ALEKS PPL Practice Questions (2026) — Free Sample Questions with Answers

Answer: 1

1. 1. Parentheses: $\frac{5}{6}-\frac{1}{4}=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}$.
2. 2. Divide: $\frac{7}{12}÷\frac{7}{8}=\frac{7}{12}\cdot \frac{8}{7}=\frac{8}{12}=\frac{2}{3}$.
3. 3. Add: $\frac{2}{3}+\frac{1}{3}=1$.
4. 4. Answer: $1$.
5. 5. Common error: adding $\frac{7}{8}+\frac{1}{3}$ before dividing, which gives $\frac{7}{12}÷\frac{29}{24}=\frac{14}{29}$.

The order of operations applies to fractions exactly as to whole numbers: grouping, then division, then addition.

Answer: 66 π ft²

1. 1. Surface area: $SA=2\pi r^2+2\pi rh$.
2. 2. The radius is half the diameter: $r=3$ ft.
3. 3. Two bases: $2\pi \cdot 3^2=18\pi$.
4. 4. Curved side: $2\pi \cdot 3\cdot 8=48\pi$.
5. 5. Total: $18\pi +48\pi =66\pi {\text{ ft}}^2$ (type 66).
6. 6. Common errors: using the diameter as the radius ($72\pi +96\pi =168\pi$) or including only one base ($57\pi$).

A cylinder's surface is two circles plus a rectangle that wraps around (width = circumference 2πr, height = h).

Answer: 21.5 cm²

1. 1. Area of the square: $10^2=100{\text{ cm}}^2$.
2. 2. The circle's diameter equals the side of the square, so its radius is $5$ cm.
3. 3. Area of the circle: $\pi r^2\approx 3.14\cdot 25=78.5{\text{ cm}}^2$.
4. 4. Shaded area: $100-78.5=21.5{\text{ cm}}^2$.
5. 5. Common errors: using the radius 10 (circle area 314, larger than the square) or using $\pi d=31.4$, the circumference, instead of the area.

For a shaded region, subtract the inner figure's area from the outer one; an inscribed circle has diameter equal to the square's side.

Answer: (-4,4]

1. 1. Subtract 3 from all three parts: $-8<2x\le 8$.
2. 2. Divide all three parts by 2: $-4<x\le 4$.
3. 3. In interval notation: $(-4,4]$.
4. 4. Common errors: swapping the brackets ($[-4,4)$) or subtracting 3 from only the middle part.

Apply each step to all three parts of a compound inequality; "<" gives a parenthesis and "≤" gives a bracket.

Answer: 8 pounds

1. 1. Let $x$ = pounds of $9 beans; then $20-x$ = pounds of $14 beans.
2. 2. Total value: $9x+14(20-x)=12\cdot 20$.
3. 3. Simplify: $9x+280-14x=240$, so $-5x+280=240$.
4. 4. $-5x=-40$, so $x=8$ pounds (and 12 pounds of the $14 beans).
5. 5. Check: $9(8)+14(12)=72+168=240$. Common error: answering 12, the amount of the $14 beans.

In a value mixture, (price × amount) for each part adds up to (price × amount) for the blend.

Answer: All real numbers

1. 1. Distribute: $6x-12+2=6x-10$.
2. 2. Combine: $6x-10=6x-10$.
3. 3. Subtract $6x$ from both sides: $-10=-10$, a true statement with no variable left.
4. 4. The equation is true for every $x$: all real numbers.
5. 5. Common errors: answering $x=0$ because the $x$-terms cancelled, or typing no solution (that is the case when the leftover statement is false).

When the variable cancels, a true statement (like −10 = −10) means every real number works; a false one (like −4 = −1) means no solution.

Answer: -2/3

1. Use the two marked points $(-2,4)$ and $(4,0)$.
2. Rise: $0-4=-4$. Run: $4-(-2)=6$.
3. Slope $m=\frac{\text{rise}}{\text{run}}=\frac{-4}{6}=-\frac{2}{3}$.
4. Answer: $-2/3$. Common errors: $-3/2$ (run over rise) and $2/3$ (dropping the sign; the line falls from left to right, so the slope is negative).

Slope is rise over run between two points on the line; a line that falls from left to right has a negative slope.

Answer: (2x - 3)(3x + 5)

1. AC method: $a\cdot c=6(-15)=-90$. Find two numbers with product $-90$ and sum $1$: $10$ and $-9$.
2. Split the middle term: $6x^2+10x-9x-15$.
3. Group: $2x(3x+5)-3(3x+5)$.
4. Factor out $(3x+5)$: $(3x+5)(2x-3)$.
5. Answer: $(2x-3)(3x+5)$. Check: $6x^2+10x-9x-15=6x^2+x-15$. ✓ Common error: $(2x+3)(3x-5)$, which expands to $6x^2-x-15$ (middle sign wrong).

For ax² + bx + c with a > 1, find two numbers that multiply to ac and add to b, split the middle term, and factor by grouping; always check by expanding.

Answer: (1 - sqrt(7))/3, (1 + sqrt(7))/3

1. Identify $a=3$, $b=-2$, $c=-2$.
2. Discriminant: $b^2-4ac=(-2{)}^2-4(3)(-2)=4+24=28$.
3. Formula: $x=\frac{-(-2)\pm \sqrt{28}}{2(3)}=\frac{2\pm \sqrt{28}}{6}$.
4. Simplify the radical: $\sqrt{28}=\sqrt{4\cdot 7}=2\sqrt{7}$, so $x=\frac{2\pm 2\sqrt{7}}{6}$.
5. Divide every term by 2: $x=\frac{1\pm \sqrt{7}}{3}$.
6. Answer: $\frac{1-\sqrt{7}}{3},\frac{1+\sqrt{7}}{3}$. Common errors: $\frac{-1\pm \sqrt{7}}{3}$ (using $b$ instead of $-b$) and $\frac{2\pm \sqrt{28}}{6}$ (correct but not simplified).

In x = (−b ± √(b² − 4ac)) / 2a, be careful with −b when b is negative, simplify the radical, then reduce only if every term shares the factor.

Answer: 0

1. Factor the right denominator: $x^2-1=(x-1)(x+1)$, so the LCD is $(x-1)(x+1)$ and $x\ne 1,-1$.
2. Multiply every term by the LCD: $x(x+1)-2(x-1)=2$.
3. Expand: $x^2+x-2x+2=2$, so $x^2-x=0$.
4. Factor: $x(x-1)=0$, so $x=0$ or $x=1$.
5. Reject $x=1$: it makes $x-1=0$. Check $x=0$: $\frac{0}{-1}-\frac{2}{1}=-2$ and $\frac{2}{-1}=-2$. Answer: $x=0$.
6. Common errors: answering $0,1$ (keeping the extraneous root) and $x^2-x-4=0$ from writing $-2(x-1)$ as $-2x-2$.

After clearing denominators, every candidate must be checked against the excluded values; a root that makes a denominator zero is extraneous.

Answer: 3+sqrt(5)

1. Multiply numerator and denominator by the conjugate $3+\sqrt{5}$: $\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$.
2. Denominator: $3^2-(\sqrt{5}{)}^2=9-5=4$.
3. Simplify: $\frac{4(3+\sqrt{5})}{4}=3+\sqrt{5}$.
4. Common errors: $\frac{6+2\sqrt{5}}{7}$ (using $9+5=14$ in the denominator) and $\frac{4\sqrt{5}}{3\sqrt{5}-5}$ (multiplying by $\sqrt{5}$ alone, which does not clear the radical).

To rationalize $a-\sqrt{b}$, multiply by its conjugate $a+\sqrt{b}$; the denominator becomes $a^2-b$.

Answer: ln(5)/3

1. Isolate the exponential: divide by 4 to get $e^{3x}=5$.
2. Take the natural log of both sides: $3x=\ln 5$.
3. Divide by 3: $x=\frac{\ln 5}{3}$.
4. Common errors: $\frac{\ln 20}{3}$ (taking logs before dividing by 4) and $\frac{\ln 5}{3}$ written as $\ln \frac{5}{3}$ (the 3 divides the log, not the argument).

Isolate the exponential expression before taking logarithms; use ln when the base is e.

Answer: 3

1. Product rule: ${\log }_2[(x+5)(x-1)]=4$.
2. Exponential form: $(x+5)(x-1)=2^4=16$.
3. Expand and set to zero: $x^2+4x-5=16$, so $x^2+4x-21=0$.
4. Factor: $(x+7)(x-3)=0$, so $x=-7$ or $x=3$.
5. Reject $x=-7$: it makes $x+5$ and $x-1$ negative, and logs of negative numbers are undefined. Check $x=3$: ${\log }_{2}8+{\log }_{2}2=3+1=4$. Answer: $x=3$.
6. Common errors: answering $-7$ (keeping the extraneous root) and $x=6$ (treating ${\log }_2(x+5)+{\log }_2(x-1)$ as ${\log }_2(2x+4)$, so $2x+4=16$).

Combine the logs into one, rewrite in exponential form, and discard any solution that makes the argument of an original logarithm zero or negative.

Answer: 12/5

1. Pythagorean identity: ${\sin }^2\theta =1-{\left(-\frac{5}{13}\right)}^2=1-\frac{25}{169}=\frac{144}{169}$.
2. $\theta$ is in Quadrant III, where sine is negative: $\sin \theta =-\frac{12}{13}$.
3. $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{-12/13}{-5/13}=\frac{12}{5}$.
4. Common errors: $-\frac{12}{5}$ (forgetting that tangent is positive in Quadrant III) and $\frac{5}{12}$ (computing $\frac{\cos \theta }{\sin \theta }$, the cotangent).

Use sin²θ + cos²θ = 1 for the missing value, choose its sign from the quadrant, then form the requested ratio.