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SAT II Math I : SAT Subject Test in Math I

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All SAT II Math I Resources

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Example Questions

Example Question #1 : Solving Inequalities

Solve for x.

$3x+2\geq -7$

Possible Answers:

$x\geq -3$

$x\leq -3$

$x\geq 1$

$x\geq 3$

$x\geq -1$

Correct answer:

$x\geq -3$

Explanation:

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, subtract 2from each side.

$3x+2-2\geq -7-2$

$3x\geq -9$

Now, divide both sides by 2.

$x\geq -3$

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Example Question #1 : Solving Inequalities

Solve the following inequality:

3x+7<9

Possible Answers:

x<-4

x< 9

x<7

x<3

$x< \frac{2}{3}$

Correct answer:

$x< \frac{2}{3}$

Explanation:

To solve for an inequality, you solve like you would for a single variable expression and get by itself.

First, subtract from both sides to get,

3x<2 .

Then divide both sides by and your final answer will be,

$x< \frac{2}{3}$ .

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Example Question #5 : Solving Inequalities

Solve the inequality: 3-(3-2x)<10

Possible Answers:

$x>-\frac{1}{5}$

x>5

$x<\frac{1}{5}$

x<5

x<-5

Correct answer:

x<5

Explanation:

Simplify the left side.

3-3+2x<10

The inequality becomes:

2x<10

Divide by two on both sides.

$\frac{2}{2}x<\frac{10}{2}$

The answer is: x<5

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Example Question #4 : Solving Inequalities

Solve the inequality: $2x+6\leq 9x-3$

Possible Answers:

$x\geq-\frac{3}{11}$

$x\geq-\frac{9}{7}$

$x\geq\frac{9}{7}$

$x\leq-\frac{9}{7}$

$x\leq-\frac{3}{11}$

Correct answer:

$x\geq\frac{9}{7}$

Explanation:

Subtract on both sides.

$2x+6-2x\leq 9x-3-2x$

$6\leq 7x-3$

Add 3 on both sides.

$6+3\leq 7x-3+3$

$9\leq 7x$

Divide by 7 on both sides.

$\frac{9}{7}\leq \frac{7x}{7}$

$\frac{9}{7}\leq x$

The answer is: $x\geq\frac{9}{7}$

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Example Question #1 : Simplifying Expressions

Simplify the expression.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

Possible Answers:

$\small \frac{x^2y^5}{z^3}$

$\small \frac{xy^3}{z^3}$

$\small \frac{z^3}{x^2y^5}$

$\small \frac{z^3}{xy^3}$

Correct answer:

$\small \frac{xy^3}{z^3}$

Explanation:

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

Because we are only multiplying terms in the numerator, we can disregard the parentheses.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}=\frac{x^2y^2x^3y^4z}{x^4y^3z^4}$

To combine like terms in the numerator, we add their exponents.

$\small \frac{x^5y^6z}{x^4y^3z^4}$

To combine like terms between the numerator and denominator, subtract the denominator exponent from the numerator exponent.

$x^{5-4}y^{6-3}z^{1-4}=xy^3z^{-3}$

Remember that any negative exponents stay in the denominator.

$\frac{xy^3}{z^3}$

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Example Question #1 : Simplifying Expressions

Give the value of that makes the polynomial $9x^{2}+ Nx + 100$ the square of a linear binomial.

Possible Answers:

None of the other responses gives a correct answer.

N = 60

N = 30

N = 90

N = 81

Correct answer:

N = 60

Explanation:

A quadratic trinomial is a perfect square if and only if takes the form

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$9x^{2}+ Nx + 100 = (3x)^{2} + Nx + 10^{2}$ , so

A = 3x and B = 10 .

For $9x^{2}+ Nx + 100$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 3x \cdot 10 = 60x$ ,

so N = 60 . This is the correct choice.

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Example Question #3 : Simplifying Expressions

Factor:

$27Q^{3}- 270Q^{2}+900Q -1,000$

Possible Answers:

$(3Q-10)(9Q^{2}- 30 Q +100)$

The polynomial is prime.

