## Arithmetic

Part 1 of our ASVAB Math Study Guide covers arithmetic. This is a 5-part unit in our **ASVAB Free Online Study Guide**. The arithmetic topics include order of operations, absolute value, factorials, exponents, square roots and cube roots.

__Order of Operations__

The order of operations must be followed when solving an equation. You can remember the proper order using the acronym **PEMDAS**, which stands for:

**P**arentheses**E**xponents**M**ultiplication and**D**ivision**A**ddition and**S**ubtraction

These are the four ranks. When you have more than one operation of the same rank, **always work from left to right**. So multiplication and division are done at the same time, moving from left to right. Addition and Subtraction are also done at the same time, moving from left to right.

**Let’s look at an example:**

7 × 3 + (8 − 2)

Review the order of operations in PEMDAS to see what to solve first. Parentheses are first in order of operations, so we must solve what is inside them first, subtracting 2 from 8 equals 6:

7 × 3 + 6

Exponents is the next operation, but we don’t have any, so we move on to multiplication. Multiplying 7 times 3 equals 21:

21 + 6

Now we are left with just an addition problem, and can solve for the final answer:

27

**Let’s consider a second, somewhat more-challenging example:**

7 × (5 − 2)^{2} − 15 ÷ 3 × 2

Parentheses are what we must address first according to “PEMDAS”:

(5 − 2) = 3

We can now replace this in the original equation. Once we have dealt with the operations inside parentheses they should be removed:

7 × 3^{2} − 15 ÷ 3 × 2

Exponents are the next operation. We do have an exponent that must be tackled:

3^{2} = 9

Let’s update our equation:

7 × 9 − 15 ÷ 3 × 2

Multiplication and Division come next in our order of operations (“PEMDAS”). We can start with:

7 × 9 = 63

Updating our equation:

63 − 15 ÷ 3 × 2

We now turn our attention to the “string” of division and multiplication, 15 ÷ 3 × 2, so we must proceed from left to right. We start with 15 ÷ 3:

15 ÷ 3 = 5, 15 ÷ 3 × 2 becomes 5 × 2. And 5 × 2 = 10.

So 15 ÷ 3 × 2 is replaced with 10. Updating our equation:

63 − 10

Addition and Subtraction are the final step in “PEMDAS”:

63 − 10 = 53

We have our result:

7 × (5 − 2)^{2} − 15 ÷ 3 × 2 = 53

It can seem like a lot of work, but none of the steps is complicated. You just need to be methodical and organized in your approach.

__Absolute Value__

Absolute value is the *magnitude* of a number and can be thought of as its distance from zero. An absolute value cannot be negative, even if the number is negative. For example, the absolute values of both 7 and −7 are equal to 7, because they are both the same distance from 0. Absolute value is symbolized with two vertical lines, | |.

*Let’s look at a few examples:*

|5| = 5

|−20| = 20

|5 − 12| = |−7| = 7

__Factorials__

A factorial is the product of an integer and all the integers below it. Factorials are symbolized with an exclamation point. To find the factorial 6 (or 6!), you would multiply 6 × 5 × 4 × 3 × 2 × 1 to get 720.

*Let’s look at an example:*

5! = ?

5 × 4 × 3 × 2 × 1 = 120

__Exponents__

Exponents are used when multiplying a number by itself. For example, 3 squared:

3^{2
}= 3 × 3

= 9

Another example, 5 cubed:

5^{3
}= 5 × 5 × 5

= 125

A number raised to the power of 1 equals itself. For example:

6^{1} = 6

A number raised to the power of 0 equals 1. For example:

6^{0} = 1

When multiplying two powers with the same base, you can add the exponents. For example:

6^{3} × 6^{4
}= 6^{3+4
}= 6^{7}

It makes sense to add the exponents since:

6^{3} = 6 × 6 × 6 and 6^{4} = 6 × 6 × 6 × 6

6^{3} × 6^{4} = (6 × 6 × 6) × (6 × 6 × 6 × 6) = 6 × 6 × 6 × 6 × 6 × 6 × 6 = 6^{7}

__Square Roots__

Finding a square root is the inverse of taking the square of a number—in other words, square root “undoes” square.

