12 n h, where n is the number of landscapers who will be working and h is the total number of hours the job will take using n landscapers. Which of the following is the best interpretation of the number 12 in the expression?

A. The company charges $12 per hour for each landscaper.

B. A minimum of 12 landscapers will work on each job.

C. The price of every job increases by $12 every hour.

D. Each landscaper works 12 hours a day.


Question 4 is based on the following expression. 9 a, raised to the power 4, plus 12 a, squared, b squared, plus, 4 b raised to the power 4

Question 4.Which of the following is equivalent to the preceding expression?

A. parenthesis, 3 a, squared, plus 2 b squared, close parenthesis, squared

B. parenthesis, 3 a, plus 2 b, close parenthesis, raised to the power 4

C. parenthesis, 9 a, squared, plus 4 b squared, close parenthesis, squared


D. parenthesis, 9 a, plus 4 b, close parenthesis, raised to the power 4


Question 5 is based on the following equation.

the square root of 2 k squared, plus 17, end square root, minus x equals 0.

Question 5.If k is greater than 0 and x equals 7 in the preceding equation, what is the value of k ?

A. 2

B. 3

C. 4

D. 5



Question 6 refers to the following figure.

Begin figure description.

The figure presents two lines, k and ℓ, that slant upward and to the right on the x y-plane. Line ℓ is to the left of line k. Two points on line ℓ are labeled with coordinates negative 5 comma 0, and 0 comma 2. Two points on line k are labeled with coordinates 0 comma negative 4, and p comma 0.

End figure description.

Question 6.In the preceding x y-plane, line l is parallel to line k. What is the value of p ?

A. 4

B. 5

C. 8

D. 10



Question 7.

If the fraction whose numerator is x raised to the power, a, squared, and whose denominator is x raised to the power b squared, end fraction, equals, x raised to the power 16, x is greater than 1, and a, plus b equals 2, what is the value of a, minus b ?

A. 8

B. 14

C. 16

D. 18



Question 8 is based on the following formula.

n A equals 360

Question 8.The measure A, in degrees, of an exterior angle of a regular polygon is related to the number of sides, n, of the polygon by the preceding formula. If the measure of an exterior angle of a regular polygon is greater than , 50 degrees, what is the greatest number of sides it can have?

A. 5

B. 6

C. 7

D. 8


Question 9.

The graph of a line in the x y-plane has slope 2 and contains the point with

coordinates 1 comma 8. The graph of a second line passes through the points

with coordinates 1 comma 2, and 2 comma 1. If the two lines intersect at the

point with coordinates a, comma b, what is the value of a, plus b ?

A. 4

B. 3

C. negative 1

D. negative 4


Question 10.Which of the following equations has a graph in the x y-plane for which y is always greater than or equal to negative 1 ?

A. y equals the absolute value of x, end absolute value, minus 2

B. y equals x squared minus 2

C. y equals, parenthesis, x minus 2, close parenthesis, squared

D. y equals x cubed minus 2


Question 11.

Which of the following complex numbers is equivalent to the fraction 3 minus 5 i over 8 plus 2 i, end fraction? (Note: i equals the square root of negative 1.)

A. the fraction 3 over 8, minus the fraction 5 i over 2

B. the fraction 3 over 8, plus the fraction 5 i over 2

C. the fraction 7 over 34, minus the fraction 23 i over 34

D. the fraction 7 over 34, plus the fraction 23 i over 34



Question 12 is based on the following formula.

R equals, the fraction F, over N plus F, end fraction

Question 12.A website uses the preceding formula to calculate a seller’s rating, R, based on the number of favorable reviews, F, and unfavorable reviews, N. Which of the following expresses the number of favorable reviews in terms of the other variables?

A. F equals the fraction R N over R minus 1, end fraction

B. F equals the fraction R N over 1 minus R, end fraction

C. F equals the fraction N over 1 minus R, end fraction

D. F equals the fraction N over R minus 1, end fraction



Question 13.What is the sum of all values of m that satisfy 2 m squared minus 16 m, plus 8, equals 0?

