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Don’t believe anyone who tells you that the Executive Assessment (EA) is a walk in the park! While it is true that getting a score in the 99th or even 90th percentile is not a requirement for admission into your desired EMBA program, earning a respectable EA score is challenging. Earning a solid score requires dedication and time. To help you in your EA preparation, this article will provide you with representative examples of Executive Assessment Algebra questions and answers. We’ll do a short algebra concepts review for each problem we cover.
If you are currently studying for the EA, you have already discovered that the Quant section includes a comprehensive review of 21 math topics, many of which you probably haven’t seen in a long time. In this article, we focus on the topics that comprise the Algebra category. Of course, practicing the questions here is just a start. If you need more EA math help after completing what we offer in this article, please check out the Target Test Prep Online Executive Assessment Course.
Here is what we’ll cover in this article:
- An Overview of the EA Quant Section
- The Topics Tested on the Quant Section of the EA
- The EA Quant Question Types
- EA Problem-Solving (PS) Questions
- EA Data Sufficiency (DS) Questions
- Topic #1: Simplifying Algebraic Expressions
- Topic #2: Solving Linear Equations with One Unknown
- Topic #3: Solving Two Linear Equations with Two Unknowns
- Topic #4: Factoring Quadratics
- Topic #5: Exponents
- Topic #6: Inequalities
- Topic #7: Absolute Value
- Topic #8: Functions
- Summary
- What’s Next?
Let’s start by looking at what’s tested in the Quant section of the EA.
An Overview of the EA Quant Section
On the EA, the Quantitative Reasoning section is presented third, after the Integrated Reasoning (IR) and the Verbal sections. You are given 30 minutes to answer the 14 math questions in the two subsections, called panels, of the Quantitative Reasoning section.
The difficulty level of the first Quant section panel is dependent on your performance in the IR section. This may sound odd at first, but recall that IR questions have a significant Quant component. So, if you score well on IR, it’s an indication that you can handle more difficult Quant questions.
The difficulty level of the second Quant panel is based on your performance on the first Quant panel.
You may think that the EA structure is a bit convoluted, but this “panel-adaptive” difficulty level allows for the EA to assess your skills quite accurately, even though the exam length is short and the number of questions is minimal.
KEY FACT:
The Executive Assessment is “panel-adaptive,” allowing for a short exam with relatively few questions to be highly accurate.
The Topics Tested on the Quant Section of the EA
The good news about the topics tested in the EA Quant section is that they are largely those you studied in high school. However, the bad news is that it’s been many years since high school! But, as you progress with your EA Quant prep, those concepts and techniques will come back to you.
However, you will discover that the math questions you see on the Executive Assessment are not the same types of questions you saw in high school. You may be familiar with the multiple-choice Problem-Solving question type, but the Data Sufficiency question type is unique to the EA and its companion exam, the GMAT. We will cover these question types in more detail later in this article.
TTP PRO TIP:
The topics tested on the EA are similar to those you saw in high school, but the question types might not be.
GMAC, the organization that owns and administers the EA, has divided the math coverage into two major categories: Arithmetic and Algebra.
An Overview of the Executive Assessment Quant Topics
Here is a list of the 21 topics covered in the Quant section of the EA.
- Arithmetic:
- Properties of Integers
- Fractions
- Decimals
- Real Numbers
- Ratio and Proportion
- Percents
- Powers and Roots of Numbers
- Descriptive Statistics
- Sets
- Counting Methods
- Discrete Probability
- Algebra:
- Simplifying Algebraic Expressions
- Equations
- Solving Linear Equations with One Unknown
- Solving Two Linear Equations with Two Unknowns
- Solving Equations by Factoring
- Solving Quadratic Equations
- Exponents
- Inequalities
- Absolute Value
- Functions
KEY FACT:
There are 2 major categories comprising 21 topics on the EA Quant section.
You should note that geometry, logarithms, and complex numbers are not on the topics list. In this article, we will concentrate on Algebra practice problems. But before we cover those, let’s become more comfortable with the Quant question types.
The EA Quant Question Types
In the Executive Assessment Quant section, you will encounter two types of Quant questions: Problem Solving (PS) and Data Sufficiency (DS) problems. Of the 14 questions in the Quant section of the EA, roughly 8 will be PS questions and 6 will be DS questions.
