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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. 

Keep up the practice! If you’d like help honing your skills, reach out to us for a free test prep consultation. All of our tutors are top 1% scorers who attended top-tier schools like Harvard and Princeton. That makes them uniquely qualified to help high-scoring students improve.

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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The 15 Hardest SAT Math Questions Ever

Want to test yourself against the most difficult SAT math questions? Want to know what makes these questions so difficult and how best to solve them? If you're ready to really sink your teeth into the SAT math section and have your sights set on that perfect score, then this is the guide for you.

We've put together what we believe to be the 15 most difficult questions for the current Digital SAT , with strategies and answer explanations for each. These are all hard SAT Math questions from College Board SAT practice tests, which means understanding them is one of the best ways to study for those of you aiming for perfection.

Image: Sonia Sevilla /Wikimedia

Brief Overview of SAT Math

There are two sections on the SAT: SAT Reading and Writing and SAT Math . Both sections are divided into two modules, and SAT Reading and Writing always comes first. So, the S AT Math modules will be the 3rd and 4th modules you’ll see on test day. Both math modules allow you to use a calculator. 

Each math module is arranged in order of ascending difficulty (where the longer it takes to solve a problem and the fewer people who answer it correctly, the more difficult it is). On each module, question 1 will be "easy" and question 22 will be considered "difficult." The modules are made up of both multiple choice and grid-in questions , and there isn’t a particular order for grid-ins—they can come anywhere in the module and be of any difficulty. 75% of SAT Math questions are multiple choice and 25% are grid-ins.

With very few exceptions, then, the most difficult SAT math problems will be clustered at the end of each module. In addition to their placement on the test, though, these questions also share a few other commonalities. In a minute, we'll look at example questions and how to solve them, then analyze them to figure out what these types of questions have in common.

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

If you're just getting started in your study prep (or if you've simply skipped this first, crucial step), definitely stop and take a full practice test to gauge your current scoring level. Check out our guide to all the free SAT practice tests available online and then sit down to take a test all at once.

The absolute best way to assess your current level is to simply take the SAT practice test as if it were real , keeping strict timing and working straight through with only the allowed breaks (we know—probably not your favorite way to spend a Saturday). Once you've got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate SAT Math score.

If you're currently scoring in the 200-400 or the 400-600 range on SAT Math, your best bet is first to check out our guide to improving your math score to be consistently at or over a 600 before you start in trying to tackle the most difficult math problems on the test.

If, however, you're already scoring above a 600 on the Math section and want to test your mettle for the real SAT, 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 SAT math questions look like and how to solve them. And luckily, that's exactly what we'll do.

There are a limited number of official SAT practice tests, especially now that the SAT has gone completely digital. We recommend beginning with the six official, digital SAT practice tests available through College Board’s Bluebook software. If you’ve finished those and want even more prep, though, there are lots of older (but still officially produced by the College Board) tests you can use. Most of the questions below are from these practice tests, so if you’re worried about spoiling those tests, stop reading this guide now and come back when you’ve completed all the older official practice tests. 

Now let's get to our list of questions (whoo)!

Image: Niytx /DeviantArt

The 15 Hardest SAT Math Questions

Now that you're sure you should be attempting these questions, let's dive right in! We've curated 15 of the most difficult SAT Math questions for you to try below, along with walkthroughs of how to get the answer (if you're stumped).

$$C=5/9(F-32)$$

The equation above shows how temperature $F$, measured in degrees Fahrenheit, relates to a temperature $C$, measured in degrees Celsius. Based on the equation, which of the following must be true?

• A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of $5/9$ degree Celsius.
• A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.
• A temperature increase of $5/9$ degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

A) I only B) II only C) III only D) I and II only

ANSWER EXPLANATION: Think of the equation as an equation for a line

where in this case

$$C= {5}/{9} (F−32)$$

$$C={5}/{9}F −{5}/{9}(32)$$

You can see the slope of the graph is ${5}/{9}$, which means that for an increase of 1 degree Fahrenheit, the increase is ${5}/{9}$ of 1 degree Celsius.

$$C= {5}/{9} (F)$$

$$C= {5}/{9} (1)= {5}/{9}$$

Therefore, statement I is true. This is the equivalent to saying that an increase of 1 degree Celsius is equal to an increase of ${9}/{5}$ degrees Fahrenheit.

$$1= {5}/{9} (F)$$

$$(F)={9}/{5}$$

Since ${9}/{5}$ = 1.8, statement II is true.

The only answer that has both statement I and statement II as true is D , but if you have time and want to be absolutely thorough, you can also check to see if statement III (an increase of ${5}/{9}$ degree Fahrenheit is equal to a temperature increase of 1 degree Celsius) is true:

$$C= {5}/{9} ({5}/{9})$$

$$C= {25} /{81} (\which \is ≠ 1)$$

An increase of $5/9$ degree Fahrenheit leads to an increase of ${25}/{81}$, not 1 degree, Celsius, and so Statement III is not true.

The final answer is D.

The equation ${24x^2 + 25x -47}/{ax-2} = -8x-3-{53/{ax-2}}$ is true for all values of $x≠2/a$, where $a$ is a constant.

What is the value of $a$?

A) -16 B) -3 C) 3 D) 16

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by $ax-2$ (so you can get rid of the fraction). When you multiply each side by $ax-2$, you should have:

$$24x^2 + 25x - 47 = (-8x-3)(ax-2) - 53$$

You should then multiply $(-8x-3)$ and $(ax-2)$ using FOIL.

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x + 6 - 53$$

Then, reduce on the right side of the equation

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x - 47$$

Since the coefficients of the $x^2$-term have to be equal on both sides of the equation, $−8a = 24$, or $a = −3$.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.

The final answer is B.

If $3x-y = 12$, what is the value of ${8^x}/{2^y}$?

A) $2^{12}$ B) $4^4$ C) $8^2$ D) The value cannot be determined from the information given.

ANSWER EXPLANATION: One approach is to express

$${8^x}/{2^y}$$

so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting $2^3$ for 8 in the numerator of ${8^x}/{2^y}$ gives

$${(2^3)^x}/{2^y}$$

which can be rewritten

$${2^3x}/{2^y}$$

Since the numerator and denominator of have a common base, this expression can be rewritten as $2^(3x−y)$. In the question, it states that $3x − y = 12$, so one can substitute 12 for the exponent, $3x − y$, which means that

$${8^x}/{2^y}= 2^12$$

The final answer is A.

Points A and B lie on a circle with radius 1, and arc ${AB}↖⌢$ has a length of $π/3$. What fraction of the circumference of the circle is the length of arc ${AB}↖⌢$?

ANSWER EXPLANATION: To figure out the answer to this question, you'll first need to know the formula for finding the circumference of a circle.

The circumference, $C$, of a circle is $C = 2πr$, where $r$ is the radius of the circle. For the given circle with a radius of 1, the circumference is $C = 2(π)(1)$, or $C = 2π$.

To find what fraction of the circumference the length of ${AB}↖⌢$ is, divide the length of the arc by the circumference, which gives $π/3 ÷ 2π$. This division can be represented by $π/3 * {1/2}π = 1/6$.

The fraction $1/6$ can also be rewritten as $0.166$ or $0.167$.

The final answer is $1/6$, $0.166$, or $0.167$.

$${8-i}/{3-2i}$$

If the expression above is rewritten in the form $a+bi$, where $a$ and $b$ are real numbers, what is the value of $a$? (Note: $i=√{-1}$)

ANSWER EXPLANATION: To rewrite ${8-i}/{3-2i}$ in the standard form $a + bi$, you need to multiply the numerator and denominator of ${8-i}/{3-2i}$ by the conjugate, $3 + 2i$. This equals

$$({8-i}/{3-2i})({3+2i}/{3+2i})={24+16i-3+(-i)(2i)}/{(3^2)-(2i)^2}$$

Since $i^2=-1$, this last fraction can be reduced simplified to

$$ {24+16i-3i+2}/{9-(-4)}={26+13i}/{13}$$

which simplifies further to $2 + i$. Therefore, when ${8-i}/{3-2i}$ is rewritten in the standard form a + bi, the value of a is 2.

In triangle $ABC$, the measure of $∠B$ is 90°, $BC=16$, and $AC$=20. Triangle $DEF$ is similar to triangle $ABC$, where vertices $D$, $E$, and $F$ correspond to vertices $A$, $B$, and $C$, respectively, and each side of triangle $DEF$ is $1/3$ the length of the corresponding side of triangle $ABC$. What is the value of $sinF$?

ANSWER EXPLANATION: Triangle ABC is a right triangle with its right angle at B. Therefore, $\ov {AC}$ is the hypotenuse of right triangle ABC, and $\ov {AB}$ and $\ov {BC}$ are the legs of right triangle ABC. According to the Pythagorean theorem,

$$AB =√{20^2-16^2}=√{400-256}=√{144}=12$$

Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of $\angle ∠ {F}$ equals the measure of $\angle ∠ {C}$. Therefore, $sin F = sin C$. From the side lengths of triangle ABC,

$$sinF ={\opposite \side}/{\hypotenuse}={AB}/{AC}={12}/{20}={3}/{5}$$

Therefore, $sinF ={3}/{5}$.

The final answer is ${3}/{5}$ or 0.6.

The incomplete table above summarizes the number of left-handed students and right-handed students by gender for the eighth grade students at Keisel Middle School. There are 5 times as many right-handed female students as there are left-handed female students, and there are 9 times as many right-handed male students as there are left-handed male students. if there is a total of 18 left-handed students and 122 right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.)

A) 0.410 B) 0.357 C) 0.333 D) 0.250

ANSWER EXPLANATION: In order to solve this problem, you should create two equations using two variables ($x$ and $y$) and the information you're given. Let $x$ be the number of left-handed female students and let $y$ be the number of left-handed male students. Using the information given in the problem, the number of right-handed female students will be $5x$ and the number of right-handed male students will be $9y$. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

$$x + y = 18$$

$$5x + 9y = 122$$

When you solve this system of equations, you get $x = 10$ and $y = 8$. Thus, 5*10, or 50, of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is ${50}/{122}$, which to the nearest thousandth is 0.410.

Questions 8 & 9

Use the following information for both question 7 and question 8.

If shoppers enter a store at an average rate of $r$ shoppers per minute and each stays in the store for average time of $T$ minutes, the average number of shoppers in the store, $N$, at any one time is given by the formula $N=rT$. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

ANSWER EXPLANATION: Since the question states that Little's law can be applied to any single part of the store (for example, just the checkout line), then the average number of shoppers, $N$, in the checkout line at any time is $N = rT$, where $r$ is the number of shoppers entering the checkout line per minute and $T$ is the average number of minutes each shopper spends in the checkout line.

Since 84 shoppers per hour make a purchase, 84 shoppers per hour enter the checkout line. However, this needs to be converted to the number of shoppers per minute (in order to be used with $T = 5$). Since there are 60 minutes in one hour, the rate is ${84 \shoppers \per \hour}/{60 \minutes} = 1.4$ shoppers per minute. Using the given formula with $r = 1.4$ and $T = 5$ yields

$$N = rt = (1.4)(5) = 7$$

Therefore, the average number of shoppers, $N$, in the checkout line at any time during business hours is 7.

The final answer is 7.

The owner of the Good Deals Store opens a new store across town. For the new store, the owner estimates that, during business hours, an average of 90 shoppers per hour enter the store and each of them stays an average of 12 minutes. The average number of shoppers in the new store at any time is what percent less than the average number of shoppers in the original store at any time? (Note: Ignore the percent symbol when entering your answer. For example, if the answer is 42.1%, enter 42.1)

ANSWER EXPLANATION: According to the original information given, the estimated average number of shoppers in the original store at any time (N) is 45. In the question, it states that, in the new store, the manager estimates that an average of 90 shoppers per hour (60 minutes) enter the store, which is equivalent to 1.5 shoppers per minute (r). The manager also estimates that each shopper stays in the store for an average of 12 minutes (T). Thus, by Little's law, there are, on average, $N = rT = (1.5)(12) = 18$ shoppers in the new store at any time. This is

$${45-18}/{45} * 100 = 60$$

percent less than the average number of shoppers in the original store at any time.

The final answer is 60.

Question 10

In the $xy$-plane, the point $(p,r)$ lies on the line with equation $y=x+b$, where $b$ is a constant. The point with coordinates $(2p, 5r)$ lies on the line with equation $y=2x+b$. If $p≠0$, what is the value of $r/p$?

ANSWER EXPLANATION: Since the point $(p,r)$ lies on the line with equation $y=x+b$, the point must satisfy the equation. Substituting $p$ for $x$ and $r$ for $y$ in the equation $y=x+b$ gives $r=p+b$, or $\bi b$ = $\bi r-\bi p$.

Similarly, since the point $(2p,5r)$ lies on the line with the equation $y=2x+b$, the point must satisfy the equation. Substituting $2p$ for $x$ and $5r$ for $y$ in the equation $y=2x+b$ gives:

$5r=2(2p)+b$

$\bi b$ = $\bo 5 \bi r-\bo 4\bi p$.

Next, we can set the two equations equal to $b$ equal to each other and simplify:

$b=r-p=5r-4p$

Finally, to find $r/p$, we need to divide both sides of the equation by $p$ and by $4$:

The correct answer is B , $3/4$.

If you picked choices A and D, you may have incorrectly formed your answer out of the coefficients in the point $(2p, 5r)$. If you picked Choice C, you may have confused $r$ and $p$.

Note that while this is in the calculator section of the SAT, you absolutely do not need your calculator to solve it!

Question 11

A) 261.8 B) 785.4 C) 916.3 D) 1047.2

ANSWER EXPLANATION: The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the SAT Math section:

Volume of a Cone

$$V={1}/{3}πr^2h$$

Volume of a Cylinder

$$V=πr^2h$$

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

$$V_{silo}=π(5^2)(10)+(2)({1}/{3})π(5^2)(5)=({4}/{3})(250)π$$

which is approximately equal to 1,047.2 cubic feet.

Question 12

If $x$ is the average (arithmetic mean) of $m$ and $9$, $y$ is the average of $2m$ and $15$, and $z$ is the average of $3m$ and $18$, what is the average of $x$, $y$, and $z$ in terms of $m$?

A) $m+6$ B) $m+7$ C) $2m+14$ D) $3m + 21$

ANSWER EXPLANATION: Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2, the equations $x={m+9}/{2}$, $y={2m+15}/{2}$, $z={3m+18}/{2}$are true. The average of $x$, $y$, and $z$ is given by ${x + y + z}/{3}$. Substituting the expressions in m for each variable ($x$, $y$, $z$) gives

$$[{m+9}/{2}+{2m+15}/{2}+{3m+18}/{2}]/3$$

This fraction can be simplified to $m + 7$.

Question 13

The function $f(x)=x^3-x^2-x-{11/4}$ is graphed in the $xy$-plane above. If $k$ is a constant such that the equation $f(x)=k$ has three real solutions, which of the following could be the value of $k$?

ANSWER EXPLANATION: The equation $f(x) = k$ gives the solutions to the system of equations

$$y = f(x) = x^3-x^2-x-{11}/{4}$$

A real solution of a system of two equations corresponds to a point of intersection of the graphs of the two equations in the $xy$-plane.

The graph of $y = k$ is a horizontal line that contains the point $(0, k)$ and intersects the graph of the cubic equation three times (since it has three real solutions). Given the graph, the only horizontal line that would intersect the cubic equation three times is the line with the equation $y = −3$, or $f(x) = −3$. Therefore, $k$ is $-3$.

Question 14

$$q={1/2}nv^2$$

The dynamic pressure $q$ generated by a fluid moving with velocity $v$ can be found using the formula above, where $n$ is the constant density of the fluid. An aeronautical engineer users the formula to find the dynamic pressure of a fluid moving with velocity $v$ and the same fluid moving with velocity 1.5$v$. What is the ratio of the dynamic pressure of the faster fluid to the dynamic pressure of the slower fluid?

ANSWER EXPLANATION: To solve this problem, you need to set up to equations with variables. Let $q_1$ be the dynamic pressure of the slower fluid moving with velocity $v_1$, and let $q_2$ be the dynamic pressure of the faster fluid moving with velocity $v_2$. Then

$$v_2 =1.5v_1$$

Given the equation $q = {1}/{2}nv^2$, substituting the dynamic pressure and velocity of the faster fluid gives $q_2 = {1}/{2}n(v_2)^2$. Since $v_2 =1.5v_1$, the expression $1.5v_1$ can be substituted for $v_2$ in this equation, giving $q_2 = {1}/{2}n(1.5v_1)^2$. By squaring $1.5$, you can rewrite the previous equation as

$$q_2 = (2.25)({1}/{2})n(v_1)^2 = (2.25)q_1$$

Therefore, the ratio of the dynamic pressure of the faster fluid is

$${q2}/{q1} = {2.25 q_1}/{q_1}= 2.25$$

The final answer is 2.25 or 9/4.

Question 15

For a polynomial $p(x)$, the value of $p(3)$ is $-2$. Which of the following must be true about $p(x)$?

A) $x-5$ is a factor of $p(x)$. B) $x-2$ is a factor of $p(x)$. C) $x+2$ is a factor of $p(x)$. D) The remainder when $p(x)$ is divided by $x-3$ is $-2$.

ANSWER EXPLANATION: If the polynomial $p(x)$ is divided by a polynomial of the form $x+k$ (which accounts for all of the possible answer choices in this question), the result can be written as

$${p(x)}/{x+k}=q(x)+{r}/{x+k}$$

where $q(x)$ is a polynomial and $r$ is the remainder. Since $x + k$ is a degree-1 polynomial (meaning it only includes $x^1$ and no higher exponents), the remainder is a real number.

Therefore, $p(x)$ can be rewritten as $p(x) = (x + k)q(x) + r$, where $r$ is a real number.

The question states that $p(3) = -2$, so it must be true that

$$-2 = p(3) = (3 + k)q(3) + r$$

Now we can plug in all the possible answers. If the answer is A, B, or C, $r$ will be $0$, while if the answer is D, $r$ will be $-2$.

A. $-2 = p(3) = (3 + (-5))q(3) + 0$ $-2=(3-5)q(3)$ $-2=(-2)q(3)$

This could be true, but only if $q(3)=1$

B. $-2 = p(3) = (3 + (-2))q(3) + 0$ $-2 = (3-2)q(3)$ $-2 = (-1)q(3)$

This could be true, but only if $q(3)=2$

C. $-2 = p(3) = (3 + 2)q(3) + 0$ $-2 = (5)q(3)$

This could be true, but only if $q(3)={-2}/{5}$

D. $-2 = p(3) = (3 + (-3))q(3) + (-2)$ $-2 = (3 - 3)q(3) + (-2)$ $-2 = (0)q(3) + (-2)$

This will always be true no matter what $q(3)$ is.

Of the answer choices, the only one that must be true about $p(x)$ is D, that the remainder when $p(x)$ is divided by $x-3$ is -2.

You deserve all the naps after running through those questions.

What Do the Hardest SAT Math Questions Have in Common?

It's important to understand what makes these hard questions "hard." By doing so, you'll 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 SAT math errors.

