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 25 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!
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 don’t do well on the first set of 22 questions, the second set will be easier–but your overall math score will be negatively affected. (Sounds confusing? See our guide to adaptive testing on the SAT here.)
You can use your calculator on both sections of the test, as well as the built-in DESMOS tool in the Bluebook app. (New to DESMOS? Here’s how to use it to save time on the SAT.)
The SAT covers the following math material:
Algebra: 35% of test. Linear equations and inequalities and their graphs and systems.
Problem Solving and Data Analysis: 15% of test. Ratios, proportions, percentages, and units; analyzing graphical data, probabilities, and statistics.
Advanced Math: 35% of test. Identifying and creating equivalent expressions; quadratic and nonlinear equations/functions and their graphs.
Geometry and Trigonometry: 10% of test. Area, volume, and angles, working with shapes including triangles and circles.
You’ll see both multiple-choice and open-ended questions on the test.
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 high score—and it is! It’s a very high score.
MIT
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.
Source: IPEDS.Note: Some of these schools are test-optional, and others don’t release test scorespublicly.
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 44 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.
You might notice the differences in formatting across some of the questions below. That’s because some are from the latest, digital SAT, and some are taken from the paper format of the test. All the example questions are useful for the digital test, though: the math hasn’t changed!
SAT Problem #1: Substitution solutions
In the given equation, c is a positive constant. Which of the following is one of the solutions to the given equation?
You might be tempted to try and solve for c in terms of x, or to try to get rid of the square root signs. But don’t fall for that! You’ll lose a lot of time.
Look at the answer choices for a clue: you might not have to get rid of the square roots signs at all. And since three of the four options include c2, you might not even need to get rid of the exponent.
Instead, substitution can help you deal with complicated equations like this. It makes them simple enough to work with, without all the hassle of actually simplifying them.
Before we do that, let’s make the equation as simple as possible by combining like terms. Take this thorny part of the equation:
…and try to get it all on one side of the equation by subtraction.
We can make things even clearer by combining the two fractions:
Now we’re ready for the substitution part. Let’s take
…and label it a.
Now our equation becomes a whole lot simpler:
We can simplify the left side to just a, giving us a = 39.
Then, we substitute the value of a back in:
Squaring both sides gives us :
Now we just need to solve for x. Adding c2 to both sides of the equation gives us:
Then, taking the square root results in:
Of these two solutions, one is among the answer choices:
CHOICE D.
SAT Problem #2: Percentages with Multiple Variables
The number a is 110% greater than the number b. The number b is 90% less than 47. What is the value of a?
This question can seem confusing, because there’s just so much going on: we’re dealing with two different unknowns and a couple of percentages.
The key is to put all of these relationships into just one equation — and get rid of the variable we’re not interested in.
What we’re ultimately looking for is the value of a. So let’s set up an equation that will give us a.
We know that a is 110% greater than b. In other words, the value of a is 100% the value of b, plus another 110% of b.
That makes 100% + 100% + 10% = 210% of b.
a = 2.1b
Remember, we don’t care what the value of b actually is. It’s just a stepping stone on the way to finding a.
So, we can substitute the value of b into the equation above.
That value is 90% less than 47. That’s another way of saying that b is 10% of 47.
Think about it — if 47 is 100%, and b is 90% less than that, then b is going to be 10% of 47.
We can write that as b = 0.10(47)
That works out to b = 4.7
Substituting that back into our equation for a, we get:
a = 2.1(4.7) = 9.87
So our final answer is:
a = 9.87
If you want, you can skip the step of solving for b and plug its value straight into the equation for a:
a = 2.1(0.10(47)) = 9.87
It’s up to you. If you tend to lose points from sloppy mistakes or going too fast, then writing out each step can be helpful. On the other hand, if time management is your big challenge on the SAT, then saving time by cutting down on steps might help. You know your own pitfalls and strong points best!
SAT Problem #3: Algebra Dressed Up as Geometry
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?
