### All GRE Math Resources

## Example Questions

### Example Question #1 : How To Multiply Even Numbers

If is even, and is odd. Which of the following must be odd?

**Possible Answers:**

**Correct answer:**

To solve, pick numbers to represent and . Let and . Now try each of the equations given:

.

Only works and is thus our answer.

### Example Question #751 : Gre Quantitative Reasoning

is even

is even

**Quantity A:**

The remainder of

**Quantity B:**

1

**Possible Answers:**

Quantity B is larger.

The relationship between the quantities cannot be determined.

The two quantities are equal.

Quantity A is larger.

**Correct answer:**

Quantity B is larger.

Begin by considering our options. For , it is either the case that:

is even and is odd, or,

is even and is even, or,

is odd and is even

Now, for , it is either the case that:

and are even, or,

and are odd

Now, for , it cannot be the case that both are odd. This means that the only viable option for is the case when both are even.

Therefore, must be even, meaning that the remainder of must be .

Therefore, quantity B is larger.

### Example Question #3 : How To Multiply Even Numbers

is even

is even

Therefore, which of the following must be true about ?

**Possible Answers:**

It could be either even or odd.

It must be even.

It must be odd.

**Correct answer:**

It could be either even or odd.

Recall that when you multiply by an even number, you get an even product.

Therefore, we know the following from the first statement:

is even or is even or both and are even.

For the second, we know this:

Since is even, therefore, can be either even or odd. (Regardless of what it is, we can get an even value for .)

Based on all this data, we can tell nothing necessarily about . If is even, then is even, even if is odd. However, if is odd while is even, then will be even.

### Example Question #531 : Arithmetic

In a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

**Possible Answers:**

**Correct answer:**

In a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

To start, let's calculate the total philosophers:

**Ockham:** * <Number following Durandus>, or

**Aquinas: *** <Number following Ockham>, or

**Scotus: ** divided by , or

Therefore, the total number is:

### Example Question #5 : How To Multiply Even Numbers

Assume and are both even whole numbers.

What is a possible solution for ?

**Possible Answers:**

**Correct answer:**

When two even numbers are multiplied, they must equal an even number. Also, since both variables are said to be even whole numbers, the answer must fit the requirement that its factors are two even numbers multiplied by one another. The only answer that fits both requirements is which can be factored into the even whole numbers and .

### Example Question #101 : Integers

Which of the following integers has an even integer value for all positive integers and ?

**Possible Answers:**

**Correct answer:**

There are certain patterns that can be used to predict whether the product or sum of numbers will be odd or even. The sum of two odd numbers is always even, as is the sum of two even numbers. The sum of an odd number and an even number is always odd. In multiplication the product of two odd numbers is always odd. While the product of even numbers, as well as the product of odd numbers multiplied by even numbers is always even. So for this problem we need to find scenarios where the only possibile answers are even. can only result in even numbers no matter what positive integers are used for and , because must can only result in even products; the same can be said for . The rules provide that the sum of two even numbers is even, so is the answer.

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