A. negative 8

B. negative 4 times the square root of 3

C. 4 times the square root of 3

D. 8


Question 14.A radioactive substance decays at an annual rate of 13 percent. If the initial amount of the substance is 325 grams, which of the following functions f models the remaining amount of the substance, in grams, t years later?

A. f of t equals 325, parenthesis, 0.87, close parenthesis, raised to the power t

B. f of t equals 325, parenthesis, 0.13, close parenthesis, raised to the power t

C. f of t equals 0.87, parenthesis, 325, close parenthesis, raised to the power t

D. f of t equals 0.13, parenthesis, 325, close parenthesis, raised to the power t


Question 15.

The expression for the fraction whose numerator is 5 x minus 2, and whose denominator is x plus 3, is equivalent to which of the following?

A. the fraction 5 minus 2, over 3, end fraction

B. 5 minus the fraction 2 over 3

C. 5 minus the fraction, 2, over x plus 3, end fraction

D. 5 minus the fraction, 17, over x plus 3, end fraction



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 three and one half must be gridded as 3.5 or

seven slash two. (If three, one, slash, two, is entered into the grid, it

will be interpreted as thirty one halves, not three and one half.)

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

Begin skippable figure description.

Example 1: If your answer is a fraction such as seven-twelfths, it should be recorded as follows. Enter 7 in the first space, the fraction bar (a slash) in the second space, 1 in the third space, and 2 in the fourth space. All four spaces would be used in this example.

Example 2: If your answer is a decimal value such as 2.5, it could be recorded as follows. Enter 2 in the second space, the decimal point in the third space, and 5 in the fourth space. Only three spaces would be used in this example.

End skippable figure description.


Example 3


Example 3: Acceptable ways to record two-thirds are: 2 slash 3, .666, and .667.



Example 4

Note: You may start your answers in any column, space permitting. Columns you don’t need to use should be left blank.


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 2 in the second space, 0 in the third space, and 1 in the fourth space. Alternatively, you may record 2 in the first space, 0 in the second space, and 1 in the third space. Spaces not needed should be left blank.



Question 16.The sales manager of a company awarded a total of $3000 in bonuses to the most productive salespeople. The bonuses were awarded in amounts of $250 or $750. If at least one $250 bonus and at least one $750 bonus were awarded, what is one possible number of $250 bonuses awarded?


Question 17 is based on the following equation.

2 x, parenthesis, 3 x plus 5, close parenthesis, plus 3, parenthesis, 3 x plus 5, close parenthesis, equals, a, x squared, plus b x, plus c

Question 17.In the preceding equation, a, b, and c are constants. If the equation is true for all values of x, what is the value of b ?





The figure presents two triangles formed by two intersecting line segments and two horizontal line segments. The two intersecting line segments intersect at a point labeled B. The upper triangle is labeled C D B, where side C D is horizontal and C is to the left of D. The lower triangle is labeled A E B, where side A E is horizontal and A is to the left of E. In the upper triangle, side D B is labeled 5, and in the lower triangle, side A B is labeled 10 and side B E is labeled 8.


Question 18.

In the preceding figure, line segment A E is parallel to line segment C D, and segment A D intersects segment C E at B. What is the length of segment C E ?





The figure presents a circle in the x y-plane. The horizontal axis is labeled x, the vertical axis is labeled y, and the origin is labeled O. The circle intersects the x-axis to the right of the origin at a point labeled B. A line is drawn from the origin upward and to the right to a point on the circle labeled A. The coordinates of point A are the square root of 3 comma 1.


Question 19.In the preceding x y-plane, O is the center of the circle, and the measure of

angle A O B is the fraction pi over a, radians. What is the value of a ?



Question 20 is based on the following system of equations. a, x plus b y, equals 12

2 x plus 8 y, equals 60

Question 20.In the preceding system of equations, a and b are constants. If the system has

infinitely many solutions, what is the value of the fraction a, over b ?


