ALEKS Math Practice Test 2026: 25 Problems With Worked Solutions

A Full ALEKS Math Practice Test, With Every Step Worked Out

The other ALEKS guides on this site explain the test format, the 0-to-100 score, the retake rules, and how to build a study plan. This one does the thing none of them do: it hands you 25 real ALEKS-style problems and works every single one out, step by step, then names the exact wrong-answer trap that quietly costs students the point.

Here is the key thing to internalize before you start. ALEKS PPL is open response and adaptive. McGraw Hill's official ALEKS PPL page says the assessment reaches a placement in 25 questions or fewer across hundreds of topics: you type an exact answer instead of picking A, B, C, or D, and the software climbs or drops in difficulty based on what you get right. So the fastest way to prepare is not to memorize facts about the test. It is to solve the kinds of problems it asks and rehearse producing the exact answer — an integer, fraction, interval, coordinate pair, expression, or exact value.

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How to use this test: cover the solution, solve on paper, and write the exact answer you would type into ALEKS. Then check the worked steps and read the trap. On the real assessment there are no options to eliminate, so training your hand to produce and enter the clean form is half the score. A correct value entered in the wrong form (a rounded decimal where ALEKS wants an exact fraction) reads as wrong.

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Tier 1 — Arithmetic and Proportional Reasoning

Foundation for every placement. Weak arithmetic here silently lowers every score above it, and many of these items give you no on-screen calculator.

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 always measured against the original amount: 20/80 = 0.25 = 25%.

Trap. Dividing by the new price instead of the original: 20/60 ≈ 33%. Percent decrease is decrease ÷ original, not decrease ÷ 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. Setting the proportion up upside down, 3/18 = x/7, which gives x = 21/18 ≈ $1.17. Keep the same quantity (pounds) in matching positions when you write the proportion.

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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Tier 2 — Core Algebra (the College-Algebra Boundary)

These skills decide most college-algebra placements: linear equations and inequalities, lines, systems, exponents, polynomials, and factoring.

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. It balances.

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 always flips 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, (4 − (−2))/(−7 − 5) = 6/−12 = −1/2. Slope is the change in y over the change in x, in that order.

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

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

Trap. Finding x = 3 and stopping. ALEKS wants the full ordered pair; a system's answer is both coordinates, not just one variable.

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: writing 2x^6 instead of 4x^6. The exponent outside the parentheses applies to the 2 as well as the x.

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

Solution. Use FOIL: 2x · x = 2x^2, 2x · 5 = 10x, −3 · x = −3x, −3 · 5 = −15. Combine the middle terms: 10x − 3x = 7x. Result: 2x^2 + 7x − 15.

Trap. Multiplying only the first and last terms to get 2x^2 − 15 and dropping the middle term. 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. Those are +7 and −2. So x^2 + 5x − 14 = (x + 7)(x − 2). Check the middle term: 7x − 2x = 5x. Correct.

Trap. Getting the signs backward with (x − 7)(x + 2), which produces a middle term of −5x. The signs must reproduce both the product (−14) and the sum (+5).

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

Solution. Factor into (x − 5)(x + 2) = 0 (since −5 and +2 multiply to −10 and add to −3). Set each factor to zero: x − 5 = 0 or x + 2 = 0, so x = 5 or x = −2.

Trap. Reporting only x = 5, or writing x = 5 and x = 2 with the wrong sign. A factored equation gives one solution per factor, and the sign is the opposite of the number in the parentheses.

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Tier 3 — Advanced Algebra and Functions (the Precalculus Boundary)

Rational expressions, radicals, function notation, the quadratic formula, and absolute value. This tier is where a precalculus placement is decided.

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

Solution. Factor both parts. The numerator is a difference of squares: (x − 3)(x + 3). The denominator factors as (x + 3)(x − 2). Cancel the common factor (x + 3): the result is (x − 3)/(x − 2). The restrictions come from the original denominator being nonzero: x ≠ −3 and x ≠ 2.

Trap. Dropping the x ≠ −3 restriction after (x + 3) cancels. The value that was excluded before cancelling is still excluded, so it must be stated. Also, never cancel individual terms, only whole factors.

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

Solution. Multiply every term by (x − 3): x = 3 + 2(x − 3). Expand: x = 3 + 2x − 6, so x = 2x − 3. Then −x = −3, giving x = 3. But x = 3 makes the original denominator zero, so it is extraneous. The equation has no solution.

Trap. Reporting x = 3 without checking it against the domain. Any candidate that makes an original denominator zero must be rejected, even when the algebra produces it cleanly.

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

Solution. Simplify each radical first: √50 = √(25 · 2) = 5√2 and √18 = √(9 · 2) = 3√2. They now have the same radical part, so add the coefficients: 5√2 + 3√2 = 8√2.

Trap. Adding under the radical to get √68, or √(50 + 18). You can only combine radicals after simplifying to a common radicand, and even then you add the coefficients, not the numbers inside.

