GA Milestones EOC Practice Questions (2026) — Free Sample Questions with Answers

Answer: D

$b^2$: Can be rational. If $b=\sqrt{5}$, then $b^2=5$.

$b\sqrt{2}$: Can be rational. If $b=\sqrt{8}$, then $b\sqrt{2}=\sqrt{16}=4$; a product of two irrational numbers is not always irrational.

$b+\sqrt{3}$: Can be rational. If $b=4-\sqrt{3}$, then $b+\sqrt{3}=4$; a sum of two irrational numbers is not always irrational.

$ab$: Correct. A nonzero rational number times an irrational number is always irrational: if $ab$ were rational, then $b=\frac{ab}{a}$ would be a quotient of rational numbers, which is rational.

Closure rules: rational ± rational and rational × rational stay rational; rational + irrational and nonzero rational × irrational are always irrational; two irrationals can combine to a rational.

Answer: B

1. 1. Gym: $\sqrt{(-4-1{)}^2+(5-2{)}^2}=\sqrt{34}\approx 5.83$.
2. 2. Library: $\sqrt{(-2-1{)}^2+(-2-2{)}^2}=\sqrt{25}=5$.
3. 3. Pool: $\sqrt{(5-1{)}^2+(6-2{)}^2}=\sqrt{32}\approx 5.66$.
4. 4. Store: $\sqrt{(7-1{)}^2+(2-2{)}^2}=\sqrt{36}=6$.
5. 5. The smallest is 5, the library. Check: $25<32<34<36$, so comparing the squared distances gives the same order.

The gym: $\sqrt{(-5{)}^2+3^2}=\sqrt{34}\approx 5.83$ blocks, farther than the library.

The library: Correct. $\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{25}=5$ blocks, the shortest of the four distances.

The pool: $\sqrt{4^2+4^2}=\sqrt{32}\approx 5.66$ blocks, close but still farther than 5.

The store: Counting blocks along the grid ($6+0=6$ versus $3+4=7$ for the library) makes the store look closest, but its straight-line distance is 6 blocks, the farthest.

Straight-line distance comes from the distance formula; counting grid blocks measures a different path and can change which location is closest.

Answer: C

1. 1. $AB$ is horizontal: $5-(-3)=8$.
2. 2. $AC=\sqrt{(1-(-3){)}^2+(4-(-2){)}^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$.
3. 3. $BC=\sqrt{(1-5{)}^2+(4-(-2){)}^2}=\sqrt{16+36}=2\sqrt{13}$.
4. 4. Perimeter $=8+2\sqrt{13}+2\sqrt{13}=8+4\sqrt{13}\approx 22.42$.
5. 5. Check: each slanted side ($\approx 7.21$) is longer than its rise of 6 and shorter than $4+6=10$.

$8+2\sqrt{13}$: Finds $AC=2\sqrt{13}$ but leaves out side $BC$, which has the same length.

20: Counts each slanted side as 6 units, its vertical rise only; slanted sides need the distance formula.

$8+4\sqrt{13}$: Correct. $AB=8$, and $AC=BC=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}$, so the perimeter is $8+4\sqrt{13}\approx 22.42$ units.

28: Adds the horizontal and vertical changes ($4+6=10$) for each slanted side instead of using the distance formula.

Horizontal and vertical sides can be counted on the grid, but slanted sides must be found with the distance formula.

Answer: C

1. 1. Model: $f(w)=400(1.25{)}^w$.
2. 2. $1.25^6\approx 3.8147$.
3. 3. $400\times 3.8147\approx 1,525.9\approx 1,526$.
4. 4. Check by continuing the table: 976.56 (week 4), 1,220.70 (week 5), 1,525.88 (week 6). ✓

1,000: Adds 100 followers each week, $400+6(100)$, a linear model.

1,221: Uses the exponent $w-1$: $400(1.25{)}^5\approx 1,221$ is the week-5 value.

1,526: Correct. Using the factor from Part A, $f(w)=400(1.25{)}^w$, so $f(6)=400(1.25{)}^6\approx 400(3.815)\approx 1,526$.

1,907: Uses the exponent 7, $400(1.25{)}^7\approx 1,907$, counting week 0 as the first week.

Part B builds on Part A: the factor found there gives the model $f(w)=a\cdot b^w$ with $a$ the week-0 value.

Answer: C

1. 1. $4=2^2$ and $8=2^3$.
2. 2. $4^{x+1}=2^{2(x+1)}$ and $8^x=2^{3x}$.
3. 3. $2(x+1)=3x\Rightarrow 2x+2=3x\Rightarrow x=2$.
4. 4. Check: $4^3=64$ and $8^2=64$. ✓

$x+1=x$: Sets the exponents equal without first writing both sides with the same base; $4$ and $8$ are different bases.

$2x+1=3x$: Writes $4^{x+1}$ as $2^{2x+1}$; the 2 must multiply the whole exponent: $2(x+1)$.

