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

Answer: A

1. 1. $y\le 2x-1$: solid line, shade below.
2. 2. $y>-x+2$: dashed line, shade above.
3. 3. The overlap is the region to the right of $(1,1)$, below the solid line and above the dashed line.
4. 4. Test $(3,1)$: $1\le 5$ true and $1>-1$ true, so it is in the region.

Graph A: Correct. Solid $y=2x-1$ shaded below, dashed $y=-x+2$ shaded above, overlapping to the right of $(1,1)$.

Graph B: Shades the right region, but $y=-x+2$ is solid; points on it are not solutions of the strict inequality.

Graph C: Shades below $y=-x+2$, the wrong half-plane for $y>-x+2$.

Graph D: Shades the right region, but the boundary types are swapped: $y=2x-1$ should be solid and $y=-x+2$ dashed.

Graph each inequality with the correct boundary (solid for $\le$, dashed for $>$) and half-plane; the solution set is the region where the shadings overlap (A-REI.D.12).

Answer: 11

1. 1. $45b+70(10)\le 1200\Rightarrow 45b\le 500$.
2. 2. $b\le 11.\overline{1}$.
3. 3. The greatest whole number is 11.
4. 4. Check: $45(11)+700=1195\le 1200$; $45(12)+700=1240>1200$.

Substitute the known value, solve the inequality, and then apply the context: a number of boxes must be a whole number, so round down to stay within the limit (A-CED.A.3).

Answer: A

1. 1. First month: $45. Remaining months: $m-1$ at $30.
2. 2. $C(m)=45+30(m-1)=30m+15$.
3. 3. Check: $C(1)=45$; $C(3)=45+60=105$, and $30(3)+15=105$.

$C(m)=30m+15$: Correct. $45+30(m-1)=30m+15$; it gives $45 for one month and $105 for three.

$C(m)=45+30m$: Charges $30 for the first month as well, so $C(1)=75$.

$C(m)=45m+30$: Swaps the two charges, making $45 the monthly rate.

$C(m)=30(m-1)$: Counts only the months after the first and drops the $45.

When the first period has a different price, the remaining $m-1$ periods get the regular rate: $C=45+30(m-1)$, which simplifies to $30m+15$ (LEAP.I.A1.2).

Answer: C

1. 1. Let $P=2x+1$ and $Q=x-4$.
2. 2. $P-Q=2x+1-x+4=x+5$.
3. 3. $P+Q=3x-3$.
4. 4. So the expression is $(x+5)(3x-3)$.
5. 5. Check by expanding both: $(4x^2+4x+1)-(x^2-8x+16)=3x^2+12x-15$, and $(x+5)(3x-3)=3x^2+12x-15$.

$(x-3)(3x-3)$: Does not distribute the negative to $-4$: computes $P-Q$ as $2x+1-x-4=x-3$.

$3x^2-15$: Squares each term separately, $(2x{)}^2+1-(x^2+16)$, dropping both middle terms.

$(x+5)(3x-3)$: Correct. $(P-Q)(P+Q)$ with $P-Q=x+5$ and $P+Q=3x-3$.

$(x+5)(3x+5)$: Gets $P-Q$ right but adds $x+4$ instead of $x-4$ for $P+Q$.

Treat each binomial as a single quantity: $P^2-Q^2=(P-Q)(P+Q)$, and subtract the whole binomial Q, distributing the negative sign (A-SSE.A.2).

Answer: C

1. 1. Rewrite the right side: $2x-10=2(x-5)$.
2. 2. Move it left: $(x-5{)}^2-2(x-5)=0$.
3. 3. Factor out $(x-5)$: $(x-5)(x-5-2)=(x-5)(x-7)=0$.
4. 4. $x=5$ or $x=7$. Check: $x=5$ gives $0=0$; $x=7$ gives $4=4$.

The only solution is $x=7$.: Divides both sides by $x-5$, which loses the solution $x=5$ (where that factor is 0).

The only solution is $x=5$.: Notices that $x=5$ makes both sides 0 but never factors, so misses $x=7$.

The solutions are $x=5$ and $x=7$.: Correct. $(x-5{)}^2-2(x-5)=(x-5)(x-7)=0$.

The equation has no real solutions.: Expands $(x-5{)}^2$ as $x^2+25$, giving $x^2-2x+35=0$, whose discriminant is negative.

Never divide both sides by an expression that can equal zero; instead move everything to one side and factor the common factor, here $(x-5)$ (LEAP.I.A1.5).

Answer: A

1. 1. $a=2$, $b=-6$, $c=1$.
2. 2. Discriminant: $(-6{)}^2-4(2)(1)=36-8=28$.
3. 3. $x=\frac{-(-6)\pm \sqrt{28}}{2(2)}=\frac{6\pm \sqrt{28}}{4}$.
4. 4. Check: $\frac{6+\sqrt{28}}{4}\approx 2.82$; $2(2.82{)}^2-6(2.82)+1\approx 15.9-16.9+1\approx 0$.

$\frac{6\pm \sqrt{28}}{4}$: Correct. $-b=6$, the discriminant is $36-8=28$, and the whole numerator is divided by $2a=4$.