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

Solution. Work from the inside out. First g(3) = 3^2 = 9. Then f(9) = 2(9) − 1 = 17.

Trap. Reversing the order and computing g(f(3)): f(3) = 5, then g(5) = 25. Composition is not commutative, so f(g(3)) and g(f(3)) usually differ. Evaluate the inner function first.

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

Solution. With a = 1, b = 4, c = 1, the quadratic formula gives x = (−4 ± √(16 − 4))/2 = (−4 ± √12)/2. Simplify √12 = 2√3: x = (−4 ± 2√3)/2. Divide both terms by 2: x = −2 ± √3.

Trap. Dividing only the radical term by 2 to get −4 ± √3. Every term in the numerator, including the −4, must be divided by the denominator.

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

Solution. An absolute value equals 7 when the inside is 7 or −7. Case one: 2x − 3 = 7, so 2x = 10 and x = 5. Case two: 2x − 3 = −7, so 2x = −4 and x = −2. Solutions: x = 5 and x = −2.

Trap. Solving only the positive case and giving x = 5 alone. An absolute-value equation almost always yields two cases; you have to set the inside equal to both the positive and the negative value.

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Tier 4 — Precalculus and Calculus Readiness

Logarithms, exponentials, inverse functions, trigonometry, and coordinate geometry. This tier is what pushes a placement from precalculus up to calculus-ready.

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

Solution. Ask: 2 to what power gives 32? Since 2^5 = 32, log₂(32) = 5. A logarithm returns the exponent.

Trap. Treating log₂(32) as 32 ÷ 2 = 16, or answering 2^5 = 32. The log is asking for the exponent, not a quotient or the original number.

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

Solution. Write 125 as a power of 5: 125 = 5^3. With equal bases, set the exponents equal: 2x = 3, so x = 3/2.

Trap. Forgetting the coefficient on the exponent and stopping at 2x = 3 → x = 3, or trying 125 ÷ 5. Match the bases first, then solve the exponent equation, including the factor of 2.

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

Solution. Write y = 3x − 5, swap x and y, then solve for y: x = 3y − 5, so 3y = x + 5 and y = (x + 5)/3. The inverse is f⁻¹(x) = (x + 5)/3.

Trap. Confusing the inverse function with the reciprocal and writing 1/(3x − 5). The inverse undoes the function; it is not one over the function.

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

Solution. From the 30-60-90 triangle (or the unit circle), sin(60°) = √3/2. Enter the exact radical, not a decimal approximation.

Trap. Typing the rounded decimal 0.87 when ALEKS wants the exact value √3/2, or confusing it with cos(60°) = 1/2. On exact-value items, a rounded decimal scores as wrong.

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

Solution. First find the hypotenuse with the Pythagorean theorem: √(3^2 + 4^2) = √25 = 5. Then sin θ = opposite/hypotenuse = 3/5.

Trap. Using 3/4 (opposite over adjacent, which is tan θ) or 4/5. Sine is opposite over hypotenuse, so you must find the hypotenuse before you can write it.

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

Solution. Area = πr^2 = π(5)^2 = 25π. Leave the answer in terms of π for an exact result.

Trap. Using the circumference formula 2πr = 10π, or rounding to 78.5 when an exact answer is requested. Area uses r^2; the exact form keeps π rather than a decimal.

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) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

Trap. Forgetting to square the differences and just adding, or subtracting the coordinates once. The distance formula is the Pythagorean theorem applied to the horizontal and vertical gaps, so both differences must be squared.

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How the Four Tiers Map to Your Placement

Because ALEKS is adaptive, you may never see the top tier if the lower tiers are shaky. That is not a glitch: once the software confirms you are not calculus-ready, it stops climbing. Use this map to aim your study time at the tier that actually decides your course.

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Why Rehearsing the Exact Answer Matters as Much as the Math

Every worked solution above ends with a specific value in a specific form: 11/24, x < −4, 8√2, (3, 1), −2 ± √3, 25π. That is not decoration. On a multiple-choice quiz you can back into an answer by testing the options; on ALEKS there are no options, so the exact form is the answer. Three habits protect those points:

• Predict the answer type before you solve — integer, fraction, interval, ordered pair, expression, or exact value. It stops you from typing a decimal where a fraction is required.
• Write restrictions and both cases. Rational expressions need domain restrictions from the original denominator; absolute-value and quadratic problems usually have two solutions; even roots need nonnegative radicands.
• Check by substitution. Put your answer back into the original problem, especially after clearing denominators, squaring, or taking a log, to catch extraneous solutions like the one in Problem 14.

Official Resources

• McGraw Hill ALEKS PPL — official product page confirming the open-response adaptive format, placement in 25 questions or fewer, hundreds of topics from basic math through Calculus I readiness, up to five attempts, and 12-month learning-module access.
• ALEKS Placement Problem Types (PDF) — official reference for the math topic families ALEKS can assess, from signed numbers and linear equations through logarithms and trigonometry.

Bottom Line

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