Stop.

If you finish before time is called, you may check your work on this section only. Do not turn to any other section.

Answers and explanations for questions 1 through 20 are provided in the next section of this document.


Answers and Explanations for Questions 1 through 20

Explanation for question 1.Choice C is correct. Subtracting 6 from each side of 5 x plus 6 equals 10 yields 5 x equals 4. Dividing both sides of 5 x equals 4 by

5 yields x equals four fifths. The value of x can now be substituted into the

expression 10 x plus 3, giving 10, parenthesis, four fifths, close parenthesis, plus 3, equals 11.

Alternatively, the expression 10 x plus 3 can be rewritten as 2, parenthesis, 5 x plus 6, close parenthesis, minus 9, and 10 can be substituted for

5 x plus 6, giving 2, parenthesis, 10, close parenthesis, minus 9, equals 11.

Choices A, B, and D are incorrect. Each of these choices leads to 5 x plus 6, which is not equal to 10, contradicting the given equation, 5 x plus 6 equals 10. For example, choice A is incorrect because if the value of

10 x plus 3 were 4, then it would follow that x equals 0.1, and the value of 5 x plus 6 would be 6.5, not 10.


Explanation for question 2.Choice B is correct. Multiplying each side of x plus y equals 0 by 2 gives

2 x plus 2 y equals 0. Then, adding the corresponding sides of 2 x plus 2 y equals 0 and 3 x minus 2 y equals 10 gives

5 x equals 10. Dividing each side of 5 x equals 10 by 5 gives x equals 2. Finally, substituting 2 for x in x plus y equals 0 gives

2 plus y equals 0, or y equals negative 2. Therefore, the solution to the given system of equations is parenthesis, 2 comma negative 2, close parenthesis.

Alternatively, the equation x plus y equals 0 can be rewritten as x equals negative y, and substituting x for negative y in 3 x minus 2 y equals 10 gives 5 x equals 10, or x equals 2. The value of y can then be found in the same way as before.

Choices A, C, and D are incorrect because when the given values of x and y are substituted into x plus y equals 0 and 3 x minus 2 y equals 10, either one or both of the equations are not true. These answers may result from sign errors or other computational errors.


Explanation for question 3.Choice A is correct. The price of the job, in dollars, is calculated using the expression 60 plus 12 n h, where 60 is a fixed price and 12 n h depends on the number of landscapers, n, working the job and the number of hours, h, the job takes those n landscapers. Since n h is the total number of hours of work done when n landscapers work h hours, the cost of the job increases by $12 for each hour a landscaper works. Therefore, of the choices given, the best interpretation of the number 12 is that the company charges $12 per hour for each landscaper.

Choice B is incorrect because the number of landscapers that will work each job is represented by n in the equation, not by the number 12.

Choice C is incorrect because the price of the job increases by 12 n dollars each hour, which will not be equal to 12 dollars unless n equals 1.

Choice D is incorrect because the total number of hours each landscaper works is equal to h. The number of hours each landscaper works in a day is not provided.


Explanation for question 4.Choice A is correct. If a polynomial expression is in the form

parenthesis, x, close parenthesis, squared, plus 2, parenthesis, x, close parenthesis, times, parenthesis, y, close parenthesis, plus,

parenthesis, y, close parenthesis, squared, then it is equivalent to parenthesis, x plus y, close parenthesis, squared. Because

9 a, raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4, equals, parenthesis, 3 a, squared, close parenthesis, squared, plus, 2, parenthesis, 3 a, squared, close parenthesis, times, parenthesis, 2 b squared, close parenthesis, plus, parenthesis, 2 b squared, close parenthesis, squared, it can be rewritten as

parenthesis, 3 a, squared, plus 2 b squared, close parenthesis, squared.

