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PSAT Math : PSAT Mathematics

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Example Questions

Example Question #1201 : Psat Mathematics

$Simplify: \frac{\sqrt[3]{81}}{\sqrt[3]{3}}$

Possible Answers:

$\frac{\sqrt[3]{81}}{\sqrt[3]{3}} = \frac{3}{1}= 3$

$\frac{\sqrt[3]{9\star 9}}{\sqrt[3]{9}}= \sqrt[3]{9}$

$\sqrt[3]{81\star 3}= \sqrt[3]{243} = 6.24$

$\sqrt[3]{\frac{81}{3}}= \sqrt[3]{27}= 3$

$\frac{81}{3}= 27$

Correct answer:

$\sqrt[3]{\frac{81}{3}}= \sqrt[3]{27}= 3$

Explanation:

$\frac{\sqrt[3]{81}}{\sqrt[3]{3}} = \sqrt[3]{\frac{81}{3}}= \sqrt[3]{27}= 3$

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Example Question #1202 : Psat Mathematics

Rationalize the denominator:

$\frac{1}{2-\sqrt{3}}$

Possible Answers:

$2+\sqrt{3}$

$\sqrt{3}$

$2-\sqrt{3}$

$\frac{2+\sqrt{3}}{7}$

Correct answer:

$2+\sqrt{3}$

Explanation:

The conjugate of $2-\sqrt{3}$ is $2+\sqrt{3}$ .

Now multiply both the numerator and the denominator by $2+\sqrt{3}$

and you get:

$\frac{2+\sqrt{3}}\left ({2-\sqrt{3}} \right )\left ( 2+\sqrt{3} \right )$

$\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right ) =2^{2}-\left ( \sqrt{3} \right )^{2} =4-3 = 1$

Hence we get

$\frac{2+\sqrt{3}}{1} = 2+\sqrt{3}$

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Example Question #1203 : Psat Mathematics

Solve for :

$3^{x-5} = \frac{1}{27}$

Possible Answers:

Correct answer:

Explanation:

$3^{x-5}=\frac{1}{27}=3^{-3}$

x-5=-3

x=2

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Example Question #1204 : Psat Mathematics

Solve for .

$\log _{5}x=2$

Possible Answers:

125

Correct answer:

Explanation:

$log_5{x} = 2$

$x = 5^{2} = 25$

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Example Question #1205 : Psat Mathematics

If,

$log_a{120}= log_b{120}$

What does a=?

Possible Answers:

a=10

a=b

$\frac{a}{b}= 120$

ab=10

b=10

Correct answer:

a=b

Explanation:

If $log_a{x} = log_b{x}$ ,

then a=b .

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Example Question #16 : Exponents And Rational Numbers

$\sqrt{x+3} -x = 1$

Possible Answers:

x=4

x=13

x=9

x=1

x=0

Correct answer:

x=1

Explanation:

From the equation in the problem statement

$\sqrt{x+3} = x + 1$

Now squaring both sides we get

$x+3=x^{2}+2x+1$ this is a quadratic equation which equals

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

and the factors of this equation are

$\left ( x+2 \right )\left ( x-1 \right )$

This gives us $x=-2\ or\ x=1$ .

However, if we plug these solutions back into the original equation, does not create an equality. Therefore, x=-2 is an extraneous solution.

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Example Question #5 : How To Find A Rational Number From An Exponent

For some positive integer , if x^3=27 , what is the value of (x+2)^3 ?

Possible Answers:

125

Correct answer:

125

Explanation:

If x^3=27 , then must equal 3 (Note that cannot be -3 because you need it to be positive.

Now, plug x=3 into the new equation (x+2)^3 :

=(3+2)^3

=5^3

=125

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Example Question #551 : Algebra

Factor 2x² - 5x – 12

Possible Answers:

(x – 4) (2x – 3)

(x + 4) (2x + 3)

(x - 4) (2x + 3)

(x + 4) (2x + 3)

Correct answer:

(x - 4) (2x + 3)

Explanation:

Via the FOIL method, we can attest that x(2x) + x(3) + –4(2x) + –4(3) = 2x² – 5x – 12.

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Example Question #111 : Exponents

x > 0.

Quantity A: (x+3)(x-5)(x)

Quantity B: (x-3)(x-1)(x+3)

Possible Answers:

Quantity A is greater

The two quantities are equal

The relationship cannot be determined from the information given

Quantity B is greater

Correct answer:

Quantity B is greater

Explanation:

Use FOIL:

(x+3)(x-5)(x) = (x² - 5x + 3x - 15)(x) = x³ - 5x² + 3x² - 15x = x³ - 2x² - 15x for A.

(x-3)(x-1)(x+3) = (x-3)(x+3)(x-1) = (x² + 3x - 3x - 9)(x-1) = (x² - 9)(x-1)

(x² - 9)(x-1) = x³ - x² - 9x^₊ 9 for B.

The difference between A and B:

(x³ - 2x² - 15x) - (x³ - x² - 9x^₊ 9) = x³ - 2x² - 15x - x³ + x² + 9x - 9

= - x² - 4x - 9. Since all of the terms are negative and x > 0:

A - B < 0.

Rearrange A - B < 0:

A < B

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Example Question #1206 : Psat Mathematics

Solve for all real values of $\small x$ .

x^3+5x^2-10x=2x^2

Possible Answers:

$2,\ -5$

$0,\ 2,\ 5$

$2,\ 5$

$0,\ 2,\ -5$

Correct answer:

$0,\ 2,\ -5$

Explanation:

x^3+5x^2-10=2x^2

First, move all terms to one side of the equation to set them equal to zero.

x^3+5x^2-2x^2-10x=0

x^3+3x^2-10x=0

All terms contain an $\small x$ , so we can factor it out of the equation.

x(x^2+3x-10)=0

Now, we can factor the quadratic in parenthesis. We need two numbers that add to $\small 3$ and multiply to $\small -10$ .

$-2*5=-10\ \text{and}\ -2+5=3$

x(x-2)(x+5)=0

We now have three terms that multiply to equal zero. One of these terms must equal zero in order for the product to be zero.

$\begin{matrix} x=0 & x-2=0 &x+5=0 \\ x=0 & x=2 & x=-5 \end{matrix}$

Our answer will be $\small x=0,2,-5$ .

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PSAT Math : PSAT Mathematics

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1201 : Psat Mathematics

Example Question #1202 : Psat Mathematics

Example Question #1203 : Psat Mathematics

Example Question #1204 : Psat Mathematics

Example Question #1205 : Psat Mathematics

Example Question #16 : Exponents And Rational Numbers

Example Question #5 : How To Find A Rational Number From An Exponent

Example Question #551 : Algebra

Example Question #111 : Exponents

Example Question #1206 : Psat Mathematics

All PSAT Math Resources

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