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Hardest SAT Math Problems (updated for Digital SAT)

Bonus Material: The Hardest SAT Math Problems Quiz

Aiming for a really great score on the SAT? Wondering if your math skills are up to the challenge of the hardest problems?

If you want to be able to get a perfect score, you have to be able to solve the hardest SAT math problems.

We used our extensive test-prep experience to find the questions that many students miss. The examples below are real problems from past official SATs. 

Give each of these 16 hard math problems a try, then read our step-by-step explanations to see if you’re solving them correctly.

If you’re thinking about getting SAT tutoring to help you tackle problems like these on the real SAT, be sure to check out our list of the 15 best SAT Tutoring Services

Then, download this quiz with 20 more of the hardest real SAT problems ever to see if you’re on track for a perfect score! 

Bonus Material: 20 of the All-Time Hardest SAT Math Problems

Math on the SAT

Math accounts for half of your Total SAT Score, regardless of whether you’re taking the old paper SAT or the new Digital SAT.

On the traditional, paper SAT (which will be phased out in early 2024), the Math section comprises section 3, which contains 20 questions, is 25 minutes long and does not allow you to use a calculator; and section 4, which contains 38 questions, is 55 minutes long and does allow a calculator.

On the upcoming digital SAT (which will come into place in spring of 2024), the format is considerably different. You’ll be given two 35-minute “modules” with 22 questions in each, with the difficulty level of the second one depending on your performance on the first one. In other words, if you do really poorly on the first set of 22 questions, the second set will be easier–but your overall math score will be negatively affected. You can use your calculator on both.

Every SAT covers the following math material:

Heart of Algebra: 33% of test . Linear equations and inequalities and their graphs and systems.

Problem Solving and Data Analysis: 29% of test . Ratios, proportions, percentages, and units; analyzing graphical data, probabilities, and statistics.

Passport to Advanced Math: 28% of test . Identifying and creating equivalent expressions; quadratic and nonlinear equations/functions and their graphs.

Additional Topics in Math: 10% of test . A wide variety of topics, including geometry, trigonometry, radians and the unit circle, and complex numbers.

On the old SAT , open-ended questions came at the end of each Math section. Many students find them harder because you can’t guess or work backwards from multiple-choice options.

However, what many students don’t know is that the first 1–3 of these grid-in questions will actually be easier than the last few multiple-choice questions. 

That’s because the math questions on the SAT get increasingly difficult over the course of each section, but the difficulty level starts over again with the grid-in questions.

The savvy student will know this and skip the harder multiple-choice questions to go answer the easier grid-in questions first. Of course, if you’re aiming for a perfect score, (on most tests) you’ll have to answer every question correctly . 

But on the new Digital SAT, these open-ended questions will pop up at different points throughout both modules. You may see them in the beginning, the middle, or the end: there’s no set place for these to appear. Nor is there a set difficulty: generally, we’ve seen these questions be slightly on the easier side, but this varies significantly from test to test.

Because there’s obviously no bubble sheet on the digital SAT, you’ll simply type your answer into the text box. Be sure to look for instructions in the question about how they want the answer formatted!

To work with us for one-on-one tutoring or for our group SAT classes, schedule a free consultation with our team .

Why these problems are essential if you’re aiming at a top school

A perfect score on the SAT Math is 800. The only way to get this score is to answer every question correctly . 

In order to score a 750, you can only miss 2 or 3 questions across both math sections .

A 750 Math SAT may sound like a very high score—and it is! It’s a very high score.

But at the very best schools in the US, three quarters of the students scored a 750 Math or better.

In fact, at the Ivy League and other top schools, at least a quarter of the students had a perfect score!

The average math scores are even higher at the top engineering schools. Three quarters of the students at CalTech had a 790 or 800, and three quarters of the students at MIT had at least a 780.

In order to be a competitive applicant to these schools, your SAT Math score should be within the “middle 50%” of the students at that school—in other words, more or less an average score for that school.

So if you’re aiming at an Ivy or one of the other top schools, you can only miss 2 or 3 questions out of the 58 math questions on the whole SAT.

If that’s your goal, make sure that you understand the problems explained below, and then try our quiz of 20 more real SAT questions that rank among the hardest questions ever.

SAT Problem #1

At first glance, this looks like a geometry question, since it talks about planes and lines and points . But this is actually an algebra question, dressed up with some geometric trappings. 

The key is to realize: 

1) We don’t need to solve for p and r individually. We just need to solve for (r/p) . 

2) The points themselves (p,r) and (2p, 5r) represent X and Y values on the line itself. (For example if p = 2 and r = 3 then that’s the same thing as an x-coordinate of 2 and a y-coordinate of 3.)

So let’s take a look at it. 

First, let’s plug in the p and r points for the x and y values to see what equations we end up with. 

y = x + b becomes r = p + b

y = 2x + b becomes 5r = 2(2p) + b or 5r = 4p + b

At this point we might get a little anxious because we have three variables. 

But we have to remember we don’t need to get the value of the individual letters, just the value of the relationship between r and p. 

That’s where b actually becomes helpful. Because we can now set both equations equal to b , plug in, and then see if we can manipulate the r and p to get them to express the same relationship we want. 

So, first set both equations equal to b to get: 

b = r – p

b = 5r – 4p

And since, obviously b = b … 

r – p = 5r – 4p

Let’s now use some basic algebra to put the like variables together, so:

Now we’re nearly home. All we have to do is manipulate the problem so r/p .

So, divide both sides by  3p :

4r / 3p = 1

Then multiply both sides by 3: 

And finally divide by 4, which gives us: 

CHOICE B  

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SAT Problem #2

This is a question that can cause all sorts of problems if you forget your exponent rules—but it’s otherwise very straightforward. 

So let’s go over a few of those rules, just to get comfortable . . . and notice a pattern. I’ve included three below:

Two things to pay attention to:

First, when we divide variables with exponents, we keep the base and subtract the exponent. When we multiply variables with exponents, we keep the base and add the exponents. When we take a variable with an exponent to an additional power, we multiply the exponents. 

Second, in order to use the first two of these rules, the two numbers must have the same base . 

There is a base x on both the top and bottom of that fraction or the left and right side of that multiplication sign. 

So how does that help us here? 

Let’s forget the first half of the problem and look at the second:

We might look back at these exponent rules and throw our hands up—the top and bottom parts of this fraction don’t have the same base, so what am I supposed to do here? 

Except… 

8 and 2 actually DO have the same base. Base 2. 

Isn’t 2^3 equal to 8? 

So if we re-write the problem, plugging in 2^3 for 8, and thinking about that third exponent rule I gave you above, the equation will look like this: 

Now let’s go back to our exponent rules once more, and look at the first one. 

Because that tells us that… 

Well, hold on a second! 

We know the value of 3x – y . 

The problem tells us it’s 12.  

So we just plug in and get our answer… 

Which is CHOICE A. 

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SAT Problem #3

A question like this confuses a lot of students because they either forget how minimums and maximums work or find it hard to keep track of which numbers they are plugging in and where. 

In order to solve it, it’s helpful to think of a function as a machine . We enter an input into the machine (an x value)—it acts on it—and then it gives us an output (a y value). 

Let’s also remember that when we’re talking about minimum and maximums we’re talking about the y value when the function is at its highest and lowest point . 

With these two facts in mind, the problem is going to be much simpler, so let’s take it on in parts…

Since the question is asking us for g(k) and k represents the maximum value of f , it’s going to be helpful to first… 

Find k .  

So what is the maximum value of f , the graphed function? Well, the maximum value (as we realized earlier) is the y value when the function is at its highest. 

Looking at the graph, it looks the function is at highest when x = 4 , and more importantly, when 

Therefore, k = 3 .

Now let’s consider our functions as machines. 

When the problem asks us for g(k) , it’s telling us that k is going to act as the input (the x value for the function). So g(k) , the value after the machine acts upon the function, is going to be the output , or the y value . 

So, g(k) is the same as g(x) , except we’re plugging in our value of k , which is 3, for our x value. 

The rest is very simple. 

We go to the table and find where x = 3 , then move our finger across to see the output for that value, which is 6.

CHOICE B. 

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SAT Problem #4

A version of this question has appeared on the SAT multiple times in recent years, and it often stumps students!

Here we have something that resembles a rotated version of the logo from Star Trek, and we’re asked to find the value of a degree inside the circle, between two points of the pointed figure.

We’re given a point that represents the center of the circle, along with two degree measurements inside the triangle-like figure. 

Generally, when we’re given a figure that looks unfamiliar to us—like the figure inside the circle— it can be extremely helpful to find a way to fix it (or cut it up) so that it’s made up of parts of shapes that are more familiar . 

So looking inside this circle, how might we “fix” this figure so that it becomes a little friendlier. 

Well, if we draw a line to the center of the circle ( P ) from the edge of the circle ( A ), then this unfamiliar figure suddenly becomes two triangles. 

And with triangles, unlike the figure we were originally given, we can apply some rules . 

Rules, for example, that dictate opposite sides of the triangle that have the same length will have the same opposite angles. 

And if we look at our drawing we see that two sides of our triangle are the same length because they’re both the radius …

And so we also know that the opposite angles of those sides will be the same… 

And we’ve been given one of those angles! 

Therefore, angles ⦣ABP and ⦣PAB will be the same—both 20 degrees. Let’s fill that in. 

Now again—because we have a triangle—we can apply another rule as well. 

We know that degrees of a triangle will add up to 180 degrees. 