$2(x+1)=3x$: Correct. $4^{x+1}=(2^2{)}^{x+1}=2^{2(x+1)}$ and $8^x=(2^3{)}^x=2^{3x}$. With equal bases, the exponents are equal: $2(x+1)=3x$, so $x=2$.

$4(x+1)=8x$: Multiplies each base by its exponent; equal powers do not mean the bases times the exponents are equal.

Write both sides as powers of the same base, using $(a^m{)}^n=a^{mn}$; then set the exponents equal.

Answer: D

1. 1. $w^2+5w-104=0$.
2. 2. Factor: $(w-8)(w+13)=0$, so $w=8$ or $w=-13$.
3. 3. Reject $-13$. Width 8 ft, length 13 ft; check $8\cdot 13=104$. ✓
4. 4. Perimeter $=2(8)+2(13)=42$ feet.

8: This is the width, $w=8$ feet, not the perimeter.

13: This is the length, $8+5=13$ feet (also the size of the negative solution, $-13$).

21: Adds the width and length once, $8+13$; the perimeter counts each side twice.

42: Correct. $w^2+5w-104=0$ factors as $(w-8)(w+13)=0$; the width must be positive, so $w=8$, the length is 13, and the perimeter is $2(8+13)=42$ feet.

Solve the quadratic, reject solutions that do not make sense (a negative width), then answer the question that was asked.

Answer: Divide each term by 2: $x^2+6x-7=0$ → Add 7 to both sides: $x^2+6x=7$ → Add 9 to both sides: $x^2+6x+9=16$ → Factor the left side: $(x+3)^2=16$ → Take square roots: $x+3=\pm4$, so $x=1$ or $x=-7$

1. 1. $2x^2+12x-14=0\to x^2+6x-7=0$.
2. 2. $x^2+6x=7$.
3. 3. $x^2+6x+9=16$.
4. 4. $(x+3{)}^2=16$.
5. 5. $x+3=\pm 4$, so $x=1$ or $x=-7$. Check $x=-7$: $2(49)-84-14=0$. ✓

Divide each term by 2: $x^2+6x-7=0$: First, make the leading coefficient 1 so that half of $b$ can be squared directly.

Add 7 to both sides: $x^2+6x=7$: Second, move the constant term to the right side to leave room to complete the square.

Add 9 to both sides: $x^2+6x+9=16$: Third, add ${\left(\frac{6}{2}\right)}^2=9$ to both sides to keep the equation balanced.

Factor the left side: $(x+3{)}^2=16$: Fourth, the left side is now a perfect square, giving the form $(x-p{)}^2=q$ with $p=-3$.

Take square roots: $x+3=\pm 4$, so $x=1$ or $x=-7$: Last, take both square roots and solve; check $x=1$: $2+12-14=0$.

Completing the square: make the leading coefficient 1, isolate the $x$-terms, add $(b/2{)}^2$ to both sides, factor, then take both square roots.

Answer: B

1. 1. $(5,26)$: $31>30$ (fails the first inequality).
2. 2. $(12,14)$: $26\le 30$ and $306\ge 300$ (passes both).
3. 3. $(16,10)$: $26\le 30$ but $278<300$ (fails the second).
4. 4. $(18.5,11)$: passes both, but 18.5 washes is not a whole number.

$(5,26)$: Earns $40+390=430$ dollars, but $5+26=31$ washes is more than 30.

$(12,14)$: Correct. $12+14=26\le 30$ and $8(12)+15(14)=96+210=306\ge 300$, with whole numbers of washes.

$(16,10)$: $16+10=26$ washes is allowed, but $128+150=278$ dollars is short of $300.

$(18.5,11)$: Satisfies both inequalities ($29.5$ washes and $313), but the club cannot do half a car wash.

A combination is viable only if it satisfies every constraint in the system and is realistic in context, such as a whole number of items.

Answer: D

1. 1. Boundary: $y=-\frac{1}{2}x+3$, with $y$-intercept 3 and slope $-\frac{1}{2}$ (falls 1 for every 2 to the right).
2. 2. The symbol $<$ means the boundary is dashed.
3. 3. "$y$ less than" means shade below the line.
4. 4. Check with $(0,0)$: $0<3$ is true, so the origin is shaded in the correct graph, as in Graph D.

Graph A: Graph A has the right line and shading, but its boundary is solid; a strict inequality needs a dashed line.

Graph B: Graph B has the dashed line but shades above it, which shows $y>-\frac{1}{2}x+3$.

Graph C: Graph C has slope $+\frac{1}{2}$; its line rises from left to right, which shows $y<\frac{1}{2}x+3$.

Graph D: Correct. Graph D has a dashed boundary (strict $<$) with slope $-\frac{1}{2}$ and $y$-intercept 3, shaded below the line.

Use $<$ or $>$ for a dashed boundary and $\le$ or $\ge$ for a solid one; in slope-intercept form, $y<$ shades below and $y>$ shades above.

