Updated December 16, 2025

Hardest ACT Math Questions: 15 Examples With Solutions

Bonus Material: ACT Score Ranges for 750+ Schools

If you’re aiming to get a top score on the ACT, you have to be able to solve the hardest ACT math problems.

We’ve been preparing students for the ACT for 20 years, so we know which questions tend to trip up even the strongest students. These example questions (and detailed solution walk-throughs) will take you through some of the hardest problems you’re likely to run into on ACT Math.

Curious what ACT Math score you should be aiming for in the first place? We lined up the 25th, 50th, and 75th percentile scores at hundreds of schools so you can see what it takes to be a competitive applicant. Find the schools on your list in the free download below!

Jump to section:
What’s on the ACT Math Section?
The Hardest Types of ACT Math Questions
#1: Probability Puzzles
#2: Inverse Functions
#3: Matrix Problems 
#4: Circle Geometry 
#5: Integer Relationships
#6: Trigonometric Relationships
#7: Natural Logarithms
#8: Probability & Primes
#9: Ratios
#10: Varying Rates
#11: Real Numbers
#12: Probability Terms
#13: Volume and Surface Area
#14: Statistics 
#15: Imaginary and Complex Numbers
Next Steps

What’s on the ACT Math Section?

Here are the basics you need to know about the ACT Math section.

  • The Math section has 45 questions and 50 minutes. That means time management is a big part of preparing. (To see the full structure of the new ACT format, check out our post breaking it down.)
  • You’ll receive a Math score up to 36. It will factor into your overall ACT score, along with your Reading and English scores. If you take the optional Science section, you’ll also get a STEM score that includes your Math and Science scores. (Here’s how the ACT is scored – it’s complicated!)
  • You may use a calculator. That’s the good news. The not-so-good news is…
  • You’ll have to memorize most formulas. Unlike the SAT, the ACT does not provide an extensive formula reference sheet.
  • The questions get harder as you go along. If you’re missing a lot of points early on in the test, it’s likely that you have some math concepts you need to work on. If you’re already scoring above a 32 or so, then practicing these advanced questions can help you reach a top score.

The Hardest Types of ACT Math Questions

What are the hardest question types on ACT Math?

That’s a very personal question. You know best what your personal stronger and weaker points are.

But based on over 20 years of experience tutoring students for the ACT, some of the topics students tend to find tricky are:

  • Matrices
  • Advanced geometry
  • Probability, especially multi-step problems
  • Statistics
  • Polynomials
  • Logarithms
  • …and whatever students personally struggle with.

The ACT test makers provide this information about the makeup of the test:

TopicPercentage of questionsNumber of questions
Real and complex number systems7-10%3-5
Algebra12-15%5-7
Functions12-15%5-7
Geometry12-15%5-7
Statistics & Probability8-12%4-5
Complex problems (combining multiple areas)40-43%18-19

As you can see, about 40% of the test involves combining multiple skill areas. These are often among the harder questions that you’ll see on ACT Math.

If your ACT Math score isn’t where you want it to be, you don’t need to give up on a great school. The best way to improve your score is practice, practice, practice! And if you’re still struggling with math concepts or time management, a little tutoring support can go a long way. Check out our analysis of 75 ACT tutoring services to find one that’s the right fit for you, or get in touch with our team for a free consultation.

Now, on to the questions.

#1: Probability Puzzles

Each time Coin Z is tossed, it lands faceup or facedown. The probability of landing faceup is 4 times the probability of landing facedown. In a game, the player wins $1.00 when Coin Z lands faceup and $3 when Coin Z lands facedown. What is the expected value of each toss of Coin Z in this game, to the nearest cent?

A. $1.20

B. $1.40

C. $1.50

D. $2.00

Why this question is hard: There are multiple steps to this problem, and it’s easy to solve for the wrong thing (like the expected value of 5 tosses). But if you approach is systematically and keep in mind what you’re solving for, you can solve this question fairly easily.

First things first: what is this question asking?

We want the expected value of each toss. In reality, in this game, you’ll get either $1 or $3 for each toss. But we want the average expected value.

