mathematics practice

Overview

This page contains 25 practice problems organised by the four content domains of the SAT Mathematics section. Each problem includes a step-by-step solution, the key concept tested, and common mistakes to avoid. These problems span the difficulty range you will encounter on the digital SAT, from straightforward applications to multi-step problems requiring careful reasoning.

Algebra

Problem 1: Linear Equations

If $3(2x - 5) = 4(x + 3) - 7$ , what is the value of $x$ ?

Solution:

$3(2x - 5) = 4(x + 3) - 7$

$6x - 15 = 4x + 12 - 7$

$6x - 15 = 4x + 5$

$6x - 4x = 5 + 15$

$2x = 20$

$x = 10$

Key concept: Solving a linear equation by distributing, combining like terms, and isolating the variable.

Common mistakes:

Forgetting to distribute the 3 to both terms in $(2x - 5)$
Dropping the negative sign when moving $-15$ to the right side
Arithmetic errors in $12 - 7 = 5$

Problem 2: Systems of Equations

A store sells notebooks for $3 each and pens for $2 each. If a customer buys 14 items and spends $34 in total, how many notebooks did the customer buy?

Solution:

Let $n$ = number of notebooks and $p$ = number of pens.

$n + p = 14 \quad \text{(total items)}$

$3n + 2p = 34 \quad \text{(total cost)}$

From the first equation: $p = 14 - n$ .

Substitute into the second:

$3n + 2(14 - n) = 34$

$3n + 28 - 2n = 34$

$n + 28 = 34$

$n = 6$

The customer bought 6 notebooks.

Key concept: Solving a system of linear equations using substitution.

Common mistakes:

Setting up the equations with incorrect coefficients (confusing which item costs $3 and which costs $2)
Solving for $n$ but answering with $p$ instead
Forgetting to verify: $6 \times 3 + 8 \times 2 = 18 + 16 = 34$ ✓

Problem 3: Linear Functions

A line passes through the points $(-2, 5)$ and $(4, -7)$ . What is the equation of this line in slope-intercept form?

Solution:

$m = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2$

Using point-slope form with point $(4, -7)$ :

$y - (-7) = -2(x - 4)$

$y + 7 = -2x + 8$

$y = -2x + 1$

The equation is $y = -2x + 1$ .

Key concept: Calculating slope from two points and writing the equation in slope-intercept form.

Common mistakes:

Reversing the order of subtraction in the slope formula (getting $m = \frac{5-(-7)}{-2-4}$ gives the same result, but inconsistent order can lead to sign errors)
Making arithmetic errors when substituting into point-slope form
Forgetting to simplify the final equation

Problem 4: Inequalities

Solve the inequality $-3(2x + 1) > 4x - 11$ . Express the solution as an inequality.

Solution:

$-3(2x + 1) > 4x - 11$

$-6x - 3 > 4x - 11$

$-6x - 4x > -11 + 3$

$-10x > -8$

Dividing by $-10$ (reverse the inequality sign):

$x < \frac{4}{5}$

Key concept: Solving a linear inequality, remembering to reverse the inequality sign when dividing by a negative number.

Common mistakes:

Forgetting to reverse the inequality sign when dividing by $-10$
Sign errors when distributing $-3$ to both terms
Writing $\frac{-8}{-10}$ incorrectly (it simplifies to $\frac{4}{5}$ , not $-\frac{4}{5}$ )

Problem 5: Absolute Value

What is the sum of all integer solutions to $|2x - 3| \leq 7$ ?

Solution:

$|2x - 3| \leq 7 \implies -7 \leq 2x - 3 \leq 7$

Add 3:

$-4 \leq 2x \leq 10$

Divide by 2:

$-2 \leq x \leq 5$

The integer solutions are $-2, -1, 0, 1, 2, 3, 4, 5$ .

Sum: $-2 + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 12$

Key concept: Solving an absolute value inequality by rewriting as a compound inequality.

Common mistakes:

Writing $2x - 3 \leq 7$ only (forgetting the negative case)
Incorrectly dividing: dividing by a positive number (2) does not require reversing the inequality
Off-by-one errors when listing integer solutions (including 6 or excluding $-2$ )

Problem 6: Linear Word Problem

A tank initially contains 200 litres of water. Water is added at a constant rate of 15 litres per minute. At the same time, water drains from a leak at a constant rate of 3 litres per minute. How many minutes will it take for the tank to contain 380 litres?

