Call Now to Set Up Tutoring:

(888) 888-0446

SAT II Math II : Absolute Value

Study concepts, example questions & explanations for SAT II Math II

Create An Account

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #32 : Mathematical Relationships

Define $g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$ .

Evaluate g(25) .

Possible Answers:

1,620

1,630

940,900

370

380

Correct answer:

370

Explanation:

$g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$

$g(25)= \left | \left | 1,000- \sqrt{25}\right | - 25^{2} \right |$

$= \left | \left | 1,000-5\right | - 625 \right |$

$= \left | \left | 995 \right | - 625 \right |$

$= \left | 995 - 625 \right |$

$= \left | 370 \right |$

=370

Report an Error

Example Question #33 : Mathematical Relationships

Define $f(x) = \left | 1 - x\right |$ .

Order from least to greatest: $f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Possible Answers:

$f\left ( \frac{17}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right )$

$f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{13}{14} \right )$

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Correct answer:

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

Explanation:

$f(x) = \left | 1 - x\right |$ , or, equivalently,

$f(x) = \left | \frac{14}{14} - x\right |$

$f\left ( \frac{9}{14} \right ) = \left | \frac{14}{14} - \frac{9}{14}\right | = \left | \frac{5}{14} \right |=\frac{5}{14}$

$f\left ( \frac{13}{14} \right ) = \left | \frac{14}{14} - \frac{13}{14}\right |= \left | \frac{1}{14} \right |=\frac{1}{14}$

$f\left ( \frac{17}{14} \right ) = \left | \frac{14}{14} - \frac{17}{14} \right |= \left | - \frac{3}{14} \right |=\frac{3}{14}$

From least to greatest, the values are

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

Report an Error

Example Question #1 : Absolute Value

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

Evaluate $2\frac{1}{2} \Psi \left (- \frac{1}{4} \right )$ .

Possible Answers:

$-1\frac{1}{5}$

Undefined.

$1\frac{1}{5}$

Correct answer:

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

$2\frac{1}{2} \Psi \left (- \frac{1}{4} \right ) = \frac{\left |1- 2\frac{1}{2} \right |}{1-\left |-\frac{1}{4}\right |}$

$= \frac{\left |-1\frac{1}{2} \right |}{1- \frac{1}{4}}$

$= \frac{ 1\frac{1}{2} }{ \frac{3}{4}}$

$= \frac{ \frac{3}{2} }{ \frac{3}{4}}$

$= \frac{3}{2} \div \frac{3}{4}$

$= \frac{3}{2} \cdot \frac{4}{3}$

$= \frac{12}{6}$

Report an Error

Example Question #61 : Sat Subject Test In Math Ii

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

If $\frac{1}{2}\Psi N = 1$ , which is a possible value of ?

Possible Answers:

N =- 3

N =- 1

$N =- 1\frac{1}{2}$

N =- 2

$N =- \frac{1}{2}$

Correct answer:

$N =- \frac{1}{2}$

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$ , so

$\frac{1}{2}\Psi N = 1$

can be rewritten as

$\frac{\left |1- \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{\left | \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{ \frac{1}{2} }{1-\left |N \right |} = 1$

$\frac{1}{2} =1-\left |N \right |$

$-\frac{1}{2} =-\left |N \right |$

$\frac{1}{2} = \left |N \right |$

Therefore, either $N = \frac{1}{2}$ or $N =- \frac{1}{2}$ . The correct choice is $N =- \frac{1}{2}$ .

Report an Error

Example Question #41 : Mathematical Relationships

Define $f(x) = \left | x+ 6\right |+ \left | x-9\right |$ .

How many values are in the solution set of the equation f(x)= 7 ?

Possible Answers:

One solution

Three solutions

Infinitely many solutions

No solutions

Two solutions

Correct answer:

No solutions

Explanation:

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If x< -6 , then

x+ 6 < 0 and x-9 < 0 ,

and this part of the function can be written as

$f(x) =- \left ( x+ 6\right )- \left (x-9\right )$

f(x) =-x-6-x+9

f(x)= -2x+3

If f(x)= 7 under this definition, then

-2x+3 = 7

-2x = 4

x = -2

However, $-2 \nless -6$ , so this is a contradiction.

