 There are two sections in the SAT covering math, and they are the third and fourth sections of the test overall. The No-Calculator section is 25 minutes and has 20 total questions, while the Calculator section is 55 minutes and has 38 total questions. Each section is scored out of 400 points by converting your raw score (how many questions you got right) to a 400 point-scale. The conversion is not the same each time, however, and depends on the test you take. Each section score is summed to give you your total math score. That means that the maximum score you can get on the math portion of the SAT is 800. The median Math total score (the 50th percentile score) usually hovers between 520 and 530.

# Most Difficult SAT Math Topics

There are four official content areas that are tested throughout both sections of the SAT Math portion: Heart of Algebra, Problem Solving and Data Analysis, Passport to Advanced Math, and Additional Topics in Math. These content areas span a wide variety of topics, some easier than others. We collected a list of the ten hardest SAT Math Topics in Part 1 of this article series here. If you haven’t read that guide yet, go through it before continuing on this article because it will give you a good roadmap and understanding of these different topics and what to look out for when solving those types of problems.

Then, you can continue on this article and try your hand at answering practice problems on some of the most difficult math topics on the SAT.

# How To Approach These Practice Problems

For each of these problems, try to answer them on your own before reading the explanation. Each answer will be at the end of the explanation, so there’s no chance you can accidentally view the answer before you’ve tried it yourself! It’s important to do this because students often end up relying on answer keys when going over hard questions. While answer keys are a good tool to understand the underlying concept and one way to approach a problem, the struggle of figuring out a problem is also part of the learning.

I’ve found that students who try their best to figure out the problem first, without looking at the answer, actually understand the concept better (because they actually have to put that understanding to the test) and find out different strategies that work for them in figuring out questions that they may not immediately know how to solve. So, don’t give up quickly when solving these problems! They are on the more difficult end of the spectrum of questions you will be asked. Approaching these with just your pencil and paper (and calculator) will prepare you best for the actual SAT when that’s all you’ll have with you.

# Practice Problems And Answer Explanations

## #1: Quadratic And Polynomial Equations

For this question, you’ll have to know what factors of a function are. “Factors” are in the form “x-a” where a is the value that solves f(a) = 0. Many people say that they “factor” a function in order to find the solution.

In order to find the solutions to a function, you must set the equation to zero and factor accordingly. For example, if you were factoring the function f(x) = x^2 – 4x – 5, you would set it to zero, and find the factors: (x-5) * (x+1). The solutions are thus 5 and -1. When you solve out f(5) and f(-1), they will equal to zero. In other words, f(5) = 0 and f(-1) = 0.

Now that you know this background, you can approach each answer choice and determine whether it’s true or false.

Answer choice A posits that x – 5 is a factor of p(x). Following our knowledge of factors, this would mean that p(5) = 0. The thing is that there is no way to conclude this based on the information given to us in the problem. Thus, this answer choice is not necessarily true.

Answer choice B similarly gives us a statement, which, following the previous logic, concludes that p(2) = 0. Again, there is no way to conclude this from the given information, so this answer choice is not necessarily true.

Answer choice C is basically the same steps as A and B. Following the logic gives us the statement of p(-2) = 0, which is not necessarily true based on the information given.

Thus, Answer choice D would be the answer. Students will probably reach it by process of elimination, but just to make sure, you should understand why it is the only true statement purely based on the information given.

Remember that a solution to a problem will present the equality, f(x) = 0. But what if you plug in an “x” value that is not a solution to the problem? Then the right-hand side will be a non-zero number that represents the remainder.

If you want to understand this concept more deeply, you can read up on the Polynomial Remainder Theorem. You can also try solving a couple of example functions and seeing what happens when you plug in solutions and non-solutions.

## #2: Trigonometry

In this problem, it will be helpful to draw a diagram. You can draw the actual building and car, and connect the distances between the two to form a triangle (the ground would be the triangle base, the building would be the height, and the top of the building to the car would be the hypotenuse.

Once you draw this diagram, label it with the information you’re given. The base of the triangle would be 120 feet and the height would be 350 feet.

The angle of depression is the angle you look down from a horizontal line to see something below. It’s basically the downward inclination of a line of sight. (Note: you should know what the angle of depression AND angle of elevation means!) In this case, the angle of depression is formed by the top-most angle (between the height and the hypotenuse).

Therefore, once you get this picture down, the problem becomes a fairly straightforward trigonometry problem. You are given two lengths and are meant to find the angle. Based on the angle, the two lengths you are given are the opposite and adjacent legs of the triangle.