Answer: C

1. 1. The table doubles each step, so $f(x)=2^x$.
2. 2. $g(5)=4(25)=100$ and $f(5)=32$.
3. 3. $f(10)=1,024$ and $g(10)=400$.

$g(x)$ is greater than $f(x)$ for all $x>0$.: Fails for large $x$: $f(10)=1,024$ but $g(10)=400$.

$f(x)$ is greater than $g(x)$ for all $x>0$.: Fails at $x=5$: $f(5)=32$ but $g(5)=100$.

$g(5)$ is greater than $f(5)$, but $f(10)$ is greater than $g(10)$.: Correct. $g(5)=100>f(5)=32$, and $f(10)=2^{10}=1,024>g(10)=400$. The exponential overtakes the quadratic.

$f(5)$ is greater than $g(5)$, but $g(10)$ is greater than $f(10)$.: Reverses the comparison; the exponential falls behind first and then overtakes the quadratic.

An increasing exponential function eventually exceeds any quadratic function, even if the quadratic is greater for a while.

Answer: C

1. 1. $a=y(0)=5$.
2. 2. Each 2-unit step multiplies $y$ by 4.
3. 3. $b^2=4$, so $b=2$.
4. 4. Check: $5(2{)}^2=20$. ✓

$y=5(4{)}^x$: Uses 4, the factor for each 2-unit step in $x$, as the factor for each 1-unit step.

$y=5+7.5x$: Fits only the first two points with a line; $y$ is multiplied, not increased by a constant.

$y=5(2{)}^x$: Correct. $y$ is multiplied by 4 each time $x$ increases by 2, so the factor per unit is $b$ with $b^2=4$, $b=2$. Check: $5(2{)}^4=80$ and $5(2{)}^6=320$.

$y=2(5{)}^x$: Swaps $a$ and $b$; at $x=0$ it gives 2, not 5.

Equal factors over equal intervals: if the step in $x$ is 2, the factor per 1 unit is the square root of the factor per step.

Answer: A

1. 1. The table fits $f(x)=x^2+x$ (check: $3^2+3=12$).
2. 2. $g(4)=28$ and $f(4)=20$, so $g$ is greater at $x=4$.
3. 3. $f(5)=30<g(5)=34$; $f(6)=42>g(6)=40$.
4. 4. After $x=6$ the gap grows because $f$ increases by more than 6 per step.

$g(4)$ is greater than $f(4)$, but $f(x)$ eventually exceeds $g(x)$ as $x$ increases.: Correct. $g(4)=28>f(4)=20$, but $f$ grows by increasing amounts (2, 4, 6, 8, …) while $g$ always grows by 6. By $x=6$, $f(6)=42>g(6)=40$.

$g(x)$ stays greater than $f(x)$ for all $x\ge 0$, since $g$ always increases by 6.: Assumes the linear function stays ahead; the quadratic's increases keep growing, so it overtakes $g$.

$f(x)$ is greater than $g(x)$ for every value of $x$ greater than 0.: Fails at $x=4$: $f(4)=20$ but $g(4)=28$.

$f$ and $g$ have the same average rate of change from $x=0$ to $x=4$.: $f$: $\frac{20-0}{4}=5$; $g$: $\frac{28-4}{4}=6$, so the rates differ.

A quadratic with a positive leading coefficient eventually exceeds any linear function, because its rate of change keeps increasing.

Answer: D

1. 1. $d=\frac{11-(-1)}{5-2}=\frac{12}{3}=4$; also $\frac{27-11}{9-5}=4$.
2. 2. $a_1=a_2-d=-1-4=-5$.
3. 3. $a_n=-5+4(n-1)$.
4. 4. Check: $a_5=-5+16=11$ and $a_9=-5+32=27$.

$a_n=-1+4(n-1)$: Uses the first table value as $a_1$, but $-1$ is $a_2$.

$a_n=-5+12(n-1)$: Uses the change between table entries, 12, as $d$; those entries are 3 terms apart.

$a_n=-9+4(n-1)$: Subtracts two differences from $a_2$; $a_1$ is only one step before $a_2$.

$a_n=-5+4(n-1)$: Correct. From $n=2$ to $n=5$ the terms rise by 12 over 3 steps, so $d=4$; one step before $a_2=-1$ is $a_1=-5$.

When terms are not consecutive, divide the change in value by the change in $n$ to get $d$, then step back to $a_1$.

Answer: A

The study does not show that adding libraries would bring in more dentists.: Correct. Larger towns tend to have more of both, so town size is a likely lurking variable. A strong correlation shows the two counts rise together, not that one causes the other.

Building more libraries in a town will cause more dentists to move there.: Treats correlation as causation; the study observed towns and did not test what happens when libraries are added.

Because $r$ is less than 1, there is no real association between the two variables.: An $r$ of 0.91 shows a strong association; it does not have to equal 1.

More dentists cause towns to build more libraries, because $r$ is positive.: A positive $r$ gives the direction of the association, not a cause in either direction.

Correlation, even a strong one, does not show causation; look for a lurking variable that could affect both quantities.