So if we know one of the inner degrees of the triangle is 20, and the other is 20—the remaining angle has to be 140 degrees. (Because 180 – 40 = 140. )

We have two of these triangles, so we know the larger inner angles of both add up to 280. 

Because a circle is 360 degrees, the number of degrees “left over” when 280 is subtracted from 360 is 80. 

So X equals 80. 

There is actually a second clever way to solve this problem, involving arc measures. Can you spot it? (If not, don’t worry! Ask us how we did it here .)

SAT Problem #5

Here we have a problem that looks quite complicated—and one I find students often waste a lot of time on. They either try to plug in answers and work backwards… 

…or they waste time trying to combine the two terms on the right side of the equation and simplifying. 

It turns out the easiest way to solve this problem is by polynomial division , because we’ve already been given the answer! It’s the right-hand side of the equation: (-8x – 3) – (53 / (ax – 2)) .

That means that this is our answer to when (24x^2 + 25x – 47) is divided by ax – 2 .

So how does that help us get a value for a ? 

Well, let’s set this up as a polynomial division problem.

We’d write it as follows: 

(I’m not putting the second half of the right side of the equation on top because that’s going to be our remainder.) 

So now we have a simple question. What number divided into 24 , gives me -8 ? 

Well, that’s easy. It’s -3 , right? 

Because -3 * -8 gives me 24 . 

So a equals -3 ,   CHOICE B .

Now, you could spend time plugging in -3 for a and dividing through the rest of the problem to make sure your answer matches the one on the exam—but generally on a timed test you really shouldn’t do more work than necessary. 

In fact, by setting this up as a polynomial division problem, we’ve saved time precisely because we don’t have to complete all the work . . . just enough to get us our answer. 

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SAT Problem #6

Because the SAT is a timed test, “difficult” includes not only questions that are hard to solve, but also those that—if a few wrong decisions are made—take a long time to solve. 

Sure, you may get the right answer, but those extra seconds or minutes wasted will inevitably cost you on other questions later on the exam.

Generally speaking, you should be able to answer each question in about a minute. If you spend more than 60 seconds on a single question, you should put down your best guess and move on (and hope that you have extra time at the end to return to this question).

To that end, let’s look at this question. You’re asked to find the value of 3x – 2 , and you’re given this equation:

(⅔)(9x – 6) – 4 = (9x – 6)

Many students will immediately think: “This is totally straightforward: Solve for x and plug it back into the equation.” 

They’ll distribute the ⅔ and end up with something like this: 

6x – 4 – 4 = 9x – 6

and then go through all the algebra from there, to get… 3x = -2 . 

These students will then find that x = (-⅔) . 

A few unlucky students will then forget that they have to plug in, and they’ll choose the trap answer C. 

The lucky ones will plug the (-⅔) back into 3x – 2 and get the correct answer, -4 , A . 

However, it turns out there is actually a much quicker way to solve this problem! 

We can solve it without ever having to plug into a second equation. 

If we simply subtract (⅔)(9x-6) from both sides, we end up with… 

-4 = (⅓)(9x-6) . 

We can realize that (⅓) of 9x-6 is the same as 3x-2 . 

And, what do you know… 

-4 = 3x – 2 . 

Ready to try some of these problems on your own? Try our quiz with 20 more of the hardest real SAT problems ever to see if you could get a perfect score on the SAT Math!

SAT Problem #7

This is a question you could muscle through, but it’s going to be a lot easier if we find a few shortcuts and work from there. Remember, a hard question isn’t necessarily difficult because of the conceptual and mathematical effort it asks from you but also because of the time it might require.

So how do we save ourselves some time? 

First, let’s notice that in the answer choices none of these numbers repeat . There are eight distinct numbers in the answer choices. Therefore, if we were pressed for time we only really have to find one of the values of c , choose the corresponding answer choice, and then move on. 

Second, let’s look at the other piece of information this problem gives us besides the quadratic. 

It tell us that a + b = 8.

This should be especially helpful because we know from FOIL (and what the rest of the problem gives us) that a * b = 15 , because abx^2 is going to be equal to 15x^2. 

Because a + b = 8 and ab = 15 , we know that the values of a and b are going to be 3 and 5. 

(We don’t know which one is which, and that’s precisely why this problem has two possible values for c .)

At this point we’ve done most of the “hard” work to save time in this problem, and it hasn’t even been particularly hard!

Now all we have to do is assign one of 3 or 5 to a , assign the other to b , FOIL out the problem, and pick whichever choice corresponds to one of the values of c . 

Let’s say a = 3 and b = 5 .

It will work like this: 

(3x + 2)(5x + 7) = 15x^2 + 21x + 10x + 14 .

Which simplifies to… 

15x^2 + 31x + 14 .

Which means c = 31 .

31 only appears once in our answer choices, so the answer must be CHOICE D.  

SAT Problem #8

When you’re faced with one of these more difficult system-of-equations problems—specifically the ones that ask you for no solutions or infinite solutions —it’s going to be much, much easier to think about the problems geometrically. 

In other words, as two line equations. 

So what does it mean for two lines to have no solutions ?

Well, for two lines to have no solutions, they’d have to never intersect , correct? 

(Just like if one of these problems asks you about two lines with infinite solutions , they’re saying that the lines are the same . They’re laid on top of each other. )

In other words, they’d have to be… parallel lines . 

And parallel lines have the same… slope! 

So this question is asking you to find the correct value for the variable that gives these lines the equivalent slope . 

Obviously, the first step is to put both of these equations in slope-intercept form. We’d end up with:

y = (-a/2)x + 2

Now the rest is very simple. All we need is a value of a that makes the slopes equal, so that it solves the equation (-a/2) = 3 .

With some basic algebra, we end up with -a = 6 . This is the same as a = -6 .

So the answer is CHOICE A, -6. 

Are these problems feeling super hard for you? Want to work on more similar problems? Check out our one-on-one tutoring with Ivy-League instructors. A great experienced tutor can help you focus on the concepts that are the hardest for you until you understand them thoroughly.

SAT Problem 9

This is another type of problem that students often have conceptual difficulty with, causing them to waste much more time than they should. 

(Remember, basically every problem in the SAT math section is designed to be solved in a minute and half or less. If you’re taking three or four minutes on a math problem, you’ve probably made a mistake!)

Some students will see that (u-t) is defined but not u or t individually, so they’ll try either solving for u in terms of t (or vice versa), or they’ll try squaring (u-t) to get a solution. (Which is closer to the correct way to solve the problem, but still incorrect). 

Instead, to solve this problem we need to remember the difference of squares . 

Remember, that the difference of squares states the following… 

(x+y)(x-y) = x^2 – xy + xy – y^2 .

Which means… 

(x+y)(x-y) = x^2 – y^2 .

And doesn’t that look awfully familiar to… u^2 – t^2 ?

In fact, we can now replace  u^2 – t^2 with (u + t)(u – t) .

So the whole problem would now read: (u + t)(u – t)(u – t) . Since we know the value of (u + t) and (u – t) , this would simply be the same as (2)(5)(2) .

Which equals our answer… 

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SAT Problem #10

What makes this question confusing is that students often get thrown off by the repetition of the (⅓). 

They forget that when the ⅓ gets factored out of the parenthesis like that, it means it’s going to apply to the whole equation: both the  x^2 AND the -2 . 

Once we remember that, we can solve this problem by difference of squares . This will save us the time of having to brute force the answer choices and FOIL each one through for the different values of k. 

We’ll simply square k and subtract it from the  x^2 for each choice. 

That will give us the following four choices: 

(⅓)(x^2  – 4)

(⅓)(x^2 – 36)

(⅓)(x^2 – 2)

(⅓) (x^2 – 6)

A student might rush to choose the third answer choice, since it appears to look like the expression at the beginning of the problem, but remember what I told you at the beginning: 

We’re going to apply that ⅓ to both the x^2 AND the k ! 

If we multiply that ⅓ through, the choices suddenly look like this…  

(⅓)(x^2) – (4/3)

(⅓)(x^2) –  (12)

(⅓)(x^2) – (⅔)

(⅓)(x^2) – (2)

. . . and so the correct answer is actually the fourth choice, CHOICE D . 

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SAT Problem #11

There are not many problems on the SAT that involve knowing the equation for a circle—in fact, circle equation problems don’t show up on every test—but that’s precisely why students often find a problem like this more difficult. 

First, let’s do a quick refresher on what the numbers in the equation of a circle mean. 

Any equation for a circle is going to be in this form: 

(x – h)^2 + (y – k)^2 = r^2

Where h and k represent the coordinates of the center and r is the radius. 

Let’s apply that to our problem here… 

(x + 3)^2 + (y – 1)^2 = 25.

Remember: because in the form of the circle equation the numbers inside the parenthesis are subtracted from x and y , when they appear inside the parenthesis as positives , that indicates the coordinate point will be negative. 

Therefore the center of this circle is at point (-3, 1) .

Because the radius is expressed as r^2 , then the 25 indicates the radius will be 5 . 

So we have a circle centered on the point (-3,1) and with a radius of 5 . 

So… now what? 

How do we figure out which of these points is not inside the circle? 

First, let’s draw the circle itself and look at it. On the SAT itself, you won’t have graph paper, so just draw a rough sketch!

Of course if we’re truly flummoxed we could graph the points, eliminate what we can . . . and guess. 

But that’s not ideal, obviously! 

Instead, let’s think about what the radius means. 