KEY FACT:
Of the 14 questions in EA Quant, 8 are Problem- Solving questions and 6 are Data Sufficiency questions.
Let’s take a closer look at Problem-Solving questions.
EA Problem-Solving (PS) Questions
If you have ever answered a multiple-choice question, you are already familiar with EA Problem-Solving questions. This multiple-choice question type presents 5 answer choices: A, B, C, D, and E, and there is only one correct answer. Any concept can be tested in a PS question, including all 21 topics we listed previously.
Let’s now turn our attention to the question type that you have probably never encountered: the Data Sufficiency (DS) question.
EA Data Sufficiency (DS) Questions
The Executive Assessment will present you with 6 Data Sufficiency (DS) questions on the two panels of the Quantitative Reasoning section. Because there are only 14 Quant questions in total, being confident in answering DS questions is paramount!
Let’s compare the two question types. In a PS question, you must calculate a specific number (or variable) answer, but in a Data Sufficiency question, you must decide only whether you have sufficient information to determine a definitive answer.
A DS question usually presents information in a problem stem, a question, and two statements. Your job is to decide if exactly one statement, either statement individually, or (3) both statements together, provide enough information to answer the given question. If this seems a bit confusing, don’t worry. With practice, you can learn the logic underlying DS questions, and soon you’ll be a pro!
TTP PRO TIP:
When solving Data Sufficiency questions, you must decide whether you have sufficient information to determine an answer.
Let’s now look at the DS answer choices.
The Data Sufficiency Answer Choices
Data sufficiency answer choices are the same for every DS question. Thus, early in your EA preparation, you will want to memorize the DS answer choices. Doing so is as important as memorizing math formulas or math shortcuts.
Here are the DS answer choices:
A: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C: BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D: EACH statement ALONE is sufficient to answer the question asked.
E: Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
TTP PRO TIP:
Memorize the DS answer choices.
We will not provide an in-depth strategy or approach to answering DS questions here. Instead, refer to this TTP article to learn the tips, tricks, and strategies for efficiently solving DS questions.
Now that we know the basic facts about the Quant section and the question types we’ll encounter there, let’s begin with our overview of EA Algebra.
Topic #1: Simplifying Algebraic Expressions
An algebraic expression does not have an equal sign. For example, 3x + 7 is an expression, but 3x + 7 = 10 is an equation. Thus, this first EA algebra topic addresses simplifying algebraic expressions, not solving for x.
When you’re simplifying algebraic expressions, you’re often factoring, using the distributive rule, and/or combining like terms. Consider the following example.
Example #1: Simplifying Algebraic Expressions
x(x + 3) – (x – 7)(x + 2) is equal to which of the following?
- 6x – 14
- 12x – 14
- 8x + 14
- 9x + 14
- 12x + 14
Solution:
We first use the distributive rule and then we combine like terms:
x(x + 3) – (x – 7)(x + 2)
x^2 + 3x – (x^2 + 2x – 7x – 14)
x^2 + 3x – x^2 + 5x + 14
8x + 14
Answer: C
Our next topic is solving single-variable linear equations.
Topic #2: Solving Linear Equations with One Unknown
From the first week of Algebra I, you were solving equations with one variable. For example, to solve the equation 3x + 8 = 5x – 4 for x, recall that you get all the “x” terms on one side and the “number” terms on the other. Then, you isolate x:
3x + 8 = 5x – 4
12 = 2x
x = 6
Now, you can verify that the solution of x = 6 is true because if you substitute 6 for x into the original equation, you obtain equality:
3(6) + 8 = 5(6) – 4
26 = 26
KEY FACT:
To solve a single-variable linear equation, collect all the variable terms on one side and the numbers on the other.
Now, let’s look at an example.
Example #2: Solving Linear Equations with One Unknown
Solve for x: (9x – 5) / 4 = 4
- 11/9
- 4/3
- 13/9
- 2
- 7/3
Solution:
(9x – 5) / 4 = 4
(9x – 5) = 16 multiply each side by 4
9x = 21 add 5 to each side
x = 21/9 divide both sides by 9
x = 7/3 reduce the fraction
Answer: E
Next, let’s consider solving a system of linear equations.
