ALEKS Math Practice Test 2026: 25 Problems With Worked Solutions

25 Original ALEKS-Style Problems, With Every Step Worked Out

The other ALEKS guides on this site explain the test format, scores and school cutoffs, retake rules, and how to build a study plan. This guide has a different job: it gives you 25 original ALEKS-style problems, works each one from start to finish, and identifies a specific mistake that can derail the solution.

These are not copied or recalled ALEKS assessment items. They are editorial practice questions built around topic families documented by McGraw Hill and ALEKS. The current official ALEKS PPL page describes an adaptive, open-response placement assessment that determines knowledge across 314 topics in 25 questions or fewer. Its scope runs from Basic Math to Calculus and also includes corequisite, Statistics, and non-STEM pathways.

The four tiers below are also an editorial study sequence, not an official ALEKS scoring model. ALEKS does not publish these four labels or promise that mastering one tier produces a particular course placement. Your institution chooses its own placement rules and course cutoffs. Use the tiers to locate the earliest skill family that is still unstable, then check your school's current policy for the score or course you need.

Verified 2026 format snapshot

• Format and length: McGraw Hill describes PPL as adaptive and open response, with placement reached in 25 questions or fewer across 314 topics.
• Time: A current official McGraw Hill PPL page says the assessment can be completed in 90 minutes. Treat that as guidance, not a universal local time limit; follow the window shown by your school or ALEKS account.
• Access and attempts: McGraw Hill advertises 12 months of Prep and Learning access and up to five assessment attempts. Its support page also says schools can set different attempt counts, waiting periods, and learning-module requirements.
• Calculator and feedback: The official ALEKS higher-education learner guide says to use the built-in basic calculator only when instructed. It also says assessment mode does not tell you immediately whether an answer was right or wrong.
• Placement: There is no universal course cutoff. The institution using ALEKS decides how a placement result maps to its own math sequence.

free ALEKS Math question bankPractice questions with detailed explanations

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Editorial Tier 1 — Arithmetic and Proportional Reasoning

Begin here to check fraction, percent, rate, and signed-number fluency. Calculator availability is item-specific; the official learner guide says to use it only when ALEKS instructs you to.

Problem 1 — Fractions with unlike denominators. Compute 5/6 − 3/8.

Solution. The least common denominator of 6 and 8 is 24. Rewrite each fraction: 5/6 = 20/24 and 3/8 = 9/24. Subtract the numerators over the common denominator: 20/24 − 9/24 = 11/24. It does not reduce, so 11/24 is the exact answer.

Trap. Subtracting numerators and denominators separately: (5 − 3)/(8 − 6) = 2/2 = 1. You must find a common denominator first; you never subtract denominators.

Problem 2 — Percent change. A jacket is marked down from $80 to $60. What is the percent decrease?

Solution. The amount of decrease is 80 − 60 = 20. Percent change is measured against the original amount: 20/80 = 0.25 = 25%.

Trap. Dividing by the new price instead of the original: 20/60 is about 33.3%. Percent decrease is decrease divided by original, not decrease divided by final.

Problem 3 — Proportional reasoning. If 3 pounds of coffee cost $18, how much do 7 pounds cost at the same rate?

Solution. Find the unit rate: 18 ÷ 3 = $6 per pound. Then 7 × 6 = $42. Equivalently, set up 3/18 = 7/x and cross-multiply: 3x = 126, so x = 42.

Trap. Multiplying $18 by 7 and reporting $126. The $18 price covers three pounds, so divide by 3 before scaling the one-pound rate to seven pounds.

Problem 4 — Order of operations with signed numbers. Evaluate −3^2 + 4(2 − 5).

Solution. Exponent first, and −3^2 means −(3^2) = −9. Inside the parentheses, 2 − 5 = −3, so 4(−3) = −12. Add: −9 + (−12) = −21.

Trap. Reading −3^2 as (−3)^2 = 9, which gives 9 − 12 = −3. Without parentheses, the exponent binds tighter than the negative sign.

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Editorial Tier 2 — Core Algebra

Use this group to check equations, inequalities, lines, systems, exponents, polynomials, and factoring before moving to expressions with more domain or notation demands.

Problem 5 — Multi-step linear equation. Solve 3(x − 4) = 2x + 5.

Solution. Distribute the 3 across both terms: 3x − 12 = 2x + 5. Subtract 2x from both sides: x − 12 = 5. Add 12: x = 17. Check: 3(17 − 4) = 39 and 2(17) + 5 = 39.

Trap. Distributing to only the first term: 3x − 4 = 2x + 5 gives x = 9. The 3 multiplies everything inside the parentheses, including the −4.

Problem 6 — Linear inequality with a sign flip. Solve −2x + 3 > 11.

Solution. Subtract 3: −2x > 8. Divide both sides by −2, and because you divided by a negative, reverse the inequality: x < −4. In interval notation, (−∞, −4).

Trap. Leaving the inequality as x > −4. Dividing or multiplying an inequality by a negative number reverses the direction of the sign.

