**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.

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. There are two Math sections of the SAT: section 3, which is 25 minutes long and does not allow you to use a calculator, and section 4, which is 55 minutes long and does allow a calculator.

Every SAT covers the following math material:

**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**. A wide variety of topics, including geometry, trigonometry, radians and the unit circle, and complex numbers.

There are 58 math questions in total across the two sections. Most of the questions are multiple-choice, but 12 questions are grid-in questions where you have to write out the answer yourself:

These grid-in questions come at the end of each 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**.

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)****. **

AND…

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*

And

*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:

*3p = 4r*

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:

*4r / p = 3*

And finally divide by 4, which gives us:

*r/p = ¾*

**CHOICE B**

### 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

*y = 3*

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*valueSo, *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. **

### 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.

### 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*.

**CHOICE A**.

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 = 3x+6*

*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…

**20**.

### 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**.

Ready to try more hard problems on your own? Download our free quiz to try 20 more of the hardest ever (real) SAT problems.

### 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**.

### 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**.

We’ve tutored thousands of students and used that experience to assemble a list of 20 more problems that students frequently miss. Can you answer them correctly? Download the quiz now to find out!

### 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.

B has 60 gears.

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:

*(a/4)^2*

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**.

### 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**.

## Next steps

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 is 528, 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**

Emily graduated *summa cum laude* from Princeton University and holds an MA from the University of Notre Dame. 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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