A. 2/5
B. 3/4
C. 4/3
D. 5/2
Source:Former Test #8, Section 4
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 becomesr = 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 #4: Exponent Rules
A. 212
B. 44
C. 82
D. The value cannot be determined from the information given.
Source: Former Test #1, section 3
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 23 equal to 8?
So if we re-write the problem, plugging in 23 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 #5: Intersecting Graphs
y = 18
y = -3(x -18)2 + 15
If the given equations are graphed in the xy-plane, at how many points do the graphs of the equations intersect?
A point where the graphs of two equations intersect is a point where those equations are equal to each other. (If that doesn’t ring a bell, think of it visually: if you draw the graphs of two equations, then any point where they cross each other has to be a point where their solution (x,y) is the same.)
So what we want to know is: if we look at this as a system of two equations, how many solutions does that system have?
There are a couple of solutions to this problem.
Actually, there are three solutions — because you can also just throw both equations into DESMOS and see the answer. That’s not going to help you understand the math, though. And it’s not always the best solution, because sometimes plugging complicated equations into DESMOS can be more time-consuming than just solving the problem.
So, back to math.
Strategy #1: One solution is to visualize (or sketch) the graphs.
y = 18 is going to be a horizontal line at, well, y = 18, for any value of x.
Does the second equation ever intersect with that line? To do that, it would have to have values below y = 18 and values above y = 18.
Looking at the second equation, we can see two things about its graph:
First, it’s going to be a parabola that opens downward. That’s because the coefficient of x2 is -3, which is negative.
Second, its vertex will be at (18, 15). That’s because 18 is the value of x for which the part of the equation involving x equals 0. When x = 18, then y = 15.
Since the parabola opens downward, the vertex is the maximum of the parabola. And if the maximum occurs at y = 15…
.…then y will never equal 18. That means the two graphs will never intersect.
CHOICE D is our answer.
Strategy #2: If you’re more of an algebra person than a visual thinker, you can also solve this problem algebraically, without visualizing any graphs.
We’re looking for points where the graphs of these equations intersect — which, remember, is another way of saying points where these equations equal one another.
Setting the two equations equal to one another gives us:
18 = -3(x-18)2 + 15
Subtract 15 from both sides of the equation:
3 = -3(x-18)2
Now simplify by dividing both sides by -3:
-1 = (x-18)2
The square of a real number can’t be negative, so this system of equations has no real solutions.
That means the graphs don’t intersect at any point:
CHOICE D.
SAT Problem #6: Functions of Functions
The complete graph of the function f and a table of values for the function g are shown above. The maximum value of f is k. What is the value of g(k)?
A. 7
B. 6
C. 3
D. 0
Source: Former Test #8, Section 4
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 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.
SAT Problem #7: Angles in Circles
Point P is the center of the circle in the figure above. What is the value of x?
Source: Former Test #5, section 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 #8: Saving Time with Polynomial Division
A. -16
B. -3
C. 3
D. 16
Source: Former Test #3, Section 3
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 (24x2 + 25x – 47) is divided byax – 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 #9: Combinations of Shapes
A circle has center G, and points M and N lie on the circle. Line segments MH and NH are tangent to the circle at points M and N, respectively. If the radius of the circle is 168 millimeters and the perimeter of quadrilateral GMHN is 3,856 millimeters, what is the distance, in millimeters, between points G and H?
For more complicated geometry questions, it’s usually a good idea to sketch the situation and fill in the information you have.
The blue line is the radius of the circle, and the orange line is the one we want to determine: the distance between point G and point H.
What else do we know? We have two more facts, based on the question:
1: Line GM and line GNare equivalent to the radius of the circle, which is 168 mm. That’s because points M and N are on the circle, and G is its center.
2: The angles GMH and GNH are both right angles. We’re told that the line segments MH and NH are tangent to the circle. The angle formed at the intersection of a line from the center of the circle to a point on the circle, and a line tangent to the circle through that point, is always a right angle.
That means that the quadrilateral GMHN is made up of two right triangles — GMH and GNH.