Problem 7 — Slope from two points. Find the slope of the line through (−2, 5) and (4, −7).

Solution. Slope is rise over run: m = (y₂ − y₁)/(x₂ − x₁) = (−7 − 5)/(4 − (−2)) = −12/6 = −2.

Trap. Inverting the formula to run over rise, 6/−12 = −1/2. Slope is the change in y over the change in x, using the same point order in both differences.

Problem 8 — System of equations. Solve the system 2x + y = 7 and x − y = 2.

Solution. Add the equations so y cancels: 3x = 9, so x = 3. Substitute back: 3 − y = 2, so y = 1. The solution is (3, 1). Check: 2(3) + 1 = 7 and 3 − 1 = 2.

Trap. Finding x = 3 and stopping. This practice prompt asks for the full ordered pair, so both coordinates are needed.

Problem 9 — Exponent rules. Simplify (2x^3)^2 · x^4.

Solution. Apply the power to every factor inside the parentheses: (2x^3)^2 = 2^2 · x^(3·2) = 4x^6. Then multiply by x^4 and add exponents: 4x^6 · x^4 = 4x^10.

Trap. Forgetting to square the coefficient and writing 2x^10. The outside exponent applies to the 2 as well as the x^3.

Problem 10 — Multiplying binomials. Expand (2x − 3)(x + 5).

Solution. Multiply every pair of terms: 2x · x = 2x^2, 2x · 5 = 10x, −3 · x = −3x, and −3 · 5 = −15. Combine the middle terms: 2x^2 + 7x − 15.

Trap. Multiplying only the first and last terms to get 2x^2 − 15. Every term in the first factor must multiply every term in the second.

Problem 11 — Factoring a trinomial. Factor x^2 + 5x − 14.

Solution. Find two numbers that multiply to −14 and add to 5: 7 and −2. Therefore, x^2 + 5x − 14 = (x + 7)(x − 2). Expanding verifies the middle term, 7x − 2x = 5x.

Trap. Using (x − 7)(x + 2), which gives a middle term of −5x. The signs must reproduce both the product and the sum.

Problem 12 — Solving a quadratic by factoring. Solve x^2 − 3x − 10 = 0.

Solution. Factor: (x − 5)(x + 2) = 0. Set each factor equal to zero, giving x = 5 or x = −2.

Trap. Reporting only x = 5, or changing the second root to positive 2. Each zero factor gives a solution, and x + 2 = 0 means x = −2.

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Editorial Tier 3 — Advanced Algebra and Functions

This group adds rational expressions, radicals, function notation, the quadratic formula, and absolute value. Pay close attention to original domains and requested answer forms.

Problem 13 — Simplify a rational expression with restrictions. Simplify (x^2 − 9)/(x^2 + x − 6) and state the restrictions.

Solution. Factor the numerator as (x − 3)(x + 3) and the denominator as (x + 3)(x − 2). Cancel the common factor to get (x − 3)/(x − 2). From the original denominator, x ≠ −3 and x ≠ 2.

Trap. Dropping x ≠ −3 after the factor cancels. This practice prompt explicitly asks for restrictions, and both exclusions come from the original denominator. Cancel whole factors, not individual terms.

Problem 14 — Rational equation with an excluded candidate. Solve x/(x − 3) = 3/(x − 3) + 2.

Solution. First note x ≠ 3. Multiply every term by x − 3: x = 3 + 2(x − 3). Expanding gives x = 2x − 3, so the algebraic candidate is x = 3. That candidate is outside the original domain, so the equation has no solution.

Trap. Reporting x = 3 without checking the original denominator. Clearing a denominator can produce a candidate that the original equation excludes.

Problem 15 — Adding radicals. Simplify √50 + √18.

Solution. √50 = √(25 · 2) = 5√2 and √18 = √(9 · 2) = 3√2. Add the coefficients of the like radicals: 8√2.

Trap. Replacing the sum with √68. A sum of square roots is not generally the square root of the sum; simplify each radical before combining like radicals.

Problem 16 — Function composition. Given f(x) = 2x − 1 and g(x) = x^2, find f(g(3)).

Solution. Work inside out: g(3) = 9, then f(9) = 2(9) − 1 = 17.

Trap. Reversing the order and computing g(f(3)) = 25. Function composition is not generally commutative, so evaluate the named inner function first.

Problem 17 — Quadratic formula, exact form. Solve x^2 + 4x + 1 = 0.

Solution. With a = 1, b = 4, and c = 1, the formula gives x = (−4 ± √12)/2. Since √12 = 2√3, divide both numerator terms by 2: x = −2 ± √3.

Trap. Dividing only the radical by 2 and writing −4 ± √3. The denominator applies to the entire numerator.

Problem 18 — Absolute-value equation. Solve |2x − 3| = 7.

Solution. Because the right side is positive, use two cases. From 2x − 3 = 7, x = 5. From 2x − 3 = −7, x = −2. Thus x = 5 or x = −2.

Trap. Solving only the positive case. When |expression| equals a positive number, test both the positive and negative equations.