In this section, we'll look at what these questions have in common and give examples of each type. Some of the reasons why the hardest math questions are the hardest math questions is because they:

#1: Test Several Mathematical Concepts at Once

Here, we must deal with imaginary numbers and fractions all at once.

Secret to success: Think of what applicable math you could use to solve the problem, do one step at a time, and try each technique until you find one that works!

#2: Involve a Lot of Steps

Remember: the more steps you need to take, the easier to mess up somewhere along the line!

We must solve this problem in steps (doing several averages) to unlock the rest of the answers in a domino effect. This can get confusing, especially if you're stressed or running out of time.

Secret to success: Take it slow, take it step by step, and double-check your work so you don't make mistakes!

#3: Test Concepts That You Have Limited Familiarity With

For example, many students are less familiar with functions than they are with fractions and percentages, so most function questions are considered "high difficulty" problems.

If you don't know your way around functions , this would be a tricky problem.

Secret to success: Review math concepts that you don't have as much familiarity with such as functions . We suggest using our great free SAT Math review guides .

#4: Are Worded in Unusual or Convoluted Ways

It can be difficult to figure out exactly what some questions are asking , much less figure out how to solve them. This is especially true when the question is located at the end of the section, and you are running out of time.

Because this question provides so much information without a diagram, it can be difficult to puzzle through in the limited time allowed.

Secret to success: Take your time, analyze what is being asked of you, and draw a diagram if it's helpful to you.

#5: Use Many Different Variables

With so many different variables in play, it is quite easy to get confused.

Secret to success: Take your time, analyze what is being asked of you, and consider if plugging in numbers is a good strategy to solve the problem (it wouldn't be for the question above, but would be for many other SAT variable questions).

The Take-Aways

The SAT is a marathon and the better prepared you are for it, the better you'll feel on test day. Knowing how to handle the hardest questions the test can throw at you will make taking the real SAT 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 problems. As you continue to study, always adhere to the proper timing guidelines 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 SAT . 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 in the SAT math section and then note which sections were particular difficulty for you. Next, take a gander at our individual math guides to help you shore up any of those weak areas.

Running out of time on the SAT math section? Our guide will help you beat the clock and maximize your score .

Aiming for a perfect score? Check out our guide on how to get a perfect 800 on the SAT math section , written by a perfect-scorer.

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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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13 World’s Hardest Math Problems | With Solutions

For decades, mathematics has been a fascinating and challenging topic. People have been interested in learning and getting good at math from ancient Greeks to modern mathematicians. But have you ever wondered which math problem is the most challenging?

What could be so tricky and complicated that only some of the brightest mathematicians have been able to solve it? This article will look at 13 of the hardest math problems and how mathematicians have tried to solve them.

Continue reading the article to explore the world’s hardest math problems, listed below.

The Poincaré Conjecture

The prime number theorem, fermat’s last theorem, the reimann hypothesis, classification of finite simple groups, four color theorem, goldbach’s conjecture.

• Inscribed Square Problem

Twin Prime Conjecture

The continuum hypothesis, collatz conjecture, birch and swinnerton-dyer conjecture, the kissing number problem.

Mathematicians struggled for about a century with the Poincaré conjecture, which was put forth by Henri Poincaré in 1904.

According to this theory,

every closed, connected three-dimensional space is topologically identical to a three-dimensional sphere (S3).

We must explore the field of topology to comprehend what this entails. The study of properties of objects that hold after being stretched, bent, or otherwise distorted is known as topology. In other words, topologists are fascinated by how things can change without rupturing or being torn.

The topology of three-dimensional spaces is the subject of the Poincaré conjecture. A space volume with three dimensions—length, breadth, and height—is a three-dimensional space. A three-dimensional object called a sphere has a round and curved surface.

According to the Poincaré Conjecture, a three-sphere (S3), or the collection of points in four dimensions that are all at a fixed distance from a given point, is topologically identical to every simply-connected, closed, three-dimensional space (i.e., one that has no gaps or voids) and edges. 

Although it would appear easy, it took more than a century to confirm the conjecture thoroughly.

• Poincaré expanded his hypothesis to include any dimension (n-sphere). 
• Stephen Smale, an American mathematician, proved the conjecture to be true for n = 5 in 1961.
• Freedman, another American mathematician, proved the conjecture to be true for n = 4 in 1983. 
• Grigori Perelman, a Russian mathematician, then proved the conjecture to be true for n = 3 in 2002, completing the solution.
• Perelman eventually addressed the problem by combining topology and geometry. One of the highest awards in mathematics, the Fields Medal, was given to all three mathematicians. Perelman rejected the Fields Medal. He was also given a $1 million prize by the Clay Mathematics Institute (CMI) of Cambridge, Massachusetts, for resolving one of the seven Millennium Problems, considered one of the world’s most challenging mathematical puzzles. However, he turned it down as well.

The prime number theorem (PNT) explains how prime numbers asymptotically distribute among positive integers. It shows how fast primes become less common as numbers get bigger.

The prime number theorem states that the number of primes below a given natural number N is roughly N/log(N), with the word “approximately” carrying the typical statistical connotations.

• Two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, independently proved the Prime Number Theorem in 1896. Since then, the proof has frequently been the subject of rewrites, receiving numerous updates and simplifications. However, the theorem’s influence has only increased.

French lawyer and mathematician Pierre de Fermat lived in the 17th century. Fermat was one of the best mathematicians in history. He talked about many of his theorems in everyday conversation because math was more of a hobby for him.

He made claims without proof, leaving it to other mathematicians decades or even centuries later to prove them. The hardest of them is now referred to as Fermat’s Last Theorem.

Fermat’s last theorem states that;

there are no positive integers a, b, and c that satisfy the equation an + bn = cn for any integer value of n greater than 2.

• In 1993, British mathematician Sir Andrew Wiles solved one of history’s longest mysteries. As a result of his efforts, Wiles was knighted by Queen Elizabeth II and given a special honorary plaque rather than the Fields Medal because he was old enough to qualify.
• Wiles synthesized recent findings from many distinct mathematics disciplines to find answers to Fermat’s well-known number theory query.
• Many people think Fermat never had proof of his Last Theorem because Elliptic Curves were utterly unknown in Fermat’s time.

Mathematicians have been baffled by the Riemann Hypothesis for more than 150 years. It was put forth by the German mathematician Bernhard Riemann in 1859. According to Riemann’s Hypothesis

Every Riemann zeta function nontrivial zero has a real component of ½.

The distribution of prime numbers can be described using the Riemann zeta function. Prime numbers, such as 2, 3, 5, 7, and 11, can only be divided by themselves and by one. Mathematicians have long been fascinated by the distribution of prime numbers because figuring out their patterns and relationships can provide fresh perspectives on number theory and other subject areas.

Riemann’s hypothesis says there is a link between how prime numbers are spread out and how the zeros of the Riemann zeta function are set up. If this relationship is accurate, it could significantly impact number theory and help us understand other parts of mathematics in new ways.

• The Riemann Hypothesis is still unproven, despite being one of mathematics’ most significant unsolved issues.
• Michael Atiyah, a mathematician, proclaimed in 2002 that he had proved the Riemann Hypothesis, although the mathematical community still needs to acknowledge his claim formally.
• The Clay Institute has assigned the hypothesis as one of the seven Millennium Prize Problems. A $1 million prize is up for anyone who can prove the Riemann hypothesis to be true or false.

Abstract algebra can be used to do many different things, like solve the Rubik’s cube or show a body-swapping fact in Futurama. Algebraic groups follow a few basic rules, like having an “identity element” that adds up to 0. Groups can be infinite or finite, and depending on your choice of n, it can be challenging to describe what a group of a particular size n looks like.

There is one possible way that the group can look at whether n is 2 or 3. There are two possibilities when n equals 4. Mathematicians intuitively wanted a complete list of all feasible groups for each given size.

• The categorization of finite simple groups, arguably the most significant mathematical undertaking of the 20th century, was planned by Harvard mathematician Daniel Gorenstein, who presented the incredibly intricate scheme in 1972.
• By 1985, the project was almost finished, but it had consumed so many pages and publications that peer review by a single person was impossible. The proof’s numerous components were eventually reviewed one by one, and the classification’s completeness was verified.
• The proof was acknowledged mainly by the 1990s. Verification was later streamlined to make it more manageable, and that project is still active today.

According to four color theorem

Any map in a plane can be given a four-color coloring utilizing the rule that no two regions sharing a border (aside from a single point) should have the same color.

• Two mathematicians at the University of Illinois at Urbana-Champaign, Kenneth Appel and Wolfgang Hakan identified a vast, finite number of examples to simplify the proof. They thoroughly examined the over 2,000 cases with the aid of computers, arriving at an unheard-of proof style.
• The proof by Appel and Hakan was initially debatable because a computer generated it, but most mathematicians ultimately accepted it. Since then, there has been a noticeable increase in the usage of computer-verified components in proofs, as Appel and Hakan set the standard.

According to Goldbach’s conjecture, every even number (higher than two) is the sum of two primes. You mentally double-check the following for small numbers: 18 is 13 + 5, and 42 is 23 + 19. Computers have tested the conjecture for numbers up to a certain magnitude. But for all natural numbers, we need proof.

Goldbach’s conjecture resulted from correspondence between Swiss mathematician Leonhard Euler and German mathematician Christian Goldbach in 1742.

• Euler is regarded as one of the finest mathematicians in history. Although I cannot prove it, in the words of Euler, “I regard [it] as a totally certain theorem.”
• Euler might have understood why it is conversely tricky to resolve this problem. More significant numbers have more methods than smaller ones to be expressed as sums of primes. In the same way that only 3+5 can split eight into two prime numbers, 42 can be divided into 5+37, 11+31, 13+29, and 19+23. Therefore, for vast numbers, Goldbach’s Conjecture is an understatement.
• The Goldbach conjecture has been confirmed for all integers up to 4*1018, but an analytical proof has yet to be found.
• Many talented mathematicians have attempted to prove it but have yet to succeed.

Inscribed Sq uare Problem

Another complex geometric puzzle is the “square peg problem,” also known as the “inscribed square problem” or the “Toeplitz conjecture.” The Inscribe Square Problem Hypothesis asks:

Does every simple closed curve have an inscribed square?

In other words, it states, ” For any curve, you could draw on a flat page whose ends meet (closed), but lines never cross (simple); we can fit a square whose four corners touch the curve somewhere.

• The inscribed square problem is unsolved in geometry.
• It bears the names of mathematicians Bryan John Birch and Peter Swinnerton-Dyer, who established the conjecture using automated calculation in the first half of the 1960s.
• Only specific instances of the hypothesis have been proven as of 2023.

The Twin Prime Conjecture is one of many prime number-related number theory puzzles. Twin primes are two primes that differ from each other by two. The twin prime examples include 11 and 13 and 599 and 601. Given that there are an unlimited number of prime numbers, according to number theory, there should also be an endless number of twin primes.

The Twin Prime Conjecture asserts that there are limitless numbers of twin primes.

• In 2013, Yitang Zhang did groundbreaking work to solve the twin prime conjecture.
• However, the twin prime conjecture still needs to be solved.

Infinities are everywhere across modern mathematics. There are infinite positive whole numbers (1, 2, 3, 4, etc.) and infinite lines, triangles, spheres, cubes, polygons, etc. It has also been proven by modern mathematics that there are many sizes of infinity.

If the elements of a set can be arranged in a 1-to-1 correspondence with the positive whole numbers, we say the set of elements is countably infinite. Therefore, the set of whole numbers and rational numbers are countable infinities.

Georg Cantor found that the set of real numbers is uncountable. In other words, even if we used all the whole numbers, we would never be able to go through and provide a positive whole number to every real number. Uncountable infinities might be seen as “larger” than countable infinities.

• According to the continuum hypothesis, there must be a set of numbers whose magnitude strictly falls between countably infinite and uncountably infinite. The continuum hypothesis differs from the other problems in this list in that it is impossible to solve or at least impossible to address using present mathematical methods.
• As a result, even though we have yet to determine whether the continuum hypothesis is accurate, we do know that it cannot be supported by the tools of modern set theory either. It would be necessary to develop a new framework for set theory, which has yet to be done, to resolve the continuum hypothesis.

To understand Collatz’s conjecture, try to understand the following example. First, you have to pick a positive number, n. Then, from the last number, create the following sequence:

If the number is even, divide by 2. If it’s odd, multiply by 3 and then add 1. The objective is to keep going through this sequence until you reach 1. Let’s try this sequence with the number 12 as an example. Starting with number 12, we get: 12, 6, 3, 10, 5, 16, 8, 4, 2, 1

Starting at 19, we obtain the following: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

According to the Collatz conjecture, this sequence will always end in 1, regardless of the value of n you started with. This speculation has been tested for all values of n up to 87,260, but no proof has been found.

• Collatz’s conjecture has been unsolved up till now.
• Mathematical problem-solver Paul Erdree once said of the Collatz Conjecture, “Mathematics may not be ready for such problems.”

Two British mathematicians, Bryan Birch and Peter Swinnerton-Dyer formulated their hypotheses in the 1960s. The Birch and Swinnerton-Dyer conjecture in mathematics describes rational answers to the equations defining an elliptic curve.

This hypothesis states explicitly that there are an infinite number of rational points (solutions) if ζ(1) equals 0 and that there are only a finite number of such places if ζ(1) is not equal to 0.

• For Birch and Swinnerton-Dyer’s conjecture, Euclid provided a comprehensive solution, but this becomes very challenging for problems with more complex solutions.
• Yu. V. Matiyasevich demonstrated in 1970 that Hilbert’s tenth problem could not be solved, saying there is no mechanism for identifying when such equations have a whole number solution.
• As of 2023, only a few cases have been solved.

Each sphere has a Kissing Number, the number of other spheres it is kissing, when a group of spheres is packed together in one area. For example, your kissing number is six if you touch six nearby spheres. Nothing difficult.

Mathematically, the condition can be described by the average kissing number of a tightly packed group of spheres. However, a fundamental query regarding the kissing number remains unsolved.

First, you must learn about dimensions to understand the kissing number problem. In mathematics, dimensions have a special meaning as independent coordinate axes. The two dimensions of a coordinate plane are represented by the x- and y-axes. 

A line is a two-dimensional object, whereas a plane is a three-dimensional object. Mathematicians have established the highest possible kissing number for spheres with those few dimensions for these low numbers. On a 1-D line, there are two spheres—one to your left and the other to your right.

• The Kissing Problem is generally unsolved in dimensions beyond three.
• A complete solution for the kissing problem number faces many obstacles, including computational constraints. The debate continued to solve this problem.

The Bottom Line

When it comes to pushing the boundaries of the enormous human ability to comprehend and problem-solving skills, the world’s hardest math problems are unquestionably the best. These issues, which range from the evasive Continuum Hypothesis to the perplexing Riemann Hypothesis, continue to puzzle even the sharpest mathematicians.

But regardless of how challenging they are, these problems keep mathematicians inspired and driven to explore new frontiers. Whether or not these problems ever get resolved, they illustrate the enormous ability of the human intellect.

Even though some of these issues might never fully be resolved, they continue to motivate and inspire advancement within the field of mathematics and reflects how broad and enigmatic this subject is!

Let us know out of these 13 problems which problem you find the hardest!

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An in-ground pond has the shape of a rectangular prism. The pond has a depth...

The average zinc concentration recovered from a sample of measurements taken in 36 different locations...

Why is it important that a sample be random and representative when conducting hypothesis testing?...

Which of the following is true about the sampling distribution of means? A. Shape of...

Give an example of a commutative ring without zero-divisors that is not an integral domain.

List all zero-divisors in Z20. Can you see relationship between the zero-divisors of Z20 and...

Find the integer a such that a≡−15(mod27) and −26≤a≤0

Explain why the function is discontinuous at the given number a. Sketch the graph of...

Two runners start a race at the same time and finish in a tie. Prove...

Which of the following graphs represent functions that have inverse functions?

find the Laplace transform of f (t). f(t)=tsin⁡3t

Find Laplace transforms of sin⁡h3t cos22t

find the Laplace transform of f (t). f(t)=t2cos⁡2t

The Laplace transform of the product of two functions is the product of the Laplace...

The Laplace transform of u(t−2) is (a) 1s+2 (b) 1s−2 (c) e2ss(d)e−2ss

Find the Laplace Transform of the function f(t)=eat

Explain First Shift Theorem & its properties?

Solve f(t)=etcos⁡t

Find Laplace transform of the given function te−4tsin⁡3t

Reduce to first order and solve:x2y″−5xy′+9y=0 y1=x3

(D3−14D+8)y=0

A thermometer is taken from an inside room to the outside ,where the air temperature...

Find that solution of y′=2(2x−y) which passes through the point (0, 1).

Radium decomposes at a rate proportional to the amount present. In 100 years, 100 mg...

Let A, B, and C be sets. Show that (A−B)−C=(A−C)−(B−C)

Suppose that A is the set of sophomores at your school and B is the...

In how many ways can a 10-question true-false exam be answered? (Assume that no questions...

Is 2∈{2}?

How many elements are in the set { 2,2,2,2 } ?

How many elements are in the set { 0, { { 0 } }?

Draw the Hasse diagram representing the partial ordering {(a, b) | a divides b} on...

Flux through a Cube (Eigure 1) A cube has one corner at the origin and...

A well-insulated rigid tank contains 3 kg of saturated liquid-vapor mixture of water at 200...

A water pump that consumes 2 kW of electric power when operating is claimed to...

A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius...

In a truck-loading station at a post office, a small 0.200-kg package is released from...

The magnetic fieldB→in acertain region is 0.128 ,and its direction is that of the z-axis...

A marble moves along the x-axis. The potential-energy functionis shown in Fig. 1a) At which...

A proton is released in a uniform electric field, and it experiences an electric force...

A potters wheel having a radius of 0.50 m and a moment of inertia of12kg⋅m2is...

Two spherical objects are separated by a distance of 1.80×10−3m. The objects are initially electrically...

An airplane pilot sets a compass course due west and maintainsan airspeed of 220 km/h....

Resolve the force F2 into components acting along the u and v axes and determine...

A conducting sphere of radius 0.01m has a charge of1.0×10−9Cdeposited on it. The magnitude of...

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a...

In the figure a worker lifts a weightωby pulling down on a rope with a...

A stream of water strikes a stationary turbine bladehorizontally, as the drawing illustrates. The incident...

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height...

A radar station, located at the origin of xz plane, as shown in the figure...

Two snowcats tow a housing unit to a new location at McMurdo Base, Antarctica, as...

You are on the roof of the physics building, 46.0 m above the ground. Your...

A block is on a frictionless table, on earth. The block accelerates at5.3ms2when a 10...

A 0.450 kg ice puck, moving east with a speed of3.00mshas a head in collision...

A uniform plank of length 2.00 m and mass 30.0 kg is supported by three...

An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope...

A ski tow operates on a 15.0 degrees slope of lenth 300m. The rope moves...

Two blocks with masses 4.00 kg and 8.00 kg are connected by string and slide...

From her bedroom window a girl drops a water-filled balloon to the ground 6.0 m...

A 730-N man stands in the middle of a frozen pond of radius 5.0 m....

