Free SAT Math Prep Questions, Problems & Tests

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Hard SAT Math Questions

SAT Math prep for the hardest problems helps you confidently tackle challenging questions. By practicing at this level, you’ll unlock the highest possible scores.

An equation of a circle in the xy-plane with radius 2 and center at (5, −4) is x² + y² + ax + by = c. What is the value of c ?

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Hint: The equation of a circle in standard form is (x − h)² + (y − k)² = r².

Explanation

Correct Answer: -37

The equation of a circle in standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the length of the radius.

The given circle has radius 2 and center at (5, −4), so plug in r = 2 and (h, k) = (5, −4) to get an equation of the circle in standard form. Then rewrite the equation in the form x² + y² + ax + by = c to identify the value of c.

(x − h)² + (y − k)² = r²	Standard form
(x − 5)² + (y − (−4))² = 2²	Plug in (h, k) = (5, −4) and r = 2
(x − 5)² + (y + 4)² = 4	Simplify

Now expand the perfect squares (x − 5)² and (y + 4)² and then rearrange the equation to the form x² + y² + ax + by = c.

(x − 5)² + (y + 4)² = 4
x² − 10x + 25 + y² + 8y + 16 = 4	Expand (x − 5)² and (y + 4)²
x² + y² − 10x + 8y + 41 = 4	Rearrange and combine constant terms on left
x² + y² − 10x + 8y = −37	Subtract 41 from both sides

Compare the resulting equation to the given equation to identify c.

Things to remember:

The equation of a circle in standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the length of the radius.
To expand (a + b)², use the identity (a + b)² = a² + 2ab + b² or rewrite the expression (a + b)² as (a + b)(a + b) and multiply.

In a warehouse, there are 7 shelves of cardboard boxes and 8 shelves of plastic boxes. Each shelf of cardboard boxes has 12 boxes that weigh 25 pounds or more and 8 boxes that weigh less than 25 pounds. Each shelf of plastic boxes has 11 boxes that weigh 25 pounds or more and 9 boxes that weigh less than 25 pounds. A box from one of these shelves will be selected at random. What is the probability of selecting a plastic box, given that the box is 25 pounds or heavier?

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Hint : To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

Explanation

Correct Answer: 22/43

The question asks for the probability of selecting a plastic box, given that the box is 25 pounds or heavier.

To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

The total number of possible outcomes is the total number of boxes that are 25 pounds or heavier.
The number of desired outcomes is the number of those boxes that are plastic.

The total boxes that weigh 25 pounds or more is the sum of the numbers of plastic boxes and cardboard boxes that weigh 25 pounds or more.

It is given that there are 8 shelves with 11 plastic boxes on each shelf that weigh 25 pounds or more and that there are 7 shelves with 12 cardboard boxes on each shelf that weigh 25 pounds or more.

Multiply each number of shelves by the number of boxes on each shelf to see that there are 8 ∙ 11 = 88 plastic boxes and there are 7 ∙ 12 = 84 cardboard boxes that weigh 25 pounds or more.

P = 88 / 88 + 84

Now simplify the probability.

$P = \frac{88}{88 + 84}$
$P = \frac{88}{172}$	Add in denominator
$P = \frac{22}{43}$	Simplify 88 / 172 = 4 . 22 / 4 . 43

The probability of selecting a plastic box, given that the box is 25 pounds or heavier, is 22 / 43.

Things to remember:
To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

The table shows three values of x and their corresponding values of y, where y = f(x) + 6 and f is a quadratic function. What is the y-coordinate of the y-intercept of the graph of y = f(x) in the xy-plane?

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Hint: For any function f(x), the graph of y = f(x − h) + k is the graph of y = f(x) transformed by a horizontal shift of h units and a vertical shift of k units.

The equation of a quadratic function in vertex form is y = a(x − h)² + k, where (h, k) is the vertex of the parabola and a is the scale factor.

Explanation

Correct Answer: 1075

The equation y = f(x) + 6 is in terms of f(x), so it is the result of a transformation to the function f(x). The constant 6 is added to the output f(x), so it represents a vertical shift 6 units up.

Each row of the table represents an (x, y) point on the graph of y = f(x) + 6. The question asks for the y-intercept of f(x), so apply the transformation to the given table to get a table of values for y = f(x).

Every point on the graph of y = f(x) + 6 is 6 units above a point on the graph of y = f(x), so subtract 6 from each y-value to get a table that represents points on the graph of y = f(x).

It is given that f(x) is a quadratic function, so its graph is a parabola. A vertical axis of symmetry passes through the vertex and is located halfway between any two points on a parabola that have the same y-value.

Notice that two points from the table have the same y-value (4). Therefore, the point with an x-value halfway between 17 and 21 is the vertex (h, k). This point is in the table (19, −8).

Plug (h, k) = (19, −8) into the vertex form equation y = a(x − h)² + k to see that an equation for the graph of f(x) is y = a(x − 19)² − 8. Plug in a different point from the table to solve for a. Choose (x, y) = (17, 4).

y = a(x − 19)² − 8
4 = a(17 − 19)² − 8	Plug in x = 17 and y = 4
4 = a(−2)² − 8	Subtract inside parentheses
4 = 4a − 8	Apply the exponent
12 = 4a	Add 8 to both sides
3 = a	Divide by 4 on both sides

Plug a = 3 into y = a(x − 19)² − 8 to see that an equation for f(x) is y = 3(x − 19)² − 8. To find the y-coordinate of the y-intercept, plug in x = 0 and solve for y.

y = 3(x − 19)² − 8
y= 3(0 − 19)² − 8	Plug in x = 0
y = 3(−19)² − 8	Subtract inside parentheses
y = 3(361) − 8	Apply the exponent
y = 1,075	Simplify

Note: It is also possible to use the given table of values to see that the equation of the transformed function is y = 3(x − 19)² − 2 and has y-intercept 1,081. Then subtract 6 to get the y-coordinate of the y-intercept of f(x).

Things to remember:

For any function f(x), the graph of y = f(x − h) + k is the graph of y = f(x) transformed by a horizontal shift of h units and a vertical shift of k units.
A vertical axis of symmetry passes through the vertex (h, k) of a parabola and is located halfway between any two points on a parabola that have the same y-value.
The equation of a quadratic function in vertex form is y = a(x − h)² + k, where (h, k) is the vertex of the parabola and a is the scale factor.

$\frac{13}{20} y - \frac{11}{20} x = \frac{10}{13} - \frac{13}{20} y$

$\frac{11}{10} x + \frac{13}{10} = r y + \frac{33}{10}$

In the system of equations shown, r is a constant. If the system has no solution, what is the value of r ?

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Hint : A system of linear equations has no solution if the same expression of x and y is equal to two different numbers.

Explanation

Correct Answer: 26/10

It is given that the system has no solution, so its graph must consist of lines that do not intersect. Therefore, the lines must be parallel. Parallel lines have the same slope but different y-intercepts.

Rewrite the given system of linear equations in slope-intercept form y = mx + b, where m is the slope and b is the y-intercept. Isolate the y-term in the first equation.

Rewrite the second equation to align like terms, then compare the two equations to identify the slopes and y-intercepts.

Set the slopes equal and solve for r to find the value of r that makes the slopes the same.

11 / 26 = 11 / 10r	Set slopes equal
( 11 / 26 )r = ( 11 / 10r )r	Multiply both sides by r
26 / 11 ( 11 / 26 )r = ( 11 / 10 ) 26 / 11	Multiply both sides by 26 / 11 , the reciprocal of 11 / 26
r = 26 / 10	Simplify

For the y-intercepts to be different and the slopes to be the same, the value of r must be 26 / 10 .

Note: It is possible to plug r = 26 / 10 into the second equation to verify that the y-intercepts are different (graph).

Things to remember:

A system of linear equations has no solution if its graph consists of two parallel lines because parallel lines never intersect.
Parallel lines have the same slope but different y-intercepts.

Alternate Method :

A system has no solution if the same expression of x and y is equal to two different numbers.

Rewrite the given system of linear equations in standard form so that the constant terms are isolated.

Now each equation is written in standard form. Notice that $- \frac{11}{20} x$ is a multiple of $- \frac{11}{10} x$ . Multiply the first equation by 2 so that the x-terms are the same ( $(- \frac{11}{10} x)$ ) and $- \frac{11}{10} x$ .

The x-terms are the same ( $(- \frac{11}{10} x)$ and $- \frac{11}{10} x$ ) and the constant terms are different ( $(\frac{20}{13} \neq - 2)$ ), so identify the value of r that makes the y-terms the same.

Therefore, the value of r must be $\frac{26}{10}$ for the system to have no solution.

Things to remember:
A system has no solution if the same expression of x and y is equal to two different numbers.

In triangle LMN, the measure of angle L is 30°, the length of LM is 10 units and the length of LN is 8 units. What is the area, in square units, of triangle LMN?

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Hint : The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

Explanation

Correct Answer: 20

The formula for the area of a triangle is $A = \frac{1}{2} bh$ , where b is the length of the base and h is the height.

Draw triangle LMN and label it with the given information. Let LM with a given length of 10 be the base of triangle and notice the height of the triangle is unknown.

The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

Draw a line segment from vertex N down and perpendicular to base LM . Label the new point of intersection P. The length of NP is the height of triangle LMN.

Triangle LNP is a right triangle with a given angle of 30°, so it is a 30°-60°-90° triangle.

In a 30°-60°-90° triangle, the hypotenuse is twice the length of the shorter leg. Compare the side lengths of the general triangle to triangle LNP to see that hypotenuse LN must be twice the height NP.

The length of the hypotenuse is 8, so set 2x equal to 8 and divide by 2 to find the length of the shorter leg x.

2x = 8	Length of hypotenuse of special right triangle
x = 4	Divide by 2 on both sides

The length of the shorter leg is 4, so the height of triangle LMN is 4. Plug the length of the base (10) and the height (4) of into the formula for the area of a triangle and simplify.

The area of triangle LMN, in square units, is 20.

Things to remember:

The formula for the area of a triangle is $A = \frac{1}{2} bh$ , where b is the length of the base and h is the height.
A 30°-60°-90° triangle has the following ratio of side lengths:

The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

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Frequently Asked Questions (FAQs)

Do these SAT math practice questions include detailed explanations?

Yes. Each SAT math problem features comprehensive explanations so you understand the reasoning behind correct and incorrect answers and avoid repeating the same mistakes on future questions.

Do these SAT math practice tests match the latest SAT format?

Yes. Our SAT math prep questions mirror the current SAT Math section format: multiple-choice and student-produced response questions that reflect the official SAT math exam structure. Students can use the full-length practice exam in our SAT Prep Course to match the 2 modules format or create custom ones focused on key topics like the sample SAT math practice tests we’ve included above.

How many SAT math practice tests are included in this product?

We provide one full-length SAT practice test, plus you can generate unlimited personalized math practice tests. To sharpen your skills, focus on weak areas, use custom tags, or revisit questions you previously missed.

How do these SAT math practice questions compare to the official SAT in difficulty?

Our SAT math problems are designed to match—or even exceed—the official SAT in difficulty, including hard questions, so you’ll be fully prepared for test day.

Do you offer additional SAT math prep resources, such as videos or tutoring?

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Yes. You can study SAT math practice questions on a computer, tablet, or smartphone, and your progress will be saved across devices for seamless SAT math prep.

How should I time myself during SAT math practice?

On the real SAT math exam, you’ll have 70 minutes for the Math section (35 minutes per module). Practice SAT math problems under these conditions or with smaller timed sets to build pacing and accuracy.

What concepts are covered on the SAT Math section?

You’ll encounter SAT math questions testing algebra, problem-solving, data analysis, geometry, trigonometry, and math in real-world contexts. Both calculator and no-calculator SAT math problems are required.

How many questions are on the SAT Math section?

There are 44 SAT math questions total, split across 2 modules. You’ll answer multiple-choice and student-produced response SAT math practice questions that mirror sample SAT math questions from official tests.

What's a good score on the SAT Math section?

SAT math test scores range from 200 to 800. Competitive colleges often look for 650–780 in Math, but your target should align with the requirements of your chosen schools. Regular SAT math practice with challenging SAT math practice problems helps achieve higher scores.