And these right triangles are identical, because they share their right angle and two sides (GM and GN are both equal to the circle’s radius, and GH is obviously the same in both triangles).
That means their third side must also be the same: MH = HN
So, how does that get us closer to figuring out GH?
Looking back at the question, we also know that the perimeter of GMHN is 3,856 millimeters. We can write that as an equation:
3,856 =GM + MH + HN + GN
We can substitute in the information we already know: GM and GN are the same as the circle’s radius, 168 millimeters. And we also know that MH = HN, so we can substitute one of them with the other to simplify the equation.
3,856 =168 + MH + MH + 168
That turns into:
3,856 =336 + 2MH
Subtracting 336 from both sides, we get:
3,520 = 2MH
Then divide by 2:
1,760 = MH
Here’s everything we know so far:
Now, the final step: to solve for GH, we can use the Pythagorean theorem. After all, GMH is a right triangle.
GM2 + MH2 = GH2
Add in what we know:
1682 + 1,7602 = GH2
Solving for GH gives us ±1,768. We’re talking about a length, so the answer must be positive. That means GH = 1,768 millimeters.
CHOICE D is our answer.
SAT Problem #10: Efficient Algebra
Based on the equation above, what is the value of 3x – 2?
A. -4
B. -4/5
C. -2/3
D. 4
Source: Former Test #10, Section 4
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 #11: Value of a Constant
If (ax + 2)(bx + 7) = 15x2 + cx + 14 for all values of x, and a + b = 8, what are the two possible values for c?
A. 3 and 5
B. 6 and 35
C. 10 and 21
D. 31 and 41
Source: Former Test #1, Section 3
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 abx2 is going to be equal to 15x2.
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) = 15x2 + 21x + 10x + 14.
Which simplifies to…
15x2 + 31x + 14.
Which means c = 31.
31 only appears once in our answer choices, so the answer must be CHOICE D.
SAT Problem #12: Translating Word Problems
For an electric field passing through a flat surface perpendicular to it, the electric flux of the electric field through the surface is the product of the electric field’s strength and the area of the surface. A certain flat surface consists of two adjacent squares, where the side length, in meters, of the larger square is 3 times the side length, in meters, of the smaller square. An electric field with strength 29.00 volts per meter passes uniformly through this surface, which is perpendicular to the electric field. If the total electric flux of the electric field through this surface is 4,640 volts • meters, what is the electric flux, in volts • meters, of the electric field through the larger square?
First things first: you don’t need to know any physics for this question! Don’t panic if some of phrases like “electric flux” are unfamiliar. This is just a question about area and ratios.
The most important skill with wordy questions like this is being able to translate the text into numbers, equations, and visuals.
We’re told that “the electric flux of the electric field through the surface is the product of the electric field’s strength and the area of the surface.” We can turn each of those long phrases into a one-letter variable — f for flux, s for strength and a for area.
f = s • a
Now let’s visualize the situation. We’ll let x be the side length of the smaller square.
Don’t lose sight of what we’re after: what we want to know is the size of the flux through the larger square. And what we have is the flux through the whole surface.
Since we’re given the electric strength of the field, you might be tempted to plus that into our formula, f = s • a, and try to solve for f.
That won’t work, though: we can only figure out the area in terms of x, so we’ll never be able to solve for f that way.
To get rid of that x, we’ll need to use the ratio of the two areas — the one we have the flux for (the total area), and the one we want the flux for (the area of the larger square).
The area of the smaller square is x2. The area of the bigger square is (3x)2, which is 9x2.
So the total area is x2 + 9x2 = 10x2.
You can plug this into f = s • a to get:
4,640 = s(10x2)
Time to plus in s =29.00, solve for x, and then put that back into the formula for the larger square — right?
You could do that. But you would lose a lot of valuable time.
In fact, the value of s is just distracting extra information — you don’t need it to solve the problem.
We know that the flux is directly proportional to the area. We also know that because the area of the smaller square is x2, and the area of the bigger square is 9x2, the bigger square accounts for 9/10 of the total area.
That means that the bigger square also gets 9/10 of the total electric flux.