(Why would that be useful? Because you can then use it to estimate how much you could expect to get from, say, 10 or 30 tosses.)

Let’s call p the probability that the coin lands facedown.
Then probability of landing faceup =4p.

Because the coin must land either faceup or facedown – there are no other options – The total probability of these options is 1. We can express that like this:

p + 4p = 1

5p = 1

p = 0.2

So now we know:

  • P(faceup) = 4 • 0.2 = 0.8
  • P(facedown) = 0.2

So what does this mean for the payout? We have a 0.8 chance – 80% – of getting $1. And we have a 0.2 chance – 20% – of getting $3.

Expected value =(0.8 • 1.00) + (0.2 • 3.00)

= 0.8 + 0.6

= 1.4

That’s answer choice C.

#2: Inverse Functions

Why this question is hard: You need to know the concept of inverse functions and how it applies to roots. If you know that, this question is very doable – but it also looks scarier than it is, which can shake students’ confidence.

A helpful way to think about finding the inverse function is that you’re swapping the x and y variables.

Write this out (on scratch paper or in your head) as 

Now swap the x and y variables.

The first step is to cube both sides of the equation:

Then subtract 2 from both sides:

And finally, divide by 4:

Now that you have x expressed in terms of y, you just need to swap the x and y variables back to get the inverse of the original function:

That’s answer choice B.

If you want more guidance learning these skills and how to use them on the test, give us a call about ACT tutoring. We’ve been tutoring for the ACT for over 20 years, and we can help you adjust to the brand-new test format.

#3: Matrix Problems

A party supply store sells 2 types of jumbo popcorn bags: salty and sweet. The numbers of bags sold on Thursday and Friday last week are given in matrix A; the selling price per bag and the profit per bag are given in matrix B. Price and profit are in dollars. What is the total profit for the sale of the 2 types of popcorn sold on Thursday and Friday?

A. $150.00

B. $158.50

C. $164.00

D. $172.25

Why this question is hard: A lot of students are intimidated by matrix problems, but these are actually very solvable once you know a couple of basic matrix rules.

First, let’s figure out what everything in the matrices represents.

Matrix A:

  • Rows: Thursday, Friday
  • Columns: salty, sweet

Matrix B:

  • Rows: salty, sweet
  • Columns: price, profit

Once you know what each matrix shows you, you don’t even need to know matrix rules specifically. From here on out, it’s just simple multiplication and addition.

What we’re looking for is the total profit for both days combined.

The first matrix gives us the number of bags sold each day. The second one gives us the price for each bag and – importantly for this question – the profit made on each bag.

What’s the total number of salty bags sold? 

From the left column of matrix A, it’s 14 + 20 = 34.

And what’s the profit on salty bags? Multiply that total by the salty profit per bag in matrix B: 

 34 • 2.25 = $76.50.

Do the same for the sweet bags: the total sold is 26 + 24 = 50.

And the profit on sweet bags is 50 • 1.75 = $87.50.

Finally, add these together to get the total profit for both days:

76.50 + 87.50 = $164.00.

That’s answer choice C.

#4: Circle Geometry

The figure shows △LMN a right isosceles triangle, inscribed in a circle with center O and radius 12 cm. What is the length, in centimeters, of arc MN shown as the thick curved line?

A. 24π

B. 18π

C. 12π

D. 6π

Why this question is hard: it involves multiple steps and requires you to remember properties of both circles and triangles.

A helpful start to any geometry problem is to write down what you know or can immediately figure out. It’s often more than you think!

  • Triangle LMN is a right isosceles triangle, so angle M is 90º and LM = MN.
  • The diameter of the circle is the hypotenuse of the triangle and equals 2 • 12 = 24 cm.
  • Because LMN is an isosceles triangle, angle L = angle N.

Let’s create 2 triangles within LMN by drawing a line from O to M that is perpendicular to LN.

Because OM is perpendicular to LN, angle MON will be 90º.

Now we can use an important property of circles to solve the problem.

