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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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

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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.

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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 $9.32\times10^{-2}$ .

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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.

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Example Question #1111 : Ssat Upper Level Quantitative (Math)

Simplify by rationalizing the denominator:

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

Possible Answers:

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

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

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

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

$- 32-8\sqrt{6}$

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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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 $6+\sqrt{3}$ :

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

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

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

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

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

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

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

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

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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 $\sqrt{14}+\sqrt{5}$ :

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

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

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

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

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

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

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

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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.

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

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

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

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

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

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

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

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

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Example Question #2 : How To Divide Square Roots

Simplify by rationalizing the denominator:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #1112 : Ssat Upper Level Quantitative (Math)

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

Possible Answers:

$\frac{\sqrt2}{4}$

$\frac{1}{8}$

$2\sqrt2$

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: $2\sqrt2$

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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 #1111 : Ssat Upper Level Quantitative (Math)

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 #2 : How To Divide Square Roots

Example Question #1112 : Ssat Upper Level Quantitative (Math)

All SSAT Upper Level Math Resources

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