The total flux is 4,640 volts • meters, so the flux through the bigger square is
• 4,640 = 0.9 • 4,640 = 4,176 volts • meters.
On the SAT, where you have a very short amount of time to spend on each question, saving even a minute can make a big difference. So before you start plugging in numbers, ask yourself what information you really need!
SAT Problem #13: Systems with No Solutions
-3x + y = 6
ax + 2y = 4
In the system of equations above, a is a constant. For which of the following values of a does the system have no solution?
A. -6
B. -3
C. 3
D. 6
Source: Former Test #10, Section 3
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.
The first step is to put both of these equations in slope-intercept form. We’d end up with:
y = 3x+6
y = –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 – = 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.
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SAT Problem #14: Systems with Infinite Solutions
In the system of equations below, a and c are constants.
x + y =
ax + y = c
If the system of equations has an infinite number of solutions (x,y), what is the value of a?
This question is kind of the opposite of the previous one. Instead of a system of equations with no solutions, now we’re looking at one with infinite solutions.
This means that the graphs of the two equations intersect an infinite number of times…
…which means that they’re the same line. The graphs are basically lying on top of each other, which is why, at any point imaginable, they have the same value for x and y.
To solve questions like this, 1. get both equations into the same format and then 2. solve for the variable you’re looking for.
In this problem, Step 1 has already been done for us: both equations are already in the form of kx + ly = m.
And because the equations must equal each other, we can use the ratios of the coefficients we already have to work out the one that’s missing (a).
In the first equation, the coeffiecient of y is . In the second equation, it’s 1 (which isn’t written).
What’s the ratio of the coefficient of y in the first equation to the coefficient of y in the second?
Just put them in like terms: 1 = , so the coefficient of y in the second equation is 3 times the coefficient of y in the first equation.
That same ratio must be true for the coefficient of x in both equations. (Why? Because these equations equal each other — that’s what it means for a system of equations to have infinitely many solutions.)
So, a must be 3 times the coefficient of x in the first equation, which is . That means that:
a = 3() =
CHOICE D.
SAT Problem #15: Difference of Squares
If u + t = 5 and u – t = 2, what is the value of (u – t)(u2 – t2)?
Source: Former Test #10, Section 3
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) = x2 – xy + xy – y2.
Which means…
(x+y)(x-y) = x2 – y2.
And doesn’t that look awfully familiar to… u2 – t2?
In fact, we can now replace u2 – t2 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 #16: Difference of Squares and Value of a Constant
The expression x2 – 2 can be rewritten as (x – k)(x + k), where k is a positive constant. What is the value of k?
A. 2
B. 6
C. 2
D. 6
Source: Former Test #7, section 3
What makes this question confusing is that students often get thrown off by the repetition of the
They forget that when the gets factored out of the parentheses like that, it means it’s going to apply to the whole equation: both the x2 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 x2for each choice.
That will give us the following four choices:
()(x2 – 4)
()(x2 – 36)
()(x2– 2)
()(x2 – 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 to both the x2 AND the k!
If we multiply that ⅓ through, the choices suddenly look like this…
()(x2) – ()
()(x2) – (12)
()(x2) – ()
()(x2) – (2)
. . . and so the correct answer is actually the fourth choice, CHOICE D.
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SAT Problem #17: Equation for a Circle
A circle in the xy-plane has equation (x + 3)2 + (y – 1)2 = 25. Which of the following points does NOT lie in the interior of the circle?
A. (-7, 3)
B. (-3, 1)
C. (0, 0)
D. (3, 2)
Source: Former Test #7, section 4
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 = r2
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 r2, 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 #18: Possible Values of a Constant
One of the factors of 2x3 + 42x2 + 208x is (x + b), where b is a positive constant. What is the smallest possible value of b?
It can be easy to feel overwhelmed by an open-ended question like this, with no answer choices to use as hints. But your first question is always:
What do I need to end up with?
And then:
What do I already know?
We want to know the value of b, and we know that one of the factors of this equation is going to be in the form (x + b).