The radius demarcates the boundaries of the circle from the center. 

In other words, any points with a distance less-than-the-radius away from the center will lie within the circle. 

And any points more-than-the-radius distance from the center will lie outside of it. 

(Any points exactly-the-radius distance from the center will lie on the circle itself.) 

So all we have to do is find the point that is more than 5 units away from our center, and that will be our answer. 

To do this requires the distance formula. 

Remember, the distance formula is

A quick note: if you ever forget the distance formula, simply plot the two points on a graph, make a triangle with the distance between the two points and the hypotenuse, and use the Pythagorean Theorem to find the length of the hypotenuse, like this: 

Going back to our problem, let’s plug each of the points in along with our radius to the equation. (I’ll include the second point here, although since that’s our center we need not actually bother with it when we’re going through the problem.) We end up with: 

√(-3 – (-7))^2 +(1-(3))^2) = √20

√(-3 – (-3))^2 +(1-(1))^2) = √0

√(-3 – (0))^2 +(1-(0))^2) = √10

√(-3 – (3))^2 +(1-(2))^2) = √37

Only the square root of 37—choice D—is an answer that is larger than five. 

So that’s our correct choice, D . 

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SAT Problem #12

More circles! Let’s recall how the equation for a circle looked. It’s… 

(x – h)^2 + (y – k)^2 = r^2 

What the problem gives us, unfortunately, does not resemble that equation… 

…so our goal is to get the equation in the problem to look like a normal equation for a circle. 

Once we do this, we’ll just have to take the square root of whatever is on the right side of the equation, and that will give us our answer.

But how? 

We need to do something called completing the square . 

For the SAT, this concept is slightly obscure—it’s one you may see only once (or not at all) on a given exam. It makes the question a bit more difficult. 

Completing the square is normally a process reserved for solving a quadratic equation, but if you look closely at the way this problem is set up – 

2x^2 – 6x + 2y^2 + 2y = 45

we see that what we really have here are two quadratic equations, so we just have to complete the square twice. 

First we have to get rid of the coefficient in front of the x and y squared, so we have to divide through by 2 . 

This gives us  x^2 – 3x + y^2 + y = 22.5 .

Now we’re reading to complete the square!

Let’s deal with the x terms first. We have to think of what number, if we had it here in the equation, would allow us to factor x^2 – 3x into something of the form (x – z)^2 , where z is a constant. If we think about it, we realize that z has to be half of b . In this case, that means half of -3 , so -1.5 .

When we pop that into our setup, we get (x – 1.5)^2 . If we FOIL this out, however, we see that we get x^2 – 3x + 2.25 .

So it turns out that in order to be able to rewrite our expression in the form we want, we need to add 2.25 to our equation. As always in algebra, we do the same thing to both sides, so now we have:

x^2 – 3x + 2.25 + y^2 + y = 22.5 + 2.25.

Now we do the same thing for the y terms! Again, we need to add something to the equation so that we could rewrite the y part of the expression in the form (y – z)^2 . To get this number, we take half of the b term and square it: 1 divided by 2 , then squared, so 0.5^2 or 0.25.

Again, we have to add this number to both sides of the equation. Now we’ve got:

x^2 – 3x + 2.25 + y^2 + y + 0.25 = 22.5 + 2.25 + 0.25.

We can factor and rewrite this like:

( x – 1.5)^2 + (y + 0.5)^2 = 25.

Alright, now this is finally in the right format for the equation for a circle!

The final step is to use this equation to find the radius.

We know that the equation for a circle is (x – h)^2 + (y – k)^2 = r^2  . Fortunately this works out really nicely, since 25 is just 5^2. The radius must be 5, CHOICE A .

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SAT Problem #13

This question involves a number of moving parts and thus can be a little overwhelming for students to follow. 

It asks us to find, based on the rotation of the first gear, the rotation of the third. 

I find many students trip up on this problem by making two errors that are simple to fix, but relatively common. They fail to take the problem step by step… and they fail to write down their work as they track through the material. 

With that in mind, let’s work through the problem. 

Because gears A and C do not connect directly, but instead through gear B, we should first try to figure out the rotational relationship between A and B (at 100 rpm) before applying that to B and C. 

Because B is larger than A (and has more gears), A is going to rotate fully multiple times before B rotates once. 

How many times? Here it’s helpful to consider a ratio. 

A has 20 gears. 

B has 60 gears. 

So A is going to have to rotate three times before B rotates once . (20 goes into 60 three times.) 

Therefore, the ratio of rotation between A and B is 3 : 1 . 

Let’s write that down and then apply the same method to figure out the ratio between B and C. 

C has 10 gears. 

Here B only has to rotate a sixth of its distance for C to rotate once, so the ratio of rotation between B and C is 1 : 6 .

Now we take the number of RPMs the problem gives us, start with the gear on the left and multiply through with our ratios. 

So if Gear A rotates 100 times RPMS per minute, Gear B will rotate a third of that distance… 

So we divide 100 by 3. 

Because we know Gear C rotates six times as fast as Gear B, we then take our answer and multiply it by 6. 

So we get (100)(⅓)(6) . 

Which gives us 200 rpm. 

CHOICE C. 

SAT Problem #14

This question appears complicated—and students often get tripped up trying to either plug in numbers (which can be time consuming) or by searching for an equation that explains the relationship between the surface area and perimeter of the cube itself. 

This is especially tempting because while the question gives us the equation for the entire surface area of the cube, it only asks for the perimeter of one of the cube’s faces. 

However… 

If we think about the properties of a cube, this question actually becomes quite simple. 

First, let’s draw a cube.  

Again, the equation the problem gives us is for the entire surface area of the cube: 6(a/4)^2 .

But when we look at the cube, we may notice that it has, in fact, six faces. 

Therefore, each face would have one sixth of the surface area of the entire cube. 

So by dividing the equation by six, we get the surface area for one face of the cube, which is: 

But the question asks for the perimeter of one face of the cube. 

Let’s examine the drawing of the cube one more time. 

What shape is each cube face? It’s a square. 

And because each side of a square (let’s call each side x ) is equal to the other, the area of the square is going to be x^2, or the length of the side times itself. 

Well, wait a moment… 

If we go back to our equation for the surface area of ONE face of the cube, (a/4)^2 , we might notice that it’s in the same form as the equation for area of the square, except instead of x being squared, it’s (a/4) . 

And if we replace the x with (a/4) , we find that each side of the square is equivalent to (a/4) . 

Which makes finding the perimeter of this square quite simple, because it has four sides. 

So we merely add the four sides together: 

(a/4) + (a/4) + (a/4) + (a/4) . . .

which equals a . 

Which in this case is CHOICE B . 

Want more practice? We collected 20 more of the hardest SAT math problems. Download the quiz and take it with a 25-minute timer to mimic the real test!

SAT Problem #15

We have a lot of variables in this question, so it’s easiest to try to incorporate the extra piece of information we’re given, b = c – (½) , as best we can and then try to simplify the problem and solve from there. 

So how can we do that? 

The problem tells us b = c – (½) , which can also be expressed as b – c = -(½) .

(Once we put the b and c together on one side, it becomes easier to replace them together with a number). 

So what’s the best way to manipulate these two equations so that we’ll have b – c , which we can then replace with the (-½) and be left with x and y ? 

Because let’s remember that the problem does not ask us to solve for x and y individually. 

Just their relationship. 

So once we’re left with x and y as our only two variables, we should be able to make good progress. 

Anyhow, looking back over these two equations it seems the easiest way to be left with b – c is to… 

…subtract the bottom equation from the top one. 

When we do so, we’re left with the following: 

(3x – 3y) + (b – c) = (5x – 5y) + (-7 – (-7))

We replace b – c with -½  

And then combine like terms to get… 

(-½) = (5x – 3x) – (5y + 3y)

(-½) = 2x – 2y

Divide through by 2 … 

-¼ = x – y

Or x = y – (¼)

So our an answer is x = y – ¼ , CHOICE A .

Download 20 more of the hardest problems ever

SAT Problem #16

There are a few ways to solve this problem. The easiest one is simply to know the “remainder theorem.”

I don’t want to get too sidetracked with details, but remainder theorem states that when polynomial g(x) is divided by (x – a) , the remainder is g(a) .

In other words, when p(x) is divided by  (x-3) here, the remainder would be p(3) , which, according to the information we’re given, is -2 . 

That leads to CHOICE D . 

But what if, like many students, you don’t know the remainder theorem? (It’s pretty obscure and there’s a good chance you won’t see a problem about it on the entire exam.) 

Let’s look at an alternative way to solve the problem. 

If p(3) equals -2 , let’s imagine a function where that might be the case. 

We could do as simple one, like y = 3x – 11 , or a more complex one, like y = x^2 + 3x – 20 .

Either way, if I plug 3 into either of these functions for x , I get -2 as a y value. 

I should also notice immediately that (x – 5) , (x – 2) , and (x + 2) are not factors of either of them.

Clearly choices A, B, and C are not things that must be true. 

This also, by process of elimination, leads to CHOICE D. 

But just to check, let’s divide x – 3 into one of these functions – say 3x – 11 – and see what happens: 

The x goes into 3x three times – and three times (x-3) equals 3x – 9 .

When I subtract 3x – 9 from 3x – 11 , I get -2 , which is my remainder. 

Which points us, again, to CHOICE D .

Test your knowledge with 20 more problems

If these problems feel really hard, don’t panic—you can still do well on the SAT without answering every question correctly. 

