Call Now to Set Up Tutoring:

(888) 888-0446

SSAT Upper Level Math : Basic Addition, Subtraction, Multiplication and Division

Study concepts, example questions & explanations for SSAT Upper Level Math

Create An Account

All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #9 : How To Multiply

Fill in the circle to yield a true statement:

$6 \times \bigcirc = 5 \mod 12$

Possible Answers:

The other answer choices are incorrect.

Correct answer:

The other answer choices are incorrect.

Explanation:

The problem is asking for a number whose product with 6 yields a number congruent to 5 in modulo 12 arithmetic - that is, a number which, when divided by 12, yields remainder 6.

However, 6 multiplied by any odd number yields a number which, when divided by 12, yields remainder 6, as is demonstrated using our choices:

$6 \times 3 = 18 ; 18 \div 12 = 1 \textrm{ R }6$

$6 \times 5 = 6 ; 30 \div 12 = 2 \textrm{ R }6$

$6 \times 7 = 42 ; 42 \div 12 = 3 \textrm{ R }6$

$6 \times 9 = 54 ; 54 \div 12 = 4 \textrm{ R }6$

Each of the choices yields a product congruent to 6 modulo 12, so none of them is a correct choice.

Report an Error

Example Question #51 : Expressions & Equations

Write in scientific notation.

Possible Answers:

$9.32\times10^{2}$

$9.32\times10^{-1}$

$9.32\times10$

$9.32\times10^{-2}$

$.932\times10$

Correct answer:

$9.32\times10^{-2}$

Explanation:

Scientific notation is used to simplify exceptionally complex numbers and to quickly present the number of significant figures in a given value. The value is converted to an exponent form using base ten, such that only a single-digit term with any given number of decimal places is used to represent the significant figures of the given value. Non-significant zeroes can be omitted from the leading term, and represented only in the base ten exponent.

The given number has three significant figure ( 932 ) so we write out number as 9.32

You must move the decimal place two places to the right, or in other words multiply by 100 . When you move the decimal place to the right, you multiply the number by $10^{-2}$ , so it is $9.32\times10^{-2}$ .

Report an Error

Example Question #1 : How To Divide Square Roots

Divide:

$18\textrm{ lbs }2\textrm{ oz } \div 5$

Possible Answers:

$3\textrm{ lbs }10 \textrm{ oz }$

$3\textrm{ lbs }12 \textrm{ oz }$

$3\textrm{ lbs }6 \textrm{ oz }$

$3\textrm{ lbs }10 \textrm{ oz }$

$3\textrm{ lbs }8 \textrm{ oz }$

Correct answer:

$3\textrm{ lbs }10 \textrm{ oz }$

Explanation:

We can write 18 pounds, 2 ounces as just ounces as follows:

$18\textrm{ lbs }2\textrm{ oz } = (18 \times 16 + 2) \textrm{ oz } = 290 \textrm{ oz }$

Divide:

$290 \div 5 = 58$

Convert 58 ounces to pounds and ounces as follows:

$58 \div 16 = 3 \textrm{ R }10$

That is, 3 pounds 10 ounces.

Report an Error

Example Question #31 : Basic Addition, Subtraction, Multiplication And Division

Simplify by rationalizing the denominator:

$\frac{16}{4-\sqrt{6}}$

Possible Answers:

$\frac{ 32-8\sqrt{6}}{5}$

$- 32-8\sqrt{6}$

$\frac{ 32-8\sqrt{6}}{11}$

$\frac{ 32+8\sqrt{6}}{11}$

$\frac{ 32+8\sqrt{6}}{5}$

Correct answer:

$\frac{ 32+8\sqrt{6}}{5}$

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator, which is $4+\sqrt{6}$ :

$\frac{16}{4-\sqrt{6}}= \frac{16(4+\sqrt{6})}{(4-\sqrt{6}) (4+\sqrt{6})}$

$= \frac{16(4+\sqrt{6})}{4^{2}-(\sqrt{6})^{2} }$

$= \frac{16(4+\sqrt{6})}{16-6}$

$= \frac{16(4+\sqrt{6})}{10}$

$= \frac{8(4+\sqrt{6})}{5}$

$= \frac{ 32+8\sqrt{6}}{5}$

Report an Error

Example Question #32 : Basic Addition, Subtraction, Multiplication And Division

Simplify by rationalizing the denominator:

$\frac{\sqrt{6}}{6-\sqrt{3}}$

Possible Answers:

$\frac{12-2\sqrt{3}}{13}$

$\frac{-\sqrt{2}+ 2 \sqrt{6}}{13}$

$\frac{-\sqrt{2}+ 2 \sqrt{6}}{11 }$

$\frac{\sqrt{2}+ 2 \sqrt{6}}{11 }$

$\frac{12+2\sqrt{3}}{11}$

Correct answer:

$\frac{\sqrt{2}+ 2 \sqrt{6}}{11 }$

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator, which is $6+\sqrt{3}$ :

$\frac{\sqrt{6}}{6-\sqrt{3}}$

$= \frac{\sqrt{6}(6+\sqrt{3})}{(6-\sqrt{3})(6+\sqrt{3})}$

$= \frac{6\cdot \sqrt{6}+\sqrt{3} \cdot \sqrt{6}}{6^{2}-(\sqrt{3})^{2}}$

$= \frac{6 \sqrt{6}+ \sqrt{18}}{36-3 }$

$= \frac{6 \sqrt{6}+ \sqrt{9} \cdot \sqrt{2}}{3 3 }$

$= \frac{6 \sqrt{6}+3\sqrt{2}}{3 3 }$

$= \frac{2 \sqrt{6}+\sqrt{2}}{11 }$

$= \frac{\sqrt{2}+ 2 \sqrt{6}}{11 }$

Report an Error

Example Question #3 : How To Divide Square Roots

Simplify by rationalizing the denominator:

$\frac{12}{\sqrt{14}-\sqrt{5}}$

Possible Answers:

$\frac{4 \sqrt{14}+4\sqrt{5}}{3}$

None of the other responses is correct.

$\frac{4 \sqrt{14}-4\sqrt{5}}{3}$

$\frac{12 \sqrt{14}-12\sqrt{5}}{19}$

$\frac{12 \sqrt{14}+12\sqrt{5}}{19}$

Correct answer:

$\frac{4 \sqrt{14}+4\sqrt{5}}{3}$

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{14}+\sqrt{5}$ :

$\frac{12}{\sqrt{14}-\sqrt{5}}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{(\sqrt{14}-\sqrt{5})(\sqrt{14}+\sqrt{5})}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{(\sqrt{14})^{2}-(\sqrt{5})^{2} }$

$= \frac{12(\sqrt{14}+\sqrt{5})}{14-5}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{9}$

$= \frac{4(\sqrt{14}+\sqrt{5})}{3}$

$= \frac{4 \sqrt{14}+4\sqrt{5}}{3}$

Report an Error

Example Question #5 : How To Divide Square Roots

Simplify by rationalizing the denominator:

$\frac{6}{\sqrt{5}+\sqrt{7}}$

Possible Answers:

$\frac{ \sqrt{7}- \sqrt{5}}{12}$

The correct answer is not among the other responses.

$\frac{ \sqrt{7}- \sqrt{5}}{3}$

$\frac{ \sqrt{7}+ \sqrt{5}}{12}$

$\frac{ \sqrt{7}+ \sqrt{5}}{3}$

Correct answer:

The correct answer is not among the other responses.

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}-\sqrt{7}$ :

$\frac{6}{\sqrt{5}+\sqrt{7}}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{(\sqrt{5}+\sqrt{7})(\sqrt{5}-\sqrt{7})}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{(\sqrt{5})^{2}-(\sqrt{7})^{2}}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{5-7}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{-2}$

$= -3(\sqrt{5}-\sqrt{7} )$

$= 3\sqrt{7}-3\sqrt{5}$

This is not among the given choices.

Report an Error

Example Question #4 : How To Divide Square Roots

Simplify by rationalizing the denominator:

$\frac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}$

Possible Answers:

$\frac{ 5\sqrt {3}+3 \sqrt{5}}{8 }$

$\frac{ 5\sqrt {3}+3 \sqrt{5}}{2 }$

$5\sqrt {3}+3 \sqrt{5}$

$\frac{ 5\sqrt {3}-3 \sqrt{5}}{2 }$

$\frac{ 5\sqrt {3}-3 \sqrt{5}}{8 }$

Correct answer:

$\frac{ 5\sqrt {3}+3 \sqrt{5}}{2 }$

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}+\sqrt{3}$

$\frac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}$

$=\frac{\sqrt{15}( \sqrt{5}+\sqrt{3})}{\left (\sqrt{5}-\sqrt{3} \right )( \sqrt{5}+\sqrt{3})}$

$=\frac{ \sqrt{15} \cdot \sqrt{5}+ \sqrt{15} \cdot \sqrt{3}}{ (\sqrt{5})^{2}-(\sqrt{3} )^{2} }$

