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

15 free PreACT questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

English (4)Math (4)Reading (3)Science (4)

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English

Question 1

Flying above the grass, a pattern of flashes is produced by a male [19] — Which choice makes the sentence most grammatically acceptable?

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Answer: B

NO CHANGE: Dangling modifier: "Flying above the grass" seems to describe "a pattern of flashes," which cannot fly.

Flying above the grass, a male produces a pattern of flashes: Correct. The modifier is followed immediately by "a male," the one doing the flying, and the active verb is clearer.

A pattern of flashes, flying above the grass, is produced by a male: The modifier still attaches to "a pattern of flashes."

Flying above the grass, the flashes of a male form a pattern: The modifier now attaches to "the flashes," not to the male.

Put the noun that performs the action of an introductory -ing phrase right after the comma.

Question 2

Question 10 asks about the preceding passage as a whole. For the sake of logic and cohesion, Paragraph 2 should be placed:

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Answer: B

where it is now.: Paragraph 2 opens with "a faster method" and refers to "a whole crew," but neither the slow method nor the crew has been described yet.

after Paragraph 3.: Correct. Paragraph 3 describes the slow, crew-based way of smoothing the ice; Paragraph 2's "Determined to find a faster method" and "a job that had taken a whole crew more than an hour" respond to it. Paragraph 3's "Running a rink, however" also follows directly from the opening of the rink in Paragraph 1.

after Paragraph 4.: Paragraph 4 begins "The new machine soon found a famous customer," which must come after the machine (the Model A) has been introduced.

after Paragraph 5.: Paragraph 5 describes the machine today and works as the conclusion; placing the 1940s invention after it breaks the time order.

Look for words that point back ("a faster method," "the new machine") and place each paragraph after the one that sets them up.

Question 3

Question 20 asks about the preceding passage as a whole. The writer is considering adding the following sentence to the third paragraph:

I counted thirty-two of them on my own back wheel.

If the writer were to add this sentence, it would most logically be placed at:

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Answer: C

Point A in the third paragraph.: At Point A, "them" would have no antecedent at all; the paragraph has not yet mentioned spokes.

Point B in the third paragraph.: At Point B, "them" would refer to the spoke wrench, and the spokes have not been introduced yet.

Point C in the third paragraph.: Correct. Placed right after "dozens of thin spokes," the sentence gives a specific count, and "them" clearly refers to the spokes.

Point D in the third paragraph.: At Point D, the sentence would interrupt the explanation of what happens when a spoke loosens and how a mechanic fixes it.

Place an added sentence right after the sentence containing its pronoun's antecedent and the idea it develops.

Question 4

Convinced at last, the valley became the centerpiece of the maps that Heezen and Tharp published together. [8] — Which choice makes the sentence most grammatically acceptable?

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Answer: C

NO CHANGE: Dangling modifier: "Convinced at last" seems to describe "the valley," which cannot be convinced.

Heezen, convinced at last, the valley became the centerpiece of the maps that he and Tharp published together.: Places "Heezen" first but leaves him without a verb, so the sentence's grammar breaks down.

Convinced at last, Heezen made the valley the centerpiece of the maps that he and Tharp published together.: Correct. The opening modifier "Convinced at last" is followed immediately by "Heezen," the person who was convinced.

Convinced at last, the maps that Heezen and Tharp published together made the valley their centerpiece.: Still dangling: the modifier now seems to describe "the maps."

An introductory modifier must be followed directly by the noun it describes.

Math

Question 5

Malik is painting the 4 walls of a room. Each wall is 16 feet long and 9 feet high. The walls contain one door that is 3 feet by 7 feet and two windows that are each 5 feet by 4 feet, none of which will be painted. He will apply 2 coats of paint, and each can covers 350 square feet. What is the fewest number of cans he must buy?

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Answer: C

Wall area: $4 \times 16 \times 9 = 576$ sq ft.
Openings: door $3 \times 7 = 21$ , windows $2 \times 5 \times 4 = 40$ ; total 61 sq ft.
Paintable: $576 - 61 = 515$ sq ft; two coats: $1, 030$ sq ft.
Cans: $1, 030 \div 350 \approx 2.94$ , rounded up to 3.

1: Paints one coat and rounds $515 \div 350 \approx 1.47$ down; both coats are needed and partial cans must be rounded up.

