MCAS Practice Questions (2026) — Free Sample Questions with Answers

Answer: A finished mural still depends on people to keep caring for it over time.: Text 2; A selection process was designed so that reputation would not decide who was chosen.: Text 1; People who made an artwork as amateurs argued that it still deserved a place.: Text 3; Decisions made before any paint is applied can decide how long a mural lasts.: Text 2

When several texts share a topic, decide which one actually develops an idea, not just which one mentions a related word; competitions and restoration appear in more than one text (RI.9-10.1).

Answer: D

He spoke several languages and helped patrons who did not speak English.: Reads "translator" literally; no one in the excerpt needs a different language.

He explained the library in simple words so that patrons could search on their own.: He does "explain the catalog in words a ten-year-old could follow," but the paragraph centers on identifying what patrons need.

He corrected patrons who used the wrong titles for books and reference works.: Reverses his behavior: he "never laughed at anyone" and calls the name "the easy part."

He could work out what patrons needed even when they could not name it.: Correct. He turns a vague description, "something about a lighthouse ... and a dog," into the right book on the shelf (paragraph 4) (RI.9-10.4).

When a writer uses a word figuratively, ask which feature of the literal meaning (turning one language into another) is being applied.

Answer: The author remembers a boy who left the library without the book he wanted. (Text 1, paragraph 4): Personal anecdote; A survey found that libraries without fines had a higher rate of items never returned. (Text 2, paragraph 5): Research finding; An economist explains that a small fine works as a signal that a deadline is real. (Text 2, paragraph 3): Expert opinion; Library records showed that on-time returns stayed almost the same after fines ended. (Text 1, paragraph 2): Research finding; The author recalls forty people waiting for twelve copies of one novel. (Text 2, paragraph 2): Personal anecdote

Classify evidence by its source: one person's experience (anecdote), a qualified person's judgment (expert opinion), or data from records or surveys (research) (RI.9-10.8).

Answer: A

to encourage readers to rethink a long-held assumption about city streams: Correct. The question "what if the water was never the problem, and the pipe was?" challenges the old view of streams "as problems to be hidden" (RI.9-10.6).

to argue that every city should daylight its buried streams as soon as possible: Overstatement: the paragraph says the answer "may be different" in each city and on each block.

to summarize the costs and benefits described in paragraphs 4 through 9: The paragraph does not restate the costs or benefits; it raises a new way of thinking.

to explain how artists use painted lines to show where streams flow underground: The painted lines are described in paragraph 11; paragraph 12 uses them only as a starting point.

A closing question often reframes the topic for the reader; ask what belief the question invites readers to reconsider.

Answer: B

Children often understand their parents' fears better than the parents do.: The speaker does grasp "what it costs," but the poem centers on the mother's change, not on who understands better.

Facing a long-held fear can change a belief that once seemed fixed.: Correct. The mother's inherited belief that "the water took what it wanted" gives way to "it gives back whatever you give it," and the map of fear is "coming apart at the folds" (lines 32–36) (RL.9-10.2).

Learning a new skill is easiest with a patient and kind instructor.: Too narrow: the instructor is one detail in the mother's larger change.

The fears people inherit from their families can never be fully overcome.: Reverses the ending, in which the old fear is softening and coming apart.

Check a theme against the poem's ending; the final image often shows where the central idea lands.

Answer: D

Young people who leave home soon forget the family they leave behind.: Neither text shows the departing sibling forgetting anyone; Emil takes the lamp and Katrin asks for letters.

Older siblings grieve more about leaving than younger siblings grieve about staying.: Both texts focus on the younger sibling's feelings; neither compares the two siblings' grief.

Families grow closer when they share the work of preparing for a journey.: Fits part of Text 2, where the household helps, but in Text 1 the packing is already done and the help is a pretext.

People facing a separation often express their feelings through objects and small actions.: Correct. The lighthouse lamp, the list and the offered room in Text 1, and the re-wrapped lunch, the pencil and the letter in Text 2, carry feelings no one says aloud (RL.9-10.2).

A shared idea must be supported in each text; reject an option that fits only one.

Answer: C

She moves from boredom with the shop to a firm plan to become a piano tuner herself.: Overstatement: she grows to value the work, but the excerpt never mentions a career plan.

She moves from resentment of her mother to resentment of Abel for hiding his hearing loss.: Reverses her feeling toward Abel; learning of his loss makes her "throat tighten" (paragraph 14), and she agrees to help him.

She moves from seeing the work as a punishment to finding meaning in a task she shares with Abel.: Correct. At first she has been "sentenced" to the shop (paragraph 3); by the end she hears a quiet that means "something more than the strings" (paragraph 21) after working beside Abel (RL.9-10.3).