Topic #3: Solving Two Linear Equations with Two Unknowns
You may see this topic referred to as “a system of linear equations” in some textbooks or search engines. No matter the name, the goal is to solve for the values of the two variables, if they exist. An example of a linear system might be:
y = 3x (equation 1)
y + x = 12 (equation 2)
The most straightforward way to solve this system of equations is by the substitution method, whereby we substitute an expression for one variable into the other equation.
Now, looking at our two equations, we see that y is already isolated in equation 1. So, we can substitute 3x (from equation 1) for y (in equation 2). This gives us:
3x + x = 12
Next, we can simplify and determine x:
4x = 12
x = 3
Next, let’s look at a more representative problem that you might encounter on the EA.
Example #3: Solving Two Linear Equations with Two Unknowns
Using the system of linear equations below, what is the value of x + y?
5x + y = 9 (equation 1)
2x – 5y = 36 (equation 2)
- -9
- -3
- 0
- 3
- 9
Solution:
Unlike in the previous problem, we will have to do some work to isolate a variable. Looking at our equations, let’s isolate y in equation 1.
Let’s do that below by subtracting 5x from both sides of the equation:
y = 9 – 5x (equation 3)
Next, we can substitute 9 – 5x for y in equation 2. This gives us:
2x – 5(9 – 5x) = 36
2x – 45 + 25x = 36
27x = 81
x = 3
Now, to determine the value of y, we substitute 3 for x in equation 3.
y = 9 – 5x
y = 9 – 5(3)
y = -6
Thus, the value of x + y is 3 + (-6) = -3
Answer: B
KEY FACT:
We can use the substitution method when solving a system of linear equations.
Note that there are other aspects of systems of linear equations, such as the elimination method and determining the number of solutions to a linear system. While these are beyond the scope of this article, you can find the needed background at our TTP Executive Assessment course.
Now, let’s move on to factoring quadratic equations.
Topic #4: Factoring Quadratics
The most common type of nonlinear equation you will need to factor is a quadratic equation. A quadratic equation is one in which the highest power of the variable is 2. The following are all quadratic equations:
x^2 + 3x + 2 = 0 10q^2 = 100q 7c^2 + 5 = 10c – 12
The ability to factor a quadratic equation is a skill you must have. Let’s review the steps in factoring a quadratic equation, using x^2 + 8x + 15 = 0 as an example.
First, know that you will set up the factored form of x^2 + 8x + 15 = 0 with two parenthesized expressions: (x + d) (x + e), where d and e are numbers. The key is to know that the product (d)(e) is equal to 15, and d + e = 8.
Next, think of the integers that multiply to yield 15. They are:
1 and 15 3 and 5 -1 and -15 -3 and -5.
Of these candidates, we know their sum must be 8. The only pair that has a sum of 8 is 3 and 5, so these are the two values for d and e.
Thus, we can express x^2 + 8x + 15 = 0 in factored form as (x + 3)(x + 5) = 0.
KEY FACT:
A quadratic equation of the form x^2 + bx + c can be factored into (x + d)(x + e) by using the facts that (d)(e) = c and d + e = b.
Now we must determine the values of x that make the equation equal to 0. We use the zero product property to solve the equation:
x + 3 = 0 or x + 5 = 0
From these two equations, we see that x = -3 or x = -5.
TTP PRO TIP:
Use the zero product property to solve a factored quadratic.
Let’s try an example factoring question.
Example #4: Factoring Quadratics
Grace sells chicken wraps from her food truck. She has determined that her profit equation is
P = [(x)(35 – x)] / 5,
where x is the number of wraps she sells during her morning shift. Assuming she sells more than 15 wraps, how many chicken wraps must she sell in order to make a profit of $50?
- 25
- 35
- 40
- 45
- 55
Solution:
We set up the profit equation with P = 50. Then we solve the equation for x.
P = [(x)(35 – x)] / 5
50 = [(x)(35 – x)] / 5 substitute P = 50
50 = (35x – x^2) / 5 distributive law
250 = 35x – x^2 multiply both sides by 5 to clear the fraction
x^2 – 35x + 250 = 0 move all terms to the left side
(x – 25)(x – 10) = 0 factor the quadratic
x = 25 or x = 10 determine the 2 solutions
We are told that Grace sold more than 15 wraps. Therefore, the only solution is x = 25 for the number of wraps she sold.
Answer: A
Next, let’s consider the Algebra topic of exponents.