2: Rounds $1, 030 \div 350 \approx 2.94$ down; 2 cans cover only 700 sq ft, less than the 1,030 needed.

3: Correct. Paintable area is $576 - 61 = 515$ sq ft; 2 coats need 1,030 sq ft, and $1, 030 \div 350 \approx 2.94$ , so he must buy 3 cans.

4: Forgets to subtract the door and windows: $2 \times 576 = 1, 152$ sq ft needs 4 cans.

In multistep area problems, subtract what is not covered, multiply by the number of layers, and round up when buying whole units.

Question 6

In $△ A B C$ below, $D$ is on $\overline{A B}$ , $E$ is on $\overline{A C}$ , and $\overline{D E}$ is parallel to $\overline{B C}$ . If $A D = 6$ , $D B = 4$ , and $D E = 9$ , what is the length of $\overline{B C}$ ?

Triangle ABC with DE parallel to BC

Figure not drawn to scale.

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Answer: D

Parallel lines make $△ A D E \sim △ A B C$ (AA).
$A B = A D + D B = 6 + 4 = 10$ .
$\frac{6}{10} = \frac{9}{B C}$ , so $6 \cdot B C = 90$ and $B C = 15$ .

6: Uses the ratio $\frac{D B}{A D} = \frac{4}{6}$ ; the triangles compare $A D$ with the whole side $A B$ .

13: Adds $D B$ to $D E$ ; lengths in similar figures scale by a ratio, not by adding.

13.5: Uses $\frac{A D}{D B} = \frac{6}{4}$ ; $D B$ is only part of side $A B$ , not a side of $△ A B C$ .

15: Correct. $△ A D E \sim △ A B C$ , so $\frac{A D}{A B} = \frac{D E}{B C}$ : $\frac{6}{10} = \frac{9}{B C}$ , giving $B C = 15$ .

A segment parallel to one side of a triangle cuts off a similar triangle; match corresponding sides of the small and large triangles, using whole sides.

Question 7

If $lo g_{2} (x + 6) = 5$ , what is the value of $x$ ?

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Answer: C

Rewrite in exponential form: $2^{5} = x + 6$ .
$32 = x + 6$ .
$x = 26$ . Check: $lo g_{2} 32 = 5$ .

4: Multiplies 2 by 5 instead of raising 2 to the 5th power.

19: Computes $5^{2} = 25$ instead of $2^{5} = 32$ ; the base of the log is the base of the power.

26: Correct. $lo g_{2} (x + 6) = 5$ means $2^{5} = x + 6$ , so $32 = x + 6$ and $x = 26$ .

38: Adds 6 to 32 instead of subtracting it.

$lo g_{b} y = k$ means $b^{k} = y$ ; convert to exponential form, then solve.

Question 8

A school play sold 150 tickets for a total of $1,480. Adult tickets cost $12 each, and student tickets cost $8 each. How many adult tickets were sold?

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Answer: A

Let $a$ = adult tickets and $s$ = student tickets: $a + s = 150$ and $12 a + 8 s = 1, 480$ .
Substitute $s = 150 - a$ : $12 a + 8 (150 - a) = 1, 480$ , so $4 a + 1, 200 = 1, 480$ .
$4 a = 280$ , so $a = 70$ . Check: $12 (70) + 8 (80) = 840 + 640 = 1, 480$ .

70: Correct. With $a + s = 150$ and $12 a + 8 s = 1, 480$ , substituting $s = 150 - a$ gives $4 a + 1, 200 = 1, 480$ , so $a = 70$ (and $s = 80$ ).

74: Divides $1,480 by $20, as if every buyer paid for one adult and one student ticket.

75: Assumes the 150 tickets were split evenly, ignoring the revenue equation.

80: This is the number of student tickets, not adult tickets.

Write one equation for the count and one for the value, then substitute; reread the question to report the variable it asks for.

Reading

Question 9

Based on paragraphs 3 and 6, a wood frog from a southern population that entered winter with unusually large glycogen stores would most likely:

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Answer: C

be protected from freezing entirely, no matter how cold the winter became.: The frog's strategy is to control freezing, not prevent it (paragraph 3); more glycogen does not stop ice from forming.

need less glucose than a northern frog to survive the same freeze.: Unsupported: the passage links glycogen to how much glucose a frog can make, not to how much it needs.

make more glucose when frozen than most frogs in its population.: Correct. The liver converts glycogen into glucose when freezing begins (paragraph 3), and paragraph 6 says larger glycogen stores supply more material for producing glucose, so this frog could make more glucose than its typical neighbors.

lose some of its ability to tolerate low oxygen while it is frozen.: Paragraph 5 does not link oxygen tolerance to glycogen stores.

