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This is one of the higher-difficulty questions that you will find on the executive assessment, but honestly, the test tries to shock and awe you into thinking something is harder than it really is.

$$f(X)=\frac{x+1}{x}$$

The way this is written, it is basically a function that dictates operations performed on some variable, in this case \(x\).

It is read as \(f\) of \(x\). \(f(x)=\) Notation same as y in the Coordinate plane. The coordinate plane is a concept that can be tested in this exam even though there is no plane geometry.

What it means is that if you plug in a value for \(x\), you are guaranteed to get some value because you are going to have to do \(\frac{x + 1}{x}\) no matter what the value of \(x\) is.

If you have nested functions, you have to process from the innermost parentheses outward because that is simply applying parenthesis as the first step in the order of operations.

Given that \(f(x)=\frac{x+1}{x}\), what s \(f(f(2))\).

What we are going to do is find the \(f(2)\), then apply the function again to the result.

$$f(2)=\frac{2+1}{2}=1.5\ \text{or}\ \frac{3}{2}$$

$$f(1.5)=\frac{1.5+1}{1.5}=1.667\ \text{or}\ \frac{5}{3}$$

This is a very basic application of functions.

A function must have a domain. A domain refers to acceptable inputs that will produce real values in the function. Note that for the Executive Assessment, there are no non-real values, everything must be representable on the number line.

So the domain of this particular function is any value where \(x\neq0\). Because if \(x=0\) then we will be dividing by 0, and any number divided by \(0=\infty\) which is a non-real value.

- You can’t divide by zero
- You can’t square root a value less than zero. If you were to square root a negative, you will end up with an imaginary number because we know that any square root is supposed to produce two values that are the same. To have a negative product you have to have two numbers, a positive and a negative. Therefore, you cannot square root a negative number.

They are unique Arithmetic or Algebraic operations that are defined by the exam.

They are not necessarily defined rules of math. They are defined proprietarily by the test and the problem that you are working through, in other words, they are, follow the instructions kind of questions.

You just need to follow the format as presented in a similar method to functions.

- Stay calm. These problems are intended to shock and awe you and make you overreact. Instead, just remember that they have to tell you what is necessary to solve. So just read through the problem. Recognize that initial shock and awe may be the most difficult part of the problem.
- Note that that symbol or function may or may not be predicated on any defined rules of mathematics.
- Remember to apply the rules of mathematics to any symbols as dictated by arithmetic or algebra. For instance, the order of operations must apply, recognize that \({a}\times{b} ={b} \times{a} = {ab}\). All these rules will apply but work through the problem very carefully until you reach the answer.

If £ represents a unique digit in the equation \(5∆∆ + ∆∆ = ∆32\), what is the value of ∆

- Stay calm
- Begin with the known information.

We know that we have 5 hundred something + a two-digit integer equals a three-digit integer ending in 32.

If ∆ is a unique digit, it cannot be 5, 3, or 2. If it is unique, it is not already there.

The only other possible hundredth digit when you add a two-digit value to 5 hundred something is 6. Because if you are adding a two-digit number it has to be less than 100, which means you cannot, for instance, get up from 5 hundred something to seven hundred-something.

The ∆ has to be 6.

To confirm the approach, \(566 + 66 = 632\). This satisfies all conditions of this unfamiliar symbol problem even though we had no idea what £ meant at the start.

Set up your scratch pad for problem

Carefully read and define the function or symbol as presented to inform problem-solving or data-sufficiency processes.

Carefully apply that function or symbol as required by the problem

- You should use an example to illustrate the function or symbol if needed. In the previous case it wasn’t needed so we just started working on the problem but if you have any doubts as to what it means, use an example.
- Consider the logical implications of the functions or symbols once you have defined it.

Consider the best approach for solving

- Take a technical or logical approach to solving if you can.
- Consider plugging in the choices or easy values to model as a frequently viable approach because oftentimes you are going to be able to use alternative tactics and in a lot of symbols and function problems, the entire point is to plug in values because that’s how a function works.

