 Ready to tackle some of the hardest ACT math questions? This blog is for two types of students: those looking to conquer any (math) challenge and those who don’t love math but are facing their fears and doing what it takes to improve their ACT score.

We are here to help. But before you jump into the questions there are some things you should know. When it comes to the hardest ACT math questions (the last ten to fifteen questions), there are two types of questions: advanced questions and complex questions.

### Types of Hardest ACT Math Questions

The first type of challenging math questions are advanced concepts that you learn in precalculus or trigonometry. These questions are usually straightforward in how they’re written, but they require more advanced knowledge of math concepts. These can be challenging because if you haven’t learned the concepts, you might not (yet) have the knowledge to solve the questions.

The other type of questions at the end of the ACT is basic concepts asked in complicated ways. These might look overwhelming, but when you take a closer look, you can see that there are relatively common math ideas disguised as complex questions. These questions are definitely possible even if you haven’t finished Algebra II.

Most of these final math questions fall into one of these categories. But they’re very rarely both advanced and complex.

### Strategies for Difficult Math Questions

When you look at the difficult ACT math questions, there are a few strategies to keep in mind.

First of all, don’t assume that just because your answer is one of the options that it is correct. The ACT is written to anticipate your mistakes. So double-check your work and don’t just choose the first “answer” you see from your calculations.

Don’t get stuck. The last ten are more challenging than questions at the beginning of the test, but question 54 isn’t necessarily easier than 57. If you get stuck on a question, move on to see if you understand anything in the next few. Look for the familiar concepts and do your best to make it through to the end.

Keep practicing. Especially when it comes to the complex questions, practice helps you determine what the question is asking. And it will help you build confidence as you tackle these difficult math questions.

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Ready to start practicing? Take a look at these four questions, two advanced concepts and two with complicated wording, and give them your best shot! Don’t forget to check your answers.

The first two questions cover advanced math concepts that you typically learn in precalculus and trigonometry. Give them a try and see if you can solve for the correct answers. Solution Steps for Question 60:
To find the element in the upper left of the product matrix, multiply the upper left element of the first matrix (3) by the upper left element of the second matrix (w) and ADD that product to the product of the upper right element of the first matrix (5) and the lower left element of the second matrix (y) for a result of 3w + 5y.  Only Answer J has this correctly expressed in the upper left position. Solution Steps for Question 54:
The domain of a function is simply the values of x that “work” for the function.  In this case, looking at the graph shows that there is no value for the function when x = 1 or when x = -2.  This can also be solved by looking at the equation itself: any value of x for which the denominator will equal zero is NOT part of the domain.  The equation x2+x-2 =0 is true when x = 1 or when x = -2.  Both methods result in Answer J.

### Complex Questions (Basic Concepts)

These math questions do look complicated, but they have basic concepts you can figure out with some organization and step-by-step thinking. Solution Steps for Question 57:
Area of a rectangle = LW
New Length = 70%L = 0.70L
New Width = 115%W = 1.15W
Area of new rectangle = (0.70L)(1.15W)
Area of new rectangle = 0.805LW = 80.5%LW = 80.5% of Area of the original rectangle
Question asks for how much SMALLER the area is: 100% – 80.5% = 19.5% SMALLER: Answer B is correct Solution Steps for Question 58:
Find how many cartons can be placed in a box: 172/12.6 = 13.65 cartons, but you can’t have a partial carton.  Therefore, each box can hold a maximum of 13 cartons.  Subtract 13 multiple times from the total of 84 cartons to find how many cartons will be in the partially-filled box:

84 – 13 = 71
71 – 13 = 58
58 – 13 = 45
45 – 13 = 32
32 – 13 = 19
19 – 13 = 6

6 cartons, each weighing 12.6 ounces, has a total weight of 75.6 ounces, which is Answer K.

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Keep practicing math questions to prepare for the difficult questions you might encounter on the ACT