When applying a passage to a new case, choose the conclusion that follows directly from the stated relationship and goes no further.

Question 10

Both authors would most likely agree that:

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Answer: A

some children have more scheduled activities than is good for them.: Correct. Passage A says children's days of "practices, lessons, and screens" leave "little room" for ideas (paragraph 2); Passage B grants that "some families have filled their children's calendars beyond what is healthy" (paragraph 4).

unstructured time usually does teenagers more harm than good.: This goes beyond Passage B and contradicts Passage A, which praises unstructured time.

structured programs should replace most of children's free time.: Passage A resists filling "every quiet hour," and Passage B rejects scheduling "from dawn to midnight."

most children today grow up without access to books at home.: Passage B says not every child has "books on the shelf," which is far weaker than "most"; Passage A never addresses it.

A "both authors would agree" answer must be supported by a sentence in each passage; check A and B separately.

Question 11

Which of the following best describes the main focus of the passage?

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Answer: C

A church restores a valuable old piano that had been neglected for eleven years.: The piano's age and condition (paragraph 6) are setting details; the story is about the narrator, not the restoration of the instrument.

A teenager resents her parents for arranging a summer job without asking her first.: Too narrow: the parents appear only in paragraph 2, and the narrator's resentment is not what the story develops.

A narrator comes to see the sense in her aunt's puzzling advice.: Correct. The narrator files "It stops arguing" under USELESS ADVICE (paragraph 3), struggles (paragraphs 4–5), then listens "the way I had watched her listen" and hears the strings settle (paragraph 7); crossing out USELESS (paragraph 8) shows she now understands the advice.

An aunt and her niece argue repeatedly about the right way to tune a piano.: There is only one brief exchange (paragraph 5), and Aunt Ruth answers it by laughing and taking the hammer, not by arguing.

A main-focus answer must cover the whole passage, including how it ends; choices built on a single paragraph are too narrow.

Science

Question 12

According to Student 2, the oak rings for 1986–1988 were narrow mainly because:

Each year, a tree adds a layer of wood called a growth ring; a wide ring means the tree grew well that year. Researchers took core samples from 30 oak trees in a forest and measured the average ring width for 1984 through 1990 (Table 1). The rings for 1986, 1987 and 1988 were unusually narrow. Three students explain the narrow rings.

Table 1
Year	Average oak ring width (mm)
1984	2.1
1985	2.0
1986	0.9
1987	0.7
1988	0.8
1989	1.9
1990	2.1

Three explanations

Student 1: From 1986 through 1988 the forest had a drought. With little water in the soil, the trees closed the pores in their leaves to save water, which reduced photosynthesis and left less sugar for growth. A drought affects every tree species in the forest, although trees on wet, low-lying soil would be affected less than trees on dry ridges.

Student 2: From 1986 through 1988 a leaf-eating caterpillar population exploded in the forest. The caterpillars stripped much of the oaks' foliage, and with fewer leaves the trees carried out less photosynthesis and grew less. These caterpillars feed only on oaks and a few related broadleaf trees, so the forest's pines would not have been affected. The damage would not depend on soil moisture.

Student 3: In 1986, 1987 and 1988, hard frosts struck in late spring, just after the oaks had opened their new leaves. The frost killed the young leaves, and the trees used stored energy to grow replacement leaves, so photosynthesis was reduced during the early growing season. Pines, which keep their needles all year and were not opening new leaves at that time, would not have been affected. Frosts are more frequent at the higher, colder elevations of the forest.

Show answer & explanation

Answer: A

caterpillars removed many of the oaks' leaves.: Correct. Student 2 says caterpillars stripped much of the oaks' foliage, so the trees photosynthesized and grew less.

the soil held too little water for the oaks.: This is Student 1's drought explanation.

late spring frosts killed the oaks' new leaves.: This is Student 3's explanation.

the oaks grew more slowly as they aged.: No student mentions age, and the rings returned to normal width in 1989 and 1990.

