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SAT II Math I : Mathematical Relationships

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

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Example Questions

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Example Question #1 : Absolute Value

Solve: $-3\left |2x+1 \right | = -3$

Possible Answers:

0,1

$\pm1$

$\pm\frac{1}{2}$

$\textup{There is no solution.}$

-1,0

Correct answer:

-1,0

Explanation:

Divide both sides by negative three.

$\frac{-3\left |2x+1 \right |}{-3} = \frac{-3}{-3}$

$\left |2x+1 \right |=1$

Since the lone absolute value is not equal to a negative, we can continue with the problem. Split the equation into its positive and negative components.

2x+1 = 1

-(2x+1) = 1

Evaluate the first equation by subtracting one on both sides, and then dividing by two on both sides.

2x+1-1= 1-1

2x=0

x=0

Evaluate the second equation by dividing a negative one on both sides.

$\frac{-(2x+1) }{-1}= \frac{1}{-1}$

2x+1 = -1

Subtract one on both sides.

2x+1 -1 = -1 -1

2x=-2

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{-2}{2}$

x=-1

The answers are: -1,0

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Example Question #71 : Sat Subject Test In Math I

What property of arithmetic is demonstrated below?

$m \angle A \cdot 2 = 2 \cdot m \angle A$

Possible Answers:

Commutative

Transitive

Associative

Distributive

Reflexive

Correct answer:

Commutative

Explanation:

The statement shows that two numbers can be multiplied in either order to yield the same product. This is the commutative property of multiplication.

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Example Question #72 : Sat Subject Test In Math I

What property of arithmetic is demonstrated below?

If $m \angle 3 = 90$ and $m \angle 4 = 90$ , then $m \angle 3 = m \angle 4$

Possible Answers:

Commutative

Symmetric

Associative

Transitive

Reflexive

Correct answer:

Transitive

Explanation:

The statement demonstrates that two expressions both equal to a third expression are equal to each other. This is the transitive property of equality.

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Example Question #73 : Sat Subject Test In Math I

What property of arithmetic is demonstrated below?

$1 \cdot m \angle 1 = m \angle 1$

Possible Answers:

Reflexive

Symmetric

Identity

Inverse

Commutative

Correct answer:

Identity

Explanation:

The idea that one multiplied by any number yields the latter number as the product is the identity property of multiplication.

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Example Question #1 : Ratios And Proportions

Convert $\small \frac{3}{8}$ into a percentage.

Possible Answers:

None of the other answers are correct.

Correct answer:

Explanation:

Set up a proportion:

$\frac{3}{8}=\frac{x}{100}$

Cross-multiply:

8x=300

Isolate by dividing by 8:

$x=\frac{300}{8}$

Finally, simplify:

$\frac{300}{8}=\frac{75}{2}=37\frac{1}{2}=37.5$

The fraction is therefore equivalent to .

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Example Question #1 : Ratios And Proportions

Convert the following fraction into a percent.

$\small \frac{7}{12}$

Possible Answers:

$\small \small \small 58.33$ %

63.26 %

$\small .58$ %

$\small 171.43$ %

.63 %

Correct answer:

$\small \small \small 58.33$ %

Explanation:

Set up a proportion:

$\small \frac{7}{12}=\frac{x}{100}$

Cross multiply:

$\small 12x=700$

Isolate $\small x$ by dividing both sides by 12:

$\small x = \frac{700}{12}$

Simplify:

$\small \frac{700}{12} = \frac{350}{6} = \frac{175}{3} = 58.33$

(Hint: If you are unable to use long division to solve the fraction $\small \frac{175}{3}$ , take a look at your answer choices and see if you can narrow down your choices. You know that value is going to be greater than one, since 175 is greater than 3. You also know that dividing any number by 3 is going to result in a decimal of .33, .66, or no decimal at all. With these two pieces of information, you should be able to isolate the correct answer without actually solving for it like I did in the last step.)

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Example Question #1 : Ratios And Proportions

Solve for :

$\frac{8}{11} = \frac{11x}{8}$

Possible Answers:

$\frac{64}{121}$

$\frac{121}{64}$

$\frac{1}{x}$

Correct answer:

$\frac{64}{121}$

Explanation:

To solve this proportion, cross multiply.

$\frac{8}{11} = \frac{11x}{8}$

64=121x

From here to solve, divide both sides by 121.

$x=\frac{64}{121}$

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Example Question #4 : Ratios And Proportions

If for every $\frac{1}{2}$ inch on a map it represents miles, how many inches will be needed for a distance of 275 miles?

Possible Answers:

5.5

4.5

6.5

Correct answer:

5.5

Explanation:

The ratio of inches to miles is

$\frac{0.5}{25}$

so you can set up a proportion of $\frac{0.5}{25}=\frac{x}{275}$ .

You then cross multiply to get

25x= 137.5 .

Then you solve for the variable by dividing by which gives us

137.5/25=5.5 .

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Example Question #1 : Ratios And Proportions

In a geometric sequence the ratio of the first term to the third term is 9:1. What is the ratio of the second term to the fifth term?

Possible Answers:

27:1

81:1

9:1

3:1

5:2

Correct answer:

27:1

Explanation:

Recall that in a geometric sequence there is a constant ratio between any pair of consecutive terms:

$r \cdot a_{n} =a_{n+1}$

If the ratio of the first term to the third term is 9:1 then the ratio of two consecutive terms is a number that when multiplied by itself is 9:

r^2 = 9

r=3

Therefore the common ratio for this sequence is 3.

There are three terms between term 2 and 5 so the common ratio will be applied to the second term three times to get the fifth term:

$r^3 \cdot a_{2} = a_{5}$

Since for this sequence

$3^{3} = 27$

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Example Question #1 : Matrices

Define $A = \begin{bmatrix} 5& 10\\ -6 & -12 \end{bmatrix}$ .

Give $A^{-1}$ .

Possible Answers:

$A^{-1} = \begin{bmatrix} -\frac{1}{24}& - \frac{1}{12}\\ \frac{1}{20} & -\frac{1}{10 }\end{bmatrix}$

$A^{-1}$ is not defined.

$A^{-1} = \begin{bmatrix} -12& -10\\ 6 & -5\end{bmatrix}$

$A^{-1} = \begin{bmatrix} 5& -10\\ 6 & -12 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -\frac{1}{10}& - \frac{1}{12}\\ \frac{1}{20} & -\frac{1}{24}\end{bmatrix}$

Correct answer:

$A^{-1}$ is not defined.

Explanation:

The inverse of a 2 x 2 matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ , if it exists, is the matrix

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

First, we need to establish that the inverse is defined, which it is if and only if determinant $AD - BC \neq 0$ .

Set A = 5, B = 10, C = -6, D =-12 , and check:

$AD - BC = 5 \cdot \left (-12 \right ) - 10 \cdot \left ( -6\right ) =-60 - (-60 )= -60 + 60 = 0$

The determinant is equal to 0, so does not have an inverse.

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All SAT II Math I Resources

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SAT II Math I : Mathematical Relationships

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

All SAT II Math I Resources

Example Questions

Example Question #1 : Absolute Value

Example Question #71 : Sat Subject Test In Math I

Example Question #72 : Sat Subject Test In Math I

Example Question #73 : Sat Subject Test In Math I

Example Question #1 : Ratios And Proportions

Example Question #1 : Ratios And Proportions

Example Question #1 : Ratios And Proportions

Example Question #4 : Ratios And Proportions

Example Question #1 : Ratios And Proportions

Example Question #1 : Matrices

All SAT II Math I Resources

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