Solution:

Net rate of change: $15 - 3 = 12$ litres per minute.

Amount needed: $380 - 200 = 180$ litres.

$12t = 180$

$t = \frac{180}{12} = 15$

It takes 15 minutes.

Key concept: Modelling a real-world scenario with a linear equation using rate of change.

Common mistakes:

Forgetting to account for the leak (using 15 litres/min instead of the net 12 litres/min)
Using the initial amount instead of the difference ( $380 - 200$ )
Arithmetic errors in $180 \div 12$

Advanced Math

Problem 7: Quadratic Equations

What are the solutions to $2x^2 - 5x - 3 = 0$ ?

Solution:

Factor: find two numbers that multiply to $2 \times (-3) = -6$ and add to $-5$ .

The numbers are $-6$ and $1$ .

$2x^2 - 6x + x - 3 = 0$

$2x(x - 3) + 1(x - 3) = 0$

$(2x + 1)(x - 3) = 0$

$2x + 1 = 0 \implies x = -\frac{1}{2}$

$x - 3 = 0 \implies x = 3$

The solutions are $x = -\frac{1}{2}$ and $x = 3$ .

Key concept: Factoring a quadratic by splitting the middle term (AC method).

Common mistakes:

Incorrect factor pair (e.g., $-2$ and $3$ multiply to $-6$ but add to $1$ , not $-5$ )
Sign errors during grouping
Forgetting that $2x + 1 = 0$ gives $x = -1/2$ , not $x = 1/2$

Problem 8: Polynomial Operations

If $f(x) = x^3 + 2x^2 - 5x + 1$ and $g(x) = x - 3$ , what is the remainder when $f(x)$ is divided by $g(x)$ ?

Solution:

By the Remainder Theorem, the remainder is $f(3)$ :

$f(3) = 3^3 + 2(3)^2 - 5(3) + 1$

$f(3) = 27 + 2(9) - 15 + 1$

$f(3) = 27 + 18 - 15 + 1$

$f(3) = 31$

The remainder is 31.

Key concept: The Remainder Theorem — the remainder of $P(x) \div (x - c)$ equals $P(c)$ .

Common mistakes:

Evaluating $f(-3)$ instead of $f(3)$ (the theorem applies to $x - c$ , so $c = 3$ , not $-3$ )
Arithmetic errors in computing $3^3$ or $2 \times 9$
Attempting full polynomial long division (correct but slower and more error-prone)

Problem 9: Exponential Functions

A population of bacteria doubles every 3 hours. If the initial population is 500, which expression gives the population after $t$ hours?

A) $500 \cdot 2t$

B) $500 \cdot 2^{3t}$

C) $500 \cdot 2^{t/3}$

D) $500 \cdot 3^{t/2}$

Solution:

The population doubles every 3 hours. After $t$ hours, the number of doubling periods is $\frac{t}{3}$ .

$P(t) = 500 \cdot 2^{t/3}$

Correct answer: C

Key concept: Writing exponential growth functions from a doubling time.

Common mistakes:

Choosing A, which is linear growth, not exponential
Choosing B, which confuses the exponent (doubling 3 times per period instead of once every 3 hours)
Choosing D, which swaps the base and the exponent coefficient

Problem 10: Function Composition

If $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ , what is $f(g(4))$ ?

Solution:

First evaluate $g(4)$ :

$g(4) = 4^2 - 3 = 16 - 3 = 13$

Then evaluate $f$ at this result:

$f(13) = 2(13) + 1 = 26 + 1 = 27$

$f(g(4)) = 27$

Key concept: Function composition — evaluating the inner function first, then using its output as the input to the outer function.

Common mistakes:

Evaluating $f(4)$ first instead of $g(4)$ (computing $g(f(4))$ instead of $f(g(4))$ )
Arithmetic errors: $4^2 = 16$ , not $8$
Forgetting to apply $f$ to the result of $g(4)$

Problem 11: Nonlinear Equations

If $x^2 + y^2 = 25$ and $x + y = 7$ , what is the value of $xy$ ?