  • A full circle always has 360º
  • The arc length or sector area of a piece of the circle is always directly proportional to what fraction of the circle it represents

That means that we can set up this relationship between the length of arc MN – which is part of the circumference of the circle – and the circumference of the entire circle.

Now we just need to find the circle’s circumference. Remembering that circumference = 2π • radius (and that we’re given the radius as 12π cm), we can plug in:

Circumference = 2π • 12 = 24π

Now we can plug that back into the equation we set up for arc length:

Simplifying:

Therefore: 

MN = 6π

That’s answer choice D.

Questions like this require you to have a very solid grasp of ACT math skills. If you need to brush up on the test content or improve your test-taking strategy, an experienced tutor can be a big help.

#5: Integer Relationships

The sum of 3 positive integers is 360, and the ratio of the integers is  5:3:2. What is the value of the smallest of the integers?

A. 36

B. 72

C. 80

D. 120

Why this question is hard: There are multiple steps. You might get tripped up by solving for x in the equation below and assuming that that’s the solution (which is given as a trap answer among the answer choices!), but the answer you’re looking for is 2x.

What do we already know?

We have three positive numbers that add up to 360. Their ratio is 5 : 3 : 2. We need to find the smallest of the three numbers.

A ratio of 5 : 3 : 2 means:

  • If the first number is 5 parts,
  • the second is 3 parts,
  • and the third is 2 parts.

Let’s call 1 part = x

Then the numbers are:

5x, 3x, 2x

Now we can set up a simple algebra problem. Since the numbers add up to 360:

5x + 3x + 2x = 360

10 x = 360

x = 36

Back to the original question: we’re looking for the smallest number, which we know is 2x. 

2x = 2 • 36 = 72

That gives us answer choice B.

#6: Trigonometric Ratios

For an angle with measure  α in a right triangle,  sin α = 3/5 and tanα = 3/4. What is the value of cosα?

A. 1/5

B. 1/2

C. 5/4

D. 4/5

Why this question is hard: Some students spend a lot of time on a question like this trying to figure out the value of α and then get from there to cos α. Don’t waste your time!

You can find cos α just by using trig ratios. Here’s how.

Remember SohCahToa? It tells us that:

sin α = opposite / hypotenuse

cos α = adjacent / hypotenuse

tan α = opposite / adjacent

Now we can express the information we got in the question like this:

sin α = 3/5 = opposite / hypotenuse

tan α = 3/4 = opposite / adjacent

And we want cos α, which is adjacent / hypotenuse. So let’s express cos α  in terms of sin α and tan α:

cos α = adjacent / hypotenuse

=  ( opposite / hypotenuse) / (opposite / adjacent) 

 = sin α / tan α

Filling in the values of sin α and tan α:

cos α = sin α / tan α 

= (3/5) / (3/4) 

= ⅗ • 4/3 

= 4/5

That’s answer choice D.

#7: Natural Logarithms

The real solution of the equation 5ex = 25 is: 

Why this question is hard: To answer this question, you need to be familiar with e and natural logarithms. It’s a question where there’s no way around simply knowing the content. 

A quick recap: e is a constant, and ln is the inverse function of ex, defined as the logarithm with base e. (If this sounds like gibberish, bookmark this page for a minute and come back to this question after refreshing your logarithm knowledge.)

Start by simplifying the equation:

Now, how do you solve for x?

Knowing that ln is the logarithm with base e, we can get x out of the exponent by taking the natural logarithm of both sides of this equation.

The natural log of ex is simply x, so we end up with:

That matches choice A.

#8: Probability & Primes

30 marbles labeled 1 through 30 were each put into a jar. One marble will be randomly drawn from this jar. What is the probability that this marble will have a prime number written on it?

A. 1/3

B. 1/4

C. 1/2

D. 11/30

Why this question is hard: It has multiple steps. You could also be tripped up by counting 1 as a prime number.

First, we need to figure out how many marbles have a prime number on them.

Then, we need to figure out the probability that one of these marbles will be taken from the jar.

Let’s start with the first step. Remember, primes are numbers greater than 1 that are only divisible by themselves and by 1.

There’s no shortcut for this – you just have to count. To reduce the chance of making a mistake, start at 1 and count up to 30, making a note of each prime you run into.