So let’s find the factors of the equation, see which one is in the form we need, and then take the value of b from there.
At first glance, this equation looks hard to factor. So let’s simplify it as much as we can. 2x goes into all the coefficients here:
2x3 + 42x2 + 208x becomes 2x(x2+ 21x+ 104)
Okay, that didn’t allow us to simply too much–but it’s enough! Because now, we can factorize the equation between the parentheses:
2x(x2+ 21x+ 104) becomes 2x(x + 8)(x + 13)
And there we go: we have all the factors of the equation — 2, x, (x + 8), and (x + 13)— and two of these factors are already in the form (x + b).
(x + b) must be either (x + 8) or (x + 13), making b either 8 or 13.
So which one is it? The question asks for the smallest possible value of b, so b must be 8.
SAT Problem #19: Circle on a Graph
In the xy-plane, the graph of 2x2 – 6x + 2y2 + 2y = 45 is a circle. What is the radius of the circle?
A. 5
B. 6.5
C. √40
D. √50
Source: Former Test #6, section 4
More circles! Let’s recall how the equation for a circle looked. It’s…
(x – h)2 + (y – k)2 = r2
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 –
2x2 – 6x + 2y2 + 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 x2 – 3x + y2 + 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 x2 – 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 x2 – 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:
x2 – 3x + 2.25 + y2 + 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 0.52 or 0.25.
Again, we have to add this number to both sides of the equation. Now we’ve got:
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 = r2. Fortunately this works out really nicely, since 25 is just 52. The radius must be 5, CHOICE A.
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SAT Problem #20: Ratios, Step by Step
A gear ration r:s is the ratio of the number of teeth of two connected gears. The ratio of the number of revolutions per minute (rpm) of two gear wheels is s:r. In the diagram below, Gear A is turned by a motor. The turning of Gear A causes Gears B and C to turn as well.
If Gear A is rotated by the motor at a rate of 100 rpm, what is the number of revolutions per minute for Gear C?
A. 50
B. 110
C. 200
D. 1,000
Source: Former Test #6, section 4
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 #21: Area and Perimeter
The surface areas of a cube is 6()2, where a is a positive constant. Which of the following gives the perimeter of one face of the cube?
A.
B. a
C. 4a
D. 6a
Source: Former Test #5, section 4
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()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:
()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 x2, 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, ()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 .
And if we replace the x with , we find that each side of the square is equivalent to .
Which makes finding the perimeter of this square quite simple, because it has four sides.
So we merely add the four sides together:
+ + + . . .
which equals a.
Which in this case is CHOICEB.
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SAT Problem #22: Multiple Percentages
A sample of a certain alloy has a total mass of 50.0 grams and is 50.0% silicon by mass. The sample was created by combining two pieces of different alloys. The first piece was 30.0% silicon by mass and the second piece was 80.0% silicon by mass. What was the mass, in grams, of the silicon in the second piece?
A. 9.0
B. 16.0
C. 20.0
D. 30.0
First things first: let’s translate all these words into numbers and variables we can work with.
Let’s call the mass of the first piece a and the mass of the second piece b.
The answer to this question is going to be the mass of silicon in the second piece — which is equal to the value of b times the mass percentage of silicon in the second piece.
We know that the total mass of the combined pieces is 50.0 grams, and 50.0% of that is silicon.
That means two things:
1: There was a total of 0.500(50.0) = 25.0 grams of silicon in both pieces combined.
And 2: The mass of both pieces combined is 50.0 grams.
With those two pieces of information, we can set up two equations.
1:a + b = 50.0
That reflects the total mass of both pieces.
2:0.300a + 0.800b = 25.0
That reflects the total mass of silicon in both pieces. After all, we have 25.0 grams of silicon in total, and we know the mass percentage of silicon in both pieces. The mass percentage times the total mass of each piece, combined, has to equal the total amount of silicon.
Now we can solve this system of equations for b— that’s the information we’re after.
