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PSAT Math : Arithmetic

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Example Questions

Example Question #1 : How To Find The Square Of An Integer

How many integers from 20 to 80, inclusive, are NOT the square of another integer?

Possible Answers:

$\displaystyle 59$

$\displaystyle 58$

$\displaystyle 56$

$\displaystyle 60$

$\displaystyle 57$

Correct answer:

$\displaystyle 57$

Explanation:

First list all the integers between 20 and 80 that are squares of another integer:

5² = 25

6² = 36

7²= 49

8² = 64

In total, there are 61 integers from 20 to 80, inclusive. 61 – 4 = 57

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Example Question #5 : How To Find The Square Of An Integer

Let the universal set $\displaystyle U$ be the set of all positive integers.

Let $\displaystyle A$ be the set of all multiples of 3; let $\displaystyle B$ be the set of all multiples of 7; let $\displaystyle C$ be the set of all perfect square integers. Which of the following statements is true of 243?

Note: $\displaystyle A'$ means "the complement of $\displaystyle A$", etc.

Possible Answers:

$243 \in A' \cap B \cap C$ $\displaystyle 243 \in A' \cap B \cap C$

$243 \in A \cap B \cap C$ $\displaystyle 243 \in A \cap B \cap C$

$243 \in A \cap B' \cap C'$ $\displaystyle 243 \in A \cap B' \cap C'$

$243 \in A \cap B' \cap C$ $\displaystyle 243 \in A \cap B' \cap C$

$243 \in A' \cap B \cap C'$ $\displaystyle 243 \in A' \cap B \cap C'$

Correct answer:

$243 \in A \cap B' \cap C'$ $\displaystyle 243 \in A \cap B' \cap C'$

Explanation:

$243 \div 3 = 81$ $\displaystyle 243 \div 3 = 81$, so 243 is divisible by 3. $243 \in A$ $\displaystyle 243 \in A$.

$243 \div 7= 34 \textup{ R }5$ $\displaystyle 243 \div 7= 34 \textup{ R }5$, so 243 is not divisible by 7. $243 \notin B$ $\displaystyle 243 \notin B$ - that is, $243 \in B'$ $\displaystyle 243 \in B'$.

$15 = \sqrt{225}< \sqrt{243} < \sqrt{256} = 16$ $\displaystyle 15 = \sqrt{225}< \sqrt{243} < \sqrt{256} = 16$, 243 is not a perfect square integer. $243 \notin C$ $\displaystyle 243 \notin C$ - that is, $243 \in C'$ $\displaystyle 243 \in C'$.

Since 243 is an element of $\displaystyle A$, $\displaystyle B'$, and $\displaystyle C'$, it is an element of their intersection. The correct choice is that

$243 \in A \cap B' \cap C'$ $\displaystyle 243 \in A \cap B' \cap C'$

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Example Question #162 : Arithmetic

Consider the inequality:

$x< x^{5}< x^{4}$ $\displaystyle x< x^{5}< x^{4}$

Which of the following could be a value of $\displaystyle x$?

Possible Answers:

$x=-1$ $\displaystyle x=-1$

$x=-\frac{4}{3}$ $\displaystyle x=-\frac{4}{3}$

There is no possible value for $x$ $\displaystyle x$

$x=1$ $\displaystyle x=1$

$x=-\frac{3}{4}$ $\displaystyle x=-\frac{3}{4}$

Correct answer:

$x=-\frac{3}{4}$ $\displaystyle x=-\frac{3}{4}$

Explanation:

Notice how $x^4$ $\displaystyle x^4$ is the greatest value. This often means that $x$ $\displaystyle x$ is negative as $(-1)^n=-1$ $\displaystyle (-1)^n=-1$ when $\dpi{100} n$ $\displaystyle \dpi{100} n$ is odd and $(-1)^n=1$ $\displaystyle (-1)^n=1$ when $\dpi{100} n$ $\displaystyle \dpi{100} n$ is even.

Let us examine the first choice, $x=-\frac{3}{4}$ $\displaystyle x=-\frac{3}{4}$

$x^5=-\frac{3^5}{4^5}=-\frac{243}{1024}> -\frac{3}{4}$ $\displaystyle x^5=-\frac{3^5}{4^5}=-\frac{243}{1024}> -\frac{3}{4}$

This can only be true of a negative value that lies between zero and one.

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Example Question #1 : How To Find The Square Of An Integer

$\sqrt{y^2-63}=9$ $\displaystyle \sqrt{y^2-63}=9$

In the equation above, if $\displaystyle y$ is a positive integer, what is the value of $\displaystyle y$?