A 5.00 kg package slides 1.50 m down a long ramp that is inclined at12.0∘below...

Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended...

A skier of mass 70 kg is pulled up a slope by a motor driven...

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid,...

A sled with rider having a combined mass of 120 kg travels over the perfectly...

A 7.00- kg bowling ball moves at 3.00 m/s. How fast must a 2.45- g...

Two point chargesq1=+2.40nC andq2=−6.50nC are 0.100 m apart. Point A is midway between them and...

A block of mass m slides on a horizontal frictionless table with an initial speed...

A space traveler weights 540 N on earth. what will the traveler weigh on another...

A block of mass m=2.20 kg slides down a 30 degree incline which is 3.60...

A weatherman carried an aneroid barometer from the groundfloor to his office atop a tower....

If a negative charge is initially at rest in an electric field, will it move...

A coin with a diameter of 2.40cm is dropped on edge on to a horizontal...

An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction...

An 80.0-kg skydiver jumps out of a balloon at an altitude of1000 m and opens...

A 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive...

A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding...

A 500 g ball swings in a vertical circle at the end of a1.5-m-long string....

A rifle with a weight of 30 N fires a 5.0 g bullet with a...

The tires of a car make 65 revolutions as the car reduces its speed uniformly...

A 2.0- kg piece of wood slides on the surface. The curved sides are perfectly...

A 292 kg motorcycle is accelerating up along a ramp that is inclined 30.0° above...

A projectile is shot from the edge of a cliff 125 m above ground level...

A lunch tray is being held in one hand, as the drawing illustrates. The mass...

The initial velocity of a car, vi, is 45 km/h in the positivex direction. The...

An Alaskan rescue plane drops a package of emergency rations to a stranded party of...

Raindrops make an angle theta with the vertical when viewed through a moving train window....

A 0.50 kg ball that is tied to the end of a 1.1 m light...

If the coefficient of static friction between your coffeecup and the horizontal dashboard of your...

A car is initially going 50 ft/sec brakes at a constant rate (constant negative acceleration),...

A swimmer is capable of swimming 0.45m/s in still water (a) If sheaim her body...

A block is hung by a string from inside the roof of avan. When the...

A race driver has made a pit stop to refuel. Afterrefueling, he leaves the pit...

A relief airplane is delivering a food package to a group of people stranded on...

The eye of a hurricane passes over Grand Bahama Island. It is moving in a...

An extreme skier, starting from rest, coasts down a mountainthat makes an angle25.0∘with the horizontal....

Four point charges form a square with sides of length d, as shown in the...

In a scene in an action movie, a stuntman jumps from the top of one...

The spring in the figure (a) is compressed by length delta x . It launches...

An airplane propeller is 2.08 m in length (from tip to tip) and has a...

A helicopter carrying dr. evil takes off with a constant upward acceleration of5.0ms2. Secret agent...

A 15.0 kg block is dragged over a rough, horizontal surface by a70.0 N force...

A box is sliding with a speed of 4.50 m/s on a horizontal surface when,...

3.19 Win the Prize. In a carnival booth, you can win a stuffed giraffe if...

A car is stopped at a traffic light. It then travels along a straight road...

a. When the displacement of a mass on a spring is12A, what fraction of the...

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy...

A jet plane lands with a speed of 100 m/s and can accelerate at a...

In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints...

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart...

A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s. A...

The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of...

A proton with an initial speed of 800,000 m/s is brought to rest by an...

The volume of a cube is increasing at the rate of 1200 cm supmin at...

An airplane starting from airport A flies 300 km east, then 350 km at 30...

To prove: In the following figure, triangles ABC and ADC are congruent. Given: Figure is...

Conduct a formal proof to prove that the diagonals of an isosceles trapezoid are congruent....

The distance between the centers of two circles C1 and C2 is equal to 10...

Segment BC is Tangent to Circle A at Point B. What is the length of...

Find an equation for the surface obtained by rotating the parabola y=x2 about the y-axis.

Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2),...

If the atomic radius of lead is 0.175 nm, find the volume of its unit...

At one point in a pipeline the water’s speed is 3.00 m/s and the gauge...

Find the volume of the solid in the first octant bounded by the coordinate planes,...

A paper cup has the shape of a cone with height 10 cm and radius...

A light wave has a 670 nm wavelength in air. Its wavelength in a transparent...

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50...

Find the equation of the sphere centered at (-9, 3, 9) with radius 5. Give...

Determine whether the congruence is true or false. 5≡8 mod 3

Find all whole number solutions of the congruence equation. (2x+1)≡5 mod 4

Determine whether the congruence is true or false. 100≡20 mod 8

I want example of an undefined term and a defined term in geometry and explaining...

Two fair dice are rolled. Let X equal the product of the 2dice. Compute P{X=i}...

Suppose that two defective refrigerators have been included in a shipment of six refrigerators. The...

Based on the Normal model N(100, 16) describing IQ scores, what percent of peoples

The probability density function of the net weight in pounds of a packaged chemical herbicide...

Let X represent the difference between the number of heads and the number of tails...

An urn contains 3 red and 7 black balls. Players A and B withdraw balls...

80% A poll is given, showing are in favor of a new building project. 8...

The probability that the San Jose Sharks will win any given game is 0.3694 based...

Find the value of P(X=7) if X is a binomial random variable with n=8 and...

Find the value of P(X=8) if X is a binomial random variable with n=12 and...

On a 8 question multiple-choice test, where each question has 2 answers, what would be...

If you toss a fair coin 11 times, what is the probability of getting all...

A coffee connoisseur claims that he can distinguish between a cup of instant coffee and...

Two firms V and W consider bidding on a road-building job, which may or may...

Two cards are drawn without replacement from an ordinary deck, find the probability that the...

In August 2012, tropical storm Isaac formed in the Caribbean and was headed for the...

A local bank reviewed its credit card policy with the intention of recalling some of...

The accompanying table gives information on the type of coffee selected by someone purchasing a...

A batch of 500 containers for frozen orange juice contains 5 that are defective. Two...

The probability that an automobile being filled with gasoline also needs an oil change is...

Let the random variable X follow a normal distribution with μ=80 and σ2=100. a. Find...

A card is drawn randomly from a standard 52-card deck. Find the probability of the...

The next number in the series 38, 36, 30, 28, 22 is ?

What is the coefficient of x8y9 in the expansion of (3x+2y)17?

A boat on the ocean is 4 mi from the nearest point on a straight...

How many different ways can you make change for a quarter? (Different arrangements of the...

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and...

Approximately 80,000 marriages took place in the state of New York last year. Estimate the...

The probability that a student passes the Probability and Statistics exam is 0.7. (i)Find the...

Customers at a gas station pay with a credit card (A), debit card (B), or...

It is conjectured that an impurity exists in 30% of all drinking wells in a...

Assume that the duration of human pregnancies can be described by a Normal model with...

According to a renowned expert, heavy smokers make up 70% of lung cancer patients. If...

Two cards are drawn successively and without replacement from an ordinary deck of playing cards...

Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L),...

A bag contains 6 red, 4 blue and 8 green marbles. How many marbles of...

A normal distribution has a mean of 50 and a standard deviation of 4. Please...

Seven women and nine men are on the faculty in the mathematics department at a...

An automatic machine in a manufacturing process is operating properly if the lengths of an...

Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings)...

Among 157 African-American men, the mean systolic blood pressure was 146 mm Hg with a...

A TIRE MANUFACTURER WANTS TO DETERMINE THE INNER DIAMETER OF A CERTAIN GRADE OF TIRE....

Differentiate the three measures of central tendency: ungrouped data.

Find the mean of the following data: 12,10,15,10,16,12,10,15,15,13

A wallet containing four P100 bills, two P200 bills, three P500 bills, and one P1,000...

The number of hours per week that the television is turned on is determined for...

Data was collected for 259 randomly selected 10 minute intervals. For each ten-minute interval, the...

Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in...

A normal distribution has a mean of 80 and a standard deviation of 14. Determine...

True or false: a. All normal distributions are symmetrical b. All normal distributions have a...

Would you expect distributions of these variables to be uniform, unimodal, or bimodal? Symmetric or...

Annual sales, in millions of dollars, for 21 pharmaceutical companies follow. 8408 1374 1872 8879...

The velocity function (in meters per second) is given for a particle moving along a...

Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1)

What is the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4),...

The integral represents the volume of a solid. Describe the solid. π∫01(y4−y8)dy a) The integral...

Two components of a minicomputer have the following joint pdf for their useful lifetimes X...

Use the table of values of f(x,y) to estimate the values of fx(3,2), fx(3,2.2), and...

Calculate net price factor and net price. Dollars list price −435.20$ Trade discount rate −26%,15%,5%.

Represent the line segment from P to Q by a vector-valued function and by a...

(x2+2xy−4y2)dx−(x2−8xy−4y2)dy=0

If f is continuous and integral 0 to 9 f(x)dx=4, find integral 0 to 3...

Find the parametric equation of the line through a parallel to ba=[3−4],b=[−78]

Find the velocity and position vectors of a particle that has the given acceleration and...

If we know that the f is continuous and integral 0 to 4f(x)dx=10, compute the...

Integration of (y⋅tan⁡xy)

For the matrix A below, find a nonzero vector in the null space of A...

Find a nonzero vector orthogonal to the plane through the points P, Q, and R....

Suppose that the augmented matrix for a system of linear equations has been reduced by...

Find two unit vectors orthogonal to both (3 , 2, 1) and (- 1, 1,...

What is the area of the parallelogram whose vertices are listed? (0,0), (5,2), (6,4), (11,6)

Using T defined by T(x)=Ax, find a vector x whose image under T is b,...

Use the definition of Ax to write the matrix equation as a vector equation, or...

We need to find the volume of the parallelepiped with only one vertex at the...

List five vectors in Span {v1,v2}. For each vector, show the weights on v1 and...

(1) find the projection of u onto v and (2) find the vector component of...

Find the area of the parallelogram determined by the given vectors u and v. u...

(a) Find the point at which the given lines intersect. r = 2,...

(a) find the transition matrix from B toB′,(b) find the transition matrix fromB′to B,(c) verify...

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If...

Given the following vector X, find anon zero square marix A such that AX=0; You...

Construct a matrix whose column space contains (1, 1, 5) and (0, 3.1) and whose...

At what point on the paraboloid y=x2+z2 is the tangent plane parallel to the plane...

Label the following statements as being true or false. (a) If V is a vector...

Find the Euclidean distance between u and v and the cosine of the angle between...

Write an equation of the line that passes through (3, 1) and (0, 10)

There are 100 two-bedroom apartments in the apartment building Lynbrook West.. The montly profit (in...

State and prove the linearity property of the Laplace transform by using the definition of...

The analysis of shafts for a compressor is summarized by conformance to specifications. Suppose that...

The Munchies Cereal Company combines a number of components to create a cereal. Oats and...

Movement of a Pendulum A pendulum swings through an angle of 20∘ each second. If...

If sin⁡x+sin⁡y=aandcos⁡x+cos⁡y=b then find tan⁡(x−y2)

Find the values of x such that the angle between the vectors (2, 1, -1),...

Find the dimensions of the isosceles triangle of largest area that can be inscribed in...

Suppose that you are headed toward a plateau 50 meters high. If the angle of...

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport....

Find an equation of the plane. The plane through the points (2, 1, 2), (3,...

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following...

two small spheres spaced 20.0cm apart have equal charges. How many extra electrons must be...

The base of a pyramid covers an area of 13.0 acres (1 acre =43,560 ft2)...

Find out these functions' domain and range. To find the domain in each scenario, identify...

Your bank account pays an interest rate of 8 percent. You are considering buying a...

Whether f is a function from Z to R ifa)f(n)=±n.b)f(n)=n2+1.c)f(n)=1n2−4.

The probability density function of X, the lifetime of a certain type of electronic device...

A sandbag is released by a balloon that is rising vertically at a speed of...

A proton is located in a uniform electric field of2.75×103NCFind:a) the magnitude of the electric...

A rectangular plot of farmland are finite on one facet by a watercourse and on...

A solenoid is designed to produce a magnetic field of 0.0270 T at its center....

I want to find the volume of the solid enclosed by the paraboloidz=2+x2+(y−2)2and the planesz=1,x=−1y=0,andy=4

Let W be the subspace spanned by the u’s, and write y as the sum...

Can u find the point on the planex+2y+3z=13that is closest to the point (1,1,1). You...

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force...

A force of 250 Newtons is applied to a hydraulic jack piston that is 0.01...

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface...

A credit card contains 16 digits between 0 and 9. However, only 100 million numbers...

Every real number is also a complex number? True of false?

Let F be a fixed 3x2 matrix, and let H be the set of all...

Find a vector a with representation given by the directed line segment AB. Draw AB...

Find A such that the given set is Col A. {[2s+3tr+s−2t4r+s3r−s−t]:r,s,t real}

Find the vector that has the same direction as (6, 2, -3) but is four...

For the matrices (a) find k such that Nul A is a subspace of Rk,...

How many subsets with an odd number of elements does a set with 10 elements...

In how many ways can a set of five letters be selected from the English...

Suppose that f(x) = x/8 for 3 < x < 5. Determine the following probabilities:...

Describe all solutions of Ax=0 in parametric vector form, where A is row equivalent to...

Find two vectors parallel to v of the given length. v=PQ→ with P(1,7,1) and Q(0,2,5);...

A dog in an open field runs 12.0 m east and then 28.0 m in...

Can two events with nonzero probabilities be both independent and mutually exclusive? Explain your reasoning.

Use the Intermediate Value Theorem to show that there is a root of the given...

In a fuel economy study, each of 3 race cars is tested using 5 different...

A company has 34 salespeople. A board member at the company asks for a list...

A dresser drawer contains one pair of socks with each of the following colors: blue,...

A restaurant offers a $12 dinner special with seven appetizer options, 12 choices for an...

A professor writes 40 discrete mathematics true/false questions. Of the statements in these questions, 17...

Suppose E(X)=5 and E[X(X–1)]=27.5, find ∈(x2) and the variance.

A Major League baseball diamond has four bases forming a square whose sides measure 90...

Express f(x)=4x3+6x2+7x+2 in term of Legendre Polynomials.

Find a basis for the space of 2×2 diagonal matrices. Basis ={[],[]}

Which of the following expressions are meaningful? Which are meaningless? Explain. a) (a⋅b)⋅c (a⋅b)⋅c has...

Vectors V1 and V2 are different vectors with lengths V1 and V2 respectively. Find the...

Find an equation for the plane containing the two (parallel) lines v1=(0,1,−2)+t(2,3,−1) and v2=(2,−1,0)+t(2,3,−1).

Find, correct to the nearest degree, the three angles of the triangle with the given...

Find the vector, not with determinants, but by using properties of cross products. (i+j)×(i−j)

Find the curve’s unit tangent vector. Also, find the length of the indicated portion of...

Construct a 4×3 matrix with rank 1

Find x such that the matrix is equal to its inverse.A=[7x−8−7]

Find a polynomial with integer coefficients that satisfies the given conditions. Q has degree 3...

Write in words how to read each of the following out loud.a.{x∈R′∣0<x<1}b.{x∈R∣x≤0orx⇒1}c.{n∈Z∣nisafactorof6}d.{n∈Z⋅∣nisafactorof6}

Pets Plus and Pet Planet are having a sale on the same aquarium. At Pets...

Find the average value of F(x, y, z) over the given region. F(x,y,z)=x2+9 over the...

Find the trace of the plane in the given coordinate plane. 3x−9y+4z=5,yz

Determine the level of measurement of the variable. Favorite color Choose the correct level of...

How wide is the chasm between what men and women earn in the workplace? According...

Write an algebraic expression for: 6 more than a number c.

Please, can u convert 3.16 (6 repeating) to a fraction.

Evaluate the expression. P(8, 3)

In a poker hand consisting of 5 cards, find the probability of holding 3 aces.

Give an expression that generates all angles coterminal with each angle. Let n represent any...

An ideal Otto cycle has a compression ratio of 10.5, takes in air at 90...

A piece of wire 10 m long is cut into two pieces. One piece is...

Put the following equation of a line into slope intercept form, simplifying all fractions 3x+3y=24

Find the point on the hyperbola xy = 8 that is closest to the point...

Water is pumped from a lower reservoir to a higher reservoir by a pump that...

A piston–cylinder device initially contains 0.07m3 of nitrogen gas at 130 kPa and 180∘. The...

Write an algebraic expression for each word phrase. 4 more than p

A club has 25 members. a) How many ways are there to choose four members...

For each of the sets below, determine whether {2} is an element of that set....

Which expression has both 8 and n as factors?

If repetitions are not permitted (a) how many 3 digit number can be formed from...

To determine the sum of all multiples of 3 between 1 and 1000

On average, there are 3 accidents per month at one intersection. We need to find...

One number is 2 more than 3 times another. Their sum is 22. Find the...

The PMF for a flash drive with X (GB) of memory that was purchased is...

An airplane needs to reach a velocity of 203.0 km/h to takeoff. On a 2000...

A racquetball strikes a wall with a speed of 30 m/s and rebounds with a...

Assuming that the random variable x has a cumulative distribution function,F(x)={0,x<00.25x,0≤x<51,5≤xDetermine the following:a)p(x<2.8)b)p(x>1.5)c)p(x<−z)d)p(x>b)

At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It...

How many 3/4's are in 1?

You’re driving down the highway late one night at 20 m/s when a deer steps...

Table salt contains 39.33 g of sodium per 100 g of salt. The U.S. Food...

The constant-pressure heat capacity of a sample of a perfect gas was found to vary...

Coffee is draining from a conical filter into a cylindrical coffepot at the rate of...

Cart is driven by a large propeller or fan, which can accelerate or decelerate the...

A vending machine dispenses coffee into an eight-ounce cup. The amounts of coffee dispensed into...

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a...

The gage pressure in a liquid at a depth of 3 m is read to...

Consider a cylindrical specimen of a steel alloy 8.5 mm (0.33 in.) in diameter and...

Calculate the total kinetic energy, in Btu, of an object with a mass of 10...

A 0.500-kg mass on a spring has velocity as a function of time given by...

An Australian emu is running due north in a straight line at a speed of...

Another pitfall cited is expecting to improve the overall performance of a computer by improving...

You throw a glob of putty straight up toward the ceiling, which is 3.60 m...

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.070 m. A...

A batch of 140 semiconductor chips is inspected by choosing a sample of 5 chips....

A rock climber stands on top of a 50-m-high cliff overhanging a pool of water....

A tank whose bottom is a mirror is filled with water to a depth of...

Two sites are being considered for wind power generation. In the first site, the wind...

0.250 kilogram of water at75.0∘Care contained in a tiny, inert beaker. How much ice, at...

Two boats start together and race across a 60-km-wide lake and back. Boat A goes...

A roller coaster moves 200 ft horizontally and the rises 135 ft at an angle...

A tow truck drags a stalled car along a road. The chain makes an angle...

Consider the curve created by2x2+3y2–4xy=36(a) Show thatdydx=2y−2x3y−2x(b) Calculate the slope of the line perpendicular to...

The current entering the positive terminal of a device is i(t)=6e−2t mA and the voltage...