Choice B is incorrect. The expression parenthesis, 3 a, plus 2 b, close parenthesis, raised to the power 4, is equivalent to the product

of parenthesis, 3 a, plus 2 b, close parenthesis, times, parenthesis, 3 a, plus 2 b, close parenthesis, times, parenthesis, 3 a, plus 2 b, close parenthesis, times, parenthesis, 3 a, plus 2 b, close parenthesis.

This product will contain the term 4, parenthesis, 3 a, close parenthesis, cubed, times, parenthesis, 2 b, close parenthesis, equals,

216 a, cubed, b. However, the given polynomial, 9 a, raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4, does not contain the term 216 a, cubed, b. Therefore,

9 a, raised to the power 4, plus, 12 a, squared,


b squared, plus 4 b, raised to the power 4, is not equal to, parenthesis, 3 a, plus 2 b, close parenthesis, raised to the power 4.

Choice C is incorrect. The expression parenthesis, 9 a, squared, plus 4 b squared, close parenthesis, squared, is equivalent to the product

parenthesis, 9 a, squared, plus 4 b squared, close parenthesis, times, parenthesis, 9 a, squared, plus 4 b squared, close parenthesis.

This product will contain the term parenthesis, 9 a, squared close parenthesis, times, parenthesis, 9 a, squared, close parenthesis, equals 81 a,

raised to the power 4. However, the given polynomial, 9 a, raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4, does not contain the term 81 a, raised to the power 4. Therefore,

9 a, raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4, is not equal to, parenthesis, 9 a, squared, plus 4 b squared, close parenthesis, squared.

Choice D is incorrect. The expression parenthesis, 9 a, plus 4 b, close parenthesis, raised to the power 4, is equivalent to the product

parenthesis, 9 a, plus 4 b, close parenthesis, times, parenthesis, 9 a, plus 4 b, close parenthesis, times, parenthesis, 9 a, plus 4 b, close parenthesis, times, parenthesis, 9 a, plus 4 b, close parenthesis.

This product will contain the term parenthesis, 9 a, close parenthesis, times, parenthesis, 9 a, close parenthesis, times, parenthesis, 9 a, close parenthesis, times, parenthesis, 9 a, close parenthesis, equals 6,561 a,

raised to the power 4. However, the given polynomial, 9 a,


raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4,

does not contain the term 6,561 a, raised to the power 4. Therefore,

9 a, raised to the power 4, plus 12 a, squared, b squared, plus 4 b, raised to the power 4, does not equal, parenthesis, 9 a, plus 4 b close parenthesis, raised to the power 4.



Choice C is correct. Since the square root of 2 k squared, plus 17, end square root, minus x, equals 0, and x equals 7, one can substitute 7

for x, which gives the square root of 2 k squared, plus 17, end

square root, minus 7, equals 0. Adding 7 to each side of the square root of 2 k squared, plus 17, end square root, minus 7, equals 0, gives

the square root of 2 k squared, plus 17, end square root, equals 7.

Squaring each side of the square root of 2 k squared, plus 17, end

square root, equals 7, will remove the square root symbol: parenthesis, the square root of 2 k squared, plus 17, end square root, close parenthesis, squared, equals, parenthesis, 7, close parenthesis, squared, or

2 k squared plus 17, equals 49. Then subtracting 17 from each side

of 2 k squared plus 17, equals 49 gives 2 k squared, equals, 49 minus 17, equals 32, and dividing each side of 2 k squared, equals 32 by 2 gives k squared, equals 16. Finally, taking the square root of each side of k squared, equals 16 gives k equals plus or minus 4, and since the problem states that k is greater than 0, it follows that k equals 4.

Since the sides of an equation were squared while solving the square root of 2 k squared, plus 17, end square root, minus 7, equals 0, it is possible that an extraneous root was produced. However, substituting 4 for k in

the square root of 2 k squared, plus 17, end square root, minus 7, equals 0, confirms that 4 is a solution for k:

the square root of 2,


parenthesis, 4, close parenthesis, squared, plus 17, end square root, minus 7, equals, the square root of 32 plus 17, end square root, minus 7, which equals, the square root of 49, end square root, minus 7, which equals 7 minus 7, which equals 0.

