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PSAT Math : PSAT Mathematics

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Example Questions

Example Question #1 : How To Find The Square Of A Sum

Expand:

$\left ( 10x+ \frac{1}{5}\right )^{2}$

Possible Answers:

$100x^{2} + \frac{1}{10}$

$100x^{2} + 2x + \frac{1}{10}$

$100x^{2} + 4x + \frac{1}{25}$

$100x^{2} + 2x + \frac{1}{25}$

$100x^{2} + \frac{1}{25}$

Correct answer:

$100x^{2} + 4x + \frac{1}{25}$

Explanation:

Use the perfect square trinomial pattern, setting $A = 10x , B = \frac{1}{5}$ :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$\left ( 10x + \frac{1}{5} \right )^{2} = (10x)^{2} + 2 (10x) \left ( \frac{1}{5} \right ) + \left ( \frac{1}{5} \right ) ^{2}$

$= 100x^{2} + 4x + \frac{1}{25}$

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Example Question #2 : Squaring / Square Roots / Radicals

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{2}$ ?

Possible Answers:

There is no $x^{2}$ term in the expansion of $\left (5 x^{2 } - x + 1\right )^{2}$ .

Correct answer:

Explanation:

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{2}$ is therefore 11.

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Example Question #3 : Squaring / Square Roots / Radicals

If $\left (5 x^{2 } - x + 1\right )^{2}$ is expanded, what is the coefficient of $x^{3}$ ?

Possible Answers:

There is no $x^{3}$ term in the expansion of $\left (5 x^{2 } - x + 1\right )^{2}$ .

Correct answer:

Explanation:

$\left (5 x^{2 } - x + 1\right )^{2}$

$= \left (5 x^{2 } - x + 1\right ) \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \left (5 x^{2 } - x + 1\right ) - x \left (5 x^{2 } - x + 1\right ) + 1 \left (5 x^{2 } - x + 1\right )$

$= 5 x^{2 } \cdot 5 x^{2 } - 5 x^{2 } \cdot x + 5x^{2 } \cdot 1 - x \cdot 5 x^{2 }+ x \cdot x - x \cdot 1+ 1\cdot 5 x^{2 } - 1 \cdot x + 1\cdot 1$

$= 25 x^{4 } - 5 x^{3 } + 5x^{2 } - 5 x^{3 }+ x ^{2 } - x + 5 x^{2 } - x + 1$

$= 25 x^{4 } - 5 x^{3 } - 5 x^{3 } + 5x^{2 }+ 5 x^{2 } + x ^{2 } - x - x + 1$

$= 25 x^{4 } -10 x^{3 } + 11 x^{2 } -2 x + 1$

The coefficient of $x^{3}$ is therefore 10.

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Example Question #3 : Squaring / Square Roots / Radicals

Expand:

$\left ( 10x+ 0.8\right )^{2}$

Possible Answers:

$100x^{2} + 1.6x + 6.4$

$100x^{2}+8x + 0.64$

$100x^{2} + 16x + 0.64$

$100x^{2} +0. 16x + 0.064$

$100x^{2} +0.8x+ 6.4$

Correct answer:

$100x^{2} + 16x + 0.64$

Explanation:

Use the perfect square trinomial pattern, setting A = 10x , B = 0.8 :

$(A + B)^{2} = A^{2} + 2AB + B ^{2}$

$(10x + 0.8)^{2} = \left (10x \right )^{2} + 2 \cdot 10x \cdot 0.8 + 0.8^{2}$

$(10x + 0.8)^{2} = 100x^{2} + 16x + 0.64$

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Example Question #1 : Squaring / Square Roots / Radicals

x² = 36

Quantity A: x

Quantity B: 6

Possible Answers:

The two quantities are equal

Quantity A is greater

The relationship cannot be determined from the information given

Quantity B is greater

Correct answer:

The relationship cannot be determined from the information given

Explanation:

x² = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

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Example Question #2 : Squaring / Square Roots / Radicals

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

Possible Answers:

12√5

48√77

14√2

8√14

4√14

Correct answer:

8√14

Explanation:

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

Hero2

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Example Question #2 : Squaring / Square Roots / Radicals

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

Possible Answers:

2ix^3y^4

$-4x^{2}y^{4}$

-2x^3y^4

4ix^2y^4

Correct answer:

$-4x^{2}y^{4}$

Explanation:

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

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Example Question #3 : Squaring / Square Roots / Radicals

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Possible Answers:

$10\sqrt{15}$

$10\sqrt{3}$

$3\sqrt{2}$

$4\sqrt{30}$

$4\sqrt{15}$

Correct answer:

$4\sqrt{15}$

Explanation:

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}*2\sqrt{3})+(\sqrt{5}*\sqrt{12})$

Combine terms under radicals.

$2\sqrt{3*5}+\sqrt{12*5}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

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Example Question #1 : Complex Numbers

Simplify by rationalizing the denominator:

$\frac{20 + 6i}{8i}$

Possible Answers:

$\frac{ 3 }{4} - \frac{ 5}{2}i$

$\frac{ 5}{2} - \frac{ 3 }{4}i$

$\frac{ 3 }{4} + \frac{ 5}{2}i$

$-\frac{ 3 }{4} + \frac{ 5}{2}i$

$- \frac{ 5}{2} + \frac{ 3 }{4}i$

Correct answer:

$\frac{ 3 }{4} - \frac{ 5}{2}i$

Explanation:

Multiply both numerator and denominator by :

$\frac{20 + 6i}{8i}$

$=\frac{\left (20 + 6i \right ) \cdot i}{8i \cdot i}$

$=\frac{ 20i + 6i^{2} }{8i ^{2}}$

$=\frac{ 20i + 6(-1) }{8 (-1)}$

$=\frac{ 20i -6 }{-8}$

$=\frac{ -6+ 20i }{-8}$

$=\frac{ -6 }{-8} + \frac{ 20i }{-8}$

$=\frac{ 3 }{4} - \frac{ 5}{2}i$

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Example Question #1 : How To Divide Complex Numbers

Divide:

$-100 \div 5i$

Possible Answers:

20i

20 - 20i

-20i

20 + 20 i

Correct answer:

20i

Explanation:

Rewrite in fraction form, and multiply both numerator and denominator by :

$-100 \div 5i$

$= \frac{-100}{5i}$

$= \frac{-100 \cdot i }{5i \cdot i }$

$= \frac{-100 i }{5i ^{2}}$

$= \frac{-100 i }{5 (-1)}$

$= \frac{-100 i }{-5}$

$= \frac{-100 }{-5}\cdot i$

= 20i

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PSAT Math : PSAT Mathematics

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : How To Find The Square Of A Sum

Example Question #2 : Squaring / Square Roots / Radicals

Example Question #3 : Squaring / Square Roots / Radicals

Example Question #3 : Squaring / Square Roots / Radicals

Example Question #1 : Squaring / Square Roots / Radicals

Example Question #2 : Squaring / Square Roots / Radicals

Example Question #2 : Squaring / Square Roots / Radicals

Example Question #3 : Squaring / Square Roots / Radicals

Example Question #1 : Complex Numbers

Example Question #1 : How To Divide Complex Numbers

All PSAT Math Resources

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