PSAT/NMSQT Practice Questions (2026) — Free Sample Questions with Answers

Answer: D

By denying that the Maya lowlands experienced severe drought: Text 2 accepts the drought ("Drought surely strained Maya society"); it disputes only that drought was the sole cause.

By agreeing that drought was the only significant cause of the collapse: Text 2 adds warfare and lost "political authority" as further pressures, so it does not see drought as the only cause.

By claiming that warfare between cities caused the droughts: Text 2 treats warfare and drought as separate stresses; it never suggests that fighting caused the lack of rain.

By arguing that drought contributed to the collapse but cannot fully explain it: Correct. Text 2 grants that drought "strained Maya society" but points to warfare, weakened authority, and cities that "declined before the worst droughts began," challenging the claim that "drought alone explains the collapse."

When Text 1 offers a single cause, check whether Text 2 accepts that cause, rejects it, or keeps it while adding others.

Answer: B

fragile: Fragile means easily harmed, the opposite of what the text shows: tardigrades "survived extreme cold, intense radiation, and even exposure to the vacuum of space."

resilient: Correct. The text lists conditions tardigrades can endure ("survived extreme cold, intense radiation, and even exposure to the vacuum of space"), so resilient (able to withstand or recover from hardship) is the precise fit.

sluggish: The dormant tun state is a response to drying out, not a general trait; the text's focus is survival under harsh conditions, not slowness.

conspicuous: Conspicuous means easy to notice, but the text calls tardigrades "microscopic," and nothing in it concerns how visible they are.

For Words in Context, predict a word from the evidence after the blank. Here the list of extreme conditions survived points to "able to withstand hardship."

Answer: B

Corn with clover and no fertilizer yielded more than corn without clover that received 90 kilograms of fertilizer.: This is inaccurate: condition 4 yielded 8.4 tonnes, less than condition 2’s 9.3 tonnes.

Corn with clover and half the usual fertilizer yielded about as much as corn with the full amount, but corn with clover and no fertilizer yielded far less.: Correct. Condition 3 (clover, 90 kg) yielded 11.1 tonnes, nearly matching condition 1 (no clover, 180 kg) at 11.2, so clover replaced part of the fertilizer. Condition 4 (clover, 0 kg) yielded only 8.4, so clover could not replace all of it.

Corn without clover that received 180 kilograms of fertilizer had the highest yield of the four conditions.: This is accurate, but on its own it does not show that clover can replace any fertilizer.

When 90 kilograms of fertilizer were applied, adding clover raised yield from 9.3 to 11.1 tonnes per hectare.: This is accurate and shows clover helps, but it does not address the “not all” part of the conclusion.

A two-part conclusion (“part, but not all”) needs a pair of comparisons, one for each part.

Answer: B

Local families resent wealthy newcomers who settle in their neighborhood without first being introduced.: The text never describes resentment. The families welcome the newcomer as a likely husband for “one or other of their daughters.”

Neighboring families take for granted that a rich, unmarried newcomer will marry one of their daughters.: Correct. The narrator says the “truth” that a wealthy single man wants a wife is “so well fixed in the minds of the surrounding families” that he is treated as “the rightful property” of one of their daughters, even though his “feelings or views” are “little known.”

A wealthy single man usually chooses a wife soon after he moves, guided by his neighbors’ advice.: The text says nothing about the man actually choosing a wife or taking advice. It stresses that his own “feelings or views” are unknown; the assumption belongs to the families.

Young women in the neighborhood compete openly with one another for the attention of any rich man who arrives.: No competition among the daughters is described. The text focuses on what the families assume, not on how the young women behave.

For a main-idea question, choose the option that captures the whole text, including the contrast it sets up (here, the families’ certainty versus the man’s unknown views).

Answer: B

beating, moreover,: A conjunctive adverb such as "moreover" cannot join two independent clauses with commas alone; this creates a comma splice.

beating; moreover,: Two independent clauses are joined, and "moreover" is a conjunctive adverb that begins the second clause. A semicolon must come before the conjunctive adverb, and a comma after it.

beating; moreover: The semicolon is correct, but the introductory conjunctive adverb "moreover" should be followed by a comma.

beating moreover,: Without a semicolon or period before "moreover," the two independent clauses run together.

Between two independent clauses, a conjunctive adverb (however, moreover, therefore) takes a semicolon before it and a comma after it.

Answer: B

Merian, a German-born naturalist and artist, traveled to Suriname in South America in 1699 to study and paint the insects there.: This sentence describes Merian's background and travel but never mentions the belief her work challenged.

At a time when many Europeans believed insects arose from mud, Merian documented the full life cycles of butterflies and moths.: This sentence sets Merian's documentation of "the full life cycles of butterflies and moths" against the belief that "insects arose spontaneously from mud," showing how her work challenged that belief.

Merian's 1705 book Metamorphosis insectorum Surinamensium included a total of 60 engraved plates of insects.: This sentence gives a detail about her book but does not connect her work to any belief of her time.

Merian was a naturalist and artist who was born in 1647 and who died in 1717, after traveling to Suriname.: This sentence summarizes her life dates and travel; it says nothing about the belief in spontaneous generation.