Solution:

From $x + y = 7$ , square both sides:

$(x + y)^2 = x^2 + 2xy + y^2 = 49$

Substitute $x^2 + y^2 = 25$ :

$25 + 2xy = 49$

$2xy = 24$

$xy = 12$

Key concept: Using the identity $(x + y)^2 = x^2 + 2xy + y^2$ to solve a system of equations without finding $x$ and $y$ individually.

Common mistakes:

Attempting to solve for $x$ and $y$ individually (possible but more time-consuming)
Forgetting the middle term $2xy$ when expanding $(x + y)^2$
Dividing 24 by 2 incorrectly

Problem 12: Rational Equations

If $\frac{x+1}{3} + \frac{x-2}{5} = 2$ , what is the value of $x$ ?

Solution:

Multiply through by the common denominator, 15:

$15 \cdot \frac{x+1}{3} + 15 \cdot \frac{x-2}{5} = 15 \cdot 2$

$5(x + 1) + 3(x - 2) = 30$

$5x + 5 + 3x - 6 = 30$

$8x - 1 = 30$

$8x = 31$

$x = \frac{31}{8}$

Key concept: Solving a rational equation by clearing denominators.

Common mistakes:

Incorrect common denominator (the LCD of 3 and 5 is 15)
Sign errors when distributing: $3(x - 2) = 3x - 6$ , not $3x + 6$
Arithmetic errors in $5 + (-6) = -1$

Problem 13: Quadratic in Vertex Form

The function $f(x) = x^2 - 8x + 3$ is rewritten in the form $f(x) = (x - h)^2 + k$ . What is the value of $h + k$ ?

Solution:

Complete the square:

$f(x) = x^2 - 8x + 3 = (x^2 - 8x + 16) - 16 + 3 = (x - 4)^2 - 13$

So $h = 4$ and $k = -13$ .

$h + k = 4 + (-13) = -9$

Key concept: Converting a quadratic from standard form to vertex form by completing the square.

Common mistakes:

Incorrectly completing the square: $(x - 4)^2 = x^2 - 8x + 16$ , so you must subtract 16 to keep the expression equivalent
Finding $h$ and $k$ correctly but making a sign error in the sum
Using $h = 8$ instead of $h = 4$ (forgetting that the vertex form is $(x - h)^2$ , so $h$ is half of $8$ )

Problem Solving and Data Analysis

Problem 14: Ratios and Percentages

A shirt originally priced at $80 is discounted by 25%. An additional 10% discount is then applied to the sale price. What is the final price of the shirt?

Solution:

After the first discount:

$80 \times (1 - 0.25) = 80 \times 0.75 = 60$

After the second discount:

$60 \times (1 - 0.10) = 60 \times 0.90 = 54$

The final price is $54.

Key concept: Successive percentage changes multiply rather than add.

Common mistakes:

Adding the percentages: $25\% + 10\% = 35\%$ and computing $80 \times 0.65 = 52$ (incorrect)
Applying the second discount to the original price: $80 \times 0.10 = 8$ and subtracting from $60 to get $52 (still wrong — successive discounts multiply)
Correct approach: each discount applies to the current price, not the original

Problem 15: Unit Conversions

A car’s fuel efficiency is 12 kilometres per litre. If petrol costs $1.45 per litre, what is the approximate cost per kilometre to drive this car? (Round to the nearest cent.)

Solution:

Cost per kilometre = (cost per litre) ÷ (kilometres per litre)

$\frac{1.45}{12} \approx 0.1208$

Rounded to the nearest cent: approximately $0.12 per kilometre.

Key concept: Using unit rates and dimensional analysis to convert between related quantities.

Common mistakes:

Multiplying instead of dividing (getting $17.40 per kilometre, which is clearly unreasonable)
Incorrect rounding (0.1208 rounds to 0.12, not 0.13)
Forgetting to divide by 12 (using $1.45 per kilometre)

Problem 16: Scatter Plots and Line of Best Fit

A set of data has a least-squares regression line $\hat{y} = 3.2x + 1.5$ , where $x$ represents years of experience and $\hat{y}$ represents predicted annual salary (in thousands of dollars). What is the predicted salary for someone with 8 years of experience?

Solution:

$\hat{y} = 3.2(8) + 1.5 = 25.6 + 1.5 = 27.1$

Since $\hat{y}$ is in thousands of dollars, the predicted salary is $\$ 27,100$.

Key concept: Using a regression equation to make predictions. The slope (3.2) means salary increases by approximately $3,200 per additional year of experience.