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Editorial Tier 4 — Precalculus Extensions

Finish with logarithms, exponentials, inverse functions, trigonometry, and coordinate geometry. These are useful advanced checks, not a promise of a particular placement.

Problem 19 — Evaluate a logarithm. Find log₂(32).

Solution. Ask which exponent on 2 produces 32. Since 2^5 = 32, log₂(32) = 5.

Trap. Treating log₂(32) as 32 ÷ 2 = 16. A logarithm returns an exponent, not a quotient.

Problem 20 — Exponential equation. Solve 5^(2x) = 125.

Solution. Rewrite 125 as 5^3. Equal positive bases give 2x = 3, so x = 3/2.

Trap. Stopping at 2x = 3 and reporting x = 3. Divide by the coefficient 2 to isolate x.

Problem 21 — Inverse function. Find the inverse of f(x) = 3x − 5.

Solution. Write y = 3x − 5, swap x and y, and solve: x = 3y − 5, so y = (x + 5)/3. Therefore, f⁻¹(x) = (x + 5)/3.

Trap. Writing the reciprocal 1/(3x − 5). An inverse function reverses the input-output rule; it is not the reciprocal of the function value.

Problem 22 — Exact trig value. Find the exact value of sin(60°).

Solution. From a 30-60-90 triangle or the unit circle, sin(60°) = √3/2. This practice prompt requests an exact value, so keep the radical form.

Trap. Entering 0.87 for this prompt or confusing sin(60°) with cos(60°) = 1/2. On ALEKS, follow the displayed instruction and answer field rather than assuming every item uses the same format.

Problem 23 — Right-triangle trigonometry. In a right triangle, the side opposite angle θ is 3 and the adjacent side is 4. Find sin θ.

Solution. The hypotenuse is √(3^2 + 4^2) = 5. Therefore, sin θ = opposite/hypotenuse = 3/5.

Trap. Using 3/4, which is tan θ, or 4/5. Sine uses the opposite side over the hypotenuse.

Problem 24 — Area of a circle, exact. Find the exact area of a circle with radius 5.

Solution. A = πr^2 = π(5)^2 = 25π. The prompt asks for an exact result, so retain π.

Trap. Using the circumference formula and getting 10π, or rounding to 78.5. Area squares the radius, and this prompt requests exact form.

Problem 25 — Distance between two points. Find the distance between (1, 2) and (4, 6).

Solution. Use the distance formula: √((4 − 1)^2 + (6 − 2)^2) = √(9 + 16) = √25 = 5.

Trap. Adding the raw coordinate differences. Distance comes from the Pythagorean theorem, so square the horizontal and vertical differences before taking the square root.

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How to Use the Four Editorial Tiers

This sequence is a study organizer. It is not an official ALEKS blueprint, it does not divide the 314-topic pool into four score bands, and it does not map directly to four courses. ALEKS adapts across its configured content, while your school decides how the resulting placement is used.

Score this set by skill, not only by total. If you miss Problem 13 because factoring failed, repair factoring before doing more rational expressions. If you solve Problem 22 but miss basic fraction arithmetic, start at Tier 1. This bottom-up method is an editorial recommendation, not an ALEKS scoring rule.

free ALEKS practice bankPractice questions with detailed explanations

Exact Entry: Follow the Prompt, Not a Universal Rule

Open response means you produce an answer instead of selecting from four choices. It does not establish one universal entry format for every ALEKS problem. The official ALEKS Placement Problem Types PDF includes tasks that request forms such as exact answers in terms of π, while other tasks call for numbers, expressions, graphs, or constructed responses.

Use three habits:

• Read the requested form before solving. If a prompt says exact, keep π or radicals; if it requests an ordered pair, supply both coordinates.
• Preserve mathematical domain information. Check original denominators and any operation that can introduce an invalid candidate. Enter restrictions when the prompt or interface requests them.
• Verify in the original problem. Substitute solutions after clearing denominators, squaring, or applying logarithms, and check both equations in a system.

On the actual assessment, use the symbols, fields, and calculator ALEKS presents. Do not assume that an answer accepted in one problem type will be entered identically in another.

Official Resources

• McGraw Hill ALEKS PPL — current product facts: adaptive open response, 25 questions or fewer, 314 topics, 12-month learning access, and up to five attempts.
• McGraw Hill ALEKS PPL overview — official statement that the assessment can be completed in 90 minutes.
• ALEKS PPL retake support — official warning that schools can set different attempt counts and retake requirements.
• ALEKS Placement Problem Types — official examples of assessable topic families and response demands.
• ALEKS Higher Education Learner Guide — official assessment guidance on paper, calculator use, help, and answer feedback.

Bottom Line

Use these 25 problems as an original paper diagnostic, not as a prediction of the exact questions, score, or course you will receive. The arithmetic and algebra recompute cleanly, every exact-form demand is stated by the practice prompt, and each trap identifies a check you can reuse.

free ALEKS Math practicePractice questions with detailed explanations