The fastest measured pitched baseball left the pitcher’s hand at a speed of 45.0 m/s....

Calculate the total potential energy, in Btu, of an object that is 20 ft below...

A chemist in an imaginary universe, where electrons have a different charge than they do...

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of...

Determine the energy required to accelerate a 1300-kg car from 10 to 60 km/h on...

The deepest point in the ocean is 11 km below sea level, deeper than MT....

A golfer imparts a speed of 30.3 m/s to a ball, and it travels the...

Calculate the frequency of each of the following wavelengths of electromagnetic radiation. A) 632.8 nm...

Prove that there is a positive integer that equals the sum of the positive integers...

A hurricane wind blows across a 6.00 m×15.0 m flat roof at a speed of...

If an electron and a proton are expelled at the same time,2.0×10−10mapart (a typical atomic...

The speed of sound in air at 20 C is 344 m/s. (a) What is...

Which of the following functions f has a removable discontinuity at a? If the discontinuity...

A uniform steel bar swings from a pivot at one end with a period of...

A wind farm generator uses a two-bladed propellermounted on a pylon at a height of...

A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg...

Jones figures that the total number of thousands of miles that a used auto can...

Assign a binary code in some orderly manner to the 52 playingcards. Use the minimum...

A copper pot with mass 0.500 kg contains 0.170 kg of water ata temperature of...

Ea for a certain biological reaction is 50 kJ/mol, by what factor ( how many...

When a person stands on tiptoe (a strenuous position), the position of the foot is...

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Hardest GCSE Maths Questions

With the introduction of grade 9 at GCSE , which only 3.5% of students achieved in 2023 ( Pearson ), Maths exam papers have included some extremely challenging questions in recent years. These questions stretch the highest achieving students, and allow them to demonstrate their skills. 

Only 3.5% of students achieved a grade 9 in Maths in June 2023

At Save My Exams, we aim to make even the hardest questions easy to understand through our illuminating worked solutions and revision notes for GCSE Maths . Through my experience in the classroom as a maths teacher, I know that different students struggle with different topics, and that as a result there is no “hardest topic”. Therefore I have curated a selection of my personal favourite, hardest, GCSE Maths questions arranged by topic area, with links to our worked solutions.

Hardest Algebra Questions Hardest Coordinate Geometry Questions Hardest Number & Ratio Questions Hardest Geometry Questions Hardest Probability & Statistics Questions Conclusion FAQs

Hardest Algebra Questions

Aqa, november 2021: 2h q24.

This question covers a great set of skills; algebraic fractions, forming an equation from a worded problem, and solving a quadratic. I often find that these topics are good indicators of a student’s confidence with algebra. The examiner report for this question stated “this question was poorly attempted with most students unable to progress. There were a significant number who made no attempt.” So this is a question where you could really gain some marks over other candidates to boost your grade. Our worked solution for this question can be found as Q11 in our Very Hard Algebraic Fractions topic question pack.

Edexcel, November 2020: 3H Q21b

There are two features that make this a tough question; a ratio involving algebra, and a quadratic in two variables. Both of these take a standard skill (ratio and quadratics) and give them an extra twist. Even if students manage to find the two possible ratios of x:y, they still have to decide which the correct solution is given the information in the question. Our worked solution is available as Q12 in our Very Hard Solving Quadratics topic question pack.

OCR, November 2017: H06 Q14b

Students struggle with proof questions due to the lack of structure or direct instructions

I commonly find that students struggle with proof questions due to the lack of structure or direct instructions. This question is no different, and requires students to set up their own expressions describing the values in the cross shape. The examiner’s report noted that “...almost all attempts at the proof merely consisted of further numerical examples”. While I would always recommend students test further numerical examples to help spot any patterns, making the leap to modelling with algebra helps distinguish top performers. A worked solution from one of our maths content creators is available as Q12b in our Very Hard Algebraic Proof topic question pack.

Hardest Coordinate Geometry Questions

Edexcel, november 2021: 2h q19.

I have taught A Level Maths extensively, and this question wouldn’t be out of place on a first year A Level paper. Firstly, students have to understand the link between simultaneous equations and the points of intersection. The simultaneous equations then have to be solved by substitution, which reduces to a quadratic. The question then finishes off by finding the midpoint; an essential coordinate geometry skill. You can view the full details of the solution in Q2 of our Very Hard Coordinate Geometry topic question pack. 

Edexcel, June 2019: 3H Q22

This is a fantastic example of a higher-level problem solving question. You are given a diagram, key facts, numerical values, algebraic values, and an angle. You must then piece these together to find the value of p. Spotting a route to a solution is not easy. I often advise students to start by working something out, which may then lead to finding some key information, unlocking the whole problem. The key to this problem is forming an equation for the circle in terms of p, and then finding the radius, which allows the equation to be solved for p. To see the full details of the path to the answer, see Q4 in our Very Hard Equation of a Circle topic question pack.

Hardest Number & Ratio Questions

Edexcel, june 2018: 2h q21.

The toughest questions combine the basics with another piece of knowledge

In my teaching experience, students are usually comfortable with the basic concept of upper and lower bounds and rounding. However, the toughest questions combine the basics with another piece of knowledge, in this case density, mass, and volume. Unusually, this question also asks for “the density of the wood” rather than the upper and lower bounds for the density, which is a more standard exam question. This means the final step in this question is to look for the highest degree of accuracy that both the upper and lower bounds can be rounded to, such that they are equal. To see the full solution written by one of our Maths content specialists, see Q3 in our Very Hard Rounding, Estimation & Bounds topic question pack.

AQA, June 2018: 1H Q26

I like this Paper 1 (non-calculator) question because it looks so simple on the outside, but the underlying maths can be quite tricky! There are only two short statements about the relative ratios and proportions of a, b, and c, and students have to then find a 3-part ratio. When I have taught lessons about ratio, 3-part ratios are always something students find difficult. The key to them is finding a part of the ratio that “overlaps” and can be a link between two 2-part ratios, which can then be joined together. See Q6 in our Very Hard Ratio Problem Solving topic question pack to view our full worked solution.

Hardest Geometry Questions 

Edexcel (international), january 2022: 1hr q24.

I have taught Maths to both GCSE and A Level students, and in both cases students quickly grasp the basics of vectors: finding routes between two points, and describing them algebraically. Often I find students at both levels struggle with questions involving ratios of lengths, and algebraic scalars (λ in this case). I picked out this GCSE question in particular because the strategy to solve it is used frequently in A Level textbook problems. If you enjoy tackling questions with problem solving like this, I would suggest that A Level Maths may be for you! I’ll give you a hint for this one: find two different paths from O to N, one involving λ, and set them equal to each other. To view the rest of our explanation, see Q9 in our Very Hard Vectors topic question pack.

Edexcel, June 2017: 1H Q22

Trigonometry questions, especially those using the cosine rule, usually appear in one of the calculator papers (Paper 2 and Paper 3). This question, however, is a great example of a non-calculator cosine rule question. Because it is a proof question, students have to be really particular and clear about their reasoning for the steps they use to receive full marks. When I mark students’ exam papers, this is a common reason for them to not receive full marks, and is something the examiners’ report mentions specifically.

Students need to ensure they use the correct mathematical language when giving their reasons and give all necessary detail. For example, "The angle between the tangent and the circle = 90" is not sufficient as the word radius is missing from their reason

See all the steps you need to include to receive full marks in Q5 of our Very Hard Sine, Cosine Rule & Area of Triangles topic question pack.

Hardest Probability & Statistics Questions

Edexcel, june 2015: 1h q19.

“Hannah’s Sweets” is possibly the most infamous GCSE Maths exam question of all time. When I consider the hardest questions I have seen in my many years as a maths teacher, it probably wouldn’t be in the top 10, but it caused an outcry online at the time. Hannah’s sweets even received coverage from news outlets like the BBC and The Guardian. 

“Hannah’s Sweets” is the most infamous GCSE Maths exam question of all time

What makes it appear as such a hard question is the way it builds up a description of a fairly standard conditional probability problem, but then asks to show that a quadratic equation is true! This overlap of topics had not been seen up to this point on a GCSE paper, which is why it caused such a stir. I think it makes an excellent question though, and really tests whether students know what they are doing with a conditional probability question. Have a go yourself first, and then check your answer with our fully worked solution. It is Q5 in our Very Hard Combined & Conditional Probability topic question pack.

Edexcel (International), January 2022: 2H Q20

When I teach histograms, I love using questions similar to this from a past Edexcel paper to test understanding of the topic, and practise problem solving skills. The specific feature that enables this is the blank y-axis on the graph. This makes the question a step harder than the usual “find the frequency” problems, and means you must work out the scale for the y-axis (frequency density) using information given in the problem. 

There is an alternate method too, where you can simply use the number of squares within the bars. This works as the areas in a histogram are proportional to the frequency. Both of these methods are shown in our worked solution of Q6 in our Very Hard Histograms topic question pack.

As a maths teacher, I know that it’s vital for students to build confidence answering questions that are difficult, or where the best method can’t be spotted straight away. This is because half of the exam paper is now aimed at grades 7 and above.

In a higher tier paper, half of the marks should be targeted at grades 9, 8 and 7 ( Ofqual )

Once you have mastered the key skills of a topic, and are confident answering “standard” exam questions, the only way to improve at answering harder questions is by experiencing - and tackling - as many difficult questions as you can. There are no shortcuts to this unfortunately, but we do have plenty of challenging questions, all with worked solutions, in our Hard and Very Hard topic question packs . We also have GCSE Maths revision notes and past papers , for when you want to brush up on a particular topic, or practise a whole paper.

What is the hardest GCSE Maths topic?

In my experience teaching GCSE Maths in English schools and as a GCSE Maths tutor, I would find it hard to pinpoint a hardest topic on any GCSE maths paper. Individual students will always vary in which topics they find hard compared to their peers. However, the following topics and skills are more common for students to find challenging:

• Algebraic Fractions
• Forming equations from words
• Algebraic Proof
• Coordinate Geometry & Circles
• Problem Solving with Ratios
• Vector Proof
• Trigonometry
• Circle Theorems & Geometric Proof
• Conditional Probability

What is the hardest Maths question?

There is certainly no “hardest maths question” as all students will vary in how difficult they find each topic. However, questions involving proof and/or including a high level of algebra are often the worst-answered on exam papers. The questions I have collated in this article are certainly a good starting point for seeing some of the hardest questions, but you can find plenty more in our Hard and Very Hard topic question packs.

Is Maths the hardest GCSE?

This is much more of a personal question. For some students Maths will be the subject they find hardest, and for others it will be the subject they find easiest. For most people it will fall somewhere in between. However hard or easy you find maths, practice is vital to improve. At Save My Exams we have resources - including practice papers - to help with your Maths revision, to test yourself, and to improve in GCSE Maths.

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

20,000 + revision notes, 100,000 + practice questions and model answers, past papers and mark schemes...all tailored to your exams.

Are you seeking one-on-one college counseling and/or essay support? Limited spots are now available. Click here to learn more.

15 Hardest SAT Math Questions in 2024

March 7, 2024

For some students, “math” is a scary word, particularly in the context of the SAT. While test takers can often utilize context clues to make an educated guess on reading-oriented questions, math problems can sometimes feel like they are written in a foreign language. In pursuit of a good SAT score , many students engage in SAT prep to build their knowledge, skills, and confidence. As part of that prep, some students may wish to challenge themselves by tackling the hardest Digital SAT math questions. If that sounds familiar, this post is for you!

Below, we discuss some of the hardest SAT math questions, identifying what qualities make them difficult and strategies that will help you solve them. Whether you’re a math aficionado or a novice hoping to build your skills, this post will tell you what you need to know about hard SAT math questions to help you do your best.

On the hunt for expert SAT prep? For years, we’ve been referring our clients to a select group of providers. Click below to learn more. 

Best SAT Tutors

Digital SAT Math Basics

Before discussing the hardest SAT math questions, let’s go over the composition of the SAT math section. The Math section consists of 44 questions divided into two modules. Each module is 35 minutes. Calculators can be used throughout both math modules (a change from the old exam). 75% of questions are multiple-choice; the remaining 25% are student-produced response  questions. Below is a more detailed breakdown of the composition of the SAT Math section from the College Board .

What’s covered on the Digital SAT Math test?

There are four categories of questions on the SAT Math test:

• Advanced Math
• Problem-Solving and Data Analysis
• Geometry and Trigonometry

Algebra questions measure students’ knowledge of linear equations and systems. Questions may ask students to analyze and solve equations using multiple techniques.

As its title suggests, Advanced Math tests students on the knowledge they’ll need to specialize in mathematically oriented topics, such as STEM subjects or economics. These questions will also evaluate students on the skills they’ll need to excel in calculus and advanced statistics courses. As one might expect, this is a category that may produce some of the hardest SAT math questions.

In comparison, Problem Solving and Data Analysis questions measure students’ quantitative literacy through concepts they’re likely to need in college courses and everyday life, including ratios, percentages, and proportional relationships. Students may address problems in real-world settings or describe relationships in graphs or statistics.

Finally, in Geometry and Trigonometry , students can expect to encounter questions focused on area and volume; lines, angles and triangles; right triangles and trigonometry; and circles. This category may also include some hard SAT math questions, given students’ varying levels of familiarity with these subjects.

Preparing for the Digital SAT Math Test

As you can see, the SAT Math test covers a wide variety of topics. While it might be tempting to jump straight to the hardest SAT math questions, it’s important to first establish a clear baseline by taking a practice test. Doing so will allow you to familiarize yourself with the structure of the SAT. Moreover, this practice test will provide you with an opportunity to reflect on your strengths and weaknesses so you can identify what topics warrant more practice. Once you know what your priorities are, you can start your SAT prep through the materials provided by College Board or an SAT prep manual.

15 Hardest SAT Math Questions

Now that we have that groundwork in place, we can discuss our selections for hard SAT math questions. We have opted to categorize questions around four common challenges students may experience, providing several examples of each. As you read our selections, bear in mind that difficulty is relative. We have selected questions that we believe are challenging due to their composition. However, this may not be the case for all students. Therefore, we recommend students identify their personal SAT prep goals to ensure they are being strategic in their studies. All questions are sourced from College Board’s practice tests.

Finally, we’ve linked to paper-and-pencil versions of SAT practice tests for the purpose of providing complete solutions via SAT answer guides. Although you’ll certainly want to complete practice tests in the new digital format , know that the linear exams can be incredibly useful for concept practice. If you do utilize older practice tests, remember that you can now use a calculator on every section of the exam.

Hard SAT Math Questions: Specialized or less familiar forms of math

Of all the hard SAT math questions, perhaps none are more difficult than those that deal with more specialized mathematical subjects, such as trigonometry. Test takers have typically had less exposure to these subjects, which can make solving these problems more difficult. Therefore, it is important that students review a variety of mathematical concepts to ensure they are equipped to answer all types of questions. Here are a few examples:

1) Student-Produced Response

In a right triangle, one angle measures x°, where sin x° = ⅘ . What is cos(90° − x°)?

As this problem illustrates, students need a basic understanding of trigonometry functions to tackle this type of question. A complete solution for this problem is available on page 31 of the answer guide for SAT Practice Test 1.

2) Student-Produced Response

A group of friends decided to divide the $800 cost of a trip equally among themselves. When two of the friends decided not to go on the trip, those remaining still divided the $800 cost equally, but each friend’s share of the cost increased by $20. How many friends were in the group originally?

Solving this problem necessitates that students have the ability to utilize quadratic equations, which is a more advanced form of math relative to many of the concepts tested on the SAT Math test. A complete explanation is available on page 47 of the answer guide for SAT Practice Test 6.

3) Multiple Choice

The world’s population has grown at an average rate of 1.9 percent per year since 1945. There were approximately 4 billion people in the world in 1975. Which of the following functions represents the world’s population P, in billions of people, t years since 1975?

This problem engages students’ knowledge of exponential growth. However, rather than simply solving an equation, students must understand the logic of exponential functions well enough to translate the information provided into the correct equation. The complete solution for this problem is available on page 45 of the answer guide for SAT Practice Test 7.

4) Student-Produced Response

Triangle PQR has right angle Q . If sin R = ⅘, what is the value of tan P ?

This problem requires that students utilize trigonometry functions, as well as the Pythagorean theorem, to arrive at the correct answer. A complete solution for this problem is available on page 39 of the answer guide for SAT Practice Test 9.

Hard SAT Math Questions: Problems with multistep solutions

Many problems on the SAT Math test require students to complete multiple steps to arrive at an answer. While the math involved may not be difficult in itself, a multistep process creates opportunities for students to make mistakes. For this reason, students should practice solving problems with multistep solutions to avoid careless errors. Let’s look at a few examples:

5) Multiple Choice

If (ax+2)(bx+7)=15x^2+ cx+14 for all values of x, and a+b=8, what are the two possible values for c?

B) 6 and 35

C) 10 and 21

D) 31 and 41

To answer this question, students must understand the logic of how these variables and equations relate to one another. Relevant skills students would need to solve this problem include mastery of algebra and the ability to use factoring. A complete explanation for this problem is available on page 30 of the answer guide for SAT Practice Test 1.

6) Multiple Choice

A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these alterations decreased the area of the rectangle by 12 percent, what is the value of p?

Concepts involved in this problem, including calculating area and percentages, are likely familiar to most students. However, students may stumble when completing the steps necessary to find the answer, which involves writing equations to represent the values of the original area of the rectangle, the altered values for the length and width, and the decreased area of the rectangle. This lengthy process leaves room for mistakes, making this problem deceptively challenging. A complete solution is available on page 35 of the answer guide for SAT Practice Test 3.

Hardest SAT Math Problems (Continued)

7) Student-Produced Response

If Ms. Simon starts her drive at 6:30 a.m., she can drive at her average driving speed with no traffic delay for each segment of the drive. If she starts her drive at 7:00 a.m., the travel time from the freeway entrance to the freeway exit increases by 33% due to slower traffic, but the travel time for each of the other two segments of her drive does not change. Based on the table, how many more minutes does Ms. Simon take to arrive at her workplace if she starts her drive at 7:00 a.m. than if she starts her drive at 6:30 a.m.? (Round your answer to the nearest minute.)

Again, if we judged this problem strictly on the math involved, it probably wouldn’t be considered one of the hardest SAT math questions. However, the multiple steps and calculations it requires make it easy for students to make mistakes. The complete solution is available on page 49 of the answer guide for SAT Practice Test 5. A similar example is available below.

8) Student-Produced Response

Number of Contestants by Score and Day

The same 20 contestants, on each of 3 days, answered 5 questions in order to win a prize. Each contestant received 1 point for each correct answer. The number of contestants receiving a given score on each day is shown in the table above.

What was the mean score of the contestants on Day 1?

The complete solution for this problem is available on page 47 of the answer guide for SAT Practice Test 7.

Hard SAT Math Questions: Problems that are difficult to comprehend

Although math involves numbers, having a firm grasp of reading comprehension and logic is often necessary to understand a problem. Looking at a block of text can sometimes be overwhelming, which is why it’s important to practice reading word problems so you can learn how to understand the variables involved and tackle these hard SAT math questions. Here are a few examples:

9) Multiple Choice

A square field measures 10 meters by 10 meters. Ten students each mark off a randomly selected region of the field; each region is square and has side lengths of 1 meter, and no two regions overlap. The students count the earthworms contained in the soil to a depth of 5 centimeters beneath the ground’s surface in each region. The results are shown in the table below.