If $-6 \le x \le 9$ , then

$x+ 6\ge 0$ and $x-9\le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right )- \left (x-9\right )$

f(x) =x+6-x+9

f(x) =15

This yields no solutions.

If x> 9 , then

x+ 6 > 0 and x-9 > 0 ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right ) + \left (x-9\right )$

f(x) =x+6+x-9

f(x)= 2x-3

If f(x)= 7 under this definition, then

2x-3 = 7

2x=10

x = 5

However, $5 \ngtr 9$ , so this is a contradiction.

f(x)= 7 has no solution.

Report an Error

Example Question #2 : Absolute Value

Define $f(x) = \left | x+ 5\right |+ \left | x-7\right |$ .

How many values are in the solution set of the equation f(x)= 12 ?

Possible Answers:

Three solutions

One solution

No solutions

Infinitely many solutions

Two solutions

Correct answer:

Infinitely many solutions

Explanation:

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If x< -5 , then

x+ 5 < 0 and x-7 < 0 ,

and this part of the function can be written as

$f(x) =- \left ( x+ 5\right )- \left (x-7\right )$

f(x) =- x-5 -x+7

f(x) =-2x+2

If $-5 \le x \le 7$ , then

$x+ 5 \ge 0$ and $x-7 \le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )- \left (x-7\right )$

f(x) =x+5 -x+7

f(x) =12

If x > 7 , then

x+ 5 > 0 and x-7 > 0 ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )+ \left (x-7\right )$

f(x) =x+5+x-7

f(x) =2x-2

The function can be rewritten as

$f(x)= \left\{\begin{matrix} -2x+2 & x < -5\\12 & -5 \le x \le 7 \\ 2x-2 & x > 7 \end{matrix}\right.$

As can be seen from the rewritten definition, every value of in the interval [-5, 7] is a solution of f(x)= 12 , so the correct response is infinitely many solutions.

Report an Error

Example Question #3 : Absolute Value

Consider the quadratic equation

$2x^{2} + 12 = 11x$

Which of the following absolute value equations has the same solution set?

Possible Answers:

$\left |x+ 1 \frac{1}{4} \right | =2 \frac{3}{4}$

$\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$

$\left |x+2 \frac{3}{4} \right | =1 \frac{1}{4}$

None of the other choices gives the correct response.

$\left |x- 1 \frac{1}{4} \right | =2 \frac{3}{4}$

Correct answer:

$\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$

Explanation:

Rewrite the quadratic equation in standard form by subtracting 11x from both sides:

$2x^{2} + 12 = 11x$

$2x^{2} + 12 - 11x = 11x - 11x$

$2x^{2} - 11x + 12= 0$

Solve this equation using the method. We are looking for two integers whose sum is and whose product is 2(12) = 24 ; by trial and error we find they are , . The equation becomes

$2x^{2} - 8x - 3x + 12= 0$

Solving using grouping:

$(2x^{2} - 8x )- (3x - 12)= 0$

2x (x-4)- 3 (x-4)= 0

(2x - 3) (x-4)= 0

By the Zero Product Principle, one of these factors must be equal to 0.

Either

2x- 3 = 0

2x- 3 + 3 = 0 + 3

2x= 3

$\frac{2x}{2}= \frac{3}{2}$

$x= \frac{3}{2} = 1\frac{1}{2}$

x- 4= 0

x- 4 + 4 = 0 + 4

x = 4

The given quadratic equation has solution set $\left \{ 1\frac{1}{2}, 4 \right \}$ , so we are looking for an absolute value equation with this set as well.

This equation can take the form

|x-a| = b

This can be rewritten as the compound equation

x-a = -b x-a = b

Adding to both sides of each equation, the solution set is

x =a -b and x =a + b

Setting these numbers equal in value to the desired solutions, we get the linear system

$a -b =1\frac{1}{2}$

a+ b = 4

Adding and solving for :

$a -b =1\frac{1}{2}$

$\underline{a+ b =4}$

$= 5\frac{1}{2}$

$2a \div 2 = 5\frac{1}{2} \div 2$

$a = 2 \frac{3}{4}$

Backsolving to find :

a+ b = 4

$2 \frac{3}{4}+ b = 4$

$2 \frac{3}{4}+ b - 2 \frac{3}{4}= 4 - 2 \frac{3}{4}$

$b =1 \frac{1}{4}$

The desired absolute value equation is $\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$ .