The average SAT Math score for US students in 2022 was 52 8, and you have to answer about 32 out of 58 math questions correctly to get this score. That’s only a little over half of the questions!

However, if you want a high score—or a perfect score—you’ll have to be able to answer tough questions like these. You’ll need a very high score to be a competitive applicant for Harvard, Stanford, MIT, or other highly competitive schools.

The good news is that it’s very possible to raise your math score! 

In fact, it’s typically easier to improve your SAT Math score than your Reading & Writing score. Good preparation (on your own or with a tutor ) will fill in the knowledge gaps for any concepts that might be shaky and then practice the most common problem types until they feel easy.

We’ve worked with students who were able to see a 200-point increase on the Math section alone, through lots of hard work and practice.

To see how your math skills stack up against the toughest parts of the SAT, download our quiz with 20 more of the hardest SAT math questions, taken from real tests administered in recent years.

Once you know where you stand, keep up the practice!

If you’re interested in customized one-on-one tutoring support from an expert SAT tutor who can help you understand these tough problems, schedule a free consultation with Jessica or one of our founders . Our Ivy-League tutors are top scorers themselves who can help you with these more advanced concepts and strategies.

Bonus Material: Quiz: 20 of the All-Time Hardest SAT Math Problems

Emily graduated  summa cum laude  from Princeton University and holds an MA from the University of Notre Dame. She was a National Merit Scholar and has won numerous academic prizes and fellowships. A veteran of the publishing industry, she has helped professors at Harvard, Yale, and Princeton revise their books and articles. Over the last decade, Emily has successfully mentored hundreds of students in all aspects of the college admissions process, including the SAT, ACT, and college application essay. 

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Hard Algebra Questions – Challenging Problems and Step-by-Step Solutions

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Understanding the Fundamentals of Algebra

Solving algebraic operations, linear equations and their graphs, solving quadratic equations, systems of equations, applications of algebra.

Algebra is a fundamental branch of mathematics that deals with symbols and the rules for manipulating these symbols . In its most basic form, it is about finding the unknowns or the variables within equations.

As a student progresses, the questions in algebra can evolve from simple linear equations to more complex problems involving multiple steps, functions, or advanced concepts. Advanced algebra questions often probe one’s understanding of concepts like polynomial functions, rational expressions, exponents, and logarithms.

Navigating the web for reliable and challenging algebra problems and solutions I’ve found to be quite a task, but essential for anyone looking to master the subject. Websites and apps aimed at providing mathematics resources offer a range of problems, from basic to complex, helping learners tackle various topics effectively.

Engaging with these platforms allows for practice and reinforcement of algebraic concepts, which is crucial for solidifying one’s understanding and proficiency in mathematics.

I have to admit, that algebra can sometimes seem like a puzzle with a missing piece. But it’s that moment of clarity when the pieces fall into place that makes solving algebra questions incredibly rewarding. Stick around as we explore some hard algebra questions along with their solutions – the satisfaction of cracking them is just around the corner.

Algebra can seem challenging, but by grasping some key concepts, I can work through even the tough problems.

The backbone of algebra is the equation, which is a statement that two expressions are equal, often containing an unknown quantity, usually represented by a letter like ( x ) or ( y ).

For instance, in the equation ( x + 3 = 7 ), my goal is to find the value of ( x ) that makes this statement true. To do this, I simplify the equation to find ( x ). Simplification might involve combining like terms or using inverse operations. So, in this case, I subtract 3 from both sides to isolate ( x ), yielding ( x = 4 ).

Inequalities are like equations, but instead of equality, they express a relation where one value is larger or smaller than another. This can be represented by symbols such as “>,” “<,” “≤,” or “≥.” For example, if I have ( x – 5 > 10 ), I’m looking for all values of ( x ) that make this inequality true.

Solving this, I add 5 to both sides to get ( x > 15 ).

Here’s a simple way to visualize the process of solving an inequality :

Remember, if I multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign flips. So, if I had ( -2x < 8 ), by dividing both sides by -2, it becomes ( x > -4 ).

Understanding the foundation allows me to simplify and solve equations and inequalities, providing the confidence to tackle more complex algebraic questions.

When I approach algebraic operations, I prioritize understanding the core concepts—exponents, factoring, complex numbers, and simplification.

These elements are the building blocks to solving more intricate algebra problems.

For exponents , remember that the base raised to the power of an exponent reflects how many times the base is multiplied by itself. For instance, $3^4$ means $3 \times 3 \times 3 \times 3$. When dealing with exponents, it’s essential to be familiar with the laws of exponents for simplification purposes.

Next is factoring . I look for the greatest common factor (GCF) or use techniques like the difference of squares, and the quadratic formula for trinomials. For example, factoring $x^2 – 9$ would yield $(x + 3)(x – 3)$ because it is the difference of squares.

When it comes to complex numbers , which are in the form $a + bi$, where $i$ is the imaginary unit $(i^2 = -1)$, I remember that these numbers can be added, subtracted, and multiplied just like real numbers, with special attention to the property of $i$.

Finally, simplification is a matter of combining like terms and using the distributive property where necessary. For instance, to simplify the expression $2(x + 3) + x$, I first distribute the $2$ to get $2x + 6 + x$ and combine like terms to get $3x + 6$.

By mastering these components, I set a strong foundation for tackling hard algebra questions effectively.

Solving Advanced Algebraic Equations

Advanced algebra often serves as the foundation for calculus, statistics, and even programming. Let’s take a look at some of the core types of equations and how to approach them, including linear, quadratic, and systems of equations.

I’ll start by discussing linear equations , characterized by their straight-line graphs. A basic form of a linear equation is the slope-intercept form, expressed as ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept. Converting to standard form , ( Ax + By = C ), may be necessary for some applications. To graph these equations, I find two points by plugging in values for ( x ) and then plotting those points. Connecting these will give me a straight line.

Moving on to quadratic equations , these are typically presented in the general form ( a$x^2$ + bx + c = 0 ). These equations form parabolas when graphed.

I can solve quadratic equations using methods such as factoring, completing the square, or using the quadratic formula, ( x = $\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ ). Each method has its own use cases; for example, factoring is efficient when the equation easily decomposes into binomials.

Lastly, I’ll tackle systems of equations , where I have to find the solution set of two or more equations. When solving linear systems , I can use methods such as substitution, elimination, or graphing.

Substitution involves solving one equation for a variable and then substituting that expression into the other equation, while elimination adds or subtracts equations to eliminate a variable. Graphing requires drawing lines of each equation to find their point of intersection.

These are standard approaches that I often use to find the solutions to various types of algebraic equations.

Algebra has countless applications in the real world, and one of the most practical applications is in calculating areas and perimeters of geometric shapes. Let me guide you through some examples:

Area of a Rectangle : To find the area of a rectangle, I use the formula:

$$ A = l \times w $$

where ( A ) represents the area, ( l ) is the length, and ( w ) is the width.

Perimeter of a Rectangle : Similarly, finding the perimeter is just as straightforward with the equation:

$$ P = 2(l + w) $$

where ( P ) stands for the perimeter.

Area of a Circle : When I look at circles, the area is found using:

$$ A = \pi r^2 $$

Here, ( A ) is the area and ( r ) is the radius of the circle.

Circumference of a Circle: And for the circumference, which is the perimeter of a circle, the formula is:

$$ C = 2\pi r $$

where ( C ) is the circumference.

Understanding these formulas allows me to solve real-life problems, such as calculating the amount of paint needed to cover a wall or the fencing required for a circular garden.

With the help of algebra, I can create equations from word problems and solve them; the results help me make informed decisions and solve practical problems efficiently.

In our exploration of challenging algebra questions , I’ve presented a range of problems that test our understanding and application of algebraic concepts. From the intricacies of trigonometric equations to the puzzles of algebraic word problems, the journey through these issues illuminates the beauty of mathematics.

I sincerely hope that the solutions and strategies discussed have helped shed light on the complexity and elegance of algebra. Remember, practicing these tough problems not only sharpens your skills but also prepares you for more advanced mathematical challenges.

Whether you’re preparing for exams or simply indulging in the joy of problem-solving, the resilience and adaptability gained here are invaluable.

Algebra does not always yield its secrets easily, but patience and persistence in working through these tough problems can be immensely rewarding. Keep this momentum going, continue nurturing your mathematical curiosity, and let your confidence grow with each equation you solve.

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Choose Your Test

Sat / act prep online guides and tips, the hardest act math question types.

You’ve studied and now you’re geared up for the ACT math section (whoo!). But are you ready to take on the most challenging math questions the ACT has to offer?

If you're up for the challenge, preparation starts by understanding the toughest question types you'll face on the ACT . So if you’ve got your heart set on that perfect score (or you’re just really curious to see what the most difficult questions will be), then this is the guide for you.

We’ve put together what we believe to be the most most difficult math question types the ACT has given to students in the past 10 years, complete with strategies and answer explanations for each . These are all real ACT math questions, so understanding and studying them is one of the best ways to improve your current ACT score and knock it out of the park on test day.

Let's dive in!

Brief Overview of the ACT Math Section

Like all topic sections on the ACT, the ACT math section is one complete section that you will take all at once. It will always be the second section on the test and you will have 60 minutes to completed 60 questions .

The ACT arranges its questions in order of ascending difficulty. As a general rule of thumb, questions 1-20 will be considered “easy,” questions 21-40 will be considered “medium-difficulty,” and questions 41-60 will be considered “difficult.”