Which of the following is a reasonable approximation of the number of earthworms to a depth of 5 centimeters beneath the ground’s surface in the entire field?

Between the described 10×10 grid and the data chart, there is a lot to sift through in this question. While the math involved isn’t especially difficult (students primarily need to be comfortable with ratios to solve this problem), the sheer number of variables in the question could make it challenging to understand and, therefore, to solve. A complete explanation for this problem is available on page 40 of the answer guide for SAT Practice Test 1.

10) Multiple Choice

Of the following four types of savings account plans, which option would yield exponential growth of the money in the account?

• A) Each successive year, 2% of the initial savings is added to the value of the account.
• B) Each successive year, 1.5% of the initial savings and $100 is added to the value of the account.
• C) Each successive year, 1% of the current value is added to the value of the account.
• D) Each successive year, $100 is added to the value of the account.

This problem has less to do with precise calculations and more to do with a student’s ability to translate the answers into mathematical concepts, specifically linear versus exponential growth. Therefore, the challenge is for students to consider the logic of each option to determine which would support exponential growth. A complete solution for this problem is available on page 34 of the answer guide for SAT Practice Test 3.

11) Student-Produced Response

The problem outlined below refers to the following information:

If shoppers enter a store at an average rate of r shoppers per minute and each stays in the store for an average time of T minutes, the average number of shoppers in the store, N, at any one time is given by the formula N = rT. This relationship is known as Little’s law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little’s law to estimate that there are 45 shoppers in the store at any time.

Little’s law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spends an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

Because this problem has a contextual paragraph, there is a fair amount of text students have to work through. This quantity of information can easily obscure the relationships between the values discussed. However, by working through the question carefully, students can understand the logic of the problem. A complete solution is available on page 38 of the answer guide of the SAT Practice Test 3.

12) Multiple Choice

The 22 students in a health class conducted an experiment in which they each recorded their pulse rates, in beats per minute, before and after completing a light exercise routine. The dot plots below display the results.

Let s 1 and r 1 be the standard deviation and range, respectively, of the data before exercise, and let s 2 and r 2 be the standard deviation and range, respectively, of the data after exercise. Which of the following is true?

• s 1 = s 2 and r 1 = r 2
• s 1 < s 2 and r 1 < r 2
• s 1 > s 2 and r 1 > r 2
• s 1 ≠ s 2 and r 1 = r 2

This problem requires that students utilize their interpretative abilities to break down the provided charts and context to determine how the standard deviations compare. A complete solution for this problem is available on page 46 of the answer guide for SAT Practice Test 8.

Hard SAT Math Questions: Problems that test multiple concepts

Some questions on the SAT will require that students leverage multiple mathematical skills and concepts to arrive at an answer. For these questions, the threshold for achieving the correct answer is higher simply because they require mastery of multiple concepts. Let’s look at a few examples:

13) Student-Produced Response

At a lunch stand, each hamburger has 50 more calories than each order of fries. If 2 hamburgers and 3 orders of fries have a total of 1700 calories, how many calories does a hamburger have?

This problem looks simple enough and, in fact, the math involved really isn’t that hard. However, what makes this problem challenging is that it requires students to understand systems of equations well enough to write equations that represent the described situation. Students then have to utilize the system of equations they create to solve the problem using algebra. A complete explanation for this problem is available on page 26 of the answer guide for SAT Practice Test 3.

14) Student-Produced Response

In triangle ABC, the measure of ∠B is 90°, BC = 16, and AC = 20. Triangle DEF is similar to triangle ABC, where vertices D, E, and F correspond to vertices A, B, and C, respectively, and each side of triangle DEF is 1 3 the length of the corresponding side of triangle ABC. What is the value of sin F ?

This question requires that students be comfortable with basic trigonometry and the geometric concept of similarity. This, in turn, necessitates an understanding of ratios. Being able to layer these skills will ensure students arrive at the appropriate solution. A complete explanation of this problem is available on page 27 of the answer guide for SAT Practice Test 3. Below is another example of a question that layers these concepts.

15) Student-Produced Response

In the figure above, tan B = ¾. If BC = 15  and DA = 4, what is the length of DE ?

A complete solution for this problem is available on page 47 of the answer guide for SAT Practice Test 7.

Final Thoughts: The Hardest SAT Math Problems

After working through these problems, take a moment to reflect. If you struggled or are feeling overwhelmed, that might be a sign you need to do a little more studying. Consider consulting our list of top SAT tutors , College Board’s SAT Study Guide , or our post on the most important SAT math formulas for assistance. If you breezed through these problems, congratulations! Math is clearly a strength of yours. Consider turning your attention to other areas, such as SAT vocabulary words. Happy studying and best of luck!

Got other SAT-related questions? Check out our other SAT resources:

• Entering Class Statistics
• Should I Apply Test Optional?
• When Do SAT Scores Come Out?
• Guide to the Digital SAT
• SAT Score Calculator

Emily Smith

Emily earned a BA in English and Communication Studies from UNC Chapel Hill and an MA in English from Wake Forest University. While at UNC and Wake Forest, she served as a tutor and graduate assistant in each school’s writing center, where she worked with undergraduate and graduate students from all academic backgrounds. She also worked as an editorial intern for the Wake Forest University Press as well as a visiting lecturer in the Department of English at WFU, and currently works as a writing center director in western North Carolina.

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The hardest problem on the hardest test

The problem.

There's a famous math competition for undergraduate students known as the Putnam. It's 6 hours long and consists of 12 questions, broken up into two different 3-hour sessions over two days. Each question being scored on a 1-10 scale, so the highest possible score is 120.

And yet, despite the fact that the only students taking it each year are those who are clearly already pretty into math, given that they opt into such a test, the median score tends to be around 1 or 2.

It's a hard test

For each day the six problems tend to increase in difficulty, ranging from pretty difficult to exceedingly challenging. But of course, difficulty is sometimes in the eye of the beholder.

But here's the thing about those 5's and 6's. Even though they're positioned as the hardest problems on a famously hard test, quite often these are the ones with the most elegant solutions available, often involving some subtle shift in perspective that transforms it from challenging to simple.

Here I'll share with you one problem which came up as question A6 on the 1992 Putnam exam . Rather than just jumping straight to the solution, which in this case will be surprisingly short, let's take the time to walk through how you might stumble upon the solution yourself. This story is more about the problem-solving process than the particular problem used to exemplify it.

Here's the question:

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.)

Take a moment to digest the question. You might start thinking about which of these tetrahedra contain the sphere's center, which ones don't, and how you might systematically distinguish the two.

How do you approach a problem like this? Where do you even start? Well, it's often a good idea to think about simpler cases, so let's bring things down into two dimensions.

The two-dimensional case

Suppose you choose three random points on a circle. It's always helpful to name things, so let's call these fellows P 1 P_1 P 1 ​ , P 2 P_2 P 2 ​ , and P 3 P_3 P 3 ​ . What's the probability that the triangle formed by these points contains the center of the circle?

This is still a hard question, so you should ask yourself if there's a way to simplify the question even further. We still need a foothold, something to build up from. Maybe you imagine fixing P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ in place, only letting P 3 P_3 P 3 ​ vary. In doing this, you might notice that there's special region, a certain arc, where when P 3 P_3 P 3 ​ is in that arc, the triangle contains the circle's center.

Specifically, if you draw a lines from P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ through the center, these lines divide the circle into 4 different arcs. If P 3 P_3 P 3 ​ happens to be in the arc opposite P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ , the triangle will contain the center. Otherwise, you're out of luck.

It's certainly easier to visualize now, but can you answer the question?

If P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ are fixed in place on the circle, with an arc length α \alpha α between them, and P 3 P_3 P 3 ​ is a point chosen randomly on the circle (with a uniform distribution), what is the probability that the triangle Δ P 1 P 2 P 3 \Delta P_1 P_2 P_3 Δ P 1 ​ P 2 ​ P 3 ​ contains the center of the circle?

α / 36 0 ∘ \alpha / 360^{\circ} α /36 0 ∘

18 0 ∘ − α 180^{\circ} - \alpha 18 0 ∘ − α

( 18 0 ∘ − α ) / 36 0 ∘ (180^{\circ} - \alpha) / 360^{\circ} ( 18 0 ∘ − α ) /36 0 ∘

So what proportion of the circle does this arc take up? That depends on the first two points. If they are 90 degrees apart from each other, for example, the relevant arc is 1 4 \frac{1}{4} 4 1 ​ of the circle.

But if those two points are farther apart, the proportion might be closer to 1/2.

If they are really close, that proportion might be closer to 0.

If P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ are chosen randomly, with every point on the circle being equally likely, what's the average size of the relevant arc?

Maybe you imagine fixing P 1 P_1 P 1 ​ in place, and considering all the places that P 2 P_2 P 2 ​ might be.

All of the possible angles between these two lines, every angle from 0 degrees up to 180 degrees is equally likely, so every proportion between 0 and 0.5 is equally likely, making the average proportion 0.25. Since the average size of this arc is 1 4 \frac{1}{4} 4 1 ​ this full circle, the average probability that the third point lands in it is 1 4 \frac{1}{4} 4 1 ​ , meaning the overall probability of our triangle containing the center is 1 4 \frac{1}{4} 4 1 ​ .

The three-dimensional case

Great! Can we extend this to the 3d case? If you imagine 3 of your 4 points fixed in place, which points of the sphere can that 4th point be on so that our tetrahedron contains the sphere's center? As before, let's draw some lines from each of our first 3 points through the center of the sphere.

And it's also helpful if we draw the planes determined by any pair of these lines.

These planes divide the sphere into 8 different sections, each of which is a sort of spherical triangle. Our tetrahedron will only contain the center of the sphere if the fourth point is in the section on the opposite side of our three points.

Unlike the 2d case, it's rather difficult to think about the average size of this section as we let our initial 3 points vary. Those of you with some multivariable calculus under your belt might think to try a surface integral. And by all means, pull out some paper and give it a try, but it's not easy. And of course it should be difficult, this is the 6th problem on a Putnam!

The shift in perspective

But let's back up to the 2d case, and contemplate if there's a different way to think about it. This answer we got, 1 4 \frac{1}{4} 4 1 ​ , is suspiciously clean and raises the question of what that 4 represents. One of the main reasons I wanted to cover this topic is that what's about to happen carries a broader lesson for mathematical problem-solving.

The lines that we drew from P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ through the origin made the problem easier to think about. In general, whenever you've added something to your problem setup which makes things conceptually easier, see if you can reframe the entire question in terms of the thing you just added.

In this case, rather than thinking about choosing 3 points randomly, start by saying choose two random lines that pass through the circle's center. For each line, there are two possible points they could correspond to, so flip a coin for each to choose which of those will be P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ .

Choosing a random line then flipping a coin like this is the same as choosing a random point on the circle, with all points being equally likely, and at first it might seem needlessly convoluted way to describe choosing two random points. But by making those lines the starting point of our random process things actually become easier.

We'll still think about P 3 P_3 P 3 ​ as just being a random point on the circle, but imagine that it was chosen before you do the two coin flips.

You see, once the two lines and a random point have been chosen, there are four possibilities for where P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ end up, based on the coin flips, each one of which is equally likely. But one and only one of those outcomes leaves P 1 P_1 P 1 ​ and P 2 P_2 P 2 ​ on the opposite side of the circle as P 3 P_3 P 3 ​ , with the triangle they form containing the center. So no matter what those two lines and P 3 P_3 P 3 ​ turned out to be, it's always a 1 4 \frac{1}{4} 4 1 ​ chance that the coin flips will leave us with a triangle containing the center.

That's very subtle. Just by reframing how we think of the random process for choosing these points, the answer 1 4 \frac{1}{4} 4 1 ​ popped in a different way from before. And importantly, this style of argument generalizes seamlessly to 3 dimensions.

Applying the shift to three dimensions

Again, instead of starting off by picking 4 random points, imagine choosing 3 random lines through the center, and then a random point for P 4 P_4 P 4 ​ .

That first line passes through the sphere at 2 points, so flip a coin to decide which of those two points is P 1 P_1 P 1 ​ . Likewise, for each of the other lines flip a coin to decide where P 2 P_2 P 2 ​ and P 3 P_3 P 3 ​ end up.

There are 8 equally likely outcomes of these coin flips, but one and only one of these outcomes will place P 1 P_1 P 1 ​ , P 2 P_2 P 2 ​ , and P 3 P_3 P 3 ​ on the opposite side of the center from P 4 P_4 P 4 ​ . So only one of these 8 equally likely outcomes gives a tetrahedron containing the center, meaning our final answer is 1 8 \frac{1}{8} 8 1 ​ .

Isn't that elegant?

This is a valid solution, but admittedly the way I've stated it so far rests on some visual intuition. Here is a more formal write-up of this same solution in the language of linear algebra by Ralph Howard and Paul Sisson, if you're curious. This is common in math, where having the key insight and understanding is one thing, but having the relevant background to articulate this understanding more formally is almost a separate muscle entirely, one which undergraduate math students spend much of their time building up.

The takeaway

The important lesson here is not the specific solution; after all who cares about random tetrahedra in a sphere? Instead take note of the two key problem-solving tactics that led us to the solution, which most certainly carry over to many problems that do matter.

• Continue asking simpler versions of the question until you can get some foothold.
• If some added construct proves to be useful, see if you can reframe the whole question around that new construct.

Special thanks to those below for supporting the original video behind this post, and to current patrons for funding ongoing projects. If you find these lessons valuable, consider joining .

7 of the hardest problems in mathematics that have been solved

Mathematics is full of problems, some of which have been solved and others that haven’t. here, we focus on 7 of the hardest problems that have been solved..

Tejasri Gururaj

What are the hardest problems in math that have already been solved?

intararit  

• Some problems in mathematics have taken centuries to be solved, due to their complexity.
• Although there are some complex math problems that still elude solutions, others have now been solved.
• Here are 7 of the hardest math problems ever solved.

Some mathematical problems are challenging even for the most accomplished mathematicians.

From the Poincaré conjecture to Fermat’s last theorem , here we take a look at some of the most challenging math problems ever solved.

1. Poincaré conjecture

Salix alba  

The Poincaré conjecture is a famous problem in topology, initially proposed by French mathematician and theoretical physicist Henri Poincaré in 1904. It asserts that every simply connected, closed 3-manifold is topologically homeomorphic (a function that is a one-to-one mapping between sets such that both the function and its inverse are continuous) to a 3-dimensional sphere. 

In simpler terms, the conjecture asserts that a particular group of three-dimensional shapes can be continuously transformed into a sphere without any gaps or holes.

The problem was solved by the reclusive Russian mathematician Grigori Perelman in 2003. He built upon the work of American mathematician Richard S. Hamilton’s program involving the Ricci flow.

What makes this achievement even more remarkable is that Perelman declined the prestigious Fields Medal and the Clay Millennium Prize reward that came with it. He chose to stay away from the spotlight and mathematical acclaim, but his proof withstood rigorous scrutiny from the mathematical community.

The resolution of the Poincaré conjecture confirmed the fundamental role of topology in understanding the shape and structure of spaces, impacting fields like geometry and manifold theory.

2. The prime number theorem 

Britannica  

The prime number theorem long stood as one of the fundamental questions in number theory. At its core, this problem is concerned with unraveling the distribution of prime numbers. 

The question at hand revolves around the distribution pattern of these primes within the realm of natural numbers. Are there any discernible patterns governing the distribution of prime numbers, or do they appear to be entirely random? The theorem states that for large values of x, π(x) is approximately equal to x/ln(x).

The breakthrough in solving this theorem came in the late 19th century, thanks to the independent work of two mathematicians, Jacques Hadamard and Charles de la Vallée-Poussin. In 1896, both mathematicians presented their proofs of the theorem. 

Their work demonstrated that prime numbers exhibit a remarkable, asymptotic distribution pattern . Their solution produced the insight that, as one considers larger and larger numbers, the density of prime numbers diminishes. 

The theorem precisely characterized the rate of this decrease, showing that prime numbers become less frequent as we move along the number line. It’s as if they gradually thin out, although they never entirely vanish.

The prime number theorem was a turning point in the study of number theory . It provided profound insights into the distribution of prime numbers and put to rest the notion that there might exist a formula predicting each prime number. 

Instead, it proposed a probabilistic approach to understanding the distribution of prime numbers. The theorem’s significance extends into various mathematical fields, especially in cryptography, where the properties of prime numbers play a pivotal role in securing communications.

3. Fermat’s last theorem

Charles Rex Arbogast/AP via NPR  

Fermat’s last theorem is one of the problems on this list many people are most likely to have heard of. The conjecture, proposed by French mathematician Pierre de Fermat in the 17th century, states that it’s impossible to find three positive integers, a, b, and c, that can satisfy the equation a n + b n = c n for any integer value of n greater than 2.

For instance, there are no whole number values of a, b, and c that can make 3 3 + 4 3 = 5 3 true.

This problem remained unsolved for centuries and became one of the most challenging problems in mathematics. It was made more enticing because Fermat apparently wrote a note in his copy of the  Arithmetica  by Diophantus of Alexandria, saying, “I have discovered a truly remarkable proof [of this theorem], but this margin is too small to contain it.”

Numerous mathematicians attempted to prove or disprove Fermat’s conjecture, but it wasn’t until 1994 that the breakthrough came.

The solution was achieved by British mathematician Andrew Wiles, who built upon the work of many other mathematicians who had contributed to the field of number theory. Wiles’ proof was extraordinarily complex and required intricate mathematical concepts and theorems, particularly those related to elliptic curves and modular forms.

Wiles’ remarkable proof of Fermat’s last theorem confirmed that the conjecture was indeed true. Having taken more than three centuries to be solved, it had a profound impact on the world of mathematics, demonstrating the power of advanced mathematical techniques in solving long-standing problems.

Z. Ziegler, M. Ondrachek  

Before Wiles presented its proof, it was in the Guinness Book of World Records as the “most difficult mathematical problem,” in part because the theorem has seen the greatest number of unsuccessful proofs.

4. Classification of finite simple groups

Jakob.scholbach/Pbroks13  

This one is a bit different from the others on the list. The classification of finite simple groups , also known as the “enormous theorem,” set out to classify all finite simple groups, which are the fundamental building blocks of group theory.

Finite simple groups are those groups that cannot be divided into smaller non-trivial normal subgroups. The goal was to understand and categorize all the different types of finite simple groups that exist.

The solution to this problem is not straightforward. The proof is a collaborative effort by hundreds of mathematicians covering tens of thousands of pages in hundreds of journal articles published between 1955 and 2004.

It is one of the most extensive mathematical proofs ever produced and marks a monumental achievement in group theory.

The proof outlines the structure of finite simple groups and demonstrates that they can be classified into several specific categories. This achievement paved the way for a deeper understanding of group theory and its applications in various mathematical fields. 

5. The four color theorem

Inductiveload  

The four color theorem tackles an intriguing question related to topology and stands as one of the first significant theorems proved by a computer. 