Common mistakes:

Forgetting that $\hat{y}$ is in thousands (answering $27.1 or $271 instead of $27,100)
Incorrect computation: $3.2 \times 8 = 25.6$ , not $24.6$
Confusing the slope and y-intercept

Problem 17: Probability

A bag contains 4 red marbles, 6 blue marbles, and 5 green marbles. Two marbles are drawn at random without replacement. What is the probability that both marbles are blue?

Solution:

Total marbles: $4 + 6 + 5 = 15$ .

Probability the first is blue: $\frac{6}{15} = \frac{2}{5}$ .

After removing one blue marble, 14 marbles remain, 5 of which are blue.

Probability the second is blue: $\frac{5}{14}$ .

$P(\text{both blue}) = \frac{6}{15} \times \frac{5}{14} = \frac{2}{5} \times \frac{5}{14} = \frac{10}{70} = \frac{1}{7}$

Key concept: Probability of dependent events (without replacement). Multiply the conditional probabilities.

Common mistakes:

Using replacement: $\frac{6}{15} \times \frac{6}{15} = \frac{36}{225}$ (incorrect)
Not reducing $\frac{6}{15}$ before multiplying, leading to $\frac{30}{210} = \frac{1}{7}$ (same answer but more error-prone)
Forgetting that “without replacement” changes the denominator and numerator for the second draw

Problem 18: Statistics

The five numbers $3, 7, a, 11, 15$ have a mean of 10. What is the value of $a$ ?

Solution:

$\text{Mean} = \frac{3 + 7 + a + 11 + 15}{5} = 10$

$\frac{36 + a}{5} = 10$

$36 + a = 50$

$a = 14$

Key concept: The mean is the sum of values divided by the count. Use the given mean to find the missing value.

Common mistakes:

Addition errors in $3 + 7 + 11 + 15$ (it equals 36)
Multiplying $10 \times 5 = 50$ incorrectly
Placing $a$ in the wrong position when computing the median instead (the question asks for the mean)

Problem 19: Statistics — IQR

A dataset has the following five-number summary: minimum = 12, $Q_1 = 18$ , median = 24, $Q_3 = 32$ , maximum = 45. What is the interquartile range (IQR)?

Solution:

$\text{IQR} = Q_3 - Q_1 = 32 - 18 = 14$

Key concept: The interquartile range measures the spread of the middle 50% of data.

Common mistakes:

Using maximum minus minimum ( $45 - 12 = 33$ ), which is the range, not the IQR
Using median minus $Q_1$ ( $24 - 18 = 6$ ), which is only half the IQR
Confusing $Q_1$ and $Q_3$ and computing $18 - 32 = -14$ (IQR is always non-negative)

Problem 20: Two-Way Tables

A survey of 200 students found the following distribution:

	Prefers Science	Prefers Literature	Total
Year 10	40	60	100
Year 11	50	50	100
Total	90	110	200

What fraction of the Year 10 students prefer Science?

Solution:

Year 10 total = 100. Year 10 students who prefer Science = 40.

$\frac{40}{100} = \frac{2}{5}$

Key concept: Reading a two-way table and computing a conditional probability/fraction from a specific row.

Common mistakes:

Using the total Science count: $40/90$ (this is the fraction of Science-preferring students who are in Year 10, not the question asked)
Using the total of all students: $40/200 = 1/5$
Confusing rows and columns

Geometry and Trigonometry

Problem 21: Area of Composite Shapes

A rectangle has length 12 and width 8. A semicircle is constructed on one of the longer sides as its diameter. What is the total area of the composite shape? (Use $\pi \approx 3.14$ .)

Solution:

Area of rectangle: $12 \times 8 = 96$ .

The semicircle has diameter 12, so radius $r = 6$ .

Area of semicircle: $\frac{1}{2}\pi r^2 = \frac{1}{2}(3.14)(36) = 56.52$ .

Total area: $96 + 56.52 = 152.52$ .

Key concept: Decomposing a composite shape into familiar figures and summing their areas.

Common mistakes:

Using diameter instead of radius ( $r = 6$ , not $12$ )
Forgetting the $\frac{1}{2}$ factor for a semicircle
Using the shorter side (8) as the diameter instead of the longer side (12)

Problem 22: Circle Theorems

A circle has centre $O$ . Points $A$ , $B$ , and $C$ lie on the circle such that angle $AOC = 140°$ . What is the measure of angle $ABC$ ?