The way the ACT classifies “easy” and “difficult” is by how long it takes the average student to solve a problem as well as the percentage of students who answer the question correctly. The faster and more accurately the average student solves a problem, the “easier” it is. The longer it takes to solve a problem and the fewer people who answer it correctly, the more “difficult” the problem.

(Note: we put the words “easy” and “difficult” in quotes for a reason—everyone has different areas of math strength and weakness, so not everyone will consider an “easy” question easy or a “difficult” question difficult. These categories are averaged across many students for a reason and not every student will fit into this exact mold.)

All that being said, with very few exceptions, the most difficult ACT math problems will be clustered in the far end of the test. Besides just their placement on the test, these questions share a few other commonalities. We'll take a look at example questions and how to solve them and at what these types of questions have in common, in just a moment.

But First: Should You Be Focusing on the Hardest Math Questions Right Now?

If you’re just getting started in your study prep, definitely stop and make some time to take a full practice test to gauge your current score level and percentile. The absolute best way to assess your current level is to simply take the ACT as if it were real , keeping strict timing and working straight through (we know—not the most thrilling way to spend four hours, but it will help tremendously in the long run). So print off one of the free ACT practice tests available online and then sit down to take it all at once.

Once you’ve got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate ACT score. If you’re currently scoring in the 0-16 or 17-24 range , your best best is to first check out our guides on using the key math strategies of plugging in numbers and plugging in answers to help get your score up to where you want it to. Only once you've practiced and successfully improved your scores on questions 1-40 should you start in trying to tackle the most difficult math problems on the test.

If, however, you are already scoring a 25 or above and want to test your mettle for the real ACT, then definitely proceed to the rest of this guide . If you’re aiming for perfect (or close to) , then you’ll need to know what the most difficult ACT math questions look like and how to solve them. And luckily, that’s exactly what we’re here for.

Hardest Types of ACT Math Questions

Now that you're positive that you should be trying out these difficult math questions, let’s get right to it! We've broken hard ACT questions into sections based on type. Additionally, the answers to these questions are in a separate section below, so you can go through them all at once without getting spoiled.

Type #1: Graph Interpretation

Knowing how to read and interpret graphs can be tricky, especially since that information can also require you to use other math skills, like finding points on a line or calculating angles.

To unpack a graph question, first ask yourself—what is the prompt actually asking me to do? Figuring out the core mathematical concept behind the graphed information will make it easier to answer the question.

Check out the two graph interpretation questions below. Notice that both are asking you to interpret the graph data by using other mathematical concepts! With that in mind, try solving them on your own to get practice solving these tough questions. (Remember: we've included the answers below!)

Graph Interpretation Questions

QUESTION #1:

In the figure below, line $q$ in the standard $(x,y)$ coordinate plane has equation $-2x+y=1$ and intersects line $r$, which is distinct from line $q$, at a point on the $x$-axis. The angles $∠a$ and $∠b$ formed by these lines and the $x$-axis are congruent. What is the slope of line $r$?

G. $-{1}/2$

K. Cannot be determined

First, turn our given equation for line q into proper slope-intercept form .

Now, we are told that the angles the lines form are congruent. This means that the slopes of the lines will be opposites of one another [Note: perpendicular lines have opposite reciprocal slopes, so do NOT get these concepts confused!].

Since we have already established that the slope of line $q$ is 2, line $r$ must have a slope of -2.

Our final answer is F , -2

QUESTION #2:

The graph of the equation $h=-at^2 + bt + c$, which describes how the height, $h$, of a hit baseball changes over time, $t$, is shown below.

If you alter only this equation's $c$ term, which gives the height at time $t=0$, the alteration has an effect on which of the following?

I. The $h$-intercept II. The maximum value of $h$ III. The $t$-intercept

H. III only

J. I and III only

K. I, II, and III

The equation we are given ($−at^2+bt+c$) is a parabola and we are told to describe what happens when we change $c$ (the $y$-intercept).

From what we know about functions and function translations , we know that changing the value of $c$ will shift the entire parabola upwards or downwards, which will change not only the $y$-intercept (in this case called the "$h$-intercept"), but also the maximum height of the parabola as well as its $x$-intercept (in this case called the $t$-intercept). You can see this in action when we raise the value of the $y$-intercept of our parabola.

Options I, II, and III are all correct.

Our final answer is K, I, II, and III

Type #2: Trigonometry Questions

Trigonometry questions are tough for a number of reasons. First, they require you to memorize formulas and values in order to properly interpret questions. There's also a lot of conversion involved: you have to know how (and when!) to convert radians to degrees and visa versa.

And don't forget functions! Trigonometry uses non-linear functions, which is tough because its graph isn't a line or part of a line. Because trigonometry combines so many advanced math concepts, they can trip you up.

Test your trig skills on the advanced questions below. Try to work the, without peeking at the answers!

Trigonometry Questions

The equations of the two graphs shown below are $y_1(t) = a_1\sin (b_1t)$ and $y_2(t)=a_2\cos(b_2t)$, where the constants $b_1$ and $b_2$ are both positive real numbers.

Which of the following statements is true of the constants $a_1$ and $a_2$?

A. $0 < a_1 < a_2$

B. $0 < a_2 < a_1$

C. $a_1 < 0 < a_2$

D. $a_1 < a_2 < 0$

E. $a_2 < a_1 < 0$

The position of the a values (in front of the sine and cosine) means that they determine the amplitude (height) of the graphs. The larger the a value, the taller the amplitude.

Since each graph has a height larger than 0, we can eliminate answer choices C, D, and E.

Because $y_1$ is taller than $y_2$, it means that $y_1$ will have the larger amplitude. The $y_1$ graph has an amplitude of $a_1$ and the $y_2$ graph has an amplitude of $a_2$, which means that $a_1$ will be larger than $a_2$.

Our final answer is B , $0 < a_2 < a_1$.

For x such that $0 < x <π/2$, the expression ${√{1 – \cos^2 x}/{\sin x} + {√{1 – \sin^2 x}/{\cos x}$ is equivalent to:

J. $-\tan{x}$

K. $\sin{2x}$

If you remember your trigonometry shortcuts , you know that $1−{\cos^2}x+{\cos^2}x=1$. This means, then, that ${\sin^2}x=1−{\cos^2}x$ (and that ${\cos^2}x=1−{\sin^2}x$).

So we can replace our $1−{\cos^2}x$ in our first numerator with ${\sin^2}x$. We can also replace our $1−{\sin^2}x$ in our second numerator with ${\cos^2}x$. Now our expression will look like this:

${√{\sin^2 x}/{\sin x}+{√{\cos^2 x}/{\cos x }$

We also know that the square root of a value squared will cancel out to be the original value alone (for example,$√{2^2}=2$), so our expression will end up as:

$={\sin{x}}/{\sin{x}}+{\cos{x}}/{\cos{x}}$

Or, in other words:

Our final answer is H , 2.

Type #3: Logarithms

Logarithms can be hard because they require you to differentiate between exponential and logarithmic questions and understand logarithmic notation. If you can't read the question, you can't answer it, after all!

You'll also need to understand exponents to work with logarithms since the two concepts are intertwined. Working through the following equation can help you learn how to manage these tough problems.

Logarithm Questions

What is the real value of $x$ in the equation $\log_2{24} – \log_2{3} = \log_5{x}$?

ANSWER #1: If you’ve brushed up on your log basics, you know that $\log_b(m/n)=\log_b(m)−\log_b(n)$. This means that we can work this backwards and convert our first expression into:

$\log_2(24)-\log_2(3)=\log_2(24/3)$

$=\log_2(8)$

We also know that a log is essentially asking: "To what power does the base need to raised in order to achieve this certain value?" In this particular case, we are asking: "To which power must 2 be raised to equal 8?" The answer to this question is 3, since $(2^3=8)$, so $\log_2(8)=3$

Now this expression is equal to $\log_5(x)$, which means that we must also raise our 5 to the power of 3 in order to achieve $x$. So:

$3=\log_5(x)$

Our final answer is J , 125.

Type #4: Functions

Functions are one of those question types that are tough for some students but easier for others. That's because as functions get more complex, so does finding the correct answer. If you understand how functions work, you can break down more complex problems into simpler steps.;

Try your hand at the hard function question below to test your knowledge.

Function Questions

The functions $y = \sin x \and y = \sin(x + a) + b$, for constants $a$ and $b$, are graphed in the standard $(x,y)$ coordinate plane below. The functions have the same maximum value. One of the following statements about the values of $a$ and $b$ is true. Which statement is it?

A. $a < 0$ and $b = 0$

B. $a < 0$ and $b > 0$

C. $a = 0$ and $b > 0$

D. $a > 0$ and $b< 0$

E. $a > 0$ and $b> 0$

The only difference between our function graphs is a horizontal shift , which means that our b value (which would determine the vertical shift of a sine graph) must be 0.

Just by using this information, we can eliminate every answer choice but A, as that is the only answer with $b=0$. For expediency's sake, we can stop here.

Our final answer is A , $a<0$ and $b=0$

Advanced ACT Math note : An important word in ACT Math questions is "must," as in "something must be true." If a question doesn't have this word, then the answer only has to be true for a particular instance (that is, it could be true.)