For Rhetorical Synthesis, identify the goal first ("challenged a common belief") and choose the only option that includes both the belief and the work that contradicted it.

Answer: 86

1. Notice that $10n+16=2(5n+8)$.
2. So $10n+16=2(43)=86$.
3. Alternatively: $5n=35$, so $n=7$, and $10(7)+16=86$.

Look for the target expression as a multiple of the given expression before solving for the variable.

Answer: A

1. The equation has the form $C=mh+b$, where $m$ is the rate of change and $b$ is the starting value.
2. The coefficient of $h$ is $65$, so $C$ increases by $65$ dollars for each 1-hour increase in $h$.
3. So $65$ is the charge per hour; $40$ is the fixed fee.

The charge, in dollars, for each hour of work: Correct. $65$ is the coefficient of $h$ (the slope), so each additional hour adds $\$65$ to the total charge.

The fixed fee, in dollars, charged for every job: The fixed fee is the constant term, $40$, which is charged even when $h=0$.

The total charge, in dollars, for a 1-hour job: A 1-hour job costs $65(1)+40=105$ dollars, not $65$ dollars.

The number of hours the plumber works each day: Nothing in the equation describes hours per day; $65$ multiplies $h$, so its units are dollars per hour.

In a linear model $y=mx+b$, the coefficient $m$ is the change in $y$ per one unit of $x$, and $b$ is the value of $y$ when $x=0$.

Answer: -32

1. Use factored form: $f(x)=a(x+1)(x-7)$.
2. $f(0)=a(1)(-7)=-7a=-14$, so $a=2$ and $f(x)=2(x+1)(x-7)$.
3. Since $a>0$, the minimum is at the vertex, halfway between the zeros: $x=\frac{-1+7}{2}=3$.
4. $f(3)=2(4)(-4)=-32$.
5. Common error: $-16$, from assuming $a=1$.

The vertex of a parabola lies on the axis of symmetry, midway between the zeros; find the leading coefficient from another known point.

Answer: B

1. Square both sides: $x+10=(x-2{)}^2=x^2-4x+4$.
2. Rearrange: $x^2-5x-6=0$, so $(x-6)(x+1)=0$ and $x=6$ or $x=-1$.
3. Check $x=6$: $\sqrt{16}=4$ and $6-2=4$. Valid.
4. Check $x=-1$: $\sqrt{9}=3$ but $-1-2=-3$. Extraneous.
5. Desmos: graph $y=\sqrt{x+10}$ and $y=x-2$; they intersect only at $(6,4)$.

$-1$ only: $x=-1$ is extraneous: $\sqrt{9}=3$, but $-1-2=-3$. A principal square root cannot equal a negative number.

$6$ only: Correct. Squaring gives $x^2-5x-6=0$, so $x=6$ or $x=-1$. Only $x=6$ checks: $\sqrt{16}=4=6-2$.

$-1$ and $6$: Both values solve the squared equation, but $x=-1$ fails the original equation ($3\ne -3$). Squaring can introduce extraneous solutions.

$-6$ and $1$: This factors $x^2-5x-6$ as $(x+6)(x-1)$, which is $x^2+5x-6$. Neither value satisfies the original equation.

Squaring both sides of a radical equation can create extraneous solutions, so always check each candidate in the original equation.

Answer: D

1. The condition is "chose robotics," so the denominator is the robotics total: $18+27=45$.
2. The favorable count is grade 11 and robotics: $27$.
3. $P=\frac{27}{45}=\frac{3}{5}$.

$\frac{27}{100}$: This is the probability that a student selected from all $100$ students is in grade $11$ and chose robotics.

$\frac{9}{20}$: This is $\frac{45}{100}$, the probability that a randomly selected student chose robotics.

$\frac{27}{50}$: This uses the grade $11$ total ($50$) as the denominator, which gives the probability that a grade $11$ student chose robotics.

$\frac{3}{5}$: Of the $45$ students who chose robotics, $27$ are in grade $11$: $\frac{27}{45}=\frac{3}{5}$.

For a conditional probability from a two-way table, the denominator is the total of the given group only.

Answer: D

1. Since $\overline{DE}\parallel \overline{BC}$, corresponding angles are congruent, so $△ADE\sim △ABC$.
2. $AB=AD+DB=6+4=10$.
3. $\frac{BC}{DE}=\frac{AB}{AD}$, so $\frac{BC}{9}=\frac{10}{6}$.
4. $BC=\frac{90}{6}=15$.

6: This uses the ratio $\frac{DB}{AD}=\frac{4}{6}$ as the scale factor; $DB$ is not a side of either similar triangle.

13: This adds $DB=4$ to $DE=9$; similar figures scale by multiplication, not by adding a length.

13.5: This uses the ratio $\frac{AD}{DB}=\frac{6}{4}$; the correct scale factor compares $AB=10$ with $AD=6$.

15: $△ADE\sim △ABC$, so $\frac{BC}{DE}=\frac{AB}{AD}=\frac{10}{6}$, and $BC=9\cdot \frac{10}{6}=15$.

A segment parallel to one side of a triangle cuts off a similar triangle; compare whole sides (AB), not the leftover piece (DB).