MAAP Algebra I Practice Questions (2026) — Free Sample Questions with Answers

Answer: D

1. 1. $\sqrt{2}\cdot \sqrt{8}=\sqrt{16}=4$ (rational).
2. 2. $\sqrt{3}\cdot \sqrt{6}=\sqrt{18}=3\sqrt{2}$ (irrational).
3. 3. $\frac{2}{5}\sqrt{10}$ and $4\pi$ are nonzero rationals times irrationals (irrational).

$\frac{2}{5}\cdot \sqrt{10}$: A nonzero rational number times an irrational number is irrational.

$4\pi$: $\pi$ is irrational, and a nonzero rational multiple of it is irrational.

$\sqrt{3}\cdot \sqrt{6}$: $\sqrt{3}\cdot \sqrt{6}=\sqrt{18}=3\sqrt{2}$; 18 is not a perfect square, so the product is irrational.

$\sqrt{2}\cdot \sqrt{8}$: Correct. $\sqrt{2}\cdot \sqrt{8}=\sqrt{16}=4$, so this product of two irrational numbers is rational (N-RN.3).

A nonzero rational times an irrational is always irrational, but a product of two irrationals can be rational, so simplify before deciding.

Answer: C

1. 1. $y\ge 2x-4$: on or above the solid line.
2. 2. $y<-x+2$: below the dashed line.
3. 3. The overlap is the region containing (0, 0), to the left of the intersection point (2, 0).

upper region: Test (0, 4): $4\ge -4$ is true, but $4<2$ is false. This region is above both lines, so it satisfies only $y\ge 2x-4$.

lower region: Test (2, −3): $-3<0$ is true, but $-3\ge 0$ is false. This region is below both lines, so it satisfies only $y<-x+2$.

left region: Correct. Test (0, 0): $0\ge -4$ and $0<2$ are both true. This region is above the solid line and below the dashed line, so it is the solution set (A-REI.12).

right region: Test (5, 0): $0\ge 6$ and $0<-3$ are both false. This region is below the solid line and above the dashed line.

The solution set of a system of inequalities is the overlap of the half-planes; a test point not on either line identifies it quickly.

Answer: D

1. 1. $x=4\pm \sqrt{3}$ means $x-4=\pm \sqrt{3}$.
2. 2. Square both sides: $(x-4{)}^2=3$.
3. 3. Expand: $x^2-8x+16-3=0$, so $x^2-8x+13=0$.
4. 4. Check with the formula: $x=\frac{8\pm \sqrt{64-52}}{2}=\frac{8\pm 2\sqrt{3}}{2}=4\pm \sqrt{3}$.

$x^2-8x+3=0$: Uses 3 as the constant; the product of the solutions is $(4+\sqrt{3})(4-\sqrt{3})=16-3=13$.

$x^2+8x+13=0$: Wrong sign on the x-term: this equation has solutions $-4\pm \sqrt{3}$.

$x^2-8x+19=0$: Adds 3 to 16 instead of subtracting: $(x-4{)}^2+3=0$ has no real solutions.

$x^2-8x+13=0$: Correct. $x-4=\pm \sqrt{3}$ means $(x-4{)}^2=3$, so $x^2-8x+16=3$, or $x^2-8x+13=0$ (A-REI.4b).

Solutions $h\pm \sqrt{q}$ come from $(x-h{)}^2=q$; expanding that equation gives the standard form.

Answer: A

1. 1. $-16t^2+40t+6=22\Rightarrow -16t^2+40t-16=0$.
2. 2. Divide by $-8$: $2t^2-5t+2=0$.
3. 3. $(2t-1)(t-2)=0$, so $t=0.5$ or $t=2$.
4. 4. Check $t=0.5$: $-16(0.25)+20+6=22$.

0.5: Correct. $-16t^2+40t+6=22$ gives $-16t^2+40t-16=0$, or $2t^2-5t+2=0$. Then $(2t-1)(t-2)=0$, so $t=0.5$ or $t=2$; the first time is 0.5 second (A-REI.4b).

1.25: This is the time of the maximum height, $t=\frac{-40}{2(-16)}=1.25$, not the time at 22 feet.

2: The ball is at 22 feet again at 2 seconds, on the way down; the question asks for the first time.

2.64: This is approximately when the ball hits the ground ($h=0$), not when $h=22$.

Set the expression equal to the target value, not to zero, and use the context to pick which solution answers the question.

Answer: A

1. 1. $\frac{253+s}{4}\ge 85$.
2. 2. Multiply by 4: $253+s\ge 340$.
3. 3. $s\ge 87$, and $s\le 100$.
4. 4. Check: $s=87$ gives $\frac{340}{4}=85$ ✓; $s=86$ gives 84.75 ✗.

Graph A: Correct. $\frac{84+91+78+s}{4}\ge 85$ gives $253+s\ge 340$, so $s\ge 87$. With the 100-point maximum, the scores are from 87 to 100, both included (A-REI.3).

Graph B: Graph B has an open dot at 87, but "at least 85" includes an average of exactly 85, so a score of 87 works.

Graph C: Graph C shades from 80 to 87, reversing the inequality; lower scores pull the average down.

Graph D: Graph D starts at 85, the target average, not the needed fourth score; a score of 85 gives an average of 84.5.

Write the inequality from the average, solve it, then apply the limits of the context (here, a maximum score of 100).

Answer: D

1. 1. Each option must satisfy $12t+8h=600$, $t\le 35$, $h\le 40$, with whole numbers t and h.
2. 2. (10, 60): $h>40$. (25, 37.5): not a whole number. (30, 31): $360+248=608\ne 600$.
3. 3. (34, 24): $408+192=600$ and both are within the supply.

10 T-shirts and 60 hats: Brings in $120+480=600$ dollars, but the band has only 40 hats.

25 T-shirts and 37.5 hats: Brings in $600, but the band cannot sell half a hat; counts must be whole numbers.

30 T-shirts and 31 hats: Fits the supply, but brings in $360+248=608$ dollars, not exactly $600.

34 T-shirts and 24 hats: Correct. $12(34)+8(24)=408+192=600$, both counts are whole numbers, and $34\le 35$ and $24\le 40$ (A-CED.3).

A viable solution must satisfy the equation and every constraint of the context, including limits and whole-number counts.

Answer: D

1. 1. $An=40+5n$.
2. 2. $An-5n=40$.
3. 3. $n(A-5)=40$, so $n=\frac{40}{A-5}$.
4. 4. Check: 20 shirts cost $140, so $A=7$, and $\frac{40}{7-5}=20$.

$n=\frac{40}{A+5}$: Sign error when collecting the n-terms: $An-5n=40$, so the factor is $A-5$, not $A+5$.

$n=\frac{40}{A}-5$: Divides 40 by A and then subtracts 5, as if $5n$ were not divided by n.

$n=\frac{A-5}{40}$: Inverts the last step: from $n(A-5)=40$, divide 40 by $A-5$.

$n=\frac{40}{A-5}$: Correct. Multiply by n: $An=40+5n$. Subtract $5n$: $An-5n=40$. Factor: $n(A-5)=40$, so $n=\frac{40}{A-5}$ (A-CED.4).

When the variable appears in more than one term, collect those terms on one side and factor the variable out before dividing.

Answer: A

1. 1. $C=2(x^2-12x)+190$.
2. 2. Half of $-12$ is $-6$, and $(-6{)}^2=36$: $C=2(x^2-12x+36)-2(36)+190$.
3. 3. $C=2(x-6{)}^2+118$.
4. 4. Check: $2(x^2-12x+36)+118=2x^2-24x+72+118=2x^2-24x+190$.

$2(x-6{)}^2+118$: Correct. Factor 2 from the x-terms: $2(x^2-12x)+190$. Add and subtract 36 inside: $2(x-6{)}^2-72+190=2(x-6{)}^2+118$, so the minimum cost is $118 at 6 dozen (A-SSE.3b).

$2(x-6{)}^2+154$: Subtracts 36 instead of $2\cdot 36=72$; the 36 added inside the parentheses is multiplied by 2.

$2(x+6{)}^2+118$: Sign error in the square: $2(x+6{)}^2$ has $+24x$, not $-24x$.

$2(x-12{)}^2-98$: Uses half of 24 without first factoring out 2; $2(x-12{)}^2=2x^2-48x+288$, which does not match.

To complete the square when $a\ne 1$, factor a out of the x-terms first, and subtract a times the added square outside the parentheses.

Answer: B

1. 1. Differences: 22, 31.68, 45.62 — not constant, so not linear.
2. 2. Ratios: 1.44, 1.44, 1.44 — constant, so exponential.
3. 3. The x-step is 2, so the factor 1.44 is per 2 units.

Linear, because $x$ increases by 2 each time: Equal steps in $x$ are needed to test either model; they do not make the function linear.

Exponential, because $y$ is multiplied by 1.44 over each 2-unit interval: Correct. $72÷50=1.44$, $103.68÷72=1.44$, and $149.30÷103.68\approx 1.44$: equal factors over equal intervals (F-LE.1a).

Exponential, because $y$ is multiplied by 1.44 for each 1-unit increase in $x$: The factor 1.44 is for a 2-unit change in $x$; per unit the factor is 1.2, since $1.2^2=1.44$.

Linear, because $y$ increases by 22 over each 2-unit interval: Checks only the first difference; the next differences are 31.68 and 45.62, so they are not equal.

Linear functions change by equal differences over equal intervals; exponential functions change by equal factors over equal intervals.

Answer: D

1. 1. $g(3)=f(2\cdot 3)$.
2. 2. $2\cdot 3=6$.
3. 3. From the table, $f(6)=31$.

15: Computes $f(3)+2=15$, treating $f(2x)$ as a vertical shift.

24: Computes $f(3+2)=f(5)$, treating $f(2x)$ as a horizontal shift.

26: Computes $2f(3)=26$, which is the vertical stretch $2f(x)$, not $f(2x)$.

31: Correct. $g(3)=f(2\cdot 3)=f(6)=31$ (F-BF.3).

In $f(kx)$ the input is multiplied before $f$ acts (a horizontal change); in $kf(x)$ the output is multiplied (a vertical change).

Answer: B

1. 1. Domain: $x-2\ge 0$, so $x\ge 2$.
2. 2. Starting point: $f(2)=0+1=1$, so $(2,1)$.
3. 3. More points: $f(3)=2$, $f(6)=3$; the graph rises to the right.

Graph A: Starts at $(-2,1)$; reading $\sqrt{x-2}$ as a shift left 2 reverses the horizontal shift.

Graph B: Correct. The domain is $x\ge 2$; the graph starts at $(2,1)$ and rises to the right through $(3,2)$ and $(6,3)$ (F-IF.7b).

Graph C: Starts at $(1,2)$, which swaps the horizontal and vertical shifts.

Graph D: Starts at $(2,1)$ but extends to the left, which is the graph of $\sqrt{2-x}+1$ with domain $x\le 2$.

For $\sqrt{x-h}+k$ the graph starts at $(h,k)$ and extends to the right, because the expression under the root must be at least 0.

Answer: C

1. 1. $0\le x\le 2$: $(7-3)÷2=2$.
2. 2. $0\le x\le 8$: $(18-3)÷8=1.875$.
3. 3. $2\le x\le 6$: $(17-7)÷4=2.5$.
4. 4. $4\le x\le 8$: $(18-9)÷4=2.25$.
5. 5. Greatest: $2\le x\le 6$.

$0\le x\le 2$: The rate is $\frac{7-3}{2}=2$, less than 2.5.