These are the primes you should find: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

That totals 10 numbers.

Now, what are the chances that one of these 10 marbles will be taken from the jar? There are 30 marbles total, and 10 have a prime number, so the probability is:

10/30

= ⅓

That gives us answer choice A.

#9: Ratios

A. 1/42

B. 5/27

C. 5/84

D. 2/27

Why this question is hard: This is an example of one very typical kind of “hard” ACT Math question: it’s not actually that hard. It just looks scary.

It looks like we need to solve some kind of complex expression, but actually, the question is asking for something fairly simple. First, figure out what you’re actually looking for.

We want the biggest possible value of this expression:

a/b • 1/c

We can start by simplifying that expression to:

a/bc

Because a is in the numerator, we want it to be as big as possible to maximize the value. And because b and c are in the denominator, meaning a will be divided by bc, we want those values to be as small as possible.

What’s the biggest possible value for a? And what are the smallest possible values of b and c? Let’s go back to the ranges provided in the question.

Finally, we can plug the values we’ve selected back into our expression:

That leads us to answer choice B.

If this isn’t making sense to you and you want more guided practice, think about reaching out to our team. Many of our tutors are perfect ACT scorers themselves, and they can help you with both the skills and strategy to do well on the test.

#10: Varying Rates

It took Danielle x minutes to cycle the d miles from home to work. Returning home on the same route, it took Danielle y minutes. On the way home, her average speed was 3 times her average speed on the way to work. Which of the following equations gives y in terms of x?

Why this question is hard: You might be tempted to try to solve for the variables, but you actually don’t need to (and can’t) do that – and it can cost you time. The answer choices also include trap answers that you’ll get if you make specific kinds of mistakes.

What’s this question asking for? The relationship between how long it took Danielle to walk to work (x) and how long it took her to walk home (y).

We’re told that Danielle’s average speed on the way home was 3 times her average speed on the way to work, so you might go for answer choice D. But that’s a trap! We’re not looking for the relationship in average speed; we’re looking for the relationship in time.

How can we set up an equation that puts x in relation to y? We know that the distance remained the same, and we know that the duration of the trip is the distance divided by the speed.

And because we know that the average speed on the way home was 3 times greater than on the way to work, we can express that in an equation that shows the relationship between these two average speeds:

Dividing both sides by d gives us:

That matches choice A.

#11: Real Numbers

A. 36

B. 37

C. 38

D. 39

Why this question is hard: It uses vocabulary that might throw you off, even though the concept is very straightforward. 

This question might throw you off with terms like “the real number line.” But all this really means is: how many real integers lie between these two values?

The first step is to determine the nearest integers to the fractions we’re given. You can plug them into a calculator to get a decimal answer, or you can write them as mixed numbers.

-50/3 = -16 ⅔

85/4 = 21 ¼

Now, what are the smallest and biggest integers within the range we’re looking at? If you’re a visual learner, you might want to sketch a line and plot these two numbers along it to visualize the range we’re talking about in this question.

The smallest integer in between these 2 values is -16 (remember, this is a bigger value than -16 ⅔ – watch out for that trap on this question!).

And the biggest integer between these 2 values is 21.

So, how many integers lie between those two?

  • 16 negative integers
  • 0 (forgetting 0 is another potential trap answer – 0 is also an integer!)
  • 21 positive integers

Add those together to find a total of 16 + 1 + 21 = 38 integers within the given range.

That’s choice C.

#12: Probability Terms

During a certain experiment, two events, X and Y, can each occur. Events X and Y are mutually exclusive during this experiment. Which of the following probabilities must be 0?

A. P(X)
B. P(Y)
C. P(X or Y)
D. P(X and Y)

Why this question is hard: To answer this question, you need to be comfortable with the probability terms used. If you are, then it’s fairly simple.

Mutually exclusive means that if X occurs, Y cannot occur. And if Y occurs, X cannot occur.

What do the various answer choices represent?

  • A shows the probability of event X occurring.

We know this is possible, so this probability is not zero. We can eliminate choice A.