Consider 3 squared again:

3^{2} = 3 × 3 = 9

If we wanted to get back from 9 to our original number, 3, we would take the square root of 9:

$\sqrt {9} = 3$

In other words, when we find the square root of a given number, we are figuring out what number we would need to square, or “multiply by itself,” in order to end up with the given number.

What is the square root of 16?

$\sqrt {16} = ?$

When looking for the square root of 16, ask yourself **“ What number squared (multiplied by itself) would give us 16?”**

Since 4 × 4 = 16, our answer is 4.

$\sqrt {16} = 4$

__Cube Roots__

Finding a cube root is a lot like finding a square root. Cube root is the inverse of taking the cube of number—in other words, cube root “undoes” cube.

Consider 5 cubed:

5^{3} = 5 × 5 × 5 = 125

$\sqrt[3]{125} = 5$

If we wanted to get back from 125 to our original number, 5, we would take the cube root of 125: In other words, when we find the cube root of a given number, we are figuring out what number we would need to cube, or “multiply by itself and then multiply by itself again,” in order to end up with the given number.

What is the cube root of 8?

$\sqrt[3]{8} = ?$

When looking for the cube root of 8, ask yourself *“What number cubed (multiplied by itself and then multiplied by itself again) would give us 8?”*

Since 2 × 2 × 2 = 8, our answer is 2.

$\sqrt[3]{8} = 2$

Now that you’ve read our lessons and tips for the Mathematics section of the ASVAB, put your skills to practice with the review quiz below. Try not to reference the above information and treat the questions like a real test.

## Part 1 Review Quiz:

Congratulations - you have completed .

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Question 1 |

### $4 + 7 + (8 − 2) =$

$14$ | |

$16$ | |

$17$ | |

$21$ |

Question 2 |

### $4! =$

$16$ | |

$44$ | |

$24$ | |

$12$ |

Question 3 |

### $|4 × 3 − 20| =$

$− 8$ | |

$8$ | |

$− 68$ | |

$68$ |

Start by solving the problem using PEMDAS. Begin with $4 × 3$, which equals $12$. Next, subtract $20$, which is $− 8$. This is not the final answer, however; the

*absolute value*of $− 8$ is $8$, since that's how far it is from $0$.

Question 4 |

### $12 ÷ 2 × 8 − (3 + 2)^2 =$

$−102$ | |

$−24.25$ | |

$23$ | |

$1849$ |

$12 ÷ 2 × 8 − 25 =$

Next in PEMDAS is multiplication and division. This does not mean you complete multiplication problems before division problems; rather, you complete whichever comes first in the problem. Thus, we solve $12 ÷ 2$ first to get $6$, and then multiply that by $8$ to get $48$. Our equation now looks like this:

$48 − 25 =$

Finish by completing the subtraction problem to get $23$.

Question 5 |

### $2^2 + 3^2 =$

$5$ | |

$10$ | |

$13$ | |

$625$ |

Question 6 |

### $3^2 × 3^3 =$

$3^6$ | |

$6^5$ | |

$9^6$ | |

$3^5$ |

Question 7 |

### $\sqrt{81} =$

$9$ | |

$18$ | |

$40.5$ | |

$6561$ |

$9^2 = 81$, so the answer is $9$.

Question 8 |

### $\sqrt[3]{64} =$

$2$ | |

$4$ | |

$8$ | |

$16$ |

Question 9 |

### $|24 − 13(8 ÷ 4) × 2| =$

$− 44$ | |

$44$ | |

$− 28$ | |

$28$ |

$|24 − 26 × 2|$

Next in PEMDAS is multiplication, so we solve $26 × 2$ to get $52$. Within the absolute value symbols, we have $24 − 52$, which equals $− 28$. Remembering that absolute value is the

*distance*from $0$ and not simply the number within the bars, the answer is $28$.

Question 10 |

### $\sqrt{16} − \sqrt[3]{8} =$

$0$ | |

$2$ | |

$4$ | |

$8$ |

$4 − 2$

Performing simple subtraction, we find the answer is $2$.

List |