$=\frac{ \sqrt{15 \cdot 5}+ \sqrt{15 \cdot 3}}{5-3 }$

$=\frac{ \sqrt{75}+ \sqrt{45}}{2 }$

$=\frac{ \sqrt{25} \cdot \sqrt {3}+ \sqrt{9} \cdot \sqrt{5}}{2 }$

$=\frac{ 5\sqrt {3}+3 \sqrt{5}}{2 }$

Report an Error

Example Question #4 : How To Divide Square Roots

Simplify by rationalizing the denominator:

$\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{6}}$

Possible Answers:

$\sqrt{3}+ \sqrt{2}$

$\frac{\sqrt{3}+ \sqrt{2}}{2}$

$\frac{9\sqrt{3}+4 \sqrt{2} }{6}$

$\frac{9\sqrt{3}+4 \sqrt{2} }{4}$

$\frac{9\sqrt{3}+4 \sqrt{2} }{9}$

Correct answer:

$\sqrt{3}+ \sqrt{2}$

Explanation:

Multiply both numerator and denominator by $\sqrt{6}$ :

$\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{6}}$

$= \frac{(3\sqrt{2}+2\sqrt{3})\cdot \sqrt{6}}{\sqrt{6}\cdot \sqrt{6}}$

$= \frac{3\sqrt{2}\cdot \sqrt{6}+2\sqrt{3}\cdot \sqrt{6}}{6}$

$= \frac{3\sqrt{12} +2\sqrt{18} }{6}$

$= \frac{3\cdot \sqrt{4} \cdot \sqrt{3}+2\cdot \sqrt{9}\cdot \sqrt{2} }{6}$

$= \frac{3\cdot 2 \cdot \sqrt{3}+2\cdot3 \cdot \sqrt{2} }{6}$

$= \frac{6 \cdot \sqrt{3}+6 \cdot \sqrt{2} }{6}$

$= \sqrt{3}+ \sqrt{2}$

Report an Error

Example Question #4 : How To Divide Square Roots

Divide: $\frac{\sqrt{64}}{\sqrt{8}}$

Possible Answers:

$2\sqrt2$

$\frac{\sqrt2}{4}$

$\frac{1}{8}$

Correct answer:

$2\sqrt2$

Explanation:

The values are of the same power. Divide the terms.

$\frac{\sqrt{64}}{\sqrt{8}}= \sqrt{\frac{64}{8}}= \sqrt{8} = \sqrt{4}\times \sqrt{2}= 2\sqrt2$

The correct answer is: $2\sqrt2$

Report an Error

View SSAT- Upper Level Tutors

Carneta
Certified Tutor

Langston University, Bachelor in Arts, Theater Arts.

View SSAT- Upper Level Tutors

Gabby
Certified Tutor

University of Oregon, Bachelor in Arts, Journalism. University of Oregon, Current Grad Student, Communication, General.

View SSAT- Upper Level Tutors

Liban
Certified Tutor

Emory University, Bachelor of Science, Mathematics/Economics.

All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Atlanta, ISEE Tutors in San Francisco-Bay Area, Physics Tutors in Denver, English Tutors in Houston, Reading Tutors in Dallas Fort Worth, English Tutors in Atlanta, Computer Science Tutors in Houston, Biology Tutors in Phoenix, Calculus Tutors in Chicago, GMAT Tutors in Houston

Popular Courses & Classes

SSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Houston, ISEE Courses & Classes in Houston, SSAT Courses & Classes in Denver, SSAT Courses & Classes in Phoenix, GRE Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Washington DC, ACT Courses & Classes in Los Angeles, MCAT Courses & Classes in Miami, SAT Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Phoenix, LSAT Test Prep in Seattle, GRE Test Prep in Chicago, ACT Test Prep in Houston, GRE Test Prep in Houston, ISEE Test Prep in Philadelphia, MCAT Test Prep in Miami, GMAT Test Prep in Boston, GRE Test Prep in Los Angeles, ACT Test Prep in Los Angeles

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

SSAT Upper Level Math : Basic Addition, Subtraction, Multiplication and Division

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #9 : How To Multiply

Example Question #51 : Expressions & Equations

Example Question #1 : How To Divide Square Roots

Example Question #31 : Basic Addition, Subtraction, Multiplication And Division

Example Question #32 : Basic Addition, Subtraction, Multiplication And Division

Example Question #3 : How To Divide Square Roots

Example Question #5 : How To Divide Square Roots

Example Question #4 : How To Divide Square Roots

Example Question #4 : How To Divide Square Roots

Example Question #4 : How To Divide Square Roots

All SSAT Upper Level Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: