### All SAT II Math I Resources

## Example Questions

### Example Question #1 : Other Mathematical Relationships

Multiply in modulo 6:

**Possible Answers:**

**Correct answer:**

In modulo 6 arithmetic, a number is congruent to the reainder of its division by 6.

Therefore, since and ,

.

The correct response is 0.

### Example Question #2 : Other Mathematical Relationships

Which is an example of a set that is *not* closed under addition?

**Possible Answers:**

The set of all negative integers

The set of all integers between 1 and 10 inclusive

The set of all positive even integers

The set

All of the sets given in the other responses are closed under addition.

**Correct answer:**

The set of all integers between 1 and 10 inclusive

A set is closed under addition if and only if the sum of any two (not necessarily distinct) elements of the set is also an element of the set.

is closed under addition, since

The set of all negative integers is closed under addition, since any two negative integers can be added to yield a third negative integer.

The set of all positive even integers is closed under addition, since any two positive even integers can be added to yield a third positive even integer.

The remaining set is the set of all integers between 1 and 10 inclusive. It is not closed under addition, as can be seen by this counterexample:

but

### Example Question #3 : Other Mathematical Relationships

varies directly as the square root of .

If then . To the nearest tenth, calculate if .

**Possible Answers:**

**Correct answer:**

varies directly as , which means that for some constant of variation ,

We can write this relationship alternatively as

where the initial conditions can be substituted on the left side and final conditions, on the right. We will be solving for in the equation

### Example Question #4 : Other Mathematical Relationships

varies inversely as the square of and directly as the cube of .

If and , then . Calculate if .

**Possible Answers:**

**Correct answer:**

varies inversely as and directly as the cube of . This means that for some constant of variation ,

We can write this relationship alternatively as

where the initial conditions can be substituted on the left side and final conditions, on the right. We will be solving for in the equation

### Example Question #1 : Basic Single Variable Algebra

Sarah notices her map has a scale of . She measures between Beaver Falls and Chipmonk Cove. How far apart are the cities?

**Possible Answers:**

**Correct answer:**

is the same as

So to find out the distance between the cities

### Example Question #1 : How To Find Direct Variation

If an object is hung on a spring, the elongation of the spring varies directly as the mass of the object. A 20 kg object increases the length of a spring by exactly 7.2 cm. To the nearest tenth of a centimeter, by how much does a 32 kg object increase the length of the same spring?

**Possible Answers:**

**Correct answer:**

Let be the mass of the weight and the elongation of the spring. Then for some constant of variation ,

We can find by setting from the first situation:

so

In the second situation, we set and solve for :

which rounds to 11.5 centimeters.

### Example Question #6 : Other Mathematical Relationships

Sunshine paint is made by mixing three parts yellow paint and one part red paint. How many gallons of yellow paint should be mixed with two quarts of red paint?

(1 gallon = 4 quarts)

**Possible Answers:**

**Correct answer:**

First set up the proportion:

x =

Then convert this to gallons:

### Example Question #4 : Other Mathematical Relationships

Sally currently has 192 books. Three months ago, she had 160 books. By what percentage did her book collection increase over the past three months?

**Possible Answers:**

**Correct answer:**

To find the percentage increase, divide the number of new books by the original amount of books:

She has 32 additional new books; she originally had 160.

### Example Question #5 : Other Mathematical Relationships

Find for the proportion .

**Possible Answers:**

**Correct answer:**

To find x we need to find the direct proportion. In order to do this we need to cross multiply and divide.

From here we mulitply 100 and 1 together. This gets us 100 and now we divide 100 by 4 which results in

### Example Question #4 : Basic Single Variable Algebra

On a map of the United States, Mark notices a scale of . If the distance between New York City and Los Angeles in real life is , how far would the two cities be on Mark's map?

**Possible Answers:**

**Correct answer:**

If the real distance between the two cities is , and = , then we can set up the proportional equation:

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### All SAT II Math I Resources

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