$(3Q-10)(3Q+10)^{2}$

$(3Q - 10)^{3}$

$(3Q-10)(9Q^{2}+ 30 Q +100)$

Correct answer:

$(3Q - 10)^{3}$

Explanation:

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot 90Q^{2}+3\cdot 300Q -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot 9Q^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

This can be factored out as the cube of a difference, where A = 3Q, B = 10 :

$A^{3}- 3A^{2}B+3AB^{2} -B^{3} = (A - B)^{3}$

Therefore,

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$= (3Q - 10)^{3}$

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Example Question #1 : Simplifying Expressions

Which of the following linear binomials is a factor of the polynomial $x^{3} -34x- 240$ ?

Possible Answers:

x - 8

x+6

x + 4

x- 6

x -10

Correct answer:

x - 8

Explanation:

By the factor theorem, a polynomial p (x) is divisible by the linear binomial x - c if and only if p(c) = 0 . We can use this fact to test each of the binomials by evaluating the dividend for the appropriate value of .

x + 4 : Evaluate the polynomial at x = -4 :

$x^{3} -34x- 240 = (-4)^{3} -34(-4)- 240 = -64 +136 - 240 = -170 \neq 0$

x + 6 : Evaluate the polynomial at x = -6 :

$x^{3} -34x- 240 = (-6)^{3} -34(-6)- 240 = -216+204 -240= -252 \neq 0$

x- 6 : Evaluate the polynomial at x = 6 :

$x^{3} -34x- 240 =6^{3} -34(6)- 240 = 216-204 -240= -228 \neq 0$

x- 8 : Evaluate the polynomial at x = 8 :

$x^{3} -34x- 240 =8^{3} -34(8)- 240 = 512-272 -240= 0$

x- 10 : Evaluate the polynomial at x = 10 :

$x^{3} -34x- 240 =10^{3} -34(10)- 240 = 1,000-340 -240= 420$

The dividend assumes the value of 0 at x = 8 , so of the choices given, x- 8 is the factor.

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Example Question #5 : Simplifying Expressions

What is 3t+ 7 increased by 40%?

Possible Answers:

4.2 t + 9.8

3t+47

4.2+7

1.2t+2.8

3t+ 9.8

Correct answer:

4.2 t + 9.8

Explanation:

A number increased by 40% is equivalent to 100% of the number plus 40% of the number. This is taking 140% of the number, or, equivalently, multiplying it by 1.4.

Therefore, 3t+ 7 increased by 40% is 1.4 times this, or

$1.4 \left (3t+ 7 \right ) = 1.4 \cdot 3t + 1.4 \cdot 7 = 4.2 t + 9.8$

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Example Question #2 : Simplifying Expressions

Which of the following is a prime factor of $x^{6} - 1$ ?

Possible Answers:

$x^{3} + 2x + 1$

$x^{3} +3 x + 1$

$x^{2} +3 x + 1$

$x^{2} + 1$

$x^{2} + x + 1$

Correct answer:

$x^{2} + x + 1$

Explanation:

$x^{6} - 1$ is the difference of two squares:

$x^{6} - 1 =\left ( x^{3} \right ) ^{2}- 1$

As such, it can be factored as follows:

$x^{6} - 1 =\left ( x^{3} \right ) ^{2}- 1 = \left ( x^{3} + 1 \right ) \left ( x^{3} - 1 \right )$

The first factor is the sum of cubes and the second is the difference of cubes; each can be factored further:

$x^{3} +1 = (x+1) \left ( x^{2}-x \cdot 1 + 1^{2} \right ) = (x+1) \left ( x^{2} - x + 1 \right )$

$x^{3} - 1 = (x-1) \left ( x^{2} + x \cdot 1 + 1^{2} \right ) = (x-1) \left ( x^{2} + x + 1 \right )$

Therefore,

$x^{6} - 1 = \left ( x^{3} + 1 \right ) \left ( x^{3} - 1 \right ) = (x+1) \left ( x^{2} - x + 1 \right ) (x-1) \left ( x^{2} + x + 1 \right )$

Of the choices, $x^{2} + x + 1$ appears in the prime factorization and is therefore the correct choice.

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #1 : Solving Inequalities

Example Question #1 : Solving Inequalities

Example Question #5 : Solving Inequalities

Example Question #4 : Solving Inequalities

Example Question #1 : Simplifying Expressions

Example Question #1 : Simplifying Expressions

Example Question #3 : Simplifying Expressions

Example Question #1 : Simplifying Expressions

Example Question #5 : Simplifying Expressions

Example Question #2 : Simplifying Expressions

All SAT II Math I Resources

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