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SAT II Math II : SAT Subject Test in Math II

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

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All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Ratios And Proportions

Determine the value of : $\frac{2x-2}{8} = \frac{8x-2}{3}$

Possible Answers:

$-\frac{1}{3}$

$-\frac{11}{29}$

$-\frac{5}{4}$

$-\frac{5}{29}$

$\frac{5}{29}$

Correct answer:

$\frac{5}{29}$

Explanation:

Cross multiply the fractions.

3(2x-2) = 8(8x-2)

Simplify both sides.

6x-6 = 64x-16

Subtract on both sides.

6x-6 -(6x)= 64x-16-(6x)

-6 = 58x-16

Add 16 on both sides.

-6+16 = 58x-16+16

10= 58x

Divide by 58 on both sides.

$\frac{10}{58}= \frac{58x}{58}$

Reduce both fractions.

The answer is: $x = \frac{5}{29}$

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Example Question #2 : Ratios And Proportions

Solve the proportion: $\frac{x}{3} = \frac{2}{5}$

Possible Answers:

$\frac{6}{5}$

$\frac{15}{2}$

$\frac{5}{2}$

Correct answer:

$\frac{6}{5}$

Explanation:

Cross multiply both sides.

5x = 2(3)

Simplify and solve for x.

5x=6

$\frac{5x}{5}=\frac{6}{5}$

The answer is: $\frac{6}{5}$

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Example Question #3 : Ratios And Proportions

Solve the proportion: $\frac{3x}{5} = \frac{1}{3}$

Possible Answers:

$\frac{5}{6}$

$\frac{3}{2}$

$\frac{2}{3}$

$\frac{5}{9}$

Correct answer:

$\frac{5}{9}$

Explanation:

Cross multiply both sides.

3(3x) = 1(5)

9x=5

Divide by 9 on both sides.

$\frac{9x}{9}=\frac{5}{9}$

The answer is: $\frac{5}{9}$

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Example Question #1 : Ratios And Proportions

Solve the proportion: $\frac{3x}{4} = \frac{8}{3}$

Possible Answers:

$\frac{9}{2}$

$\frac{2}{9}$

$\frac{1}{2}$

$\frac{32}{9}$

Correct answer:

$\frac{32}{9}$

Explanation:

Cross multiply the two fractions.

3(3x) = 8(4)

9x=32

Divide by nine on both sides.

$\frac{9x}{9}=\frac{32}{9}$

The answer is: $\frac{32}{9}$

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Example Question #5 : Ratios And Proportions

Solve the proportion: $\frac{2x}{9} = \frac{7}{3}$

Possible Answers:

$\frac{3}{2}$

$\frac{7}{6}$

$\frac{6}{7}$

$\frac{14}{3}$

$\frac{21}{2}$

Correct answer:

$\frac{21}{2}$

Explanation:

Cross multiply the two fractions.

(2x)(3) = (7)(9)

6x = 63

Divide by six on both sides.

$\frac{6x }{6}= \frac{63}{6}$

The answer is: $\frac{21}{2}$

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Example Question #1 : Matrices

Multiply:

$\begin{bmatrix} x&7 \\ y& 3 \end{bmatrix}\begin{bmatrix} 2\\ 7 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix}2 x&14 \\7 y& 21 \end{bmatrix}$

$\begin{bmatrix}2 x&7 \\7 y& 3 \end{bmatrix}$

$\begin{bmatrix} 2x+49\\ 2y+21 \end{bmatrix}$

$\begin{bmatrix}98x\\ 42y\end{bmatrix}$

$\begin{bmatrix} 2x+7y\\ 35 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 2x+49\\ 2y+21 \end{bmatrix}$

Explanation:

The product of a 2 x 2 matrix and a 2 x 1 matrix is a 2 x 1 matrix.