Topic #5: Exponents
It is important to review all the rules of exponents during your EA prep. Here, we consider some basic exponent properties about squaring a number, which we summarize as follows:
- Squaring a positive number yields a positive number. For example, 3^2 = 9.
- Squaring a negative number yields a positive number. For example, (-4)^2 = 16
- Squaring a positive proper fraction makes it smaller. For example, (1/2)^2 = 1/4
- Squaring a negative proper fraction makes it greater because the result is a positive proper fraction. For example, (-1/3)^2 = 1/9.
Let’s use these concepts to solve the following Data Sufficiency (DS) question about exponents. Use the following answer choices:
A: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C: BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D: EACH statement ALONE is sufficient to answer the question asked.
E: Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Example #5: Exponents
What is the value of p – q?
p^2 / q^2 = 1
p / q = 1
Solution:
Statement (1) by itself:
Let’s use some convenient numbers to test the equation.
If we let p = 1 and q = 1, then p^2 / q^2 = 1/1 = 1, and p – q = 1 – 1 = 0.
However, if we let p = -2 and q = 2, then p^2 / q^2 = 4/4 = 1, and p – q = -2 – 2 = -4.
Because we have obtained two different answers for the value of p – q, then statement one is not sufficient to answer the question. Therefore, we can eliminate answer choices A and D.
Statement (2) by itself:
We can multiply both sides of the equation by q to obtain p = q. Thus, we see that since the two variables are equal, their difference p – q must be equal to 0. Hence, we know that Statement 2 is sufficient to answer the question.
Answer: B
The next Algebra topic is inequalities.
Topic #6: Inequalities
An equation describes two expressions that are equal, but an inequality describes two expressions that are not equal. We use the “greater than” symbol, >, and the “less than” symbol, <, when we are expressing inequalities. For example:
x > 8 specifies that x is any value greater than 8.
x < -2 specifies that x is any value less than -2.
KEY FACT:
An inequality describes two expressions that are not equal.
Note that the topic of inequalities includes problems with “less than or equal to”, ≤, or “greater than or equal to”, ≥, symbols. Thus, x ≤ 14 or x ≥ -8 are also considered inequalities.
Equations and basic linear inequalities are solved identically, so there is no big jump going from equations to inequalities! The difference between solving an equation and an inequality is how the solution is stated. An equation yields a finite number of solutions, such as x = 3, whereas an inequality produces an interval, such as x > 2.
KEY FACT:
An equation has a finite number of solutions. An inequality’s solution is expressed as an interval.
There is one big difference between equations and inequalities when we are solving them. Notably, when multiplying or dividing inequalities by a negative number, we reverse the direction of the inequality sign.
Here is a basic example.
What is the solution set for the inequality -2x > 8 ?
Here, we see we have to divide by -2, but when we divide an inequality by a negative number, we must reverse the direction of the inequality sign.
So, when we divide -2x > 8 by -2, we reverse the inequality sign, obtaining:
-2x / -2 < 8 / -2
x < -4
KEY FACT:
If we multiply or divide an inequality by a negative number, we must reverse the direction of the inequality sign.
Let’s consider an example.
Example #6: Inequalities
If -5x – 3 ≥ 2x – 4, which of the following cannot be a value of x ?
- 0
- 1/8
- -1/3
- -1
- 1
Solution:
First, let’s solve for x, remembering that if we multiply or divide an inequality by a negative number, we must reverse the direction of the inequality sign.
-5x – 3 ≥ 2x – 4
-7x ≥ -1
7x ≤ 1
x ≤ 1/7
All answer choices except x = 1 are in the solution set.
Answer: E
Now, let’s turn our attention to the topic of absolute value.
Topic #7: Absolute Value
The absolute value of a number is the distance between that number and zero on the number line. We enclose an absolute value statement inside a pair of vertical bars.
For example, |50| = 50 and |-50| = 50
We can extend this to understand that if |x| = 5, then x can be either 5 or -5.
KEY FACT:
The absolute value of a number is the positive distance between that number and zero.
Whatever is inside the absolute value bars, whether negative or positive, is a positive value. So, when we are given an absolute value equation, we generally solve for the expression inside the absolute value bars for two cases: positive and negative. Let’s work through a very simple example to illustrate this two-case procedure.
If |x + 1| = 12, what are the possible values of x?