Allow yourself to model and to back-solve as the problem allows and you’ll be able to knock these problems down to pay. These problems are often a differentiator for 150+ in this exam because the exam doesn’t necessarily think that these weird functions and symbols are that difficult, but students do. So they are a great opportunity for you to separate yourself from everybody else by just remaining calm and executing the steps of the problem as outlined above.

If \(f(x)=1-x^2\), which of the following expresses the value of \(f(x)\)?

A. \(1-x^4\)

B. \(1+x^3\)

C. \(1+2x+x^4\)

D. \(1-2x+x^4\)

E. \(1-2x^2+x^4\)

We will set up our scratch pad, listing our choices ‘*a’* through ‘e’ with a line at the top to write down what the question is asking for.

\(f(x^2)=?\)

a. \(1-x^4=1-(3^2)^2=1-81=-80\)

b.

c.

d.

e.

In this case, we have relatively complex expressions so we are not going to write them down. We skip to the end and see that we are being asked for the value of \(f(x^2)\).

In general, it is best to set up the technical approach first. We know that we nest the functions.

We know that if we are looking for \(f(x^2)\), that stands in for the \(x\).

Therefore, \(f(x^2)=1-(x^2)^2=1-x^4\)

The correct answer is choice A.

But if we didn’t set up the technical approach, we can still solve the problem by just plugging in easy values. In this case, we have \(1-(x^2)\), and we have 1s and 2s and 4s, so we want to avoid these numbers that are already in the problem so let’s say \(x=3\)

If \(x=3\rightarrow x^2=9\) since \(f(x)=1-x^2\rightarrow f(x)=1-9 \rightarrow f(x) = -8\) therefore,

$$\begin{align*}f(x^2)&=1-92\\ f(x^2)&=1-81\\ f(x^2)&=-80\end{align*}$$

If we plug in our values of *x* we should be able to get a result of -80. Remember \(x=3\).

Looking at choice A, on our scratch pad, \(1-x^4=1-(3^2)^2=1-81=-80\).

On the exam go on and test the rest of them to make sure there is no duplicate. Still, if we take a look at choice B, the answer > 0 so it is obviously incorrect. The same goes for choice C. The same applies to choices D and E.

We can solve this quickly through modeling. We know -80 is our result, we just need to find out what answer choices if we plug in 3 we will end up with -80 as our result.

For any two unique positive integers \(a\) and \(b\), \(a⌂b\) represents the greatest common factor of \(a\) and \(b\). If \(x\)y\), what is the value of \(x⌂y\)?

(1) \(x=16\)

(2) \(y=17\)

We start with what we know. We know that for any two unique integers a and b, a ≠ b; a > 0; b > 0; \(a⌂b=GCF\) of \(a\) and \(b\).

\(x<y\). the question is asking for the value of \(x⌂y\).

we need to know the GCF of \(x\) and \(y\).

(i) \(x=16\)

if \(x = 16\) and \(y = 32\), the greatest common factor 16. But if \(y = 30\) then the GCF is just going to be 2. We’ve got multiple possible outcomes and therefore condition (1) by itself is not sufficient. We are left with BC&E.

(ii) \(y=17\)

if \(y = 17\), we know that \(x < y\). we look at the factors of 17. They are 1 and 17. And since \(X < 17\). The GCF of 17 and anything less than it is 1. That gives us everything that we need to solve the problem and makes condition (2) by itself sufficient. We can select choice B.

Both of these examples at first glance seem very complex, you wonder what these functions and symbols are. But that is the exam taking advantage of your assumptions and trying to make you think things are harder than they are. That is a very important skill for you to have as a manager which is why this is a common tactic of the executive assessment.

Remain calm, work through the steps of the problem, and these weird functions and symbols will become part of your path to getting the score you need to attend your target program.

Practice more of these problems on your own to improve your skills for your test date.

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