Multiply each row in the first matrix by the column matrix by multiplying elements in corresponding positions, then adding the products, as follows:

$\begin{bmatrix} x&7 \\ y& 3 \end{bmatrix}\begin{bmatrix} 2\\ 7 \end{bmatrix}$

$=\begin{bmatrix} x \cdot 2+ 7 \cdot 7 \\ y \cdot 2 + 3 \cdot 7 \end{bmatrix}$ \

$=\begin{bmatrix} 2x + 49 \\ 2 y+ 21 \end{bmatrix}$

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Example Question #2 : Matrices

Multiply:

$\begin{bmatrix} x& 5\\ -4& 3 \end{bmatrix}\begin{bmatrix} 6\\ y \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 6x+5y \\ 3y-24 \end{bmatrix}$

$\begin{bmatrix}6 x&30\\ -4y& 3 y\end{bmatrix}$

$\begin{bmatrix}6 x&5\\ -4y& 3 \end{bmatrix}$

$\begin{bmatrix} 6x-4y \\ 3y+30 \end{bmatrix}$

$\begin{bmatrix} 30xy \\ -72y \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 6x+5y \\ 3y-24 \end{bmatrix}$

Explanation:

The product of a 2 x 2 matrix and a 2 x 1 matrix is a 2 x 1 matrix.

Multiply each row in the first matrix by the column matrix by multiplying elements in corresponding positions, then adding the products, as follows:

$\begin{bmatrix} x& 5\\ -4& 3 \end{bmatrix}\begin{bmatrix} 6\\ y \end{bmatrix}$

$= \begin{bmatrix} x \cdot 6 + 5\cdot y \\ -4\cdot 6 + 3 \cdot y \end{bmatrix}$

$= \begin{bmatrix} 6x+5y \\ 3y-24 \end{bmatrix}$

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Example Question #1 : Matrices

Define matrix $A = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$

For which of the following matrix values of is the expression defined?

Possible Answers:

The expression is defined for all of the values of given in the other responses.

$B = \begin{bmatrix} 3 \end{bmatrix}$

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$

$B = \begin{bmatrix} 3& -5\\ 7& -1\\4& -7 \end{bmatrix}$

$B = \begin{bmatrix} 3 \\ 7 \\4 \end{bmatrix}$

Correct answer:

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$

Explanation:

For the matrix product to be defined, itis necessary and sufficent for the number of columns in to be equal to the number of rows in .

$A = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$ has two columns. Of the choices, only

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$ has two rows, making it the correct choice.

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Example Question #4 : Matrices

$A = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix}$

$B = \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

Calculate: A - B

Possible Answers:

$\begin{bmatrix} x+4& - x+1\\ -x+12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+13\\ -x-2 & x - 5 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+1\\ -x-2 & x - 5 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+1\\ -x-2 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

Explanation:

To subtract two matrices, subtract the elements in corresponding positions:

$A - B = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix} - \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

$= \begin{bmatrix} x+8 - 4& 7- \left ( x-6 \right )\\ 5-\left ( x - 7 \right ) & x - 7 -2 \end{bmatrix}$

$= \begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

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Example Question #1 : Matrices

Evaluate:

$\begin{vmatrix} y-3&5\\ y+3& 5 \end{vmatrix}$

Possible Answers:

$y^{2} - 34$

10y

$y^{2} + 16$

Correct answer:

Explanation:

The determinant of the matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ is

$\begin{vmatrix} A& B\\ C& D \end{vmatrix} = AD - BC$ .

Substitute A = y-3, C = y + 3 , B = D = 5 :

$\begin{vmatrix} y-3&5\\ y+3& 5 \end{vmatrix}$

= (y-3) 5 - 5 (y+3)

=(5y - 15) - (5y+15)

=5y - 5y - 15-15

= -30

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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Ratios And Proportions

Example Question #2 : Ratios And Proportions

Example Question #3 : Ratios And Proportions

Example Question #1 : Ratios And Proportions

Example Question #5 : Ratios And Proportions

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #1 : Matrices

Example Question #4 : Matrices

Example Question #1 : Matrices

All SAT II Math II Resources

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