To solve for x, we follow the two-case process. For the positive case, we drop the absolute value bars and solve the equation. In the negative case, we drop the absolute value bars and negate the entire expression inside the bars.
Case 1: The Positive Case
We simply drop the absolute value bars and leave the expression untouched, giving us:
x + 1 = 12
x = 11
Case 2: The Negative Case
We drop the absolute value bars and multiply what was inside the absolute value bars by -1. Then, we solve for the variable, giving us:
-(x + 1) = 12
-x – 1 = 12
-x = 13
x = -13
So, x can be either 11 or -13.
Let’s try an example question.
Example #7: Absolute Value
If |4y – 3| = 1, then what is the value of y?
(1) y is a positive number
(2) y > y^2
Solution:
The question stem gives us a lot of information. We can first solve the absolute value equation for two cases: when (4y – 3) is positive and when (4y – 3) is negative, as follows:
Case 1: 4y – 3 is positive
4y – 3 = 1
4y = 4
y = 1
Case 2: 4y – 3 is negative
-(4y – 3) = 1
-4y + 3 = 1
-4y = -2
y = 1/2
Thus, the two solutions are y = 1 and y = 1/2.
Statement (1) by itself:
y is a positive number
Statement (1) says that y is a positive number. Since both possible values of y are positive, statement (1) by itself is not sufficient to determine a unique value for y. Therefore, we can eliminate answer choices A and D.
Statement (2) by itself:
y > y^2
We can re-express statement (2) as y^2 < y. We know that the only way that y^2 can be less than y is if y is a positive proper fraction — that is, if y is between 0 and 1.
To illustrate this, note that if y = 1/2, then y^2 = 1/4. This shows us that y > y^2 when y is a positive proper fraction.
From the question stem, we determined that y = 1 and y = 1/2 are the two possible values of y. Statement (2) tells us that y must be a positive proper fraction. Thus, there is only one possible value of y: 1/2. Therefore, statement (2) by itself is sufficient to answer the question.
Answer: B
Our final EA Algebra topic is functions.
Topic #8: Functions
Basically, a mathematical function is like a machine that takes an input and produces an output. For example, if the input is 4 and the machine instruction is to cube the input, then the output is 4^3, or 64. Symbolically, we would summarize this mathematical operation, using function notation, as:
f(x) = x^3 For any input x, the function instruction is to cube x.
f(4) = 4^3 When x = 4, the function cubes it
f(4) = 64 The output of the function when x = 4 is 64.
We used function notation f(x) above. Recall that f(x) is pronounced “f of x.” It is simply saying that the value of our function is based on the input value x. Note that f(x) does not denote multiplication. It is, again, a notation that identifies a function.
We can use letters other than x to represent the input value. Earlier, we saw a function defined as f(m), and m acts identically to x.
Even though we usually use the letter “f’” to indicate a function (because it is the first letter of the word “function”), we may see functions expressed as g(x) or h(x).
KEY FACT:
The standard notation denoting a function is f(x), which is pronounced “f of x.” Other letters can also be used.
Let’s look at an example.
Example #8: Evaluate a Function
If f(x) = (x + 6) / (x^2 – 2), what is the value of f(-4)?
- -1/9
- 1/7
- 3/5
- 5/7
- 7
Solution:
To evaluate f(-4), we substitute -4 for x in the function f(x) = (x + 6) / (x^2 – 2):
f(-4) = (-4 + 6) / ((-4)^2 – 2)
f(-4) = 2 / (16-2)
f(-4) = 2/14 = 1/7
Answer: B
Summary
The Executive Assessment Quantitative Reasoning section tests students on 21 major math topics, including arithmetic and algebra. The section consists of 14 questions, including 8 Problem-Solving questions and 6 Data Sufficiency questions.
In this article, we covered 8 of the EA Algebra topics. They include:
- Simplifying Algebraic Expressions
- Solving Linear Equations with One Unknown
- Solving Two Linear Equations with Two Unknowns
- Solving Quadratic Equations
- Exponents
- Inequalities
- Absolute Value
- Functions
What’s Next?
The two important aspects of EA Quant success are knowledge of the 21 topics that are tested and familiarity with the question types. In this article, we have provided you with 8 example questions covering both Problem-Solving (PS) and Data Sufficiency (DS) questions.
The Quant section of the Executive Assessment can be challenging. To make you feel more comfortable, read our top 10 tips for mastering EA Quant.