SAT Math Practice Questions With Full Solutions

The fastest way to raise an SAT math score is not to do more questions — it is to do questions in the four content domains the test actually uses, and to write out a full solution for every one you miss. Below are original practice questions in each domain, worked end to end, plus the structure you need to know before you start.

What the digital SAT math section looks like

The math section is delivered in two modules of 35 minutes each, 44 questions in total. About three quarters are multiple choice and the rest are student-produced responses, where you type the answer with no options to work backwards from. The section is adaptive at the module level: your performance on the first module determines whether the second module is drawn from an easier or harder pool. That is why finishing module 1 carefully matters more than rushing to finish module 2.

A built-in graphing calculator is available on every math question, and a reference sheet with common geometry formulas sits one tap away. Both facts change how you should prepare: memorising the area of a cylinder is wasted effort, while being fluent at typing an equation into a grapher to find an intersection is worth real points. Our SAT math reference sheet lists exactly what is given and what is not.

The four content domains, with their approximate share of the section:

Domain	Share	What it covers
Algebra	~35%	Linear equations, linear functions, systems, linear inequalities
Advanced Math	~35%	Quadratics, polynomials, exponentials, radicals, nonlinear systems
Problem-Solving and Data Analysis	~15%	Ratios, rates, percentages, probability, data inference
Geometry and Trigonometry	~15%	Area and volume, lines and angles, triangles, circles, right-triangle trig

Two thirds of the section is algebra in one form or another. If you are triaging, that is the answer.

Domain 1 — Algebra

Question 1. A phone plan charges a fixed monthly fee plus a per-gigabyte rate. A month using 4 GB costs $46; a month using 9 GB costs $66. What is the cost of a month using 12 GB?

Solution. The cost model is linear: $C = mg + b$. The rate is the slope,

$m = \frac{66 - 46}{9 - 4} = \frac{20}{5} = 4$

so $4 per gigabyte. Substituting the first point: $46 = 4(4) + b$, so $b = 30$. Then $C(12) = 4(12) + 30 = 78$. $78.

The trap is answering $4 or $30 — the question asked for a total, not a parameter. Read the last line of the question again before selecting. You can check any line-through-two-points work with the slope-intercept form calculator.

Question 2. If $3x - 2y = 14$ and $x + 2y = 10$, what is the value of $x + y$?

Solution. Add the equations: $4x = 24$, so $x = 6$. Substituting, $6 + 2y = 10$ gives $y = 2$. So $x + y = 8$.

Notice that adding was chosen because the $y$ terms cancelled immediately. On the SAT, elimination usually beats substitution — the test writers construct the coefficients so that one operation clears a variable. Practise more of these with the system of equations solver.

Domain 2 — Advanced Math

Question 3. The function $f(x) = x^2 - 6x + 5$ has zeros at $x = a$ and $x = b$ with $a < b$. What is the value of $b - a$?

Solution. Factor: $x^2 - 6x + 5 = (x - 1)(x - 5)$, so $a = 1$ and $b = 5$, giving $b - a = 4$.

A faster route for uglier coefficients: the distance between the roots equals $\sqrt{\Delta}/|a|$ where $\Delta = b^2 - 4ac$ is the discriminant. Here $\Delta = 36 - 20 = 16$, so the gap is $\sqrt{16}/1 = 4$. Check factorisations with the quadratic equation solver.

Question 4. A population of bacteria doubles every 6 hours. If there are 500 cells at time zero, how many cells are there after 24 hours?

Solution. Twenty-four hours is four doubling periods, so

$P = 500 \cdot 2^{24/6} = 500 \cdot 2^4 = 500 \cdot 16 = 8000.$

Write exponential growth as $P = P_0 \cdot r^{t/k}$ where $k$ is the time for one multiplication by $r$. Almost every SAT exponential question is that formula with different words wrapped around it.

Question 5. If $\sqrt{2x + 7} = x - 4$, what is the value of $x$?

Solution. Square both sides: $2x + 7 = x^2 - 8x + 16$, so $x^2 - 10x + 9 = 0$, which factors as $(x - 1)(x - 9) = 0$, giving $x = 1$ or $x = 9$. Now check both against the original equation, because squaring can create false solutions. At $x = 1$: the left side is $3$, the right side is $-3$ — rejected. At $x = 9$: left side $5$, right side $5$ — valid. $x = 9$.

Extraneous roots are a favourite SAT trap. Any time you square both sides, checking is part of the method, not an optional extra.

Domain 3 — Problem-Solving and Data Analysis

Question 6. A shirt is discounted 20%, and then a further 10% is taken off the sale price at the register. The final price is what percentage of the original?

Solution. Multiply the factors: $0.80 \times 0.90 = 0.72$. The final price is 72% of the original, a total discount of 28% — not 30%. Successive percentage changes multiply; they never add.

Question 7. The values 4, 7, 7, 9, 13 are a data set. If the value 21 is added, which changes more, the mean or the median?

Solution. Original mean is $40/5 = 8$; original median is $7$. New set: 4, 7, 7, 9, 13, 21 with mean $61/6 \approx 10.17$ and median $(7 + 9)/2 = 8$. The mean moved by about 2.17, the median by 1. The mean changes more.

This is the single most repeated data-analysis idea on the test: outliers drag the mean and barely move the median. Try your own sets in the mean, median and mode calculator.

Domain 4 — Geometry and Trigonometry

Question 8. In right triangle $ABC$, the right angle is at $B$, $AB = 9$, and $BC = 12$. What is $\sin A$?

Solution. The hypotenuse is $AC = \sqrt{9^2 + 12^2} = \sqrt{225} = 15$. From angle $A$, the opposite side is $BC = 12$, so $\sin A = 12/15 = 4/5 = 0.8$.

Recognising the 9-12-15 triangle (a 3-4-5 scaled by 3) saves the whole calculation. The scaled Pythagorean triples worth knowing on sight: 3-4-5, 5-12-13, 8-15-17, 7-24-25. Verify any triangle with the Pythagorean theorem calculator.

Question 9. A circle has equation $x^2 + y^2 - 6x + 4y = 12$. What is its radius?

Solution. Complete the square in both variables: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$, so $(x - 3)^2 + (y + 2)^2 = 25$. The radius is $\sqrt{25} = 5$ and the centre is $(3, -2)$.

The circle equation is not on the SAT reference sheet. Memorise it, along with the completing-the-square move that converts a scattered equation into centre-radius form — our completing the square tool shows every line.

How to use practice questions properly

Doing 200 questions and glancing at the answer key produces almost no gain. What works:

1. Time yourself in modules, not in single questions. Thirty-five minutes, 22 questions, no interruptions. Pacing is a skill you can only practise at full length.
2. Keep an error log. One row per miss: the question, the domain, and — the important column — why. Misread, arithmetic slip, missing formula, or genuinely did not know the concept. After 40 rows, the pattern in that last column is your study plan.
3. Redo every missed question from scratch a week later. Not reading the solution: solving it again on blank paper. If you cannot reproduce it, you did not learn it.
4. Use official full-length practice tests for score prediction. Third-party questions are fine for drilling a skill; only official adaptive tests give a realistic score estimate.

Related: SAT math reference sheet for every formula you are given and every formula you must memorise.