Before answering viewpoint questions, summarize each student's cause in a few words so they do not blur together.

Question 13

According to the results of Study 1, as the NaCl concentration increased, the percent of seeds that germinated:

Salt and radish seed germination

Salt in soil water can slow or prevent seed germination (the sprouting of a seed). Researchers studied how sodium chloride (NaCl) affects radish seeds. In every study, 50 seeds were placed on filter paper in a covered dish and wetted with 10 mL of a solution. After 5 days, the researchers counted the seeds that had sprouted and calculated the percent germinated.

Study 1. Dishes were wetted with solutions of 0, 50, 100, 150 or 200 millimolar (mM) NaCl and kept at 22°C (Table 1).

Study 2. Dishes were wetted with 0 mM or 100 mM NaCl and kept at 12°C, 17°C, 22°C or 27°C (Table 2).

Study 3. Some seeds were primed: soaked in distilled water for 12 hours and blotted dry before being placed in a dish. Primed and unprimed seeds were wetted with 100 mM NaCl and kept at 22°C (Table 3). Seeds in Studies 1 and 2 were unprimed.

Table 1
NaCl concentration (mM)	Seeds germinated (%)
0	96
50	90
100	72
150	40
200	12

Table 2
Temperature (°C)	Germinated at 0 mM NaCl (%)	Germinated at 100 mM NaCl (%)
12	60	30
17	84	52
22	96	72
27	94	70

Table 3
Seed treatment	Seeds germinated (%)
Unprimed	72
Primed	86

Show answer & explanation

Answer: B

increased only.: This reverses the trend; germination fell from 96% to 12%.

decreased only.: Correct. Table 1 shows germination falling at every step: 96%, 90%, 72%, 40% and 12% as NaCl rose from 0 to 200 mM.

increased and then decreased.: Germination never rose; even the first step (0 to 50 mM) lowered it from 96% to 90%.

remained the same.: Germination changed by 84 percentage points across the concentrations.

Describe a trend by checking every step in the table, not just the first and last values.

Question 14

A student predicted that doubling the length of the string would double the period. Do the results of Experiment 1 support this prediction?

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Answer: C

Yes, because the period increased every time the string length increased.: The period did increase, but that alone does not show it doubled when the length doubled.

Yes, because doubling the length from 0.25 m to 0.50 m increased the period from 1.00 s to 2.01 s.: At 0.50 m the period was 1.42 s; 2.01 s is the period for 1.00 m.

No, because doubling the length from 0.50 m to 1.00 m increased the period only from 1.42 s to 2.01 s.: Correct. 2.01 s is about 1.4 times 1.42 s, not 2 times. In Table 1 the period doubled (1.00 s to 2.01 s) only when the length was quadrupled (0.25 m to 1.00 m).

No, because the period stayed the same when the string length changed.: The period changed from 1.00 s to 2.84 s across the lengths in Table 1; it stayed the same only when mass changed (Experiment 2).

To test a "doubling" claim, find two trials whose inputs differ by a factor of 2 and compare their outputs.

Question 15

According to Figure 1, on July 15 the DO concentration at a depth of 8 m was closest to which of the following?

In summer, sunlight warms the surface of a deep lake, and a layer in which temperature drops quickly with depth (the thermocline) separates warm surface water from cold bottom water. In winter, the lake is covered with ice. Scientists measured water temperature and dissolved oxygen (DO) concentration every 2 m from the surface (0 m) to the bottom (14 m) at the deepest point of the lake on July 15 and on January 15. Table 1 shows the temperatures, and Figure 1 shows the DO concentrations.

Table 1
Depth (m)	July water temperature (°C)	January water temperature (°C)
0	24	1
2	23	3
4	22	4
6	15	4
8	10	4
10	8	4
12	7	4
14	7	4

July January

Show answer & explanation

Answer: B

1.6 mg/L: This is the July value at 10 m, one point too deep.

3.0 mg/L: Correct. On the July line in Figure 1, the point at 8 m lies midway between the 2 and 4 mg/L gridlines, at 3.0 mg/L.

5.6 mg/L: This is the July value at 6 m, one point too shallow.

9.6 mg/L: This is the January value at 8 m; it reads the wrong line.

Find the depth on the x-axis, go up to the line for the right month, then read across to the y-axis.

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