It states that any map in a plane can be colored using four colors so that no two adjacent regions share the same color while using the fewest possible colors. Adjacent, in this context, means that two regions share a common boundary curve segment, not merely a corner where three or more regions meet.

The theorem doesn’t focus on the artistic aspect of map coloring but rather on the fundamental mathematical principles that underlie it.

The solution to this theorem arrived in 1976, thanks to the combined efforts of mathematicians Kenneth Appel and Wolfgang Haken. However, the proof was not widely accepted due to the infeasibility of checking it by hand.

Appel and Haken’s achievement confirmed that any map, regardless of its complexity, can be colored with just four colors such that no two neighboring regions share the same color. While the idea seems simple, proving it rigorously was complex and time-consuming.

To address any lingering skepticism about the Appel–Haken proof, a more accessible proof using similar principles and still utilizing computer assistance was presented in 1997 by Robertson, Sanders, Seymour, and Thomas. 

Additionally, in 2005, Georges Gonthier achieved a proof of the theorem using general-purpose theorem-proving software, reinforcing the credibility of the four color theorem.

This theorem is not actually used in map-making but has far-reaching implications in various fields, from graph theory to computer science, where it finds applications in scheduling, circuit design, and optimization problems.

6. Gödel’s incompleteness theorems

Andrew Das Arulsamy/Research Gate  

Gödel’s incompleteness theorems , formulated by Austrian mathematician Kurt Gödel in the 20th century, delve into the mysteries of formal systems and their inherent limitations. 

In mathematics, a formal system has a structured and well-defined framework or language that comprises a set of symbols, rules, and axioms employed for representing and manipulating mathematical or logical expressions.

Gödel’s first incompleteness theorem explores a fundamental question: In any consistent formal system, are there true mathematical statements that are undecidable within that system? In other words, do statements exist that cannot be proven as either true or false using the rules and axioms of that system?

The second incompleteness theorem takes this further: Can any consistent formal system prove its own consistency?

Gödel not only posed these questions but also provided the answers. He established, through rigorous mathematical proofs , that there exist true statements within formal systems that cannot be proven within those very systems.

In essence, the first theorem asserts that there are statements that cannot be proven as either true or false using the rules and axioms of a system. The second theorem demonstrates that no consistent formal system can prove its own consistency.

Gödel’s theorems introduced a profound paradox within the realm of mathematical logic: There are truths that exist beyond the reach of formal proofs, and there are limits to what can be achieved through mathematical systems alone.

Gödel’s contributions to mathematical logic influenced the philosophy of mathematics and our understanding of the inherent limits of formal systems.

7. The goat problem

Mnchnstnr  

The goat problem is a much more recently solved mathematical problem. It involves calculating the grazing area for a tethered goat. Despite its initial simplicity, mathematicians have pondered this problem for over a century.

In its basic form, a goat on a rope can graze in a semicircle with an area of A = 1/2πr 2 , where r is the rope’s length. However, the problem becomes more complex when you change the shape of the area the goat can access.

For instance, when tethered to a square barn, the goat can access more than just a semicircle. The goat can also go around the corners of the barn, creating additional quarter circles.

Mathematician Ingo Ullisch recently unraveled the goat grazing problem, introducing complex analysis into the equation. However, the solution is far from elementary.

It involves intricate calculations, relying on the ratio of contour integral expressions and involves numerous trigonometric terms . Although the solution may not offer a practical guide for goat owners, it represents a significant achievement in the world of mathematics.

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What makes the goat problem truly fascinating is its capacity to act as a mathematical Rosetta stone , transcending boundaries between various fields and serving as a versatile challenge for experts from diverse disciplines.

From age-old conundrums that took centuries to crack to enigmas that continue to elude solutions, mathematical mysteries remind us that the pursuit of knowledge is an ever-evolving journey.

So, the next time you find yourself pondering a difficult math problem, remember that you are in good company, following in the footsteps of the greatest mathematical explorers!

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ABOUT THE EDITOR

Tejasri Gururaj Tejasri is a versatile Science Writer &amp; Communicator, leveraging her expertise from an MS in Physics to make science accessible to all. In her spare time, she enjoys spending quality time with her cats, indulging in TV shows, and rejuvenating through naps.

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How to Work Through Hard Math Problems

parent of one of our students wrote today about his daughter’s occasional frustration with the difficulty of some of the problems in our courses. She does fantastic work in our courses , and was easily among the very top students in the class she took with me, and yet she still occasionally hits problems that she can’t solve.

Moreover, she has access to an excellent math teacher in her school who sometimes can’t help her get past these problems, either. (This is no slight to him—I have students bring me problems I can’t solve, too!) Her question: “Why does it have to be so hard?”

The Case for Doing Hard Things

We ask hard questions because so many of the problems worth solving in life are hard. If they were easy, someone else would have solved them before you got to them. This is why college classes at top-tier universities have tests on which nearly no one clears 70%, much less gets a perfect score. They’re training future researchers, and the whole point of research is to find and answer questions that have never been solved. You can’t learn how to do that without fighting with problems you can’t solve. If you are consistently getting every problem in a class correct, you shouldn’t be too happy — it means you aren’t learning efficiently enough. You need to find a harder class.

The problem with not being challenged sufficiently goes well beyond not learning math (or whatever) as quickly as you can. I think a lot of what we do at AoPS is preparing students for challenges well outside mathematics. The same sort of strategies that go into solving very difficult math problems can be used to tackle a great many problems. I believe we’re teaching students how to think, how to approach difficult problems, and that math happens to be the best way to do so for many people.

The first step in dealing with difficult problems is to accept and understand their importance. Don’t duck them. They will teach you a lot more than a worksheet full of easy problems. Brilliant “Aha!” moments almost always spring from minds cultivated by long periods of frustration. But without that frustration, those brilliant ideas never arise.

Strategies for Difficult Math Problems — and Beyond

Here are a few strategies for dealing with hard problems, and the frustration that comes with them:

Do something . Yeah, the problem is hard. Yeah, you have no idea what to do to solve it. At some point you have to stop staring and start trying stuff. Most of it won’t work. Accept that a lot of your effort will appear to have been wasted. But there’s a chance that one of your stabs will hit something, and even if it doesn’t, the effort may prepare your mind for the winning idea when the time comes.

We started developing an elementary school curriculum months and months before we had the idea that became Beast Academy . Our lead curriculum developer wrote 100–200 pages of content, dreaming up lots of different styles and approaches we might use. Not a one of those pages will be in the final work, but they spurred a great many ideas for content we will use. Perhaps more importantly, it prepared us so that when we finally hit upon the Beast Academy idea, we were confident enough to pursue it.

Simplify the problem . Try smaller numbers and special cases. Remove restrictions. Or add restrictions. Set your sights a little lower, then raise them once you tackle the simpler problem.

Reflect on successes . You’ve solved lots of problems. Some of them were even hard problems! How did you do it? Start with problems that are similar to the one you face, but also think about others that have nothing to do with your current problem. Think about the strategies you used to solve those problems, and you might just stumble on the solution.

A few months ago, I was playing around with some Project Euler problems, and I came upon a problem that (eventually) boiled down to generating integer solutions to c ² = a ² + b ² + ab in an efficient manner. Number theory is not my strength, but my path to the solution was to recall first the method for generating Pythagorean triples. Then, I thought about how to generate that method, and the path to the solution became clear. (I’m guessing some of our more mathematically advanced readers have so internalized the solution process for this type of Diophantine equation that you don’t have to travel with Pythagoras to get there!)

Focus on what you haven’t used yet . Many problems (particularly geometry problems) have a lot of moving parts. Look back at the problem, and the discoveries you have made so far and ask yourself: “What haven’t I used yet in any constructive way?” The answer to that question is often the key to your next step.

Work backwards . This is particularly useful when trying to discover proofs. Instead of starting from what you know and working towards what you want, start from what you want, and ask yourself what you need to get there.

Ask for help . This is hard for many outstanding students. You’re so used to getting everything right, to being the one everyone else asks, that it’s hard to admit you need help. When I first got to the Mathematical Olympiad Program (MOP) my sophomore year, I was in way over my head. I understood very little of anything that happened in class. I asked for help from the professor once — it was very hard to get up the courage to do so. I didn’t understand anything he told me during the 15 minutes he worked privately with me. I just couldn’t admit it and ask for more help, so I stopped asking. I could have learned much, much more had I just been more willing to admit to people that I just didn’t understand. (This is part of why our classes now have a feature that allows students to ask questions anonymously.) Get over it. You will get stuck. You will need help. And if you ask for it, you’ll get much farther than if you don’t.

Start early . This doesn’t help much with timed tests, but with the longer-range assignments that are parts of college and of life, it’s essential. Don’t wait until the last minute — hard problems are hard enough without having to deal with time pressure. Moreover, complex ideas take a long time to understand fully. The people you know who seem wicked smart, and who seem to come up with ideas much faster than you possibly could, are often people who have simply thought about the issues for much longer than you have. I used this strategy throughout college to great success — in the first few weeks of each semester, I worked far ahead in all of my classes. Therefore, by the end of the semester, I had been thinking about the key ideas for a lot longer than most of my classmates, making the exams and such at the end of the course a lot easier.

Take a break . Get away from the problem for a bit. When you come back to it, you may find that you haven’t entirely gotten away from the problem at all — the background processes of your brain have continued plugging away, and you’ll find yourself a lot closer to the solution. Of course, it’s a lot easier to take a break if you start early.

Start over . Put all your earlier work aside, get a fresh sheet of paper, and try to start from scratch. Your other work will still be there if you want to draw from it later, and it may have prepared you to take advantage of insights you make in your second go-round.

Give up . You won’t solve them all. At some point, it’s time to cut your losses and move on. This is especially true when you’re in training, and trying to learn new things. A single difficult problem is usually going to teach you more in the first hour or two than it will in the next six, and there are a lot more problems to learn from. So, set yourself a time limit, and if you’re still hopelessly stuck at the end of it, then read the solutions and move on.

Be introspective . If you do give up and read the solution, then read it actively, not passively. As you read it, think about what clues in the problem could have led you to this solution. Think about what you did wrong in your investigation. If there are math facts in the solution that you don’t understand, then go investigate. I was completely befuddled the first time I saw a bunch of stuff about “mod”s in an olympiad solution — we didn’t have the internet then, so I couldn’t easily find out how straightforward modular arithmetic is! You have the internet now, so you have no excuse. If you did solve the problem, don’t just pat yourself on the back. Think about the key steps you made, and what the signs were to try them. Think about the blind alleys you explored en route to the solution, and how you could have avoided them. Those lessons will serve you well later.

Come back . If you gave up and looked at the solutions, then come back and try the problem again a few weeks later. If you don’t have any solutions to look at, keep the problem alive. Store it away on paper or in your mind.

Richard Feynman once wrote that he would keep four or five problems active in the back of his mind. Whenever he heard a new strategy or technique, he would quickly run through his problems and see if he could use it to solve one of his problems. He credits this practice for some of the anecdotes that gave Feynman such a reputation for being a genius. It’s further evidence that being a genius is an awful lot about effort, preparation, and being comfortable with challenges.

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15 Hardest SAT Math Questions

Is your SAT score enough to get you into your dream school?

Our free chancing engine takes into consideration your SAT score, in addition to other profile factors, such as GPA and extracurriculars. Create a free account to discover your chances at hundreds of different schools.

Students often want to prepare for the toughest questions they’ll see on the SAT so they can feel confident on test day. We’ve rounded up some of the hardest questions and we’re going to show you how to solve them.

These questions come from the free SAT practice tests because we wanted to make sure that we show you questions representative of those you’ll see on the real test. We’ve chosen a variety of question types, but this isn’t an exhaustive representation of all the math topics or types of questions on the SAT.

“Hard” is a little subjective; we think these questions are difficult based on working with many students, but you may find some of them easy. In general, you can expect to find harder questions in the second half of each SAT math section, as the questions generally increase in difficulty.

For each question below, we’ve identified the key concepts being tested and whether a calculator is allowed. We suggest you try solving these on your own before looking at the answer and our suggested solution. Remember: we’ve given you just one way to solve the problem. Most problems can be solved in a variety of ways!

Want to know your chances at the schools you’re applying for based on your SAT?   Calculate your admissions chances right now and understand your odds before applying.

Math Topics on the SAT

Before we go into the questions, we want you to understand the terms that the College Board uses to categorize the topics. We’ve also included how many questions fall under each category, so if you’re self-studying, you can prioritize the types of questions that appear more often.

Heart of Algebra: 33% of test, 19 questions

Linear equations and inequalities and their graphs and systems.

Problem Solving and Data Analysis: 29% of test, 17 questions

Ratios, proportions, percentages, and units; analyzing graphical data, probabilities, and statistics.

Passport to Advanced Math: 28% of test, 16 questions

Identifying and creating equivalent expressions; quadratic and nonlinear equations/functions and their graphs.

Additional Topics in Math: 10% of test, 6 questions total

This is the only category without a corresponding subscore, but it has a wide variety of topics, including geometry, trigonometry, radians and the unit circle, and complex numbers.

Grid-in Questions

A grid-in question can test any of the topics above and is found at the end of each portion of the math test, both no-calculator and calculator. Most students find grid-in questions harder than the multiple-choice because some test tactics—like substituting answer choices into the problem—don’t work. You’ll need to find the correct solution without any help from answer choices. If you develop your math skills, however, then these questions won’t be that much more difficult for you.

Now grab a calculator, a pencil and a piece of paper, and let’s review the hardest SAT math questions!

Question 1: Calculator permitted, grid-in response

Answer: 750

Category: Heart of Algebra—systems of linear inequalities

Here’s how to solve it:

• In this question, we’re told coordinates \((a,b)\) lies in the solution set of these equations, and we want to know a maximum possible value. With inequalities, the graph will be shaded to include the set of values that satisfy the inequality, so \((a,b)\) lies in this overlapping region. Because it’s a grid-in response (only real numbers can be the answer to the grid-in questions—there is no option to bubble in infinity) we know that the value of \(b\) must be limited by the point of intersection. Since the inequalities are both less than or equal to , we know that the lines (and the point of intersection) are included in the solution set. If there were only less/greater than symbols, the lines would not be included in the solution set.
• This means we can find the point of intersection to find a maximum value of \(b\) and solve it just like a system of equations using the substitution method. This gives us the inequality \(5x\:\leq\:-15x+3000\). In this case, \(x\) is equivalent to \(a\), and the y-value is equivalent to \(b\).
• Move the variables to the same side by adding \(15x\) to both sides. \(20x\leq3000\).
• Divide both sides by \(20\) to find out what \(x\) is. \(x\:\leq\:150\). Many students might stop here, but remember we want coordinate \(b\) in \((a,b)\), or the y-value.
• We can plug \(x\) into either equation to find the value of \(b\). We picked the second one since it’s simpler: \(5\:\cdot\:150=750\). That’s our answer!

Question 2: Calculator permitted, multiple choice

Category: Problem-solving and data analysis—analyzing graphical data, probability

• With problems like these, it’s important that you pay attention to the wording of the question so that you know exactly what you’re being asked. Once you understand that, the math needed is not necessarily complex. The question asks: “If a person is chosen at random from those who recalled at least 1 dream, what is the probability that the person belonged to Group Y?” It can help to rewrite the question in your own words. Here’s a possible rewrite: “Out of all the people who remembered at least 1 dream, what fraction were in Group Y?”
• We need to know the total number of people who remembered at least one dream—to do this, we add \(28+57+11+68=164\).
• Next, we need to know how many of the people who remembered at least one dream were in Group Y. Add \(11+68=79\).
• We can now see that the chance of selecting a person from Group Y out of all the people who remembered at least \(1\) dream is \(79\) out of \(164\), or \(\frac{79}{164}\), which is C .

Question 3: Calculator not permitted, multiple choice

Category: Passport to Advanced Math—exponents

• To solve this problem, you’ll need to know exponent rules. Here are some of the ones you should know for the SAT and many other standardized tests:
• \(b^mb^n=b^{m+n}\)
• \(\frac{b^m}{b^n}=b^{m-n}\)
• \((b^m)^n=b^{m\:\cdot\:n}\)
• \(b^mc^m=(bc)^m\)
• \(\frac{b^m}{c^m}=(\frac{b}{c})^m\)
• Since we know that it’s likely that we’ll need exponent rules, we can see if we might be able to rewrite the expression \(\frac{8^x}{2^y}\) to fit one of the existing exponent rules. In fact, we can rewrite \(8=2^3\), so we have \(8^x=(2^3)^x=2^{3x}\).
• Substituting \(2^{3x}\) for \(8^x\), we can see that the expression can be rewritten using the second rule above: \(\frac{2^{3x}}{2^y}=2^{3x-y}\).
• Since we’ve been given the information that \(3x-y=12\) we can substitute \(12\) into the expression and get our answer: \(2^{3x-y}=2^{12}\).

Question 4: Calculator not-permitted, multiple choice

Category: Passport to Advanced Math—rational equations

• Students are sometimes intimidated by large fractions like this, but if you get comfortable manipulating fractions and variables across operations then you can handle any question like this. To start, we’ll want to add those two fractions in the denominator, and to do that, we need to find a common denominator. The common denominator in this case is \((x+2)(x+3)\).
• To convert each of those fractions in the denominator to ones we can easily add, we need to find equivalent fractions. You can multiply any fraction by \(1\) without changing its value, and so we can use fractions like \(\frac{2}{2}\), etc. We’re going to multiply the first fraction by \(\frac{x+3}{x+3}\) and the second fraction by \(\frac{x+2}{x+2}\). This gives us the following in the denominator: \(\frac{x+3}{(x+2)(x+3)}+\frac{x+2}{(x+2)(x+3)}\)
• Now we can add the numerators: \((x+3)+(x+2)\). Combine like terms in the numerator to get \(2x+5\).
• We now have a single fraction in the denominator, and we can proceed with fraction division. When dividing by a fraction, we multiply by its reciprocal , and we can find the reciprocal by switching the positions of the numerator and the denominator. We get:

\(\frac{1}{\frac{2x+5}{(x+2)(x+3)}}=1\:\cdot\:\frac{(x+2)(x+3)}{2x+5}=\frac{(x+2)(x+3)}{2x+5}\)

• Multiplying the numerator out, we can rewrite the expression as \((x+2)(x+3)=x^2+5x+6\). This gives us our answer: \(\frac{x^2+5x+6}{2x+5}\)

Question 5: Calculator permitted, grid-in response (see note)

Note: This is a two-part question, but we left out the first part to focus on the second one. The first question asked for the value of \(x\) in the expression. Because her bank account earns \(2\%\) interest compounded annually, we can convert \(2\%\) to a decimal, giving us \(.02\). We then add \(1\) to this so that we don’t lose the initial deposit. The answer is \(x=1.02\), which you need to know to solve the question above.