Solution:

Angle $AOC = 140°$ is a central angle. The arc $AC$ has measure $140°$ .

Angle $ABC$ is an inscribed angle that subtends the same arc $AC$ .

An inscribed angle is half the measure of its intercepted arc:

$\angle ABC = \frac{140°}{2} = 70°$

Key concept: The Inscribed Angle Theorem — an inscribed angle is half the central angle that subtends the same arc.

Common mistakes:

Assuming the inscribed angle equals the central angle (answering $140°$ )
Dividing by the wrong number or misidentifying which arc is subtended
Confusing inscribed and central angles when the vertex is on vs. at the centre

Problem 23: Angle Relationships

Two parallel lines are cut by a transversal. One of the eight angles formed measures $125°$ . What is the measure of an angle adjacent to this one?

Solution:

Adjacent angles formed by intersecting lines are supplementary (they sum to $180°$ ).

$180° - 125° = 55°$

The adjacent angle measures $55°$ .

Key concept: Linear pairs formed by intersecting lines are supplementary.

Common mistakes:

Assuming adjacent angles are equal (confusing with vertical angles, which are equal)
Selecting $125°$ without recognising the question asks for the adjacent angle
Forgetting that this applies regardless of whether the lines are parallel (adjacent angles are always supplementary)

Problem 24: Right Triangle Trigonometry

In right triangle $ABC$ with right angle at $C$ , the hypotenuse $AB = 13$ and leg $BC = 5$ . What is $\sin(A)$ ?

Solution:

First find $AC$ using the Pythagorean theorem:

$AC^2 + BC^2 = AB^2$

$AC^2 + 25 = 169$

$AC^2 = 144$

$AC = 12$

For angle $A$ : opposite side is $BC = 5$ , hypotenuse is $AB = 13$ .

$\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$

Key concept: Applying SOH CAH TOA and the Pythagorean theorem in a right triangle.

Common mistakes:

Using $AC$ (adjacent) instead of $BC$ (opposite) for $\sin(A)$
Computing $\cos(A) = 12/13$ instead of $\sin(A) = 5/13$
Confusing which angle is $A$ (the angle at vertex $A$ , not at $B$ )

Problem 25: Similar Triangles

Triangle $ABC$ is similar to triangle $DEF$ with a scale factor of $2:3$ (every length in $ABC$ is $\frac{2}{3}$ of the corresponding length in $DEF$ ). If the area of $ABC$ is 24 square units, what is the area of $DEF$ ?

Solution:

The ratio of areas of similar figures equals the square of the ratio of corresponding lengths.

$\frac{\text{Area of } ABC}{\text{Area of } DEF} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

$\frac{24}{\text{Area of } DEF} = \frac{4}{9}$

$\text{Area of } DEF = 24 \times \frac{9}{4} = 6 \times 9 = 54$

The area of $DEF$ is 54 square units.

Key concept: The area ratio of similar figures is the square of the length ratio.

Common mistakes:

Using the length ratio directly ( $24 \times \frac{3}{2} = 36$ ) instead of squaring it
Squaring the wrong ratio ( $\frac{9}{4}$ vs. $\frac{4}{9}$ )
Setting up the proportion inverted: $\frac{\text{Area of } DEF}{\text{Area of } ABC} = \frac{4}{9}$

Problem 26: Volume and Surface Area

A right circular cylinder has radius 5 cm and height 12 cm. What is the total surface area of the cylinder? (Use $\pi \approx 3.14$ .)

Solution:

Total surface area = lateral area $+$ areas of two bases.

Lateral area: $2\pi rh = 2(3.14)(5)(12) = 376.8$

Area of each base: $\pi r^2 = (3.14)(25) = 78.5$

Total: $376.8 + 2(78.5) = 376.8 + 157 = 533.8$ cm².

Key concept: Surface area of a cylinder = $2\pi rh + 2\pi r^2$ .

Common mistakes:

Forgetting to include both circular bases (omitting the factor of 2 on $\pi r^2$ )
Using the diameter (10) instead of the radius (5)
Computing volume ( $\pi r^2 h$ ) instead of surface area

Problem 27: Similar Triangles and Midsegments

In triangle $PQR$ , point $M$ is the midpoint of $PQ$ and point $N$ is the midpoint of $PR$ . If $QR = 16$ , what is the length of $MN$ ?