$\frac{-6\pm \sqrt{28}}{4}$: Uses $b$ instead of $-b$ in the numerator.

$6\pm \frac{\sqrt{28}}{4}$: Divides only the radical by $2a$.

$\frac{6\pm \sqrt{44}}{4}$: Computes the discriminant as $b^2+4ac=36+8$.

In $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ the whole numerator, including $-b$, is divided by $2a$; with $b=-6$, $-b=6$ (A-REI.B.4b). Radicals need not be simplified.

Answer: B

1. 1. Elkhorn: $1060/1000=1.06$, so $E(t)=1000(1.06{)}^t$; Hartwell adds $80 a year.
2. 2. Year 1 gains: Hartwell $80, Elkhorn $60, so Hartwell leads at first.
3. 3. $t=10$: $H=1800$, $E\approx 1790.85$; $t=11$: $H=1880$, $E\approx 1898.30$. Elkhorn passes Hartwell.

The Elkhorn account gains more money than the Hartwell account every year.: In year 1 Elkhorn gains $60 and Hartwell $80, so this fails at the start.

Hartwell has more for the first several years; Elkhorn eventually has more.: Correct. Both start at $1,000; Hartwell leads in years 1 through 10, but 6% growth passes $80 a year by year 11.

The Hartwell account always has more, since it gains $80 a year and Elkhorn $60.: Uses only the first year; Elkhorn's yearly gain grows ($60, $63.60, $67.42, …).

Both accounts grow by the same percent each year, since both start at $1,000.: Hartwell grows by a constant $80, a smaller percent each year; only Elkhorn has a constant percent.

A quantity growing by a constant percent eventually exceeds one growing by a constant amount, even if the linear one starts out ahead (F-IF.C.9, F-LE.A.3).

Answer: -11

1. 1. $f(x)=2(x^2-6x)+7$.
2. 2. Complete the square: $2(x^2-6x+9)-2(9)+7$.
3. 3. $=2(x-3{)}^2-11$, so $k=-11$.
4. 4. Check: $f(3)=18-36+7=-11$. (Subtracting 9 instead of 18 gives the wrong value $-2$.)

After factoring a out of the x-terms, the number added inside the parentheses is multiplied by a, so subtract $a\cdot (b/2{)}^2$ outside (F-IF.C.8a).

Answer: A, C, E

1. 1. $\sqrt{12}\cdot \sqrt{3}=\sqrt{36}=6$.
2. 2. $\sqrt{6}+\sqrt{6}=2\sqrt{6}$.
3. 3. $\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3$.
4. 4. $5+\sqrt{20}$: rational + irrational.
5. 5. $(2-\sqrt{3})(2+\sqrt{3})=4-3=1$.
6. 6. $\frac{2}{3}\pi$: nonzero rational × irrational.

$\sqrt{12}\cdot \sqrt{3}$: Correct: $\sqrt{36}=6$.

$\sqrt{6}+\sqrt{6}$: $2\sqrt{6}$ is irrational; adding equal radicals does not remove the radical.

$\frac{\sqrt{18}}{\sqrt{2}}$: Correct: $\sqrt{9}=3$.

$5+\sqrt{20}$: Rational + irrational is irrational; $\sqrt{20}$ is not a whole number.

$(2-\sqrt{3})(2+\sqrt{3})$: Correct: the middle terms cancel, giving $4-3=1$.

$\frac{2}{3}\pi$: A nonzero rational times the irrational $\pi$ is irrational.

Simplify each expression completely before classifying it: products and quotients of square roots can be perfect squares, and conjugate products remove the radical (LEAP.I.A1.7, N-RN.B.3).

Answer: 3 6: Irrational; 2 18: Rational; (4 + 3) + (4 - 3): Rational; 5 + 5: Irrational; (1 + 2)(1 + 2): Irrational

1. Simplify each expression before classifying: products of square roots combine under one radical; opposite irrational parts cancel.
2. Rows 2 and 3 are counterexamples to the claim; rows 1, 4 and 5 are sums or products of two irrationals that happen to be irrational, which cannot prove "always".

Model full-credit response: "Mateo's claim is false. $\sqrt{2}$ and $\sqrt{18}$ are irrational, but $\sqrt{2}\cdot \sqrt{18}=\sqrt{36}=6$ is rational. Also $(4+\sqrt{3})+(4-\sqrt{3})=8$ is rational. One counterexample is enough to show a claim about 'always' is false." Rubric elements: (1) correct classification; (2) at least one counterexample for the sum or product; (3) the conclusion that the claim is false.

Answer: A

1. 1. Multiply $2x+3y=12$ by 2: $4x+6y=24$.
2. 2. The system now says $4x+6y=24$ and $4x+6y=18$; subtracting gives $0=6$, false.
3. 3. Graphically: both lines have slope $-\frac{2}{3}$, with y-intercepts 4 and 3, so they are parallel and distinct.

Doubling $2x+3y=12$ gives $4x+6y=24$, and no pair $(x,y)$ can make $4x+6y$ equal both 24 and 18.: Correct. An equivalent form of line $p$ shows that any solution would need $4x+6y$ to equal two different numbers, which is impossible, so the system is inconsistent (A-REI.C.5).