$0\le x\le 8$: This interval has the greatest total change, 15, but the rate is $\frac{15}{8}=1.875$.

$2\le x\le 6$: Correct. The rate is $\frac{17-7}{6-2}=\frac{10}{4}=2.5$, the greatest of the four (F-IF.6).

$4\le x\le 8$: The rate is $\frac{18-9}{4}=2.25$, less than 2.5.

Compare average rates, not total changes: divide each change in output by the length of its interval.

Answer: B

Graph A: The return trip takes only 8 minutes (14 to 22), the same speed as the walk out; Mateo walks home more slowly and arrives at 30 minutes.

Graph B: Correct. Rising to 0.6 mile by 8 minutes, flat from 8 to 14 minutes, then falling to 0 at 30 minutes; the 16-minute return is slower than the 8-minute walk out (F-IF.4).

Graph C: Keeps rising after the rest, which shows total distance walked, not distance from home.

Graph D: The walk out takes 16 minutes and the return 8 minutes, which reverses which part of the trip is slower.

Match each phase of the description to a graph feature: increasing, constant (horizontal) and decreasing pieces, with steepness showing speed.

Answer: A

1. 1. $f(x+2)=(x+2{)}^2-3(x+2)$.
2. 2. $(x+2{)}^2=x^2+4x+4$ and $-3(x+2)=-3x-6$.
3. 3. Combine: $x^2+4x+4-3x-6=x^2+x-2$.
4. 4. Check with $x=1$: $f(3)=9-9=0$ and $1+1-2=0$.

$x^2+x-2$: Correct. Replace every $x$ with $(x+2)$: $(x+2{)}^2-3(x+2)=x^2+4x+4-3x-6=x^2+x-2$ (F-IF.2).

$x^2-3x+2$: This is $f(x)+2$, which adds 2 to the output instead of the input.

$x^2-3x-2$: Squares $(x+2)$ as $x^2+4$, dropping the middle term $4x$.

$x^2+x+10$: Distributes $-3(x+2)$ as $-3x+6$; the correct product is $-3x-6$.

To find $f(\text{expression})$, substitute the whole expression, in parentheses, for every $x$; $f(x+2)$ is not $f(x)+2$.

Answer: B

1. 1. Condition: 10th graders, total 120.
2. 2. 10th graders who bring lunch: 54.
3. 3. $54÷120=0.45$, or 45%.

About 24.5%: This is $54÷220$, the joint relative frequency out of all students, not out of 10th graders.

45%: Correct. Among the 120 10th graders, 54 bring lunch: $54÷120=0.45=45\%$ (S-ID.5).

About 54.5%: This is $120÷220$, the percent of all students who are in grade 10.

About 64.3%: This is $54÷84$, the percent of lunch-bringers who are 10th graders; the condition is reversed.

For a conditional relative frequency, divide by the total of the group named in the condition ("of the 10th graders").

Answer: B

1. 1. Total: $4+9+6+3+2=24$, so the median is between the 12th and 13th values.
2. 2. Cumulative counts: 4 below 10, $4+9=13$ below 20.
3. 3. The 12th and 13th values are in the 10-to-20 interval.

Exactly 9 students have a commute time of 15 minutes.: The bar shows 9 students somewhere from 10 to 20 minutes; a histogram does not give individual values.

The median commute time is from 10 minutes up to 20 minutes.: Correct. The median is the mean of the 12th and 13th values; 4 values are below 10 and 13 are below 20, so both lie in the 10-to-20 interval (S-ID.1).

More than half of the students have commutes of 20 minutes or more.: Only $6+3+2=11$ of 24 students are at 20 minutes or more, which is less than half.

The longest commute time is exactly 50 minutes.: The last interval excludes 50, and the longest value could be anywhere from 40 up to 50 minutes.

A histogram shows counts in intervals, not exact values; use cumulative counts to locate the interval that contains the median.