Possible Answers:

$\displaystyle 6$

$\displaystyle 11$

$\displaystyle -12$

$\displaystyle 12$

$\displaystyle 9$

Correct answer:

$\displaystyle 12$

Explanation:

Begin by squaring both sides of the equation:

$\sqrt{y^2-63}=9$ $\displaystyle \sqrt{y^2-63}=9$

$(\sqrt{y^2-63})^2=9^2$ $\displaystyle (\sqrt{y^2-63})^2=9^2$

$\displaystyle y^2-63=81$

Now solve for y:

$\displaystyle y^2-63+63=81+63$

y^2=144 $\displaystyle y^2=144$

y=12 $\displaystyle y=12$

Note that $\displaystyle y$ must be positive as defined in the original question. In this case, $\displaystyle y$ must be 12.

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Example Question #2 : Factoring And Simplifying Square Roots

Simplify. Assume all variables are positive real numbers.

$\sqrt{16x^{5}{y^{2}}}$ $\displaystyle \sqrt{16x^{5}{y^{2}}}$

Possible Answers:

$4x^{2}y\sqrt{x}$ $\displaystyle 4x^{2}y\sqrt{x}$

$4x^{2}y$ $\displaystyle 4x^{2}y$

$4x^{5}y$ $\displaystyle 4x^{5}y$

$4x^{2}y^{2}$ $\displaystyle 4x^{2}y^{2}$

$4x^{5}y^{2}$ $\displaystyle 4x^{5}y^{2}$

Correct answer:

$4x^{2}y\sqrt{x}$ $\displaystyle 4x^{2}y\sqrt{x}$

Explanation:

$\sqrt{16x^{5}{y^{2}}}$ $\displaystyle \sqrt{16x^{5}{y^{2}}}$

The index coefficent in $\sqrt[n]{b}$ $\displaystyle \sqrt[n]{b}$ is represented by $\displaystyle n$. When no index is present, assume it is equal to 2. $\displaystyle b$ under the radical is known as the radican, the number you are taking a root of.

First look for a perfect square, $\sqrt{16}=4$ $\displaystyle \sqrt{16}=4$

Then to your Variables $\sqrt{x^{5}y^{2}}$ $\displaystyle \sqrt{x^{5}y^{2}}$

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$\sqrt{x^{5}} \rightarrow 2\cdot2=4 +1$ $\displaystyle \sqrt{x^{5}} \rightarrow 2\cdot2=4 +1$

So you would take out a $x^{2}$ $\displaystyle x^{2}$ and would be left with a $\sqrt{x}$ $\displaystyle \sqrt{x}$

$\sqrt{y^{2}} \rightarrow 2\cdot1=2$ $\displaystyle \sqrt{y^{2}} \rightarrow 2\cdot1=2$

*Dividing the radican exponent by the index - gives you the number of variables that should be pulled out.

The final answer would be $4x^{2}y\sqrt{x}$ $\displaystyle 4x^{2}y\sqrt{x}$.

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Example Question #3 : Factoring And Simplifying Square Roots

Simplify. Assume all integers are positive real numbers.

$\sqrt[3]{24x^{16}y^{12}}$ $\displaystyle \sqrt[3]{24x^{16}y^{12}}$

Possible Answers:

$2x^{5}y^{4}$ $\displaystyle 2x^{5}y^{4}$

$2x^{5}y^{4}\sqrt[3]{3x}$ $\displaystyle 2x^{5}y^{4}\sqrt[3]{3x}$

$8x^{16}y^{12}$ $\displaystyle 8x^{16}y^{12}$

$2x^{2}y^{3}\sqrt[3]{3xy}$ $\displaystyle 2x^{2}y^{3}\sqrt[3]{3xy}$

$2x^{8}y^{4}\sqrt[3]{3x^{2}}$ $\displaystyle 2x^{8}y^{4}\sqrt[3]{3x^{2}}$

Correct answer:

$2x^{5}y^{4}\sqrt[3]{3x}$ $\displaystyle 2x^{5}y^{4}\sqrt[3]{3x}$

Explanation:

$\sqrt[3]{24x^{16}y^{12}}$ $\displaystyle \sqrt[3]{24x^{16}y^{12}}$

Index of $\displaystyle 3$ means the cube root of Radican $24x^{16}y^{12}$ $\displaystyle 24x^{16}y^{12}$

Find a perfect cube in $\displaystyle 24$ $\rightarrow$ $\displaystyle \rightarrow$ $\sqrt[3]{8\cdot 3}$ $\displaystyle \sqrt[3]{8\cdot 3}$

Simplify the perfect cube, giving you $2\sqrt[3]{3}$ $\displaystyle 2\sqrt[3]{3}$.