Report an Error

Example Question #8 : Absolute Value

Solve -8 - (2 |5x - 7|) = -14 .

Possible Answers:

$-\frac{5}{2}, -7$

$\frac{7}{9}, 5$

$\frac{5}{4}, 4$

$\frac{4}{5}, 2$

$\frac{2}{5}, 3$

Correct answer:

$\frac{4}{5}, 2$

Explanation:

First we need to isolate the absolute value term. We do with using some simple algebra:

-8 - (2 |5x - 7|) = -14

- (2 |5x - 7|) = -6

|5x - 7| = 3

Now we solve two equations, one with the right side of the equation positive, one negative. Let's start with the positive:

5x - 7 = 3

5x = 10

x=2

And now the negative:

5x - 7 = -3

5x = 4

$x=\frac{4}{5}$

So our answers are:

$x=\frac{4}{5}, 2$

Report an Error

Example Question #1 : Absolute Value

Solve 7 + (5 |2x - 5|) = -8 .

Possible Answers:

No solutions

4, 1

$\frac{1}{4}$

Correct answer:

No solutions

Explanation:

First, we have to isolate the absolute value:

5 |2x - 5|= -15

|2x - 5|= -3

Let's take a look at our equation right now. It's saying that the absolute value has to be a negative number, which isn't possible. So no solutions exist.

Report an Error

Example Question #42 : Mathematical Relationships

Solve 5 - (4 |-2x - 8|) = 5 .

Possible Answers:

No solutions

Correct answer:

Explanation:

First, we need to isolate the absolute value:

4 |-2x - 8| = 0

|-2x - 8| = 0

Because the equation is set equal to , we can drop the absolute value symbols and solve normally:

-2x - 8= 0

-2x = 8

x=-4

Report an Error

View SAT Subject Test in Mathematics Level 2 Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View SAT Subject Test in Mathematics Level 2 Tutors

Samarth
Certified Tutor

Maharaja Sayajirao University of Baroda, Bachelor of Engineering, Mechanical Engineering. Concordia University-Seward, Master...

View SAT Subject Test in Mathematics Level 2 Tutors

Lilian Natalia
Certified Tutor

University of Texas Rio Grande Valley, Bachelor of Science, Mechanical Engineering. University of Texas Rio Grande Valley, Ma...

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Phoenix, GMAT Tutors in Houston, English Tutors in San Francisco-Bay Area, English Tutors in Miami, French Tutors in Houston, Physics Tutors in San Diego, SSAT Tutors in Denver, ACT Tutors in Seattle, Biology Tutors in Chicago, ISEE Tutors in New York City

Popular Courses & Classes

ACT Courses & Classes in Atlanta, ISEE Courses & Classes in New York City, SAT Courses & Classes in San Diego, SAT Courses & Classes in Boston, SSAT Courses & Classes in Houston, ACT Courses & Classes in Seattle, MCAT Courses & Classes in Seattle, Spanish Courses & Classes in San Diego, GRE Courses & Classes in Seattle, SAT Courses & Classes in Atlanta

Popular Test Prep

SAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Dallas Fort Worth, ISEE Test Prep in New York City, SSAT Test Prep in Chicago, LSAT Test Prep in Atlanta, GRE Test Prep in Chicago, SAT Test Prep in San Diego, SAT Test Prep in Philadelphia, ACT Test Prep in Seattle, MCAT Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

SAT II Math II : Absolute Value

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #32 : Mathematical Relationships

Example Question #33 : Mathematical Relationships

Example Question #1 : Absolute Value

Example Question #61 : Sat Subject Test In Math Ii

Example Question #41 : Mathematical Relationships

Example Question #2 : Absolute Value

Example Question #3 : Absolute Value

Example Question #8 : Absolute Value

Example Question #1 : Absolute Value

Example Question #42 : Mathematical Relationships

All SAT II Math II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: