FREE SAT MATH PRACTICE

Free SAT Math Practice Questions with Full Solutions

Work through SAT Math questions that test structure, reasoning, timing decisions, and common Digital SAT traps. Each question includes a full solution, a common mistake, and a method note.

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Quick SAT Math recognition tips

SAT Math often rewards recognizing the structure of a question before calculating. These tips will help you use the free questions below more effectively.

Look for structure first. Factored quadratics, exponential forms, and similar solids often have a shortcut hidden in the setup.
Do not over-expand. Many SAT questions are faster if you use roots, shifts, signs, or scale factors instead of expanding everything.
Use Desmos selectively. Desmos is powerful, but not every question needs it. Sometimes the fastest route is algebraic recognition.
Watch the wording. Phrases like “one possible value,” “no solution,” and “similar solids” change the solving route.

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Geometry and Measurement

SAT Practice Question

Geometry and Measurement Medium Multiple Choice Identify the legs of a right triangle and calculate area
A right triangle has sides of length 3√2, 5√2, and √68 units. What is the area of the triangle, in square units?
A. 8√2 + √68
B. 30√68
C. 30
D. 15

Correct answer: Choice D

15

Full solution Because the triangle is right, the longest side is the hypotenuse. Compare the side lengths: √68 is the largest. Check the legs: (3√2)2 + (5√2)2 = 18 + 50 = 68, so the legs are 3√2 and 5√2. The area is 12 × 3√2 × 5√2 = 12 × 15 × 2 = 15.
Common mistake A common mistake is multiplying the hypotenuse by one of the legs. For a right triangle, area uses the two perpendicular legs, not the hypotenuse.
Method note First identify the hypotenuse by finding the longest side. Then use Area = 12 × leg × leg.
SAT note SAT note: This is a route-choice question. A fast check is to square the sides and confirm which two sides form the legs.
Desmos note Desmos is not needed. Mental math and the Pythagorean relationship are faster.

SAT Practice Question

Geometry and Measurement Hard Free Response Use similarity scale factors for volume and surface area
The table shows the volume of two similar solids, right circular cylinder A and right circular cylinder B.

Volume (cubic units)
Right circular cylinder A32π
Right circular cylinder B864π

The radius of right circular cylinder A is 2 units. The surface area of right circular cylinder A is kπ square units, and the surface area of right circular cylinder B is nπ square units, where k and n are constants. What is the value of n − k? The surface area of a right circular cylinder with radius r and height h is 2πr2 + 2πrh.

Correct answer: 320

Full solution Cylinder A has volume 32π and radius 2. Since V = πr2h, we get 32π = π(22)h, so h = 8. The surface area of A is 2π(22) + 2π(2)(8) = 8π + 32π = 40π, so k = 40. The volume ratio from A to B is 864π32π = 27. For similar solids, volume scale factor is the cube of the linear scale factor, so the linear scale factor is 3. Surface area scales by the square of the linear scale factor, so the surface area scale factor is 9. Therefore, cylinder B has surface area 9 × 40π = 360π, so n = 360. Thus n − k = 360 − 40 = 320.
Common mistake A common mistake is multiplying surface area by 27. Volume scales cubically, but surface area scales by the square of the linear scale factor.
Method note For similar solids: linear scale factor affects length, its square affects area, and its cube affects volume.
SAT note SAT note: This is a similarity scaling question hidden inside cylinder geometry.
Desmos note Desmos is not needed. In fact, it can make this question slower. Use similarity ratios: volume scales by the cube of the linear scale factor, and surface area scales by the square of the linear scale factor.

Quadratics and Functions

SAT Practice Question

Quadratics and Functions Medium Multiple Choice Find the minimum value of a shifted quadratic function
f(x) = 3x² + 48x + 193

The function g is defined by g(x) = f(x + 7). What is the minimum value of g(x)?
A. −15
B. −8
C. 1
D. 8

Correct answer: Choice C

1

Full solution The transformation g(x) = f(x + 7) shifts the graph horizontally, but it does not change the minimum value. Find the minimum of f(x) = 3x² + 48x + 193. The vertex occurs at x = −b/(2a) = −48/(2·3) = −8. Then f(−8) = 3(64) + 48(−8) + 193 = 192 − 384 + 193 = 1. Therefore, the minimum value of g(x) is also 1.
Common mistake A common mistake is trying to substitute x + 7 into the whole expression and expand. That works, but it is slower and increases the chance of arithmetic errors.
Method note Horizontal shifts change where the minimum occurs, not the minimum value itself.
SAT note SAT note: Recognize function shifts quickly. The test often rewards avoiding unnecessary expansion.
Desmos note Can be solved using Desmos, but it can also be solved by finding the vertex and noticing that g(x) is only a horizontal transformation of f(x), so the minimum value stays the same. Desmos is more helpful here if your algebra foundation is weak or if remem

SAT Practice Question

Quadratics and Functions Hard Free Response Use signs of a factored quadratic to locate integer roots
f(x) = (x − a)(x − b)

The function f is defined by the given equation, where a and b are integer constants. If f(17) > 0, f(20) < 0, and f(23) > 0, what is one possible value of a + b?

Correct answer: 40

Full solution The expression (x − a)(x − b) changes sign between its two roots and is positive outside the roots. Since f(17) > 0, f(20) < 0, and f(23) > 0, one root must be between 17 and 20 and the other must be between 20 and 23. Because a and b are integers, possible roots are one from {18,19} and one from {21,22}. Possible sums are 18+21=39, 18+22=40, 19+21=40, or 19+22=41. A clean possible answer is 40.
Common mistake A common mistake is assuming the roots are exactly 17, 20, and 23. The signs only tell where the roots must lie.
Method note For factored quadratics with positive leading coefficient, the function is positive outside the roots and negative between them.
SAT note SAT note: Free-response questions may have multiple valid answers. Enter any accepted value.
Desmos note Desmos can help visualize the sign pattern, but interval reasoning is faster.

Geometry and Trigonometry

SAT Practice Question

Geometry and Trigonometry Hard Free Response Use unit circle coordinates to evaluate a trigonometric ratio
In the xy-plane, a unit circle with center at the origin O contains point A with coordinates (1, 0) and point B with coordinates (5√34, −3√34). If the measure of angle AOB is p radians, what is the value of cos psin p?

Correct answer: 53

Full solution Point A = (1,0) lies on the positive x-axis. Point B has coordinates (5√34, −3√34). The angle p is the angle between the ray to A and the ray to B. The cosine of this angle is based on the horizontal component, so cos p = 5/√34. The sine of the angle measure is the positive magnitude of the vertical change from the positive x-axis, so sin p = 3/√34. Therefore, cos psin p = (5√34)/(3√34) = 5/3.
Common mistake A common mistake is using the negative y-coordinate directly and getting −5/3. The question asks for the measure of angle AOB, so the angle measure is positive.
Method note Use the unit circle coordinates as cosine and sine magnitudes for the angle from the positive x-axis.
SAT note SAT note: This is mainly a coordinate-trigonometry recognition question, not a long trigonometry problem.
Desmos note Desmos is not needed.

Linear Equations

SAT Practice Question

Linear Equations Hard Free Response Find the condition for a linear equation to have no solution
5(kx − n) = −6514x − 8518

In the given equation, k and n are constants and n > 1. The equation has no solution. What is the value of k?

Correct answer: −1314

Full solution Expand the left side: 5(kx − n) = 5kx − 5n. For a linear equation to have no solution, the coefficients of x must be equal but the constants must be different. Therefore, 5k = −6514. Divide both sides by 5: k = (−6514) ÷ 5 = −6570 = −1314. Since n > 1, the constant −5n is less than −5, which is not equal to −8518, so the constants are different.
Common mistake A common mistake is trying to solve for x. But the question says the equation has no solution, so the goal is to match x-coefficients and keep constants unequal.
Method note No solution in a linear equation means same slope or coefficient, different constant.
SAT note SAT note: This is a structure question. Do not over-solve; compare the two sides.
Desmos note Desmos is not needed.

Polynomial Functions

SAT Practice Question

Polynomial Functions Medium Multiple Choice Use zeros of a monic cubic to identify the coefficient of x²
g(x) = x³ + ax² + bx + c

The function g is defined by the given equation, where a, b, and c are integer constants. The zeros of the function are −2, −7, and 6. What is the value of a?
A. −84
B. −3
C. 3
D. 84

Correct answer: Choice C

3

Full solution For a monic cubic with zeros r1, r2, and r3, the coefficient of x2 is the negative of the sum of the zeros. The zeros are −2, −7, and 6. Their sum is −2 + (−7) + 6 = −3. Therefore, a = −(−3) = 3.
Common mistake A common mistake is multiplying the roots and choosing 84 or −84. The product relates to the constant term, not the coefficient of x2.
Method note For x3 + ax2 + bx + c, use a = −(sum of zeros).
SAT note SAT note: This is much faster using root-sum structure than expanding the full polynomial.
Desmos note Desmos is not needed.

Exponential Functions

SAT Practice Question

Exponential Functions Hard Multiple Choice Connect exponential growth factor to equivalent exponent forms
For the function f, for each increase in the value of x by c, where c is a positive constant, the value of f(x) increases by a factor of 27. Which of the following equivalent forms of the function f displays 1c as a coefficient of x?
A. f(x) = 35(3)(12)x
B. f(x) = 35(33)(16)x
C. f(x) = 35(9)(14)x
D. f(x) = 35(27(13)x)12

Correct answer: Choice B

f(x) = 35(33)(16)x

Full solution If increasing x by c multiplies f(x) by 27, then a natural form is f(x) = 35(27)(1c)x. Since 27 = 33, this can be written as 35(33)(1c)x. Choice B is 35(33)(16)x, so it displays 1c as the coefficient of x, with c = 6.
Common mistake A common mistake is choosing an equivalent-looking expression without checking whether the displayed coefficient of x is actually 1c.
Method note Look for the form where the base represents the full growth factor and the exponent contains (1c)x.
SAT note SAT note: This is an exponent-structure question. The wording is more important than computation.
Desmos note Can be solved by Desmos, but it usually takes more time. The better route is to understand the structure of exponential growth questions because this pattern is common on the SAT.

SAT Practice Question

Exponential Functions Hard Multiple Choice Use substitution in an exponential function with unknown constants
The function f is defined by f(x) = ax − b, where a and b are constants. In the xy-plane, the graph of y = f(x) passes through the points (c, 5) and (2c, 137), where c is a constant. Which of the following could be the value of b?
A. 7
B. 16
C. 132
D. 142

Correct answer: Choice A

7

Full solution Let ac = t. From the point (c,5), we get t − b = 5, so t = b + 5. From the point (2c,137), we get a2c − b = 137. Since a2c = (ac)2 = t2, we have t2 − b = 137. Substitute b = t − 5: t2 − (t − 5) = 137, so t2 − t − 132 = 0. Factor: (t − 12)(t + 11) = 0. Since t = ac is positive, t = 12. Then b = t − 5 = 7.
Common mistake A common mistake is treating a2c as 2ac. It should be (ac)2.
Method note Use substitution when the exponents are related. Letting ac = t turns the problem into a quadratic.
SAT note SAT note: This is a strong example of structure over brute force.
Desmos note Desmos is not useful unless values are already known. Algebraic substitution is the clean route.

Advanced Geometry

SAT Practice Question

Advanced Geometry Hard Free Response Use intersecting diagonals and distance relationships in a kite-like quadrilateral
LMNK
Note: Figure not drawn to scale.

In quadrilateral KLMN shown, KL = 3, LM = 3, KN = 27, and MN = 27. Diagonals KM and LN intersect at point G, where GK = 1 and GM = 1. If the length of diagonal LN is √p + √w, where p and w are integers, what is the value of p + w?

Correct answer: 736

Full solution Since GK = 1 and GM = 1, point G is the midpoint of diagonal KM, so KM = 2. Also, KL = LM = 3, so point L is equidistant from K and M. Similarly, KN = MN = 27, so point N is also equidistant from K and M. Therefore, both L and N lie on the perpendicular bisector of KM, which passes through G. In right triangle KGL, KL = 3 and GK = 1, so GL = √(3² − 1²) = √8. In right triangle KGN, KN = 27 and GK = 1, so GN = √(27² − 1²) = √728. Thus LN = √8 + √728, so p + w = 8 + 728 = 736.
Common mistake A common mistake is assuming the diagram is drawn to scale or trying to estimate the diagonal visually. The note says the figure is not drawn to scale.
Method note Use equal side lengths to identify a perpendicular bisector relationship.
SAT note SAT note: This is a high-value geometry reasoning question. Draw the diagonals and use right triangles.
Desmos note Desmos is not needed.

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