In this case, the majority of the time, for a graph to shift horizontally to the left requires $a>0$. However, because $\sin(x)$ is a periodic graph, $\sin(x+a)$ would shift horizontally to the left if $a=-π/2$, which means that for at least one value of the constant $a$ where $a<0$, answer A is true. In contrast, there are no circumstances under which the graphs could have the same maximum value (as stated in the question text) but have the constant $b≠0$.

As we state above, though, on the real ACT, once you reach the conclusion that $b=0$ and note that only one answer choice has that as part of it, you should stop there. Don't get distracted into wasting more time on this question by the bait of $a<0$!

Whoo! You made it to the finish line—go you!

What Do the Hardest ACT Math Questions Have in Common?

Now, lastly, before we get to the questions themselves, it is important to understand what makes these hard questions “hard.” By doing so, you will be able to both understand and solve similar questions when you see them on test day, as well as have a better strategy for identifying and correcting your previous ACT math errors.

In this section, we will look at what these questions have in common and give examples for each type. In the next section, we will give you all 21 of the most difficult questions as well as answer explanations for each question, including the ones we use as examples here.

Some of the reasons why the hardest math questions are the hardest math questions are because the questions do the following:

#1: Test Several Mathematical Concepts at Once

EXAMPLE QUESTION:

Consider the functions $f(x) = √{x}$ and $g(x) = 7x + b$. In the standard $(x,y)$ coordinate plane, $y = f(g(x))$ passes through $(4,6)$. What is the value of $b$?

E. $4 – {7√6}$

As you can see, this question deals with a combination of functions and coordinate geometry points.

#2: Require Multiple Steps

Many of the most difficult ACT Math questions primarily test just one basic mathematical concept. What makes them difficult is that you have to work through multiple steps in order to solve the problem. (Remember: the more steps you need to take, the easier it is to mess up somewhere along the line!)

An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit?

A. $19/900$

B. $81/900$

C. $90/900$

D. $171/900$

E. $271/{1{,}000}$

Though it may sound like a simple probability question, you must run through a long list of numbers with 0 as a digit. This leaves room for calculation errors along the way.

#3: Use Concepts You're Less Familiar With

Another reason the questions we picked are so difficult for many students is that they focus on subjects you likely have limited familiarity with. For example, many students are less familiar with algebraic and/or trigonometric functions than they are with fractions and percentages, so most function questions are considered “high difficulty” problems.

The functions $y =$ sin$x$ and $y =$ sin$(x + a) + b$, for constants $a$ and $b$, are graphed in the standard $(x,y)$ coordinate plane below. The functions have the same maximum value. One of the following statements about the values of $a$ and $b$ is true. Which statement is it?

D. $a > 0$ and $b < 0$

E. $a > 0$ and $b > 0$

Many students get intimidated with function problems because they lack familiarity with these types of questions.

#4: Give You Convoluted or Wordy Scenarios to Work Through

Some of the most difficult ACT questions are not so much mathematically difficult as they are simply tough to decode. Especially as you near the end of the math section, it can be easy to get tired and misread or misunderstand exactly what the question is even asking you to find.

In the complex plane, the horizontal axis is called the real axis and the vertical axis is called the imaginary axis. The complex number $a + bi$ graphed in the complex plane is comparable to the point $(a,b)$ graphed in the standard $(x,y)$ coordinate plane.

The modulus of the complex number $a + bi$ is given by ${√{a^2+b^2}$ . Which of the complex numbers $z_1$ , $z_2$ , $z_3$ , $z_4$ , and $z_5$ below has the greatest modulus?

This question presents students with a completely foreign mathematical concept and can eat up the limited available time.

#5: Appear Deceptively Easy

Remember—if a question is located at the very end of the math section, it means that a lot of students will likely make mistakes on it. Look out for these questions, which may give a false appearance of being easy in order to lure you into falling for bait answers. Be careful!

Which of the following number line graphs shows the solution set to the inequality $|x–5|<1$ ?

This question may seem easy, but, because of how it is presented, many students will fall for one of the bait answers.

#6: Involve Multiple Variables or Hypotheticals

The more difficult ACT Math questions tend to use many different variables—both in the question and in the answer choices—or present hypotheticals. (Note: The best way to solve these types of questions—questions that use multiple integers in both the problem and in the answer choices—is to use the strategy of plugging in numbers .)

If $x$ and $y$ are real numbers such that $x > 1$ and $y < -1$, then which of the following inequalities must be true?

A. ${x/y} > 1$

B. ${|x|^2} > |y|$

C. ${x/3} – 5 > {y/3} – 5$

D. ${x^2} + 1 > {y^2} + 1$

E. ${x^{-2}} > {y^{-2}}$

Working with hypothetical scenarios and variables is almost always more challenging than working with numbers.

The Take-Aways

Taking the ACT is a long journey; the more you get acclimated to it ahead of time, the better you'll feel on test day. And knowing how to handle the hardest questions the test-makers have ever given will make taking your ACT seem a lot less daunting.

If you felt that these questions were easy , make sure not underestimate the effect of adrenaline and fatigue on your ability to solve your math problems. As you study, try to follow the timing guidelines (an average of one minute per ACT math question) and try to take full tests whenever possible. This is the best way to recreate the actual testing environment so that you can prepare for the real deal.

If you felt these questions were challenging , be sure to strengthen your math knowledge by checking out our individual math topic guides for the ACT . There, you'll see more detailed explanations of the topics in question as well as more detailed answer breakdowns.

What’s Next?

Felt that these questions were harder than you were expecting? Take a look at all the topics covered on the ACT math section and then note which sections you had particular difficulty in. Next, take a look at our individual math guides to help you strengthen any of those weak areas.

Running out of time on the ACT math section? Our guide to helping you beat the clock will help you finish those math questions on time.

Aiming for a perfect score? Check out our guide on how to get a perfect 36 on the ACT math section , written by a perfect-scorer.

Want to improve your ACT score by 4 points?

Check out our best-in-class online ACT prep classes . We guarantee your money back if you don't improve your ACT score by 4 points or more.

Our classes are entirely online, and they're taught by ACT experts . If you liked this article, you'll love our classes. Along with expert-led classes, you'll get personalized homework with thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step, custom program to follow so you'll never be confused about what to study next.

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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10 Hard Math Problems That Continue to Stump Even the Brightest Minds

Maybe you’ll have better luck.

For now, you can take a crack at the hardest math problems known to man, woman, and machine. For more puzzles and brainteasers, check out Puzzmo . ✅ More from Popular Mechanics :

• To Create His Geometric Artwork, M.C. Escher Had to Learn Math the Hard Way
• Fourier Transforms: The Math That Made Color TV Possible
• The Game of Trees is a Mad Math Theory That Is Impossible to Prove

The Collatz Conjecture

In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician Terence Tao. And while the story of Tao’s breakthrough is promising, the problem isn’t fully solved yet.

A refresher on the Collatz Conjecture : It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers (positive integers from 1 through infinity).

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Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways. But he most likely can’t adapt his methods to yield a complete solution to the problem, as Tao subsequently explained. So, we might be working on it for decades longer.

The Conjecture lives in the math discipline known as Dynamical Systems , or the study of situations that change over time in semi-predictable ways. It looks like a simple, innocuous question, but that’s what makes it special. Why is such a basic question so hard to answer? It serves as a benchmark for our understanding; once we solve it, then we can proceed onto much more complicated matters.

The study of dynamical systems could become more robust than anyone today could imagine. But we’ll need to solve the Collatz Conjecture for the subject to flourish.

Goldbach’s Conjecture

One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes.” You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for all natural numbers.

Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler , considered one of the greatest in math history. As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”

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Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.

Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math.

The Twin Prime Conjecture

Together with Goldbach’s, the Twin Prime Conjecture is the most famous in Number Theory—or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.

When two primes have a difference of 2, they’re called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many twin primes? The Twin Prime Conjecture says yes.

Let’s go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you’re ready to follow a bit of heady Number Theory.

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All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a factor of 3. Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime.

How’s your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.

The good news is that we’ve made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.

For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The closest we’ve come —given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.

The Riemann Hypothesis

Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math. It’s one of the seven Millennium Prize Problems , with $1 million reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.

There is a function, called the Riemann zeta function, written in the image above.

For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then 𝜁(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6. When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.

So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. The hypothesis is that the behavior continues along that line infinitely.

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The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.

If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.

The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer Conjecture is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain English. This Conjecture involves the math topic known as Elliptic Curves.

When we recently wrote about the toughest math problems that have been solved , we mentioned one of the greatest achievements in 20th-century math: the solution to Fermat’s Last Theorem. Sir Andrew Wiles solved it using Elliptic Curves. So, you could call this a very powerful new branch of math.

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In a nutshell, an elliptic curve is a special kind of function. They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and Number Theory.

British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s. Its exact statement is very technical, and has evolved over the years. One of the main stewards of this evolution has been none other than Wiles. To see its current status and complexity, check out this famous update by Wells in 2006.

The Kissing Number Problem

A broad category of problems in math are called the Sphere Packing Problems. They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space, like fruit at the grocery store. Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.

When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky. A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation. But a basic question about the kissing number stands unanswered.

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First, a note on dimensions. Dimensions have a specific meaning in math: they’re independent coordinate axes. The x-axis and y-axis show the two dimensions of a coordinate plane. When a character in a sci-fi show says they’re going to a different dimension, that doesn’t make mathematical sense. You can’t go to the x-axis.

A 1-dimensional thing is a line, and 2-dimensional thing is a plane. For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions. It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right. There’s proof of an exact number for 3 dimensions, although that took until the 1950s.