The lines have different y-intercepts, 4 and 3, so the two lines can never intersect.: True facts, but not enough: lines with different y-intercepts usually do intersect. The argument also needs the slopes to be equal.

The lines have slopes $-\frac{2}{3}$ and $-\frac{4}{6}$, which are different, so the lines never meet.: $-\frac{4}{6}=-\frac{2}{3}$, so the slopes are equal; lines with different slopes always intersect.

Substituting $x=0$ gives $y=4$ for line $p$ and $y=3$ for line $q$, so no point works.: Checks a single value of $x$. Two lines can disagree at $x=0$ and still meet at another $x$.

A system has no solution when elimination leaves a false statement; graphically the lines are parallel (same slope, different intercepts).

Answer: C

1. 1. If $q>0$: $x-3=\pm \sqrt{q}$ gives two different real solutions.
2. 2. If $q=0$: $x-3=0$ gives only $x=3$.
3. 3. If $q<0$: no real number squares to a negative number, so there is no real solution.
4. 4. Check with examples: $(x-3{)}^2=0$ is true only at $x=3$; $(x-3{)}^2=-4$ is true for no real $x$.

He is correct, because the $\pm$ symbol always produces two values of $x$.: Writes $\pm$ without checking the sign of $q$; when $q=0$ both signs give the same value, and when $q<0$ there is no real square root.

He is incorrect, because two real solutions occur only when $q$ is a perfect square.: Confuses real with rational: $q=11$ gives the two real solutions $3\pm \sqrt{11}$ even though 11 is not a perfect square.

He is incorrect, because $q=0$ gives one real solution and $q<0$ gives none.: Correct. A square is never negative, so $(x-3{)}^2=q$ has two real solutions for $q>0$, exactly one ($x=3$) for $q=0$, and none for $q<0$ (A-REI.B.4b).

He is incorrect, because $q<0$ gives one real solution and $q=0$ gives none.: Swaps the two cases: $q=0$ gives $x-3=0$, the single solution $x=3$; a negative $q$ gives no real solution.

Model full-credit response: "Rafael is incorrect. $(x-3{)}^2$ is never negative. If $q>0$, $x=3\pm \sqrt{q}$ gives two solutions; if $q=0$, the only solution is $x=3$; if $q<0$, there is no real solution, because no real number squared is negative. For example, $(x-3{)}^2=-4$ has no real solution." Rubric elements: (1) correct solution of Part A ($q=11$) and the completed argument; (2) the first error in Sofia's work identified and corrected; (3) a conclusion that the claim is false, supported by the three cases for $q$ or a counterexample.

Answer: 122820 bottles

1. 1. Bottles per day: $1,150\times 0.6=690$.
2. 2. Bottles per year: $690\times 178=122,820$.
3. 3. Accepted range: about $\pm 5\%$, roughly 116,700 to 128,900, so rounded estimates such as 120,000 or 123,000 count.
4. 4. Check with rounding: $1,200\times 0.6\times 180\approx 130,000$, the same size. ✓

A reasoned estimate multiplies a per-unit rate by the number of units and the number of time periods; check the size with rounded numbers.

Answer: A, C, D

1. 1. Multiplier for both markdowns: $0.70\times 0.80=0.56$.
2. 2. Final price: $90\times 0.56=50.40$; discount $=1-0.56=44\%$.
3. 3. Check step by step: $90\times 0.7=63.00$; $63.00\times 0.8=50.40$. ✓

The final price is 56% of the regular price.: Correct: $0.70\times 0.80=0.56$.

The two markdowns together equal a 50% discount.: Adds the percents; the 20% is taken from the already-reduced price, so the total discount is less than 50%.

The final price is $50.40.: Correct: $90\times 0.56=50.40$.

The two markdowns together equal a 44% discount.: Correct: paying 56% means a $100\%-56\%=44\%$ discount.

The final price is $45.00.: This is a 50% discount of $90, from adding 30% and 20%.

Taking 20% off first and then 30% off would give a lower price.: Multiplication is commutative: $0.80\times 0.70=0.70\times 0.80$, so the order does not matter.

Successive discounts multiply; the combined discount is 1 minus the product of the multipliers, never the sum of the percents.

Answer: 93 guests

1. 1. Inequality: $450+22g\le 2500$.
2. 2. $22g\le 2050$, so $g\le 93.18\dots$
3. 3. Guests are counted in whole numbers, so the greatest is 93 (rounding up to 94 breaks the budget).
4. 4. Check: $450+22(93)=2496\le 2500$ ✓; $450+22(94)=2518>2500$.

Model full-credit response (modeling, computation, interpretation): "Quarry Hall: $C=22g+450$; Willow Pavilion: $C=16g+900$. Setting them equal gives $6g=450$, so $g=75$, where both cost $2,100. For more than 75 guests Willow is cheaper because its per-guest charge is lower. With a $2,500 budget at Quarry Hall, $22g+450\le 2500$ gives $g\le 93.2$, so at most 93 guests." Rubric elements: correct equations, correct break-even and budget computations, and interpretations that respect whole-number guests.