Mathematics Study Guide for the Wonderlic Personnel Test

Algebra

Before we discuss how to address this type of problem, let’s review some facts about them in relation to the whole Wonderlic® test.

• Out of 50 questions on the test, there will only be 2 or 3 of these.
• Almost nobody finishes all 50 problems.
• You can miss quite a few of the 50 questions and still obtain a really good score.
• Each of these questions is almost guaranteed to take you more—sometimes much more—than 14 seconds to answer.

Now, in the whole scheme of things, since each question is worth the same—1 point—doesn’t it seem like this type of question would be the type to skip? We think so. Even if you are really good at math, just going through the steps necessary (creating one or multiple equations and solving them) is just too much of a time waster for just one point.

If you want to try them anyway, the questions about age and number are the quickest, involving only one equation. The information about the third kind of algebraic question (“challenging”) is included below simply for your reading pleasure. It is not recommended that you spend time on the challenging algebraic problems during the actual test, unless you finish early and go back to them.

Age Problems

If you like to muse about people’s ages for fun, you might consider these questions fun and easy to solve quickly. If so, go for it! This type of Wonderlic® question gives you information about the ages of two people and you have to find the answer to a question about the age of one person at a certain time. Here is an example:

When Anderson was 11 years old, his sister was twice his age. Now Anderson is 32 years old. How old is his sister?

42
43
50
54
64

First of all, you can’t just double his age now to find the answer; Anderson’s sister is not always going to be twice his age. But, if she was twice his age at 11, she was 11 years older and will always be 11 years older. So, just add 11 to his age now. 32 + 11 = 43, which is your answer.

Number Problems

If you simply love algebra and can make quick work of one-step algebra problems, this type of question might be one to try. It requires only one equation and a little bit of quick calculation with pencil and paper. Here’s how it could go:

There were 77 pencils in a bag. There were 10 times as many yellow pencils as red pencils. How many red pencils were in the bag?

6
7
8
9
10

Let’s call the number of red pencils (our needed amount) x. If there are ten times as many yellow pencils, yellow pencils could be called 10x. If all of them total 77, we could write:

x + 10x = 77

and solve

11x = 77

x = 7

Again, if you can do this type of calculation very quickly, even if you have to use pencil and paper, you could get a quick and correct answer.

Multi-Step Problems

This explanation is included only for those of you who actually get to the end of the 50 question test and have extra time. A good sign of a question of this type will be an extensive list of figures, no mention of finding an average, and the necessity of setting up two equations to solve.

If you remember your high school Algebra, you might remember learning about systems of equations for a month or two. There are a few ways to solve systems of equations: graphing, elimination, and substitution. For the types of problems you might see on the Wonderlic®, substitution will be the best and fastest method to solve, if you have extra time to go back and attempt multi-step problems on the test.

Here are the steps to solve a word problem requiring solving systems of equations using the substitution method.

1) Identify two variables.
2) Write two equations.
3) Isolate one of the variables.
4) Substitute.
5) Distribute and solve.

This may seem like Greek to you, so it’s best to show these steps at work with the following example.

Here is an example:

30 tickets were sold to adults and students to see the school play. Student tickets are $4 and adult tickets sell for $8. If the play grossed $160 in total, how many students bought tickets?

10
12
15
18
20

1) Identify two variables. In this case, let’s use a for the number of adult tickets sold and s for the number of student tickets sold.

2) Write two equations with those variables and the information in the problem. For this problem, the two equations would be:

\[s + a = 30\] \[4s + 8a = 160\]

3) Isolate one of the variables. To speed up the process, look at what the question is asking for and isolate the other variable in the simpler equation. Here, we need to find how many students bought tickets, so let’s isolate the other variable: a. Subtract s away and your new equation will look like \(a = 30 - s\).

4) Now comes the substitution. We just isolated a, so substitute what it equals (30 - s) for a in the second equation. This might seem confusing, but it’s just the old switcheroo. The second equation is now \(4s + 8(30 - s) =160\).

5) Distribute and solve.

\(4s + 240 - 8s = 160\) (distribute)
\(-4s + 240 = 160\) (combine 4s and -8s)
\(-4s = -80\) (subtract 240)
\(s = 20\) (divide by -4)

The answer is: 20 students bought tickets.

Note: If you went through all this trouble and then realized that the question was asking for the other amount, don’t fret. No need to restart—just look at your first equation and substitute in the value you know. 20 + a = 30. Subtract 30, and you’ll find that ten adults bought tickets.

With a little practice, this process can take maybe a minute or so to complete, which is well above the time suggested to spend on each question. This is why we must stress again: only attempt this type of problem if you’ve already finished all other problems on the test.