Therefore, you can set up the equation tan(angle) = 120/350.

Solving for this using inverse tan (use your calculator) gives you 18.9 degrees. Remember to solve for this problem in degrees, not radians!

## #3: Coordinate Geometry

The easiest way to solve this problem with a calculator is to use the graphing function. When you graph it, you will see the curve below and can clearly tell that it only touches quadrants I, II, and IV. Therefore, the answer is quadrant IV because there is no point of the curve that touches that quadrant (and therefore, there will be no coordinates that satisfy the inequality).

To solve this without a calculator, you can draw a rough sketch of the parabola using the equation given and key points you can derive. First, expand the expression to y = x^2 – 6x + 8. Thus, you can tell that the graph will be an upwards parabola. Because y is greater than or equal to the right hand side, y will be the area above the parabola.

Then, you can use the equation to find some key points: the vertex, x-intercepts, and y-intercept.

To find the vertex, you can actually use the original form the equation. In the following form (h, k) will be the vertex.

Y = a(x-h)^2 + k

Therefore, the vertex will be (3, 1). If you want to use the expanded version, you can use the vertex formula: h = – b/2a where y = ax^2 +bx +c. You will get the same answer.

To find the y-intercept, you can set the original equation’s x’s to 0. When you do this, you will find that y = 8. Therefore, the y-intercept is (0, 8).

To find the x-intercepts, you can set the original equation’s y to 0. Using the expanded version of the equation will be helpful here. You can factor the equation 0 = x^2 – 6x + 8 to be 0 = (x-2) * (x-4). Therefore, the solutions are x = 2, 4. The x-intercepts will be (2, 0), and (4, 0).

Using these four points, you can see where the parabola will roughly cross each axis and you will get a graph similar to the one in Desmos. From this, you can easily see that the parabola and its inside crosses through all quadrants except for quadrant III. Therefore, there is no point in quadrant III that will satisfy the inequality.

## #4: 3-D Geometry

You should know that the formula for the surface area of the cube is 6a^2. If you didn’t know it, you should add it to your list of formulas to memorize. But if you forget it in the heat of the moment – don’t worry. You can also derive it using logic. The area of one face of the cube with sides a will be a^2. Since there are 6 faces, the surface area will be 6a^2.

Similarly, you can derive the perimeter of one face of the cube to be 4a (4 sides to a square with each side having a length of a).

The problem gives you the specific cube’s surface area equation, and you will notice that the only thing that makes it different from the standard formula is that the a is divided by 4. Thus, each side length will be a/4. Say, each side length S = a/4.

Therefore, the perimeter of once face of this specific cube is 4S. Putting this in terms of a, it will be 4 * (a/4). The 4’s cancel out and you’re left with just a.

## #5: Complex Numbers (Imaginary Numbers)

This problem is asking you to rewrite the equation so that its denominator does not include a complex number (this will be correct form). To do this, you must use the complex conjugate. This means that you multiply the numerator and denominator by the conjugate.

The conjugate of a complex number is obtained by changing the sign of its imaginary part. Therefore, the conjugate here will be 8 – 2i.

When you multiply the denominator by 8 – 2i, you will get the following:

64 – 16i +16i – 4i^2

64 – 4i^2

64 – 4 * (-1)

64 +4 = 68

When you multiply the numerator by 8 – 2i, you will get the following:

24 – 6i – 40i +10i^2

24 – 46i + 10(-1) = 14-46i.

Therefore, the final expression you will get is (14 – 46i) / 68. If you simplify this (divide the numerator and denominator by 2), you will get (7 – 23i) / 34.

## #6: Word Problems

The trickiest part of a word problem is setting it up. You have to figure out three things: 1) What are they asking you to solve?, 2) What data is given?, 3) How can you make sense of the data to answer the question?

So first, we determine that the question is asking us for the pears produced by Type B trees. Let’s name this variable “B.” B is the unknown we will be solving for.

Second, the data given to us is the following:

• Type A trees produced 20 percent more pears than Type B trees did
• Type A trees produced 144 pears

Let’s name the variable “A” as to how many pears Type A trees produced. So A = 144.

Now that we’ve identified the problem and the data, we have to figure out how to rearrange and manipulate the data in order to find the answer. In this case, we can set up a linear equation.

B + 0.20(B) = 144

We got this because if type B trees produced 20% more, then they would have produced the same amount as type A trees, which is 144.

Solving for this, we get:

1.2B = 144

B = 120

## #7: Ratios

We are given measurements in this problem that are in different units, so the first step we have to take is to standardize the units. You can either convert all the lengths into inches, meters, or micrometers, but, in this case, it will be easiest to work with micrometers. If working with inches or meters, there will be a lot of zeros / scientific notation, which will still give you the right answer but will make you more prone to errors.

Since the layer thickness is already given to us in micrometers, we just need to convert the 1 inch length of each layer. Using the conversion given to us in the problem, we find:

1 inch = 0.254 meters = 25400 micrometers (converting from meters to micrometers is a 1:10^6 ratio, so you would have to multiply 0.0245 *10^6 to get 25400)

Thus, we know that the total height of the cube is 25400 micrometers. Since each layer is 102 micrometers, we can divide 25400 by 102 to get how many layers the printer needs to create.

25400 / 102 = 249 layers.

Each layer takes 6 seconds, so it will take a total of 249 layers * 6 seconds = 1494 seconds to print the whole cube.

The answer choices are in minutes, so dividing 1494 seconds by 60 seconds/minute, will give us 24.9 minutes, which is approximately 25 minutes.

## #8: Matrices And Determinants To find the determinant of a matrix, you should use the following formula (add it to your study guide and memorize it for the test!):

In this problem’s case:

• a = 1
• b = 2
• c = -1
• d = 0

So, plugging it into the formula, we get, |A| = 1(0) – 2(-1) = 2.

## #9: Data Analysis And Statistics

To find the probability that those that recalled at least 1 dream were in Group Y, we need to find the following data:

• Total of those overall who recalled at least 1 dream
• Total of those in Group Y who recalled at least 1 dream

To find the total of those who recalled at least 1 dream, we can add the data in columns “1 to 4” and “5 or more” in the “Total” row.

39 +125 = 164 people who recalled at least 1 dream.

To find the total of those only in Group Y who recalled at least 1 dream, we can add to the data in columns“1 to 4” and “5 or more” in the “Group Y” row.

11 + 68 = 79 people in Group Y who recalled at least 1 dream.

The probability will be the number of people in Group Y who recalled at least 1 dream “out of” (divided by) the number of people in Total who recalled at least 1 dream, which is 79/164.

## #10: Probability

The current probability of drawing a red marble is 12/32. We have to add red marbles to the bag in order to make the probability ⅗.

Let’s say that the number of red marbles that you are adding to the bag is x.

So if x = 1, you add 1 red marble, and the probability of red marbles in the bag will be 13/33.

We can set up the following equation:

(12 + x) / (32 + x) = ⅗.

Adding the marble will increase both the numerator and denominator because it will be increasing the number of red marbles in the bag, as well as the total amount of marbles in the bag overall.

You can solve for the above equation using cross multiplication. Then, it is a simple solve for x to find your answer.

5(12 + x) = 3(32 + x)

60 + 5x = 96 + 3x

2x = 36

X = 18

If you’re having trouble setting up the equation, another way to solve this problem is to use the answer choices given. It will take slightly longer, but since there are only 5 possible answer choices, it’s possible to go through each one in a reasonable amount of time. In this way, you would just add the answer choice to the numerator and denominator of 12/32, and see which one will give you a fraction that simplifies to ⅗.

For example, to try answer choice A, you would do:

(12 + 13) / (32 + 13) = 25/45 = 5/9, which is not the right answer, so you can eliminate it.

Eventually, you can eliminate all but answer choice G.

# Tips For SAT Math

For tips on how to ace the SAT Math section, as well as how to approach difficult problems, check out our guides:

These will provide you with time-tested strategies and effective study habits to prepare you best for the Math section, as well as flesh out your understanding of the section and the types of questions it asks in general.

# Prepare For SAT Math With SoFlo Tutors

Struggling to understand all these math topics and solve our example problems? SoFlo is here to help. We will help you prepare for the SAT Math section, reach your target score, and get into your dream school. All of our tutors attend top universities around the nation and scored in the 99th percentile on their SAT. They will work one-on-one on you to identify and improve your weak areas and increase your chances of getting your target score. Our tutors schedule to sessions to your availability and our costs start at \$60/hour. Check out our tutoring services to book a session!

Andie Pinga is an expert SoFlo tutor and graduated from the University of Pennsylvania with a B.A. in Economics and a minor in Anthropology. She scored a 35 on her ACT and enjoys rock climbing and playing the guitar.

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