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$ $\displaystyle x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$

$y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$ $\displaystyle y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$

The final answer would be

$2x^{5}y^{4}\sqrt[3]{3x}$ $\displaystyle 2x^{5}y^{4}\sqrt[3]{3x}$

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Example Question #1763 : Psat Mathematics

Simplify square roots. Assume all integers are positive real numbers.

Simplify as much as possible. List all possible answers.

1a. $\sqrt{12}$ $\displaystyle \sqrt{12}$

1b. $\sqrt{48}$ $\displaystyle \sqrt{48}$

1c. $\sqrt[3]{24}$ $\displaystyle \sqrt[3]{24}$

Possible Answers:

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$ and $\sqrt{3}$ $\displaystyle \sqrt{3}$ and $2\sqrt[3]{3}$ $\displaystyle 2\sqrt[3]{3}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$ and $4\sqrt{3}$ $\displaystyle 4\sqrt{3}$ and $2\sqrt[3]{3}$ $\displaystyle 2\sqrt[3]{3}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$ and $4\sqrt{3}$ $\displaystyle 4\sqrt{3}$ and $8\sqrt{3}$ $\displaystyle 8\sqrt{3}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$ and $4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

Correct answer:

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$ and $4\sqrt{3}$ $\displaystyle 4\sqrt{3}$ and $2\sqrt[3]{3}$ $\displaystyle 2\sqrt[3]{3}$

Explanation:

When simplifying radicans (integers under the radical symbol), we first want to look for a perfect square. For example, $\sqrt{12}$ $\displaystyle \sqrt{12}$ is not a perfect square. You look to find factors of $\displaystyle 12$ to see if there is a perfect square factor in $\displaystyle 12$, which there is.

1a. $\sqrt{12}=\sqrt{4\cdot 3}=\sqrt{4}\cdot\sqrt{3}= 2\sqrt{3}$ $\displaystyle \sqrt{12}=\sqrt{4\cdot 3}=\sqrt{4}\cdot\sqrt{3}= 2\sqrt{3}$

Do the same thing for $\sqrt{48}$ $\displaystyle \sqrt{48}$.

1b. $\sqrt{48}=\sqrt{16\cdot 3}=\sqrt{16}\cdot\sqrt{3}=4\sqrt{3}$ $\displaystyle \sqrt{48}=\sqrt{16\cdot 3}=\sqrt{16}\cdot\sqrt{3}=4\sqrt{3}$

1c.Follow the same procedure except now you are looking for perfect cubes.

$\sqrt[3]{24}=\sqrt[3]{8\cdot 3}=\sqrt[3]{8}\cdot\sqrt[3]{3}=2\sqrt[3]{3}$ $\displaystyle \sqrt[3]{24}=\sqrt[3]{8\cdot 3}=\sqrt[3]{8}\cdot\sqrt[3]{3}=2\sqrt[3]{3}$

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Example Question #1 : Simplifying Square Roots

Simplify

9 ÷ √3

Possible Answers:

none of these

not possible

3√3

Correct answer:

3√3

Explanation:

in order to simplify a square root on the bottom, multiply top and bottom by the root

Asatsimplifysquare_root

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Example Question #31 : Basic Squaring / Square Roots

Simplify:

√112

Possible Answers:

4√10

4√7

10√12

Correct answer:

4√7

Explanation:

√112 = {√2 * √56} = {√2 * √2 * √28} = {2√28} = {2√4 * √7} = 4√7

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Example Question #1 : Factoring And Simplifying Square Roots

Simplify:

√192

Possible Answers:

8√3

4√3

None of these

4√2

8√2

Correct answer: 8√3

Explanation:

√192 = √2 X √96

√96 = √2 X √48

√48 = √4 X√12

√12 = √4 X √3

√192 = √(2X2X4X4) X √3

= √4X√4X√4 X √3

= 8√3

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PSAT Math : Arithmetic

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : How To Find The Square Of An Integer

Example Question #5 : How To Find The Square Of An Integer

Example Question #162 : Arithmetic

Example Question #1 : How To Find The Square Of An Integer

Example Question #2 : Factoring And Simplifying Square Roots

Example Question #3 : Factoring And Simplifying Square Roots

Example Question #1763 : Psat Mathematics

Example Question #1 : Simplifying Square Roots

Example Question #31 : Basic Squaring / Square Roots

Example Question #1 : Factoring And Simplifying Square Roots

All PSAT Math Resources

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