Beyond 3 dimensions, the Kissing Problem is mostly unsolved. Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see on this chart . For larger numbers, or a general form, the problem is wide open. There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come.

The Unknotting Problem

The simplest version of the Unknotting Problem has been solved, so there’s already some success with this story. Solving the full version of the problem will be an even bigger triumph.

You probably haven’t heard of the math subject Knot Theory . It ’s taught in virtually no high schools, and few colleges. The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.

For example, you might know how to tie a “square knot” and a “granny knot.” They have the same steps except that one twist is reversed from the square knot to the granny knot. But can you prove that those knots are different? Well, knot theorists can.

✅ Up Next: The Amazing Math Inside the Rubik’s Cube

Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing. The cool news is that this has been accomplished! Several computer algorithms for this have been written in the last 20 years, and some of them even animate the process .

But the Unknotting Problem remains computational. In technical terms, it’s known that the Unknotting Problem is in NP, while we don ’ t know if it’s in P. That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an impossibly long time. For now.

If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest. On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound. Eventually, we’ll find out.

The Large Cardinal Project

If you’ve never heard of Large Cardinals , get ready to learn. In the late 19th century, a German mathematician named Georg Cantor figured out that infinity comes in different sizes. Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.

There is the first infinite size, the smallest infinity , which gets denoted ℵ₀. That’s a Hebrew letter aleph; it reads as “aleph-zero.” It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.

Next, some common sets are larger than size ℵ₀. The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀. But the reals aren’t that big; we’re just getting started on the infinite sizes.

✅ More Mind-Blowing Stuff: Mathematicians Discovered a New 13-Sided Shape That Can Do Remarkable Things

For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals. It’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.” Then, if their proof is good, that’s the new largest known cardinal. Until someone else comes up with a larger one.

Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward. There’s now even a beautiful wiki of known large cardinals , named in honor of Cantor. So, will this ever end? The answer is broadly yes, although it gets very complicated.

In some senses, the top of the large cardinal hierarchy is in sight. Some theorems have been proven, which impose a sort of ceiling on the possibilities for large cardinals. But many open questions remain, and new cardinals have been nailed down as recently as 2019. It’s very possible we will be discovering more for decades to come. Hopefully we’ll eventually have a comprehensive list of all large cardinals.

What’s the Deal with 𝜋+e?

Given everything we know about two of math’s most famous constants, 𝜋 and e , it’s a bit surprising how lost we are when they’re added together.

This mystery is all about algebraic real numbers . The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers. The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.

✅ Try It Yourself: Can You Solve This Viral Brain Teaser From TikTok?

All rational numbers, and roots of rational numbers, are algebraic. So it might feel like “most” real numbers are algebraic. Turns out, it’s actually the opposite. The antonym to algebraic is transcendental, and it turns out almost all real numbers are transcendental—for certain mathematical meanings of “almost all.” So who’s algebraic , and who’s transcendental?

The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century. You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?

Well, we do know that both 𝜋 and e are transcendental. But somehow it’s unknown whether 𝜋+e is algebraic or transcendental. Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them. So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.

Is 𝛾 Rational?

Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers.

Rational numbers can be written in the form p/q, where p and q are integers. So, 42 and -11/3 are rational, while 𝜋 and √2 are not. It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?

Meet the Euler-Mascheroni constant 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly; it looks like the image above.

✅ One More Thing: Teens Have Proven the Pythagorean Theorem With Trigonometry. That Should Be Impossible

The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.” So, it’s a combination of two very well-understood mathematical objects. It has other neat closed forms, and appears in hundreds of formulas.

But somehow, we don’t even know if 𝛾 is rational. We’ve calculated it to half a trillion digits, yet nobody can prove if it’s rational or not. The popular prediction is that 𝛾 is irrational. Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it.

Dave Linkletter is a Ph.D. candidate in Pure Mathematics at the University of Nevada, Las Vegas. His research is in Large Cardinal Set Theory. He also teaches undergrad classes, and enjoys breaking down popular math topics for wide audiences.

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Problem solving

There are N problems numbered 1..N which you need to complete. You've arranged the problems in increasing difficulty order, and the i th problem has estimated difficulty level i . You have also assigned a rating vi to each problem. Problems with similar vi values are similar in nature. On each day, you will choose a subset of the problems and solve them. You've decided that each subsequent problem solved on the day should be tougher than the previous problem you solved on that day. Also, to make it less boring, consecutive problems you solve should differ in their vi rating by at least K. What is the least number of days in which you can solve all problems?

Input Format

The first line contains the number of test cases T. T test cases follow. Each case contains an integer N and K on the first line, followed by integers v1,...,vn on the second line.

Constraints

1 <= T <= 100 1 <= N <= 300 1 <= vi <= 1000 1 <= K <= 1000

Output Format

Output T lines, one for each test case, containing the minimum number of days in which all problems can be solved.

Sample Input

Sample Output

Explanation

For the first example, you can solve the problems with rating 5 and 7 on the first day and the problem with rating 4 on the next day. Note that the problems with rating 5 and 4 cannot be completed consecutively because the ratings should differ by at least K (which is 2). Also, the problems cannot be completed in order 5,7,4 in one day because the problems solved on a day should be in increasing difficulty level.

For the second example, all problems can be solved on the same day.

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Hard Maths Quiz Questions With Answers

Are you prepared to take on the ultimate mathematical challenge? Look no further – our "Hard Maths Quiz" is designed to push your mathematical prowess to the limit! Whether you're a math enthusiast seeking a cerebral workout or a student gearing up for a challenging exam, this quiz will put your mathematical skills to the test. Our carefully curated set of difficult math quiz questions with answers covers a wide range of mathematical concepts, from advanced algebra to intricate calculus and mind-bending geometry. You'll encounter problems that require creative problem-solving and a deep understanding of mathematical principles. Why tackle the Read more hardest math quiz? Because it's an excellent way to sharpen your critical thinking and problem-solving abilities. Plus, it's a fun and engaging way to challenge yourself and see just how far your math skills can take you. So, are you ready to embark on this mathematical journey? Take the "Hard Maths Quiz" now and prove that you've got what it takes to conquer the most challenging math problems. Remember, it's not about the destination; it's about the mathematical journey itself!

Hard Maths Questions and Answers

What is 6372+5849.

Rate this question:

If there are 23 boys and 21 girls in class A, Class B has 21 boys and 22 girls. How many girls and boys are there in Class A and B combined?

73 boys and 70 girls

35 boys and 37 girls

44 boys and 43 girls

50 boys and 47 girls

40 boys and 42 girls

Your mom is in the market. She bought 22kg of fish and 24kg of chicken. You ate 4kg of chicken and 12kg of fish. How many kg was left of the 2 foods combined?

What if you divide 352 by 12, what if you divide 70 by 69, what if you divide 20 by  0.5, what if you multiply 45.5 by 2, what if you add 7778.1 into 0.0001, you are the baker. you have 253 loaves of bread and 152 donuts. you are making 13 more loaves of bread and 4 donuts, while a boy bought 7 donuts and 17 loaves of bread. how many loaves of bread and donuts do you still have, you have $30.00; you bought a flower that cost $4.50 and sold it to a lady. she gave you $1.20. then you went to the donut store and bought 3 pieces of donuts that cost $12.45. how much money do you still have.

I don't want to waste my money.

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FREE daily maths challenges

A new KS2 maths challenge every day. Perfect as lesson starters - no prep required!

25 Fun Maths Problems For KS2 And KS3 (From Easy To Very Hard!)

Fun maths problems are one of the things mathematicians love about the subject; they provide an opportunity to apply mathematical knowledge, logic and problem solving skills all at once.  In this article, we’ve compiled 25 fun maths problems, each covering various topics and question types. They’re aimed at students in KS2 & KS3. We’ve categorised them as:

Maths word problems

Maths puzzles, fraction problems, multiplication and division problems, geometry problems, problem solving questions, maths puzzles are everywhere, how should teachers use these maths problems.

Teachers could make use of these maths problem solving questions in a number of ways, such as:

• embed into a relevant maths topic’s teaching.
• settling tasks at the beginning of lessons.
• break up or extend a maths worksheet.
• keep students thinking mathematically after the main lesson has finished.

Some are based on real life or historical maths problems, and some include ‘bonus’ maths questions to help to extend the problem solving fun! As you read through these problems, think about how you could adjust them to be relevant to your students or to practise different skills. 

These maths problems can also be used as introductory puzzles for maths games such as those introduced at the following links:

• KS2 maths games
• KS3 maths games

Need more support teaching reasoning, problem solving and planning for depth ? Read here for free CPD for you and your team of teachers.

1. Home on time – easy

Type: Time, Number, Addition

A cinema screening starts at 14:35. The movie lasts for 2 hours, 32 minutes after 23 minutes of adverts. It took 20 minutes to get to the cinema. What time should you tell your family that you’ll be home?

Answer: 17:50

2. A nugget of truth – mixed

Type: Times Tables, Multiplication, Multiples, Factors, Problem Solving 

Chicken nuggets come in boxes of 6, 9 or 20, so you can’t order 7 chicken nuggets. How many other impossible quantities can you find (not including fractions or decimals)?

Answer: 1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34, 37, or 43

There is actually a theorem which can be used to prove that every integer quantity greater than 43 can be ordered.

