### All SAT Math Resources

## Example Questions

### Example Question #1 : Even / Odd Numbers

Which of the following could represent the sum of 3 consecutive odd integers, given that *d* is one of the three?

**Possible Answers:**

3*d* + 4

3*d* + 3

3*d* – 3

3*d* – 9

3*d* – 6

**Correct answer:**

3*d* – 6

If the largest of the three consecutive odd integers is *d*, then the three numbers are (in descending order):

*d*, *d* – 2, *d* – 4

This is true because consecutive odd integers always differ by two. Adding the three expressions together, we see that the sum is 3*d* – 6.

### Example Question #1 : Integers

, where and are distinct positive integers. Which of the following could be values of and ?

**Possible Answers:**

4 and 5

0 and 20

10 and 10

–10 and 30

5 and 15

**Correct answer:**

5 and 15

Since and must be positive, eliminate choices with negative numbers or zero. Since they must be distinct (different), eliminate choices where . This leaves 4 and 5 (which is the only choice that does not add to 20), and the correct answer, 5 and 15.

### Example Question #1 : Even / Odd Numbers

The sum of three consecutive odd integers is 93. What is the largest of the integers?

**Possible Answers:**

**Correct answer:**

Consecutive odd integers differ by 2. If the smallest integer is x, then

x + (x + 2) + (x + 4) = 93

3x + 6 = 93

3x = 87

x = 29

The three numbers are 29, 31, and 33, the largest of which is 33.

### Example Question #1 : Integers

Solve:

**Possible Answers:**

**Correct answer:**

Add the ones digits:

Since there is no tens digit to carry over, proceed to add the tens digits:

The answer is .

### Example Question #1 : Even / Odd Numbers

At a certain high school, everyone must take either Latin or Greek. There are more students taking Latin than there are students taking Greek. If there are students taking Greek, how many total students are there?

**Possible Answers:**

**Correct answer:**

If there are students taking Greek, then there are or students taking Latin. However, the question asks how many total students there are in the school, so you must add these two values together to get:

or total students.

### Example Question #1 : Integers

Add:

**Possible Answers:**

**Correct answer:**

Add the ones digit.

Since the there is a tens digit, use that as the carryover to the next term.

Add the tens digit including the carryover.

The hundreds digit is 7.

Combine the ones digit of each calculation in order.

The answer is:

### Example Question #1 : Integers

Add:

**Possible Answers:**

**Correct answer:**

Add the ones digit.

Carry over the one from the tens digit to the next number.

Add the tens digit with the carry over.

Carry over the one from the tens digit to the hundreds digit.

Add the hundreds digit with the carry over.

The thousands digit has no carry over. The second number has no thousands digit. This means that the thousands is one. Combine all the ones digits from each of the previous calculations.

The correct answer is:

### Example Question #1 : Even / Odd Numbers

When 8 integers are multiplied their product is negative, then at most how many of the integers can be negative?

**Possible Answers:**

7

4

5

8

1

**Correct answer:**

7

When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.

### Example Question #1531 : Sat Mathematics

If n is an odd integer, all of the following must be odd integers EXCEPT:

**Possible Answers:**

n^{2} + 2n

n^{3} + 2

n^{2}

n^{2} + 4n + 4

n^{2} +2n + 1

**Correct answer:**

n^{2} +2n + 1

Let's examine the choice n^{2}. We can rewrite n^{2} as n * n, which would be mutiplying an odd number (because n is odd) by an odd number. Multiplying two odd numbers always produces another odd number. So this can't be the correct answer.

Next, let's look at n^{3} + 2. We can rewrite this as n^{2} * n + 2. We already established that n^{2} must be odd, so then n^{2} * n must also be odd. If we then add 2 to an odd number, we still get an odd number. So we can eliminate this choice as well.

Now, let's look at the choice n^{2} + 2n. Let's factor this as n(n + 2). We know that n must be odd, and we know that n+2 must be odd. Therefore, n(n + 2) is odd, because multiplying two odd numbers gives us an odd number.

Next, we can consider n^{2} + 4n + 4, which we can factor as (n + 2)(n + 2). Once again, n + 2 is an odd number, so multiplying n + 2 by itself will also be odd.

Finally, let's analyze n^{2} + 2n + 1. We can rewrite this as (n + 1)(n + 1). Since n is odd, n+1 must be an even number. When we multiply an even number by an even number, we get an even number, so (n + 1)(n + 1) must be even, and it cannot be odd.

The answer is n^{2} + 2n + 1.

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

If n and m are both positive even integers, which of the following must be odd?

I. (n + 1)(m + 1)

II. nm + 1

III. nm + m

**Possible Answers:**

II only

II and III only

I and II only

I, II, and III

I only

**Correct answer:**

I and II only

Let us analyze I, II, and III one at a time.

Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.

Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.

We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.

Only choice I and II will always produce odd numbers.

The answer is I and II only.

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