We can rewrite equation 1 by subtracting a from both sides, which gives us a = 50.0 – b
Now we can substitute that into equation 2:
0.300(50 – b) + 0.800b = 25.0
Getting rid of the parentheses:
15.0 – 0.300b + 0.800b = 25.0
Subtracting 15.0 from both sides and adding like terms:
0.500b = 10.0
Dividing both sides by 0.500:
b = 20.0
The mass of the second piece is 20.0 grams.
If you’re feeling hasty, you might select answer choice D and move on to the next question. But be careful!
The questions asks for the mass of silicon in the second piece, not the total mass.
To get that answer, we need to take the mass we found and multiply it by the mass percentage of silicon, which the question tells us is 80.0%.
amount of silicon in the piece = (0.800)20.0 = 16.0
We end up with 16.0 grams — so CHOICE C is the answer we’re looking for.
SAT Problem #23: Relationships of Variables
3x + b = 5x – 7
3y + c = 5y – 7
In the equations above, b and c are constants. If b is c minus , which of the following is true?
A. x is y minus .
B. x is y minus .
C. x is y minus 1.
D. x is y plus .
Source: Former Test #3, section 4
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 isx = y – ¼, CHOICE A.
SAT Problem #24: Length and Surface Area
Two identical rectangular prisms each have a height of 90 centimeters (cm). The base of each prism is a square, and the surface area of each prism is K cm2. If the prisms are glued together along a square base, the resulting prism has a surface area of 92/47K cm2. What is the side length, in cm, of each square base?
Since this question mentions prisms, you might think its about volume. But don’t be fooled — this is a simple area question.
Let’s visualize what we’re dealing with: two rectangular prisms with the same height, 90 cm.
Then we glue the two prisms together along a square base, resulting in…
…a larger rectangular prism, this one with side length 90 + 90 = 180 cm.
What we want to know is the side length of one square base of this prism (which is the same as for the two smaller prisms). We can call the side length of each square base x.
What we already know is 1) the longer side length of the rectangular faces of the prisms (90cm or 180 cm), 2) the surface area of each smaller prism (K cm2) , and 3) the surface area of the larger prism (92/47 K cm2).
(Why isn’t the surface area of the larger prism just 2K cm2? Because, even though it’s made up up the two smaller prisms stuck together, we’ve lost some surface area by sticking them together.)
To get closer to x, think about what goes into the surface area of a smaller prism. Take a look at the sketch above: it’s equal to the area of two square bases (one at each end) plus four rectangular faces.
Let’s set up some equations to express the situation. Call the area of a square base s and the area of a rectangular face r. The total surface area is K, giving us K = 2s + 4r
Since each square base is, obviously, square, both sides are equal to x. That gives us s = x • x = x2. (We’ll leave out the cm2 for now, since it’s the unit for all the areas we’ll be working with.)
What about the rectangular faces? Those have one 90cm side and one side that’s shared with the square bases. So r = 90x.
Plugging this back into K = 2s + 4r:
K = 2x2 + 4(90x) = 2x2 + 360x
Now we have the area of each smaller prism in terms of x.
Next step: the area of the larger, glued-together prism.
The larger prism’s surface area is equal to that of the two smaller prisms put together — minus the two square bases that we lost by gluing them together. In equation form, that becomes:
92/47 K = 2K – 2x2
(Remember to subtract the area of two bases from the total, not just one!)
If we subtract 92/47 K from both sides, we get:
0 = 2/47 K – 2x2
Now we can simplify by multiplying both sides of the equation by 47/2. That gives us:
0 = K – 47x2
So:
K = 47x2
Almost there! We can plug this information back into K = 2x2 + 360x, which we worked out above.
47x2= 2x2 + 360x
Subtract 2x2from both sides:
45x2= 360x
Divide both sides by x:
45x = 360
Now divide both sides by 45:
x = 8 cm
And that’s our final answer.
SAT Problem #25: Polynomial Remainders
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.
Source: Former Test #1, section 4
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 = x2 + 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. A good SAT score doesn’t have to mean acing the test.
Harvard University
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!
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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