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SSAT Upper Level Math : Number Concepts and Operations

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

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All SSAT Upper Level Math Resources

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Example Questions

Example Question #7 : How To Multiply

You are asked to fill in both circles in the statement

$\bigcirc \times \bigcirc \equiv 1 \mod 12$

with the same number from the set

$\left \{ 0, 1, 2, 3,...11\right \}$

to make a true statement.

How many ways can you do this?

Possible Answers:

Two

None of the other responses is correct.

None

Six

Four

Correct answer:

Four

Explanation:

The problem is asking for a number whose square is a number congruent to 1 in modulo 12 arithmetic - that is, a number whose square, when divided by 12, yields remainder 1. This square must be odd, so the number squared must also be odd. Therefore, we need only test the odd integers. We see that:

$1^{2} = 1 ; 1 \div 12 = 0 \textrm{ R }1$

$3^{2} = 9 ; 9 \div 12 = 0 \textrm{ R }9$

$5^{2} = 25 ; 25 \div 12 = 2 \textrm{ R }1$

$7^{2} =49 ; 49 \div 12 = 4 \textrm{ R }1$

$9^{2} = 81 ; 81 \div 12 = 8 \textrm{ R }9$

$11^{2} = 121; 121 \div 12 = 10 \textrm{ R }1$

Four of the integers have squares congruent to 1 in modulo 12 arithmetic.

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Example Question #8 : How To Multiply

Multiply:

$6 \times 2 \textrm{ lbs } 5 \textrm{ oz }$

Possible Answers:

$14 \textrm{ lbs } 8 \textrm{ oz }$

$14 \textrm{ lbs }$

None of the other responses is correct.

$14 \textrm{ lbs } 6 \textrm{ oz }$

$13 \textrm{ lbs } 14 \textrm{ oz }$

Correct answer:

$13 \textrm{ lbs } 14 \textrm{ oz }$

Explanation:

We can write 2 pounds, 5 ounces as just ounces as follows:

$2 \textrm{ lbs } 5 \textrm{ oz } =\left ( 2 \times 16 + 5 \right ) \textrm{ oz } = 37 \textrm{ oz }$

Multiply:

$6 \times 2 \textrm{ lbs } 5 \textrm{ oz }$

$= 6 \times 37 \textrm{ oz }$

$= 222 \textrm{ oz }$

Divide by 16, noting quotient and remainder, to get pounds and ounces:

$222 \div 16 = 13 \textrm{ R } 14$

Therefore, the correct response is 13 pounds, 14 ounces.

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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 #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 $4+\sqrt{6}$ :

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

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

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

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

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

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

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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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SSAT Upper Level Math : Number Concepts and Operations

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #7 : How To Multiply

Example Question #8 : How To Multiply

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

All SSAT Upper Level Math Resources

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