Choices A, B, and D are incorrect because substituting any of these values for k in

the square root of 2 k squared, plus 17, end square root, minus 7, equals 0 does not yield a true statement.

Explanation for question 6.Choice D is correct. Since lines l and k are parallel, the lines have the same

slope. Line l passes through the points with coordinates negative 5

comma 0, and the point with coordinates, 0 comma 2, so its slope is

the fraction 0 minus 2, over negative 5 minus 0, end fraction, which is

two fifths. The slope of line k must also be two fifths. Since line k has slope

two fifths and passes through the points with coordinates 0 comma negative 4, and the point with coordinates, p comma 0, it follows that

the fraction negative 4 minus 0, over 0 minus p, end fraction, equals

two fifths, or the fraction 4 over p equals two fifths. Multiplying each side

of the fraction 4 over p equals two fifths by 5 p gives 20 equals 2 p, and therefore, p equals 10.

Choices A, B, and C are incorrect and may result from conceptual or calculation errors.



Choice A is correct. Since the numerator and denominator of the fraction x raised to the power a, squared, over, x raised to the power b squared, end fraction, have a common base, it follows by the laws of exponents that this expression can

be rewritten as x, raised to the power a, squared minus b squared, end

power. Thus, the equation of the fraction x raised to the power a, squared, over, x raised to the power b squared, end fraction, equals 16 can be

rewritten as x, raised to the power a, squared minus b squared, end power, equals, x raised to the power 16. Because the equivalent expressions have the common base x, and x is greater than 1, it follows that the exponents of the two expressions must also be equivalent. Hence, the equation

a, squared, minus b squared equals 16 must be true. The left-hand side of this new equation is a difference of squares, and so it can be factored:

parenthesis, a, plus b, close parenthesis, times, parenthesis, a, minus b, close parenthesis, equals 16. It is given that parenthesis, a, plus b, close parenthesis, equals 2; substituting 2 for the factor parenthesis, a, plus b, close parenthesis, gives 2, parenthesis a, minus b, close parenthesis, equals 16. Finally, dividing both sides of 2, parenthesis, a, minus b, equals 16 by 2 gives a, minus b equals 8.

Choices B, C, and D are incorrect and may result from errors in applying the laws of exponents or errors in solving the equation a, squared minus b squared equals 16.


Explanation for question 8.Choice C is correct. The relationship between n and A is given by the equation

n A equals 360. Since n is the number of sides of a polygon, n must be a

positive integer, and so n A equals 360 can be rewritten as A equals the fraction 360 over n. If the value of A is greater than 50, it follows that

the fraction 360 over n is greater than 50 is a true statement. Thus,

50 n is less than 360, or n is less than the fraction 360 over 50, which equals 7.2. Since n must be an integer, the greatest possible value of n is 7.

Choices A and B are incorrect. These are possible values for n, the number of sides of a regular polygon, if A, is greater than 50, but neither is the greatest possible value of n.

Choice D is incorrect. If A, is less than 50, then n equals 8 is the least possible value of n, the number of sides of a regular polygon. However, the question asks for the greatest possible value of n if A, is greater than 50, which is n equals 7.

Explanation for question 9.Choice B is correct. Since the slope of the first line is 2, an equation of this line can be written in the form y equals 2 x plus c, where c is the y-intercept of

the line. Since the line contains the point with coordinates, 1 comma 8, one can substitute 1 for x and 8 for y in y equals 2 x plus c, which gives

8 equals 2, parenthesis, 1 ,close parenthesis, plus c, or c equals 6. Thus, an equation of the first line is y equals 2 x plus 6. The


slope of the second line is equal to the fraction 1 minus 2, over 2 minus 1, end fraction, or negative 1. Thus, an equation of the second line can be written in the form y equals negative x plus d, where d is the y-intercept of the line. Substituting 2 for x and 1 for y gives 1 equals negative 2 plus d, or d equals 3. Thus, an equation of the second line is y equals negative x plus 3. Since a is the x-coordinate and b is the y-coordinate of the intersection point of the two lines, one can substitute a for x and b for y in the two equations, giving the system b equals 2 a, plus 6 and b equals negative a, plus 3. Thus, a can be found by solving the equation

2 a, plus 6 equals negative a, plus 3, which gives a, equals negative 1. Finally, substituting negative 1 for a into the equation b equals negative a, plus 3 gives b equals negative, parenthesis, negative 1, close parenthesis, plus 3, or b equals 4. Therefore, the value of

a, plus b is 3.

Alternatively, since the second line passes through the points with

coordinates, 1 comma 2, and the point with coordinates 2 comma 1, an equation for the second line is x plus y equals 3. Thus, the intersection

point of the first line and the second line, with coordinates, a, comma b, lies on the line with equation x plus y equals 3. It follows that a, plus b equals 3.

Choices A and C are incorrect and may result from finding the value of only a or b, but not calculating the value of a, plus b.

Choice D is incorrect and may result from a computation error in finding equations of the two lines or in solving the resulting system of equations.


Explanation for question 10.Choice C is correct. Since the square of any real number is nonnegative, every

point on the graph of the quadratic equation y equals, parenthesis, x minus 2, close parenthesis, squared in the x y-plane has a nonnegative y-coordinate. Thus, y is greater than or equal to 0 for every point on the

graph. Therefore, the equation y equals, parenthesis, x minus 2, close parenthesis, squared has a graph for which y is always greater than or equal to negative 1.

Choices A, B, and D are incorrect because the graph of each of these equations in

the x y-plane has a y-intercept at the point with coordinates 0 comma negative 2. Therefore, each of these equations contains at least one point where y is less than negative 1.


Choice C is correct. To perform the division the fraction 3 minus 5 i, over 8

plus 2 i, end fraction, multiply the numerator and denominator of the fraction 3 minus 5 i, over 8 plus 2 i, end fraction by the conjugate of the denominator, 8 minus 2 i. This gives

the fraction, parenthesis, 3 minus 5 i close parenthesis, over, parenthesis, 8 plus 2 i, close parenthesis, times, the fraction, parenthesis, 8 minus 2 i, close parenthesis, over, parenthesis, 8 minus 2 i, close parenthesis, equals, the fraction whose numerator is, 24 minus 6 i minus 40 i, plus, parenthesis, negative 5 i, close parenthesis, times, parenthesis, negative 2 i close parenthesis, and whose denominator is 8 squared minus, parenthesis 2 i, close

parenthesis, squared, end fraction. Since i squared equals negative 1, this


can be simplified to the fraction whose numerator is, 24 minus 6 i, minus 40 i minus 10, and whose denominator is 64 plus 4, end fraction, equals the fraction 14 minus 46 i over 68, which then simplifies to

the fraction 7 over 34 minus the fraction 23 i over 34.

Choices A and B are incorrect and may result from misconceptions about fractions.

For example, the fraction a, plus b over c plus d, end fraction is equal to

the fraction a, over c plus d, end fraction, plus the fraction b over

c plus d, end fraction, not the fraction a, over c plus the fraction b over d.

Choice D is incorrect and may result from a calculation error.


Choice B is correct. Multiplying each side of R equals the fraction F

over N plus F, end fraction by N plus F gives R parenthesis, N plus F, close parenthesis, equals F, which can be rewritten as

R N plus R F, equals F. Subtracting R F from each side of R N plus R F, equals F gives R N equals F minus R F,

which can be factored as R N equals F, parenthesis, 1 minus R,

close parenthesis. Finally, dividing each side of R N equals F, parenthesis, 1 minus R, close parenthesis by 1 minus R expresses F in terms

of the other variables: F equals the fraction R N over 1 minus R, end fraction.


Choices A, C, and D are incorrect and may result from calculation errors when rewriting the given equation.


Explanation for question 13.Choice D is correct. The problem asks for the sum of the roots of the quadratic equation 2 m squared minus 16 m plus 8, equals 0. Dividing each side of the equation by 2 gives m squared, minus 8 m plus 4, equals 0. If the roots of m squared minus 8 m plus 4 equals 0 are

s sub 1 and s sub 2, then the equation can be factored as

m squared minus 8 m plus 4, equals, parenthesis, m minus s sub 1, close parenthesis, times, parenthesis, m minus s sub 2, close parenthesis, equals 0. Looking at the coefficient of x on each side of

m squared minus 8 m plus 4 equals, parenthesis m minus s sub 1 close parenthesis, times parenthesis m minus s sub 2 close parenthesis gives negative 8 equals negative s sub 1, minus s sub 2, or s sub 1 plus s sub 2 equals 8.

Alternatively, one can apply the quadratic formula to either 2 m squared minus 16 m plus 8, equals 0 or m squared minus 8 m plus 4, equals 0. The quadratic formula gives two solutions, 4 minus 2

times the square root of 3 and 4 plus 2 times the square root of 3, whose sum is 8.

Choices A, B, and C are incorrect and may result from calculation errors when applying the quadratic formula or a sign error when determining the sum of the roots of a quadratic equation from its coefficients.


Explanation for question 14.Choice A is correct. Each year, the amount of the radioactive substance is reduced by 13 percent from the prior year’s amount; that is, each year, 87 percent of the previous year’s amount remains. Since the initial amount of the radioactive

substance was 325 grams, after 1 year, 325, parenthesis, 0.87, close

parenthesis, grams remains; after 2 years 325, parenthesis, 0.87, close parenthesis, times, parenthesis, 0.87, close parenthesis, equals, 325, parenthesis, 0.87 close parenthesis, squared, grams

remains; and after t years, 325, parenthesis, 0.87, close parenthesis,

raised to the power t grams remains. Therefore, the function f of t equals 325, parenthesis, 0.87, close parenthesis, raised to the power t models the remaining amount of the substance, in grams, after t years.

Choice B is incorrect and may result from confusing the amount of the substance remaining with the decay rate.

Choices C and D are incorrect and may result from confusing the original amount of the substance and the decay rate.


Explanation for question 15.Choice D is correct. Dividing 5 x minus 2 by x plus 3 gives the following long division operation:

Divide x plus 3, into 5 x minus 2, place 5 on top of the long division symbol. Multiply x plus 3 by 5, which yields 5 x plus 15. Subtract 5 x plus 15, from 5 x minus 2 which yields the remainder, negative 17.

Therefore, the expression the fraction 5 x minus 2, over x plus 3, end

fraction can be rewritten as 5, minus the fraction, 17, over x plus 3, end fraction.

Alternatively, the fraction 5 x minus 2, over x plus 3, end fraction can be

rewritten as the fraction 5 x minus 2, over x plus 3, end fraction, equals, the fraction whose numerator is, parenthesis, 5 x plus 15, close parenthesis, minus 15, minus 2, and whose denominator is x plus 3 end fraction; equals, the fraction whose numerator is 5, parenthesis, x plus 3, close parenthesis, minus 17, and whose denominator is x plus 3, end fraction, which equals 5 minus the fraction, 17, over x plus 3, end fraction.


Choices A and B are incorrect and may result from incorrectly canceling out the

x in the expression the fraction 5 x minus 2, over x plus 3, end fraction.

Choice C is incorrect and may result from finding an incorrect remainder when performing long division.