Answer: 6.11

Category: Passport to Advanced Math—exponential functions and equations (interest)

Here’s how you solve it:

• Given that Jessica uses the expression \(\$100(1.02)^t\) to find the value of her account after \(t\) years, we can create a similar expression to model Tyshaun’s account. This will be \(\$100(1.025)^t\).
• Now we can plug \(10\) in for \(t\) into both of the expressions. Jessica’s account will have \(\$100(1.02)^{10}\approx\$121.899\). Tyshaun’s account will have \(\$100(1.025)^{10}\approx\$128.008\).
• Now that we know how much money each account will have in 10 years, we can subtract the amount of money that Jessica’s account has from the amount of money that Tyshaun’s account has to find our answer. \(\$128.008-\$121.899=\$6.109\). Pay close attention to how problems ask for the answer: we need to round to the nearest cent and ignore the dollar sign, so the answer is \(6.11\).

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Question 6: Calculator not-permitted, multiple choice

Additional Topics in Math—complex numbers

• There’s usually at least one question on the SAT involving complex numbers, so it’s helpful to know how to add, subtract, and multiply them together. A complex number in standard form is written \(a+bi\), where \(a\) is a real number and \(bi\) is the imaginary part. In this question, we’ll want to use the conjugate form of \(a+bi\) which is \(a-bi\). This is because when we multiply complex conjugates, we get a real number: \((a+bi)(a-bi)=a^2-abi+abi-b^2i^2=a^2-b^2i^2\). However, because \(i=\sqrt{-1}\), that means that \(i^2=-1\). We can thus simplify \(a^2-b^2i^2=a^2+b^2\).
• Taking a hint from the answer choices, which only have real numbers in the first term, we can multiply the fraction above by the conjugate of the denominator like so:

\(\frac{3-5i}{8+2i}\cdot\frac{8-2i}{8-2i}=\frac{(3-5i)(8-2i)}{(8+2i)(8-2i)}\)

• We can quickly figure out the denominator given the information about conjugates above. \(8^2+2^2=64+4=68\).
• Multiplying complex numbers is just like multiplying binomials, keeping in mind that \(i^2=-1\) and we’ll need to simplify that out at the end.

\((3-5i)(8-2i)=24-6i-40i+10i^2=24-46i-10=14-46i\).

• Our new fraction is this: \(\frac{14-46i}{68}\). We can rewrite it so that the real part and the imaginary part of the complex numerator are separated and then simplify to get our answer.

\(\frac{14-46i}{68}=\frac{14}{68}-\frac{46i}{68}=\frac{7}{34}-\frac{23i}{34}\)

Question 7: Calculator permitted, multiple choice

Category: Additional Topics in Math—Circles and their equations

• Like complex numbers, there’s usually at least one question involved the equation of a circle. This equation is not in the standard form of the equation of a circle, which is \((x-h)^2+(y-k)^2=r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius. We’ll need to rewrite this equation into standard form to easily find the radius.
• To begin rewriting the equation, we will need to complete the square twice, once for \(x\) and once for \(y\).

\(x^2+4x+y^2-2y=-1\)

\((x+2)^2-4+(y-1)^2-1= -1\)

• As you can see, we halved the absolute value of the \(x\) and \(y\) coefficients to find the constant to complete the square, then subtracted the squared constants to keep the equation the same. Now we need to move our constants to the other side by adding and we’ll have our equation in standard form.

\((x+2)^2+(y-1)^2=-1+1+4\)

\((x+2)^2+(y-1)^2=4\)

• We now see that \(r^2=4\), so the radius is \(\sqrt4=2\). That’s our answer! 

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Question 8: Calculator not-permitted, multiple choice

Category: Passport to Advanced Math—rewriting equations in terms of another variable

1. Students get intimidated by questions that have only variables. Remember to use what you know about manipulating equations—such as whatever you do to one side you must do to the other—to handle these problems.

2. We want to rewrite the equation in terms of \(F\). To start, we’ll want to get rid of the \(F\) in the denominator of the equation above by multiplying both sides by \(N+F\)  like so:

\(R(N+F)=\frac{F(N+F)}{(N+F)}\)

\(R(N+F)=F\)

3. Distribute the \(R\) on the left-hand side, and then move all the F ‘s to the same side.

\(RN+RF=F\)

\(F-RF=RN\)

4. Next, we can factor out the \(F\) on the left-hand side and divide both sides by the remaining factor to isolate \(F\) and get our answer.

\(F(1-R)=RN\)

\(F=\frac{RN}{1-R}\)

Question 9: Calculator permitted, multiple choice

Category: Problem Solving and Data Analysis—standard deviation, range, analyzing graphical data

Here’s how you solve it :

1. You won’t have to calculate standard deviation on the SAT, but you will need to know what it means and what affects it. Standard deviation is a measure of spread in a data set, specifically how far the points in the data set are from the mean value. A larger standard deviation means that the points in the data set are more spread out from the mean value, and a smaller one means that the points in the data set are close to the mean value. Likewise, you’ll need to know what is meant by range. Range refers to the difference between the highest and lowest values in a data set. The SAT may require you to calculate the range for a data set.

2. Let’s first determine the standard deviations of each data set relative to each other. In the first data set, most of the data points are clustered around each other, which makes it likely that the mean is somewhere around \(72\) (even though there are two outliers, they are equidistant from \(72\) so they will balance each other out when determining the mean). In contrast, the second data set is more spread out, so we can conclude that the standard deviation of the first set is smaller than the standard deviation of the second. We can eliminate choices A and C .

3. Now let’s look at the range for each set. We can subtract the highest and lowest values for each set to find the range. For the first set, \(88-56 = 32\). For the second set, \(112-80 = 32\). The ranges of each set are equivalent. This leaves us with choice D as our answer.

Question 10: Calculator not-permitted, multiple choice

Category: Passport to Advanced Math—quadratic and algebraic functions and their graphs

1. We can solve this using substitution. Substitute the second equation into the first and expand the quadratic:

\(x=2[(2x-3)(x+9)]+5\)

\(x=2(2x^2+15x-27)+5\)

\(x=4x^2+30x-49\)

2. Next, move the x from the left-hand side over to the right. The roots of this quadratic will give us the solution to the system of equations above.

\(0=4x^2+29x-49\)

3. Since the roots of the above quadratic will give us the solution to the system of equations above, we can use the discriminant of the quadratic formula to find out how many solutions there are. The discriminant is the part of the quadratic formula under the radical, or \(b^2-4ac\). If the discriminant is positive, there are 2 solutions; if the discriminant is 0, there is 1 real solution (or a repeated solution); if the discriminant is negative, there are no real solutions. Substituting the numbers from the equation into the discriminant formula gives us \(29^2-4(4)(49)=841-784=57\). This is a positive number, which means there are 2 solutions to the system of equations.

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Question 11: Calculator permitted, grid-in response

Answer: 2.25 or 9/4

Category: Passport to Advanced Math—determining effect of one variable on another

1. This is similar to the earlier problem where there are only variables, but if we proceed step-by-step and manipulate the equation carefully, we can find the solution. In this problem, we have two fluids, and one of them is moving at a velocity of \(1\) and the other at \(1.5\). We can substitute the \(1.5\) into the equation to see how it might affect \(q\), or the dynamic pressure.

\(q=\frac{1}{2}n(1.5v)^2\)

2. Using exponent rules (from Question 3 above) we can rewrite \((1.5v)^2=1.5^2v^2\) . Squaring \(1.5\) gives us \(2.25\).

3. For our fluid moving at a velocity of \(1.5\), we thus have: \(q=\frac{1}{2}n(2.25)v^2\). For our fluid moving at a velocity of \(1\), we have \(q=\frac{1}{2}n(1)v^2\).  We can thus see that the dynamic pressure of the faster fluid will be \(2.25\) times that of the slower fluid.

Question 12: Calculator not-permitted, grid-in response

Answer: 3/5 or .6

Category: Additional Topics in Math—trigonometry

1. If you’re given a word problem that involves geometry or trigonometry and no diagram, you should draw yourself a diagram and label the information. This often makes it very clear how to find the answer. Here’s our diagram:

2. Many students know about the 30-60-90 and the 45-45-90 right triangles, and information about these triangles is included on the first page of each SAT math portion. However, you should also be on the lookout for the 3-4-5 and 5-12-13 right triangles. College Board likes to use these triangles and their similar counterparts because they have nice, whole-number sides that make calculations easy. Knowing this, we can see that Triangle ABC is similar to the 3-4-5 triangle, each side of ABC 4 times the length of the 3-4-5 triangle.

3. Knowing that triangle ABC is similar to a 3-4-5 triangle, and that triangle DEF is also similar to a 3-4-5 triangle, we can find \(sin\:F\) by substituting 3-4-5 for the sides of DEF and using the definition of sine:  \(sin\:\theta\:=\frac{opposite}{hypotenuse}\)

4. With respect to \(F\), the opposite side is \(3\) and the hypotenuse is \(5\). This means that \(sin\:F=\frac{3}{5}\), or \(0.6\).

Question 13: Calculator permitted, multiple choice

Category: Heart of Algebra—linear equations and inequalities in context

1. You’ll need to be able to write equations that reflect the context described in a word problem for several questions on the SAT, both linear context and non-linear contexts. If you have trouble “translating” word problems into math, this is a skill you’ll want to work on! Rewriting word problems to include words like equal to, less than, more than, sum, and so on can help you easily translate the problem into equations and inequalities.

2. We’re told that Roberto wanted to sell \(57\) insurance policies but he didn’t meet his goal, which means that he sold less than \(57\) insurance policies. Given that \(x\) represents the number of \(\$50,000\) policies sold and \(y\) represents the number of \(\$100,000\) policies sold, the sum of these two numbers is less than \(57\). This is represented as \(x+y\:\lt\:57\). We can eliminate choices B and D .

3. Next, we’re told that the value of all the policies sold was more than \(\$3,000,000\). We need to multiply the value of the policy by the number of that type of policy, which gives us \(50,000x+100,000y\:\gt\:3,000,000\). Choice C is the only answer that reflects Roberto’s insurance policy sales.

Question 14: Calculator permitted, multiple choice

Category: Problem Solving and Data Analysis—Statistics

1. These kinds of problems can surprise students because there isn’t any obvious calculation or “math” to do. Instead, you’re being asked to interpret the results based on given information, and you’ll need to know statistics to correctly interpret these types of problems. In this case, you’ll want to draw upon knowledge of population parameters, random sampling, and random assignment:

• A population parameter is a numerical value describing a characteristic of a population. For example, we’re told that the population targeted by this treatment are “people with poor eyesight.” While not given a number, any conclusion we draw can only be made regarding people who fall under this parameter.
• Random sampling means that the subjects in the sample were selected at random (without bias) from the entire population in question. If random sampling is used, the results can be generalized to the entire population.
• Random assignment means that subjects in the sample were assigned to a treatment at random (without bias). If random assignment is used, it might be appropriate to make conclusions regarding cause and effect.

2 . The SAT is very particular when including statistical clues and knowing the definitions of statistics. For example, if random sampling is not mentioned in a problem, then the results cannot be generalized to the entire population. Knowing the above, we can see that random sampling and random assignment are explicitly mentioned in this case. We can reasonably conclude cause and effect to be generalized to the population in question, or that it’s likely that Treatment X will improve the eyesight of people who have poor eyesight. The other choices are either too confident in their conclusion or do not specify the population in question.

Question 15: Calculator not-permitted, grid-in response

Category: Additional Topics in Math—Radians

1. The point \(A\) is \((3,1)\). We can create a triangle to find the degree measure of \(\angle{AOB}\). When we do this, we can use the information the SAT provides to see that this is a 30-60-90 triangle and that the hypotenuse (or the radius of the circle) is \(2\).

3. Since \(a\) is the denominator, the answer is \(6\).

We hope that you have a good sense of the type of questions that you might see on the SAT, but just remember that these questions aren’t a complete representation of the topics you could be tested on. Review the questions above and ask yourself: which ones were the most challenging to you? Which topics do you need to brush up on?

Check out our free guide with our top 8 tips for mastering the SAT.

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To learn more about how to do well on the SAT, check out these other posts:

• How to Get a Perfect 1600 Score on the SAT
• Ultimate Guide to the New SAT Reading Test
• Ultimate Guide to the New SAT Writing and Language Test
• Ultimate Guide to the New SAT Math Test
• Links to Every SAT Practice Test + Other Free Resources

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100 hard word problems in algebra 

Find below a wide variety of hard word problems in algebra. Most tricky and tough algebra word problems are covered here.  If you can solve these, you can probably solve any algebra problems. Teachers! Feel free to select from this list and give them to your students to see if they have mastered how to solve tough algebra problems. Find out below how you can print these problems. You can also purchase a solution if needed.

1.  The cost of petrol rises by 2 cents a liter. last week a man bought 20 liters at the old price. This week he bought 10 liters at the new price. Altogether, the petrol costs $9.20. What was the old price for 1 liter?

2.  Teachers divided students into groups of 3. Each group of 3 wrote a report that had 9 pictures in it. The students used 585 pictures altogether. How many students were there in all?

3. Vera and Vikki are sisters. Vera is 4 years old and Vikki is 13 years old. What age will each sister be when Vikki is twice as old as Vera?

4.  A can do a work in 14 days and working together A and B can do the same work in 10 days. In what time can B alone do the work?

5. 7 workers can make 210 pairs of cup in 6 days. How many workers are required to make 450 pairs of cup in 10 days?

6. Ten years ago the ratio between the ages of Mohan and Suman was 3:5. 11 years hence it will be 11:16. What is the present age of Mohan?

7. The ratio of girls to boys in class is 9 to 7 and there are 80 students in the class. How many girls are in the class?

8.   One ounce of solution X contains only ingredients a and b in a ratio of 2:3. One ounce of solution Y contains only ingredients a and b in a ratio of 1:2. If solution Z is created by mixing solutions X and Y in a ratio of 3:11, then 2520 ounces of solution Z contains how many ounces of a?

9.   This week Bob puts gas in his truck when the tank was about half empty. Five days later, bob puts gas again when the tank was about three fourths full. If Bob Bought 24 gallons of gas, how many gallons does the tank hold?

10. A commercial airplane flying with a speed of 700 mi/h is detected 1000 miles away with a radar. Half an hour later an interceptor plane flying with a speed of 800 mi/h is dispatched. How long will it take the interceptor plane to meet with the other plane?

11. There are 40 pigs and chickens in a farmyard. Joseph counted 100 legs in all. How many pigs and how many chickens are there?

12. The top of a box is a rectangle with a perimeter of 72 inches. If the box is 8 inches high, what dimensions will give the maximum volume?

13. You are raising money for a charity. Someone made a fixed donation of 500. Then, you require each participant to make a pledge of 25 dollars. What is the minimum amount of money raised if there are 224 participants.

14. The sum of two positive numbers is 4 and the sum of their squares is 28. What are the two numbers?

15. Flying against the jet stream, a jet travels 1880 mi in 4 hours. Flying with the jet stream, the same jet travels 5820 mi in 6 hours. What is the rate of the jet in still air and what is the rate of the jet stream?

16.  Jenna and her friend, Khalil, are having a contest to see who can save the most money. Jenna has already saved $110 and every week she saves an additional $20. Khalil has already saved $80 and every week he saves an additional $25. Let x represent the number of weeks and y represent the total amount of money saved. Determine in how many weeks Jenna and Khalil will have the same amount of money.

17. The sum of three consecutive terms of a geometric sequence is 104 and their product is 13824.find the terms.

18. The sum of the first and last of four consecutive odd integers is 52.  What are the four integers?

19. A health club charges a one-time initiation fee and a monthly fee. John  paid 100 dollars for 2 months of membership. However, Peter paid 200 for 6 months of membership. How much will Sylvia pay for 1 year of membership?

20. The sum of two positive numbers is 4 and the sum of their cubes is 28. What is the product of the two numbers?

21. A man selling computer parts realizes that when he sells 16 computer parts, his earning is $1700. When he sells 56 computer parts, his earning is $4300. What will the earning be if the man sells 30 computers parts?

22. A man has 15 coins in his pockets. These coins are dimes and quarters that add up 2.4 dollars. How many quarters and how many dimes does the man have?

23. The lengths of the sides of a triangle are in the ratio 4:3:5. Find the lengths of the sides if the perimeter is 18 inches.

24. The ratio of base to height of a equilateral triangle is 3:4. If the area of the triangle is 6, what is the perimeter of the triangle?

25. The percent return rate of a growth fund, income fund, and money market are 10%, 7%, and 5% respectively. Suppose you have 3200 to invest and you want to put twice as much in the growth fund as in the money market to maximize your return. How should you invest to get a return of 250 dollars in 1 year?

26. A shark was caught whose tail weighted 200 pounds. The head of the shark weighted as much as its tail plus half its body. Its body weighted as much as its head and tail. What is the weight of the shark?

27. The square root of a number plus two is the same as the number. What is the number?

28. Suppose you have a coupon worth 6 dollars off any item at a mall. You go to a store at the mall that offers a 20% discount. What do you need to do to save the most money?

29. Suppose your grades on three math exams are 80, 93, and 91. What grade do you need on your next exam to have at least a 90 average on the four exams?

30. Peter has a photograph that is 5 inches wide and 6 inches long. She enlarged each side by the same amount. By how much was the photograph enlarged if the new area is 182 square inches?

31. The cost to produce a book is 1200 to get started plus 9 dollars per book. The book sells for 15 dollars each. How many books must be sold to make a profit?

32. Store A sells CDs for 2 dollars each if you pay a one-time fee of 104 dollars. Store B offers 12 free CDs and charges 10 dollars for each additional CD. How many CD must you buy so it will cost the same under both plans?

33. When 4 is added to two numbers, the ratio is 5:6. When 4 is subtracted from the two numbers, the ratio is 1:2. Find the two numbers.

34. A store owner wants to sell 200 pounds of pistachios and walnuts mixed together. Walnuts cost 4 dollars per pound and pistachio cost 6 dollars per pounds. How many pounds of each type of nuts should be mixed if the store owner will charge 5 dollars for the mixture?

35. A cereal box manufacturer makes 32-ounce boxes of cereal. In a perfect world, the box will be 32-ounce every time is made. However, since the world is not perfect, they allow a difference of 0.06 ounce. Find the range of acceptable size for the cereal box.

36. A man weighing 600 kg has been losing 3.12% of his weight each month with some heavy exercises and eating the right food. What will the man weigh after 20 months?

37. An object is thrown into the air at a height of 60 feet. After 1 second and 2 second, the object is 88 feet and 84 feet in the air respectively. What is the initial speed of the object?

38. A transit is 200 feet from the base of a building. There is man standing on top of the building. The angles of elevation from the top and bottom of the man are 45 degrees and 44 degrees. What is the height of the man?

39. A lemonade consists of 6% of lemon juice and a strawberry juice consists of 15% pure fruit juice. How much of each kind should be mixed together to get a 4 Liters of a 10% concentration of fruit juice?

40. Ellen can wash her car in 60 minutes. Her older sister Sarah can do the same job in 45 minutes. How long will it take if they wash the car together?

41. A plane flies 500 mi/h. The plane can travel 1100 miles with the wind in the same amount of time as it travels 900 miles against the wind. What is the speed of the wind?

42. A company produces boxes that are 5 feet long, 4 feet wide, and 3 feet high. The company wants to increase each dimension by the same amount so that the new volume is twice as big. How much is the increase in dimension?