Solution:

The segment connecting the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length.

$MN = \frac{1}{2} \times QR = \frac{1}{2} \times 16 = 8$

Key concept: The Triangle Midsegment Theorem.

Common mistakes:

Assuming $MN = QR = 16$ (it is half, not equal)
Assuming $MN = \frac{1}{4} QR = 4$
Confusing which side $MN$ is parallel to ( $MN \parallel QR$ since $M$ and $N$ are midpoints of the other two sides)

Problem 28: Angle Relationships in Polygons

What is the sum of the interior angles of a regular octagon?

Solution:

Sum of interior angles of an $n$ -sided polygon:

$S = (n - 2) \times 180°$

For an octagon ( $n = 8$ ):

$S = (8 - 2) \times 180° = 6 \times 180° = 1080°$

Each interior angle of a regular octagon: $1080° \div 8 = 135°$ .

Key concept: The polygon interior angle sum formula.

Common mistakes:

Using $(n - 1) \times 180° = 1260°$ (incorrect formula)
Using $n \times 180° = 1440°$ (incorrect formula)
Confusing interior and exterior angles (exterior angle sum is always $360°$ )

Problem 29: Right Triangle Trigonometry — Application

A ladder leans against a wall, making a $72°$ angle with the ground. If the foot of the ladder is 4 feet from the base of the wall, how long is the ladder?

Solution:

The ground, wall, and ladder form a right triangle. The ladder is the hypotenuse.

The distance from the wall (4 ft) is the side adjacent to the $72°$ angle.

$\cos(72°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{L}$

$L = \frac{4}{\cos(72°)}$

$\cos(72°) \approx 0.309$

$L = \frac{4}{0.309} \approx 12.94$

The ladder is approximately 13 feet long.

Key concept: Choosing the correct trigonometric ratio based on which sides are known. Since we know the adjacent side and want the hypotenuse, we use cosine.

Common mistakes:

Using $\sin(72°)$ or $\tan(72°)$ instead of $\cos(72°)$ (identifying the wrong sides)
Computing $\cos(72°)$ incorrectly
Placing 4 in the numerator without the cosine function: $L = 4 \times \cos(72°) \approx 1.24$ (this would be the distance from the wall to the top of the ladder, not the ladder length)

Problem 30: Circle Equation

A circle has the equation $x^2 + y^2 - 6x + 8y - 11 = 0$ . What is the radius of this circle?

Solution:

Complete the square for both $x$ and $y$ .

Group $x$ terms and $y$ terms, move the constant:

$(x^2 - 6x) + (y^2 + 8y) = 11$

Complete the square for $x$ : $x^2 - 6x + 9 = (x - 3)^2$ (add 9 to both sides)

Complete the square for $y$ : $y^2 + 8y + 16 = (y + 4)^2$ (add 16 to both sides)

$(x - 3)^2 + (y + 4)^2 = 11 + 9 + 16 = 36$

This gives centre $(3, -4)$ and radius $r = \sqrt{36} = 6$ .

Key concept: Converting the expanded form of a circle equation to standard form by completing the square.

Common mistakes:

Sign errors when completing the square: $x^2 - 6x$ requires adding $(-3)^2 = 9$ , giving $(x - 3)^2$
Forgetting to add the same constants to the right side of the equation
Taking the square root of 36 incorrectly, or confusing radius with diameter

Summary

Domain	Problems	Key Topics
Algebra	1-6	Linear equations, systems, inequalities, absolute value
Advanced Math	7-13	Quadratics, polynomials, exponentials, function composition, rational equations
Problem Solving and Data Analysis	14-20	Percentages, unit conversions, regression, probability, statistics, two-way tables
Geometry and Trigonometry	21-30	Composite areas, circle theorems, similar triangles, trigonometry, volume, polygon angles

The most effective approach to these problems is:

Identify the concept before reaching for a formula.
Write out every step — the SAT rewards accuracy over speed.
Check your answer by substituting back or estimating whether the result is reasonable.
Review the common mistakes for each problem type — most errors fall into predictable patterns.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

Confusing terminology or concepts that appear similar but have distinct meanings.
Overlooking key assumptions or boundary conditions that limit applicability.

Mark this page as reviewed