Explanation for question 16.The correct answer is 3, 6, or 9. Let x be the number of $250 bonuses awarded, and let y be the number of $750 bonuses awarded. Since $3000 in bonuses were awarded, and this included at least one $250 bonus and one $750 bonus, it follows that 250 x plus 750 y equals 3,000, where x and y are positive integers. Dividing each side of 250 x plus 750 y equals 3,000 by 250 gives x plus 3 y equals 12, where x and y are positive integers. Since 3 y and 12 are each divisible by 3, it follows that

x equals 12 minus 3 y must also be divisible by 3. If x equals 3, then y equals 3; if x equals 6, then y equals 2; and if x equals 9, then y equals 1. If x equals 12, then y equals 0, but this is not possible, since there was at least one $750 bonus awarded. Therefore, the possible numbers of $250 bonuses awarded are 3, 6, and 9. Any of the numbers 3, 6, or 9 may be gridded as the correct answer.


The correct answer is 19. Since 2 x, parenthesis, 3 x plus 5, close parenthesis, plus 3, parenthesis, 3 x plus 5, close parenthesis, equals, a, x squared, plus b x plus c, for all values of x, the two sides of the equation are equal, and the value of b can be determined by simplifying the left-hand side of the equation and writing it in the same form as the right-hand side. Using the distributive property, the equation becomes

parenthesis, 6 x squared plus 10 x, close


parenthesis, plus, parenthesis, 9 x plus 15, close parenthesis, equals, a, x squared, plus b x plus c. Combining like terms gives 6 x squared plus 19 x plus 15, equals, a, x squared, plus b x plus c. The value of b is coefficient of x, which is 19.

Explanation for question 18.The correct answer is 12. Angles A B E and D B C are vertical angles and thus have the same measure. Since segment A E is parallel to segment C D, angles A and D are of the same measure by the alternate interior angle theorem. Thus, by the angle-angle theorem, triangle A B E is similar to triangle D B C, with vertices A, B,

and E corresponding to vertices D, B, and C, respectively. Thus, the

fraction A B over D B, equals the fraction E B over C B, or the fraction 10 over 5, equals the fraction 8 over C B. It follows that C B equals 4, and so

C E equals C B plus B E, which equals 4 plus 8, which equals 12.

Explanation for question 19.The correct answer is 6. By the distance formula, the length of radius O A is

the square root of, parenthesis, the square root of 3, close parenthesis, squared, plus 1 squared, end square root, equals the square root

of 3 plus 1, end square root, which equals 2. Thus, sine parenthesis, angle A O B, close parenthesis, equals one half. Therefore, the measure

of angle A O B is 30 degrees, which is equal to 30, parenthesis, the fraction pi over 180, close parenthesis, equals, the fraction pi over 6 radians. Hence, the value of a is 6.



The key is one fourth or .25. In order for a system of two linear equations to have infinitely many solutions, the two equations must be equivalent. Thus, the equation a, x plus b y, equals 12 must be equivalent to the equation

2 x plus 8 y, equals 60. Multiplying each side of a, x plus b y, equals 12 by 5 gives 5 a, x plus 5 b y, equals 60, which must be equivalent to 2 x plus 8 y, equals 60. Since the right-hand sides of

5 a, x plus 5 b y, equals 60 and 2 x plus 8 y, equals 60

are the same, equating coefficients gives 5 a, equals 2, or a, equals

two fifths and 5 b equals 8, or b equals eight fifths. Therefore, the

value of the fraction a, over b equals, parenthesis, two fifths, close parenthesis, divided by, parenthesis eight fifths, close parenthesis, which is equal

to one fourth. Either the fraction one fourth or its equivalent decimal, .25, may be gridded as the correct answer.

Alternatively, since a, x plus b y, equals 12 is equivalent to 2 x plus 8 y, equals 60, the equation a, x plus b y,

equals 12 is equal to 2 x plus 8 y, equals 60 multiplied on each side by the same constant. Since multiplying 2 x plus 8 y, equals 60 by a constant does not change the ratio of the coefficient of x to the coefficient of y, it

follows that the fraction a, over b equals two eighths, which equals one fourth.


Stop. This is the end of the answers and explanations for questions 1 through 20.


Download - SAT Practice Test 2 Math Test: No Calculator, for Assistive Technology

Top Related