  • B shows A shows the probability of event Y occurring.

We know this is possible, so this probability is not zero. We can eliminate choice B.

  • C shows the cumulative probability of event X or event Y occurring – basically, the probability that one of them will occur.

We’re told both events are possible, so this probability is not zero. We can eliminate choice C.

  • D shows the probability that both X and Y will occur.

We’re told that this is impossible: the events are mutually exclusive. The probability of this scenario happening is zero, so choice D is our answer.

Wondering how to set your goal score on ACT Math? Here’s the most recent data on how admitted students at over 500 schools scored, so you can see what you need at your goal schools.

#13: Volume and Surface Area

Dana has a cylindrical barrel that has an inside diameter of 12 inches and contains water. When she put marbles in the barrel, the water level rose 3 inches. What is the volume of the marbles in cubic inches?

A. 36π

B. 72π

C. 108π

D. 324π

Why this question is hard: There are several steps, and you need to remember the relevant formulas.

The first step with geometry questions is to visualize the situation.

Now it’s easier to see what you’re looking for.

The volume of the marbles is going to equal the volume of the water that they displaced. And that volume is the volume of the short cylinder between the original water level and the new water level.

Here’s where it’s important to have formulas up your sleeve on ACT Math. This is the formula for the volume of a cylinder:

The radius is the same as that of the barrel, which is 6 inches (not 12 inches! That’s the diameter). And the height is 3 inches. 

That gives us:

V = π (6 • 6) • 3

= 108 π

This is the volume of the displaced water, so its also the volume of the marbles.

That’s answer choice C.

#14: Statistics

A school randomly selected 120 students and asked them how many states they had been to in the past year. The table shows the results. Based on these data, for this population of students, what is the best estimate of the mean number of states visited in the past year?

A. 0.57

B. 2.6

C. 2.8

D. 3.2

Number of states1234
Number of students20305020

Why this question is hard: Not because you actually need a grasp of super-advanced concepts to answer it. The tricky factors aren’t really math-related: 

  • It takes a few steps to get to the answer, which can cost you time.
  • It’s easy to make a small mistake and end up with the wrong answer choice.

You’re looking for the average (mean) number of states these students visited. What do you need to know to calculate that?

  • How many students there are
  • How many states were visited

We already know there were 120 students total. What about the number of state visits?

We can figure that out by multiplying the number of students in each category by the number of states that that group visited. (Note that we don’t care if the states were the same or different – it’s just about the number of visits.)

The number of visits adds up to:

(1 • 20) + (2 • 30) + (3 • 50) + (4 • 20)

= 20 + 60 + 150 + 80

= 310

We have 310 state visits and 120 students. So the mean number of visits per student is:

310 / 120 = 2.6

That brings us to choice B.

#15: Imaginary and Complex Numbers

Given that i is the imaginary unit, which of the following numbers is equal to (5+3i)2?

Why this question is hard: This question relies on understanding imaginary numbers. If you don’t, then it’s pretty much impossible to answer correctly. But if you do, then it becomes a very easy question.

To recap, i is an imaginary number that represents √-1. That means that i2 = -1.

Start by expanding this like any other square:

Here’s where imaginary numbers come in. Since i = √-1, 9i2 = -9

That gives us choice D.

Next Steps

These are some of the hardest questions you’ll see on ACT Math – but all of them are doable. WIth enough time, hard work, and patience, you can gain confidence approaching even the hardest ACT Math questions.

If you’re aiming for a top score, tutoring support can also be helpful. We’ve been working with students preparing for the ACT for over two decades, and many of our tutors achieved perfect scores on the ACT themselves. Our students often see their scores go up by several points over just a few months of tutoring. 

Learn more about our programs here, and get in touch with our team for a free consultation on the right option for you.

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Ro

Rosamond graduated summa cum laude from Princeton. During college, she worked as a peer tutor for Italian, French, Greek, and academic writing. Since graduation, she’s continued to tutor students of all ages in French, Italian, Latin, English, and math, some of them for several years. She’s currently a graduate student studying the relationship between religion and secular law, and she works as a freelance journalist and translator.

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