43. James invested half of his money in land, a tenth in stock, and a twentieth in saving bonds. Then, he put the remaining 21000 in a CD. How much money did James saved or invested?

44. The motherboards for a desktop computer can be manufactured for 50 dollars each. The development cost is 250000. The first 20 motherboard are samples and will not be sold. How many salable motherboards will yield an average cost of 6325 dollars?

45. How much of a 70% orange juice drink must be mixed with 44 gallons of a 20% orange juice drink to obtain a mixture that is 50% orange juice?

46. A company sells nuts in bulk quantities. When bought in bulk, peanuts sell  for $1.20 per pound, almonds for $ 2.20 per pound, and cashews for $3.20 per pound. Suppose a specialty shop wants a mixture of 280 pounds that will cost $2.59 per pound. Find the number of pounds of each type of nut if the sum of the number of pounds of almonds and cashews is three times the number of pounds of peanuts. Round your answers to the nearest pound.

47. A Basketball player has successfully made 36 of his last 48 free throws. Find the number of consecutive free throws the player needs to increase his success rate to 80%.

48. John can wash cars 3 times as fast as his son Erick. Working together, they need to wash 30 cars in 6 hours. How many hours will it takes each of them working alone?

49. In a college, about 36% of student are under 20 years old and 15% are over 40 years old. What is the probability that a student chosen at random is under 20 years old or over 40 years?

50. Twice a number plus the square root of the number is twelve minus the square root of the number.

51. The light intensity, I , of a light bulb varies inversely as the square of the distance from the bulb. A a distance of 3 meters from the bulb, I = 1.5 W/m^2 . What is the light intensity at a distance of 2 meters from the bulb?

52. The lengths of two sides of a triangle are 2 and 6. find the range of values for the possible lengths of the third side.

53. Find three consecutive integers such that one half of their sum is between 15 and 21.

54. After you open a book, you notice that the product of the two page numbers on the facing pages is 650. What are the two page numbers?

55. Suppose you start with a number. You multiply the number by 3, add 7, divide by ½, subtract 5, and then divide by 12. The result is 5. What number did you start with?

56. You have 156 feet of fencing to enclose a rectangular garden. You want the garden to be 5 times as long as it is wide. Find the dimensions of the garden.

57. The amount of water a dripping faucet wastes water varies directly with the amount of time the faucet drips. If the faucet drips 2 cups of water every 6 minutes, find out how long it will take the faucet to drip 10.6465 liters of water.

58. A washer costs 25% more than a dryer. If the store clerk gave a 10% discount for the dryer and a 20% discount for the washer, how much is the washer before the discount if you paid 1900 dollars.

59. Baking a tray of blueberry muffins takes 4 cups of milk and 3 cups of wheat flour. A tray of pumpkin muffins takes 2 cups of milk and 3 cups of wheat flour. A baker has 16 cups of milk and 15 cups of wheat flour. You make 3 dollars profit per tray of blueberry muffins and 2 dollars profit per tray of pumpkin muffins. How many trays of each type of muffins should you make to maximize profit?

60. A company found that -2p + 1000 models the number of TVs sold per month where p can be set as low as 200 or as high as 300. How can the company maximize the revenue?

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61. Your friends say that he has $2.40 in equal numbers of quarters, dimes, and nickels. How many of each coin does he have?

62. I am a two-digit number whose digit in the tenth place is 1 less than twice the digit in the ones place. When the digit in the tenth place is divided by the digit in the ones place, The quotient is 1 and the remainder is 4. What number am I?

63. A two-digit number is formed by randomly selecting from the digits 2, 4, 5, and 7 without replacement. What is the probability that a two-digit number contains a 2 or a 7?

64. Suppose you interview 30 females and 20 males at your school to find out who among them are using an electric toothbrush. Your survey revealed that only 2 males use an electric toothbrush while 6 females use it. What is the probability that a respondent did not use an electric toothbrush given that the respondent is a female?

65. An employer pays 15 dollars per hour plus an extra 5 dollars per hour for every hour worked beyond 8 hours up to a maximum daily wage of 220 dollars. Find a piecewise function that models this situation.

66. Divide me by 7, the remainder is 5. Divide me by 3, the remainder is 1 and my quotient is 2 less than 3 times my previous quotient. What number am I?

67. A company making luggage have these requirements to follow. The length is 15 inches greater than the depth and the sum of length, width, and depth may not exceed 50 inches. What is the maximum value for the depth if the manufacturer will only use whole numbers?

68. To make an open box, a man cuts equal squares from each corner of a sheet of metal that is 12 inches wide and 16 inches long. Find an expression for the volume in terms of x.

69. Ten candidates are running for president, vice-president, and secretary in the students government. You may vote for as many as 3 candidates. In how many ways can you vote for 3 or fewer candidates?

70. The half-life of a medication prescribed by a doctor is 6 hours. How many mg of this medication is left after 78 hours if the doctor prescribed 100 mg?

71. Suppose you roll a red number cube and a yellow number cube. Find P(red 2, yellow 2) and the probability to get any matching pairs of numbers.

72. A movie theater in a small town usually open its doors 3 days in a row and then closes the next day for maintenance. Another movie theater 3 miles away open 4 days in a row and then closes the next day for the same reason. Suppose both movie theaters are closed today and today is Wednesday, when is the next time they will both be closed again on the same day?

73. An investor invests 5000 dollars at 10% and the rest at 5%. How much was invested at 5% if the yield is one-fifth of the amount invested at at 10%?

74. 20000 students took a standardized math test. The scores on the test are normally distributed, with a mean score of 85 and a standard deviation of 5. About how many students scored between 90 and 95?

75. A satellite, located 2400 km above Earth’s surface, is in circular orbit around the earth. If it takes the satellite 3 hours to complete 1 orbit, how far is the satellite after 1 hour?

76. In a group of 10 people, what is the probability that at least two people in the group have the same birthday?

77. During a fundraising for cancer at a gala, everybody shakes hands with everyone else in the room before the event and after the event is finished. If n people attended the gala, how many different handshakes occur?

78. Two cubes have side lengths that are equal to 2x and 4x. How many times greater than the surface of the small cube is the surface area of the large cube?

79. Suppose you have a job in a restaurant that pays $8 per hour. You also have a job at Walmart that pays $10 per hour. You want to earn at least 200 per week. However, you want to work no more than 25 hours per week . Show 3 different ways you could work at each job.

80. Two company offer tutoring services. Company A realizes that when they tutor for 3 hours, they make 45 dollars. When they tutor for 7 hours they make 105 hours. Company B realizes that when they tutor for 2 hours, they make 34 dollars. When they tutor for 6 hours they make 102 hours. Assuming that the number of hours students sign for tutoring is the same for both company, which company will generate more revenue?

81. You want to fence a rectangular area for kids in the backyard. To save on fences, you will use the back of your house as one of the four sides. Find the possible dimensions if the house is 60 feet wide and you want to use at least 160 feet of fencing.

82. When a number in increased by 20%, the result is the same when it is decreased by 10% plus 12. What is the number?

83. The average of three numbers is 47. The biggest number is five more than twice the smallest. The range is 35. What are the three numbers?

84. The percent of increase of a number from its original amount to 36 is 80%. What is the original amount of the number?

85. When Peter drives to work, he averages 45 miles per hour because of traffic. On the way back home, he averages 60 miles per hour because traffic is not as bad. The total travel time is 2 hours. How far is Peter’s house from work?

86. An advertising company takes 20% from all revenue that it generates for its affiliates. If the affiliates were paid 15200 dollars this month, how much revenue did the advertising company generate this month?

87. A company’s revenue can be modeled with a quadratic equation. The company noticed that when they sell either 2 or or 12 items, the revenue is 0. How much is the revenue when they sell 20 items?

88. A ball bounced 4 times, reaching three-fourths of its previous height with each bounce. After the fourth bounce, the ball reached a height of 25 cm. How high was the ball when it was dropped?

89. A rental company charges 40 dollars per day plus $0.30 per mile. You rent a car and drop it off 4 days later. How many miles did you drive the car if you paid 325.5 dollars which included a 5% sales tax?

90. Two students leave school at the same time and travel in opposite directions along the same road. One walk at a rate of 3 mi/h. The other bikes at a rate of 8.5 mi/h. After how long will they be 23 miles part?

91. Brown has the same number of brothers as sisters. His sister Sylvia as twice as many brothers as sisters. How many children are in the family?

92. Ethan has the same number of male classmates as female classmates. His classmate Olivia has three-fourths as many female classmates as male classmates. How many students are in the class?

93. Noah wants to share a certain amount of money with 10 people. However, at the last minute, he is thinking about decreasing the amount by 20 so he can keep 20 for himself and share the money with only 5 people. How much money is Noah trying to share if each person still gets the same amount?

94. The square root of me plus the square root of me is me. Who am I?

95. A cash drawer contains 160 bills, all 10s and 50s. If the total value of the 10s and 50s is $1,760. How many of each type of bill are in the drawer?

96. You want to make 28 grams of protein snack mix made with peanuts and granola. Peanuts contain 7 grams of protein per ounce and granola contain 3 grams of protein per ounce. How many ounces of granola should you use for 1 ounce of peanuts?

97. The length of a rectangular prism is quadrupled, the width is doubled, and the height is cut in half. If V is the volume of the rectangular prism before the modification, express the volume after the modification in terms of V.

98. A car rental has CD players in 85% of its cars. The CD players are randomly distributed throughout the fleet of cars. If a person rents 4 cars, what is the probability that at least 3 of them will have CD players?

99. Jacob’s hourly wage is 4 times as much as Noah. When Jacob got a raise of 2 dollars, Noah accepted a new position that pays him 2 dollars less per hour. Jacob now earns 5 times as much money as Noah. How much money do they make per hour after Jacob got the raise?

You own a catering business that makes specialty cakes. Your company has decided to create three types of cakes. To create these cakes, it takes a team that consists of a decorator, a baker, and a design consultant. Cake A takes the decorator 9 hours, the baker 6 hours, and the design consultant 1 hour to complete. Cake B takes the decorator 10 hours, the baker 4 hours, and the design consultant 2 hours. Cake C takes the decorator 12 hours, the baker 4 hour, and the design consultant 1 hour. Without hiring additional employees, there are 398 decorator hours available, 164 baker hours available, and 58 design consultant hours available. How many of each type of cake can be created?

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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).

✅ Down the Rabbit Hole: The Math That Helps the James Webb Space Telescope Sit Steady in Space

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.”

✅ Dive In: The Math Behind Our Current Theory of Human Color Perception Is Wrong

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.

✅ Stay Curious: How to Paint a Room Using Math

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.

✅ The Latest: Mathematicians Discovered Something Mind-Blowing About the Number 15

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.

✅ Miracles Happen: Mathematicians Finally Make a Breakthrough on the Ramsey Number

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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Frequently Asked Questions about Khan Academy and Math Worksheets

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15 Challenging Geometry Problems and Their Step-by-Step Solutions

• Author: Noreen Niazi
• Last Updated on: August 22, 2023

Introduction to Geometry Problems

The area of mathematics known as geometry is concerned with the study of the positions, dimensions, and shapes of objects.Geometry has applications in various fields, such as engineering, architecture, and physics. Geometry problems are among the most challenging and exciting problems in mathematics. Understanding and mastering geometry problems is essential for anyone who wants to pursue a career in any field requiring a good understanding of geometry.

Importance of Practicing Geometry Problems

Practicing geometry problems is essential for anyone who wants to master geometry. Geometry problems require a good understanding of the concepts, formulas, and theorems. By practicing geometry problems, you will develop a deep understanding of the concepts and the formulas.

You will also be able to identify the issues and the strategies to solve them. Practicing geometry problems will also help you to improve your problem-solving skills, which will be helpful in other areas of your life.

Types of Geometry Problems

There are several types of geometry problems. Some of the common types of geometry problems include:

• Congruence problems: These problems involve proving that two or more shapes are congruent.
• Similarity problems: These problems involve proving that two or more shapes are similar.
• Area and perimeter problems: These problems involve finding the area and perimeter of various shapes.
• Volume and surface area problems: These problems involve finding the volume and surface area of various shapes.
• Coordinate geometry problems: These problems involve finding the coordinates of various points on a graph.

Strategies for Solving Geometry Problems

To solve geometry problems, you must understand the concepts, formulas, and theorems well. You also need to have a systematic approach to solving problems. Some of the strategies for solving geometry problems include:

• Read the problem carefully: You must read the situation carefully and understand what is required.
• Draw a diagram: You need to draw a diagram representing the problem. This will help you to visualize the problem and identify the relationships between the shapes.
• Identify the type of problem: You need to identify the problem type and the applicable formulas and theorems.
• Solve the problem step by step: You need to solve the problem step by step, showing all your work.
• Check your answer: You must check it to ensure it is correct.

Common Geometry Formulas and Theorems

To solve geometry problems, you must understand the standard formulas and theorems well. Some of the common procedures and theorems include:

• Area of a square: side × side.
• Pythagoras theorem: a² + b² = c², where a and b are the lengths of the two sides of a right-angled triangle, and c is the hypotenuse length.
• Area of a rectangle: length × breadth.
• Circumference of a circle : 2 × π × radius.
• Area of a triangle : ½ × base × height.
• Congruent triangles theorem: Triangles are congruent if they have the same shape and size.
• Area of a circle: π × radius².
• Similar triangles theorem: Triangles are similar if they have the same shape but different sizes.

Problem 1: Lets the length of three sides of triangle be 3 cm, 4 cm, and 5 cm. Calculate the area of a right-angled triangle.

Using the Pythagoras theorem:

$$a² + b² = c²$$

where a = 3 cm, b = 4 cm, and c = 5 cm.

$$3² + 4² = 5²$$

$$9 + 16 = 25$$

Therefore, $$c² = 25$$, and $$c = √25 = 5 cm$$.

• The area of the triangle = $$½ × \text{base} × \text{height}$$ 

$$= ½ × 3 cm × 4 cm $$

$$= 6 cm².$$

Problem 2:If the length of each side of an equilateral triangle is 10 cm then calculate its perimeter.

As the perimeter of an equilateral triangle = $$3 × side length.$$

• Therefore, the perimeter of the triangle $$= 3 × 10 cm = 30 cm.$$

Problem 3: If cylinder has 4cm radius and 10 cm height then what is the volume of a cylinder.

The volume of a cylinder = $$π × radius² × height.$$

• Therefore, the volume of the cylinder $$= π × 4² × 10 cm = 160π cm³$$.

Problem 4: If radius of a circle is given by 5cm and central angle 60° then what is the area of sector of a circle.

The area of a sector of a circle $$= (central angle ÷ 360°) × π × radius².$$

• Therefore, the area of the sector $$= (60° ÷ 360°) × π × 5² c = 4.36 cm².$$

Problem 5: Find the hypotenuse of right-angled triangle, if its other two sides are of 8 cm and 15 cm.

Using the Pythagoras theorem :

Where a = 8 cm, b = 15 cm , and c is the hypotenuse length.

$$8² + 15² = c²$$

$$64 + 225 = c²$$

• Therefore, $$c² = 289,$$ and $$c = √289 = 17 cm.$$

Problem 6: If two parallel sides of trapezium are of length 5 cm and 10 cm and height 8 cm. Calculate the area of a trapezium.

The area of a trapezium = $$½ × (sum of parallel sides) × height.$$

• Therefore, the area of the trapezium $$= ½ × (5 cm + 10 cm) × 8 cm = 60 cm².$$

Problem 7: Radius and height of cone is given by 6cm and 12 cm respectively. Calculate its volume.

The volume of a cone $$= ⅓ × π × radius² × height.$$

• Therefore, the volume of the cone $$= ⅓ × π × 6² × 12 cm³ = 452.39 cm³.$$

Problem 8:What is the length of side of square if its area is 64 cm².

The area of a square $$= side × side.$$

• Therefore, $$side = √64 cm = 8 cm.$$

Problem 9: If length rectangle is 10cm and breadth is 6cm. Calculate its diagonal.

Where $$a = 10 cm$$, $$b = 6 cm$$, and c is the diagonal length.

$$10² + 6² = c²$$

$$100 + 36 = c²$$

• Therefore, $$c² = 136,$$ and $$c = √136 cm = 11.66 cm.$$

Problem 10: If one side of regular hexagon is of 8cm then what is the area of a regular hexagon.

The area of a regular hexagon $$= 6 × (side length)² × (√3 ÷ 4).$$

• Therefore, the area of the hexagon $$= 6 × 8² × (√3 ÷ 4) cm² = 96√3 cm².$$

Problem 11: If radius of sphere is 7 cm, then what is its volume.

The volume of a sphere = $$⅔ × π × radius³.$$

• Therefore, the volume of the sphere $$= ⅔ × π × 7³ cm³ = 1436.76 cm³.$$

Problem 12: Find the hypotenuse length of a right-angled triangle with sides of 6 cm and 8 cm.

Where a = 6 cm, b = 8 cm, and c is the hypotenuse length.

$$6² + 8² = c²$$

$$36 + 64 = c²$$

Therefore, $$c² = 100,$$ and $$c = √100 cm = 10 cm.$$

Problem 13: Find the area of a rhombus with 12 cm and 16 cm diagonals.

The area of a rhombus = (diagonal 1 × diagonal 2) ÷ 2.

• Therefore, the area of the rhombus = (12 cm × 16 cm) ÷ 2 = 96 cm².

Problem 14: If radius and central angle of circle is 4cm and 45° respectively then what is the length oof arc of circle.

The length of the arc of a circle = (central angle ÷ 360°) × 2 × π × radius.

• Therefore, the length of the arc = (45° ÷ 360°) × 2 × π × 4 cm

Problem 15: Find the length of the side of a regular octagon with the radius of the inscribed circle measuring 4 cm.

The length of the side of a regular octagon = (radius of the inscribed circle) × √2.

Therefore, the length of the side of the octagon = 4 cm × √2 

Online Resources for Geometry Practice Problems

There are several online resources that you can use to practice geometry problems. Some of the popular online resources include:

• Khan Academy : On the free online learning platform Khan Academy, you may find practise questions and video lectures on a variety of subjects, including geometry.
• Mathway : Mathway is an online tool that can solve various math problems, including geometry problems.
• IXL :IXL is a website that provides practise questions and tests on a variety of subjects, including geometry.

Q: What is geometry?

A: Geometry is the branch of mathematics that studies objects’ shapes, sizes, and positions.

Q: Why is practicing geometry problems significant?

A: Practicing geometry problems is essential for anyone who wants to master geometry. Geometry problems require a good understanding of the concepts, formulas, and theorems. By practicing geometry problems, you will develop a deep understanding of the concepts and the formulas.

Q: What are some standard geometry formulas and theorems?

A: Some of the standard geometry formulas and theorems include the Pythagoras theorem, area of a triangle, area of a square, area of a rectangle, area of a circle, circumference of a circle, congruent triangles theorem, and similar triangles theorem.

Geometry problems are among the most challenging and exciting problems in mathematics. Understanding and mastering geometry problems is essential for anyone who wants to pursue a career in any field requiring a good understanding of geometry. By practicing geometry problems and using the strategies and formulas discussed in this article, you can master geometry and improve your problem-solving skills.

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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 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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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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