3. A pet problem – mixed

Type: Number, Problem Solving, Forming and Solving Equations, Simultaneous Equations, Algebra

Eight of my pets aren’t dogs, five aren’t rabbits, and seven aren’t cats. How many pets do I have?

Answer: 10 pets (5 rabbits, 3 cats, 2 dogs)

4. The price of things – mixed

Type: lateral thinking problem

A mouse costs £10, a bee costs £15, and a spider costs £20. How much does a duck cost? Answer: £5 (£2.50 per leg)

Looking for more word problems, solutions and explanations? Read our article on word problems for primary school.

25 Fun Maths Problems - Printable

Download a printable version of these fun maths problems together with answers and mark scheme.

5. A dicey maths challenge – easy

Type: Place value, number, addition, problem solving

Roll three dice to generate three place value digits. What’s the biggest number you can make out of these digits? What’s the smallest number you can make?

Add these two numbers together. What do you get?

Answer: In most cases, 1,089.

Bonus: Who got a different result? Why?

6. PIN problem solving – mixed

Type: Logic, problem solving, reasoning

I’ve forgotten my PIN. Six incorrect attempts locks my account: I’ve used five! Two digits are displayed after each unsuccessful attempt: “2, 0” means 2 digits from that guess are in the PIN, but 0 are in the right place.

What should my sixth attempt be?

Answer: 6347

7. So many birds – mixed

Type: Triangular Numbers, Sequences, Number, Problem Solving

On the first day of Christmas my true love gave me one gift. On the second day they gave me another pair of gifts plus a copy of what they gave me on day one. On day 3, they gave me three new gifts, plus another copy of everything they’d already given me. If they keep this up, how many gifts will I have after twelve days?

Answer: 364

Bonus: This could be calculated as 1 + (1 + 2) + (1 + 2 + 3) + … but is there an easier way? What percentage of my gifts do I receive on each day?

8. I 8 sum maths questions – mixed

Type: Number, Place Value, Addition, Problem Solving, Reasoning

Using only addition and the digit 8, can you make 1,000? You can put 8s together to make 88, for example.

Answer: 888 + 88 + 8 + 8 + 8 = 1,000 Bonus: Which other digits allow you to get 1,000 in this way?

9. Quizzical – easy

Type: Fractions, Adding Fractions, Equivalent Fractions, Fractions to Percentages

4 friends entered a maths quiz. One answered \frac{1}{5} of the maths questions, one answered \frac{1}{10} , one answered \frac{1}{4} , and the other answered \frac{4}{25} . What percentage of the questions did they answer altogether?

Answer: 71%

10. Ancient problem solving – mixed

Type: Fractions, Reasoning, Problem Solving

Ancient Egyptians only used unit fractions (like \frac{1}{2} , \frac{1}{3} or \frac{1}{4} ). For \frac{2}{3} , they’d write \frac{1}{3} + \frac{1}{3} . How might they write \frac{5}{8} ?

Answer: \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} is correct. So is \frac{1}{2} + \frac{1}{8} .

Bonus: Which solution is better? Why? Can you find any more? What if subtractions are allowed?

Learn more about unit fractions here

11. everybody wants a pizza the action – hard.

An infinite number of mathematicians buy pizza. The first wants \frac{1}{2} pizza. The second wants \frac{1}{4} pizza. The third & fourth want \frac{1}{8} and \frac{1}{16} each, and so on. How many pizzas should they order?

Answer: 1 Each successive mathematician wants a slice that is exactly half of what is left:

12. Shade it black – hard

Type: Fractions, Reasoning, Problem Solving What fraction of this image is shaded black?

Answer: \frac{1}{3}

Look at the L-shaped part made up of two white and one black squares: \frac{1}{3} of this part is shaded. Zoom in on the top-right quarter of the image, which looks exactly the same as the whole image, and use the same reasoning to find what fraction of its L-shaped portion is shaded. Imagine zooming in to do the same thing again and again…

13. Giving is receiving – easy

Type: Number, Reasoning, Problem Solving

5 people give each other a present. How many presents are given altogether?

14. Sharing is caring – mixed

I have 20 sweets. If I share them equally with my friends, there are 2 left over. If one more person joins us, there are 6 sweets left. How many friends am I with?

Answer: 6 people altogether (so 5 friends!)

15. Times tables secrets – mixed

Type: Area, 2D Shape, Rectangles

Here are 77 letters:

BYHRCGNGNEOEAAHGHGCURPUTSTSASHHSBOBOREOPEEMEMEELATPEPEFADPHLTLTUT IEEOHOHLENRYTITIIAGBMTNTNFCGEIIGIG

How many different rectangular grids could you arrange all 77 letters into?

Answer: Four: 1⨉77, 77⨉1, 11⨉7 & 7⨉11. If the letters are arranged into one of these, a message appears, reading down each column starting from the top left.

Bonus: Can you find any more integers with the same number of factors as 77? What do you notice about these factors (think about prime numbers)? Can you use this system to hide your own messages?

16. Laugh it up – hard

Type: Multiples, Lowest Common Multiple, Times Tables, Division, Time

One friend jumps every \frac{1}{3} of a minute. Another jumps every 31 seconds. When will they jump together? Answer: After 620 seconds

17. Pictures of matchstick triangles – easy

Type: 2D Shapes, Equilateral Triangles, Problem Solving, Reasoning

Look at the matchsticks arranged below. How many equilateral triangles are there?

Answer: 13 (9 small, 3 medium, 1 large)

Bonus: What if the biggest triangle only had two matchsticks on each side? What if it had four?

18. Dissecting squares – mixed

Type: Reasoning, Problem Solving

What’s the smallest number of straight lines you could draw on this grid such that each square has a line going through it?

19. Make it right – mixed

Type: Pythagoras’ theorem

This triangle does not agree with Pythagoras’ theorem. 

Adding, subtracting, multiplying or dividing each of the side lengths by the same integer can fix it. What is the integer?

Answer: 3 

8 – 3 = 5

The new side lengths are 3, 4 and 5 and  32 + 42 = 52.

20. A most regular maths question – hard

Type: Polygons, 2D Shapes, tessellation, reasoning, problem-solving, patterns

What is the regular polygon with the largest number of sides that will self-tessellate?

Answer: Hexagon.

Regular polygons tessellate if one interior angle is a factor of 360°. The interior angle of a hexagon is 120°. This is the largest factor less than 180°.

21. Pleased to meet you – easy

Type: Number Problem, Reasoning, Problem-Solving

5 people meet; each shakes everyone else’s hand once. How many handshakes take place?

Person A shakes 4 people’s hands. Person B has already shaken Person A’s hand, so only needs to shake 3 more, and so on.

Bonus: How many handshakes would there be if you did this with your class?

22. All relative – easy

Type: Number, Reasoning, Problem-Solving

When I was twelve my brother was half my age. I’m 40 now, so how old is he?

23. It’s about time – mixed

Type: Time, Reasoning, Problem-Solving

When is “8 + 10 = 6” true?

Answer: When you’re telling the time (8am + 10 hours = 6pm)

24. More than a match – mixed

Type: Reasoning, Problem-Solving, Roman Numerals, Numerical Notation

Here are three matches:

How can you add two more matches, but get eight? Answer: Put the extra two matches in a V shape to make 8 in Roman Numerals:

25. Leonhard’s graph – hard

Type: Reasoning, Problem-Solving, Logic

Leonhard’s town has seven bridges as shown below. Can you find a route around the town that crosses every bridge exactly once?

Answer: No!

This is a classic real life historical maths problem solved by mathematician Leonhard Euler (rhymes with “boiler”). The city was Konigsberg in Prussia (now Kaliningrad, Russia). Not being able to find a solution is different to proving that there aren’t any! Euler managed to do this in 1736, practically inventing graph theory in the process.

Many of these 25 maths problems are rooted in real life, from everyday occurrences to historical events. Others are just questions that might arise if you say “what if…?”. The point is that although there are many lists of such problem solving maths questions that you can make use of, with a little bit of experience and inspiration you could create your own on almost any topic – and so could your students. 

For a kick-starter on creating your own maths problems, read our article on KS3 maths problem solving .

Looking for additional support and resources at KS3? You are welcome to download any of the secondary maths resources from Third Space Learning’s resource library for free. There is a section devoted to GCSE maths revision with plenty of maths worksheets and GCSE maths questions . There are also maths tests for KS3, including a Year 7 maths test , a Year 8 maths test and a Year 9 maths test Other valuable maths practice and ideas particularly around reasoning and problem solving at secondary can be found in our KS3 and KS4 maths blog articles. Try these fun maths problems for KS2 and KS3, SSDD problems , KS3 maths games and 30 problem solving maths questions . For children who need more support, our maths intervention programmes for KS3 achieve outstanding results through a personalised one to one tuition approach.

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Since 2013 these personalised one to 1 lessons have helped over 150,000 primary and secondary students become more confident, able mathematicians.

Learn how the programmes are aligned to maths mastery teaching or request a personalised quote for your school to speak to us about your school’s needs and how we can help.

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Columbia University has become the epicenter of a growing showdown between student protesters, college administrators and Congress over the war in Gaza and the limits of free speech.

Nicholas Fandos, who covers New York politics and government for The Times, walks us through the intense week at the university. And Isabella Ramírez, the editor in chief of Columbia’s undergraduate newspaper, explains what it has all looked like to a student on campus.

On today’s episode

Nicholas Fandos , who covers New York politics and government for The New York Times

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Inside the week that shook Columbia University .

The protests at the university continued after more than 100 arrests.

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