She moves from confidence in her skill to doubt that she can finish the piano on time.: Reverses the direction of the change; her early doubts (paragraphs 17–18) give way to success in paragraph 19.

For a character-change question, identify where the character starts and where she ends, then choose the option that names both accurately.

Answer: B

A pool loses 2% of its original volume of water each day.: 2% of the original volume is the same amount every day, so the volume decreases by a constant difference: a linear model.

A sample of a substance loses 2% of its current mass each hour.: Correct. Losing 2% of the current mass means the mass is multiplied by 0.98 each hour; equal factors over equal intervals define exponential decay (F-LE.A.1c).

A company hires 2 new workers for every 100 new orders that it receives.: Workers increase by a constant amount per 100 orders, so the relationship is linear.

A runner increases her weekly running distance by 2 miles each week.: A constant increase of 2 miles per week is a constant rate of change: linear growth.

Linear functions change by equal differences over equal intervals; exponential functions change by equal factors. A percent of a fixed starting amount is a constant difference, but a percent of the current amount is a constant factor.

Answer: A

1. Let $u=x+2$, so the expression is $3u^2-7u-6$.
2. Factor: $3u^2-7u-6=(3u+2)(u-3)$ (check: $-9u+2u=-7u$).
3. Substitute back: $3(x+2)+2=3x+8$ and $(x+2)-3=x-1$.
4. Check with $x=0$: $3(4)-7(2)-6=-8$, and $(8)(-1)=-8$ ✓.

$(3x+8)(x-1)$: Correct. Let $u=x+2$: $3u^2-7u-6=(3u+2)(u-3)$. Replacing $u$ gives $(3(x+2)+2)((x+2)-3)=(3x+8)(x-1)$.

$(3x+2)(x-3)$: Factors $3u^2-7u-6$ correctly but never replaces $u$ with $x+2$; this product is $3x^2-7x-6$, which is not equivalent.

$(3x+4)(x-1)$: Replaces $u$ but does not distribute the 3: $3(x+2)+2$ is $3x+8$, not $3x+4$.

$(3x-8)(x+1)$: Sign error: this expands to $3x^2-5x-8$, but the given expression is $3x^2+5x-8$.

When the same binomial appears repeatedly, treat it as a single quantity $u$, factor, then substitute back and simplify each factor.

Answer: (1) 1.020, (2) 2, (3) 1014.50

1. $1.0824^{\frac{1}{4}}\approx 1.01999\approx 1.020$ (on a calculator, raise 1.0824 to the 0.25 power).
2. Quarterly growth rate: $1.020-1=0.020$, or 2% per quarter.
3. $B(3)=800(1.0824{)}^3\approx 800(1.268134)\approx 1014.50$ dollars.
4. Check: 12 quarters at the quarterly factor give $800(1.019992{)}^{12}\approx 1014.50$ ✓.

Exponent properties let you rewrite $b^t$ as ${\left(b^{\frac{1}{n}}\right)}^{nt}$; the base $b^{\frac{1}{n}}$ is the growth factor for each $\frac{1}{n}$ of the original period, and that factor minus 1 is the rate.

Answer: D

1. Substitute $s=150-a$ into $12.50a+7.50s=1925$: $12.50a+1125-7.50a=1925$.
2. So $5a=800$ and $a=160$, giving $s=150-160=-10$.
3. A negative ticket count is impossible, so the report is wrong.
4. Check with a bound: $150\cdot 12.50=1875<1925$.

It could be accurate, because 150 tickets at an average price of about $12.83 give $1,925.: An average price above $12.50 is impossible when the highest price is $12.50, so this reasoning actually shows the report is wrong.

It could be accurate, because solving the system gives 160 adult tickets for Saturday.: Solving $a+s=150$, $12.50a+7.50s=1925$ gives $a=160$ and $s=-10$; a negative number of student tickets means there is no valid solution.

It cannot be accurate, because $1,925 is not a multiple of $7.50.: Revenue is a combination $12.50a+7.50s$, which need not be a multiple of $7.50 (for example, $12.50+7.50=20$), so this is not a valid test.

It cannot be accurate, because even 150 adult tickets would bring in only $1,875.: Correct. The greatest possible revenue from 150 tickets is $150\cdot 12.50=1875$ dollars, which is less than $1,925. Equivalently, the system's solution has $s=-10$, which is impossible.

A system that models a real situation must have a solution that makes sense in context: whole, non-negative counts. Bounding the total (all at the highest price) is a quick check.

Answer: D

1. Ratio $1:2$ means $P$ is $\frac{1}{1+2}=\frac{1}{3}$ of the way from $A$.
2. Change from $A$ to $B$: $\Delta x=8-(-4)=12$, $\Delta y=7-(-2)=9$.
3. $P=(-4+4,-2+3)=(0,1)$.
4. Check: $AP=\sqrt{4^2+3^2}=5$ and $PB=\sqrt{8^2+6^2}=10$, so $AP:PB=1:2$ ✓.

Point W: W is $(4,4)$, two-thirds of the way from $A$; it gives $AP:PB=2:1$, measuring the ratio from $B$ instead of from $A$.

Point X: X is $(2,2.5)$, the midpoint; it treats the ratio $1:2$ as the fraction $\frac{1}{2}$ instead of $\frac{1}{1+2}=\frac{1}{3}$.

Point Y: Y is $(-2,-0.5)$, one-sixth of the way from $A$; it gives $AP:PB=1:5$.

Point Z: Correct. $P$ is $\frac{1}{3}$ of the way from $A$ to $B$: $\left(-4+\frac{1}{3}(12),-2+\frac{1}{3}(9)\right)=(0,1)$, which is point Z.

To partition a segment from $A$ in the ratio $m:n$, move $\frac{m}{m+n}$ of the way from $A$ toward $B$ in both coordinates.

Answer: C

1. $\tan \theta =\frac{h-5}{60}\approx \frac{21.0}{60}=0.35$.
2. $\theta ={\tan }^{-1}(0.35)\approx 19.3°$.
3. Since $19.3°\ne 17.5°$, the claim is incorrect.

The claim is correct, because the angle of elevation and the distance are inversely proportional.: Angles are not inversely proportional to distance; the tangent of the angle is $\frac{h-5}{d}$, and the angle itself does not halve when $d$ doubles.

The claim is incorrect; doubling the distance doubles the angle, so the angle would be 70°.: Moving farther away makes the angle smaller, not larger.

The claim is incorrect; $\tan \theta =\frac{h-5}{60}$, so the angle is about 19.3°, not 17.5°.: Correct. With opposite side $h-5\approx 21.0$ ft and adjacent side 60 ft, $\tan \theta \approx 0.35$ and $\theta ={\tan }^{-1}(0.35)\approx 19.3°$. The tangent ratio halves, but the angle does not.

The claim is incorrect; $\sin \theta =\frac{h-5}{60}$, so the angle is about 20.5°.: Uses sine, but 60 feet is the adjacent side, not the hypotenuse; the ratio of opposite to adjacent is the tangent.

Trigonometric ratios, not angles, are what change proportionally with side lengths; use the inverse ratio to find an angle.

Answer: 3.7

1. Group terms: $(x^2-6x)+(y^2+10y)=-20$.
2. Complete each square: add 9 and 25 to both sides: $(x^2-6x+9)+(y^2+10y+25)=-20+9+25$.
3. $(x-3{)}^2+(y+5{)}^2=14$, so the center is $(3,-5)$ and $r=\sqrt{14}\approx 3.74$, which rounds to 3.7.
4. Check: $3.74^2\approx 14.0$ ✓.

Complete the square in $x$ and in $y$, adding the same amounts to both sides, to reach $(x-h{)}^2+(y-k{)}^2=r^2$. The radius is the square root of the right side (not 14).

Answer: B

It is reasonable, because the line of fit passes close to every point in the data.: A good fit inside the data range does not show that the trend continues far outside it.

It may not be reliable, because 20 hours is far beyond the 1 to 8 hours in the data.: Correct. The arithmetic is right ($2.38\cdot 20+1.73=49.33$), but predicting far outside the range of the data (extrapolation) assumes the linear trend continues, which the data cannot confirm; volunteers may tire or run out of litter.

It is wrong, because the line of fit gives about 10.1 bags for 20 hours, not 49.3 bags.: Divides 20 by the slope instead of multiplying: $\frac{20}{2.38}+1.73\approx 10.1$. The line does give about 49.3.

It is reasonable, because a positive slope shows that bags always increase with hours.: A positive association in the data does not guarantee the pattern holds for every value, especially far beyond the data.

Predictions from a line of fit are most trustworthy within the range of the data; extrapolating far beyond it is risky even when the fit is strong.

Answer: (1) 2/15, (2) less than

1. $P(\text{first vowel})=\frac{4}{10}$.
2. $P(\text{second vowel}\mid \text{first vowel})=\frac{3}{9}$.
3. $\frac{4}{10}\cdot \frac{3}{9}=\frac{12}{90}=\frac{2}{15}\approx 0.133$.
4. With replacement: ${\left(\frac{4}{10}\right)}^2=0.16>0.133$.

For dependent events, $P(A\text{ and }B)=P(A)\cdot P(B\mid A)$: after a draw without replacement, both the favorable count and the total decrease by 1.