Study on the flyACTPractice testScience

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ACT Science practice test

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  • 40 minutes
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Questions 1–6

Two scientists discussed why sparkling water releases carbon dioxide gas when a sealed bottle is opened.

Scientist 1 argued that carbon dioxide remains dissolved more readily under high pressure and at low temperature. Opening a bottle lowers the pressure above the liquid, allowing gas to leave the water. This scientist predicted that an unshaken warmer bottle would release more gas than an otherwise identical colder bottle.

Scientist 2 agreed that opening provides a route for gas to escape but claimed that temperature has little effect over the tested range. According to Scientist 2, shaking is the main cause of differences in gas release because bubbles begin forming on disturbed surfaces.

Researchers stored identical sealed bottles for 24 hours at 5°C, 15°C, 25°C, or 35°C. Without shaking them, the researchers opened 5 bottles at each temperature and collected the gas released during the next 2 minutes. The mean masses are shown in Table 1 and Figure 1. The bottles contained equal volumes from the same production batch and were opened with the same apparatus.

Table 1

Storagetemperature (°C)Mean mass of gasfrom unshakenbottles (g)50.8151.3251.9352.6

Figure 1

0.00.51.01.52.02.53.03.54.05152535Storage temperature (°C)Mean mass of gas(g)Unshaken bottles

Gas passed through flexible tubing into a gas-tight collection bag. The researchers determined released mass by weighing each bottle immediately before opening and again after 2 minutes; no liquid left the bottle. A separate unopened bottle at each temperature was weighed twice and showed no measurable mass change. This check indicated that ordinary scale drift was negligible compared with the observed losses.

In an additional treatment, 5 bottles stored at 15°C were shaken for 30 seconds immediately before opening. Those bottles released a mean of 2.1 g of gas. Table 2 compares that shaken treatment with the unshaken 15°C bottles from Table 1. The researchers noted that Table 1 and Figure 1 alone address the temperature predictions because none of those bottles were shaken, while Table 2 provides evidence about shaking.

Table 2

Treatment at 15°CMean mass of gas(g)Unshaken1.3Shaken for 30seconds2.1

Question 1

According to Table 1, what mean mass of gas was released from the unshaken bottles stored at 25°C?

Show answer

Answer: B — 1.9 g

The 25°C row of Table 1 gives a mean released gas mass of 1.9 g.

Question 2

Scientist 2 claimed that shaking is the main cause of differences in gas release and that temperature has little effect over the tested range. Which of the following results, if found, would most strongly support Scientist 2?

Show answer

Answer: C — The shaken 15°C bottles released more gas than the unshaken 35°C bottles.

Scientist 2 treats shaking as the main cause and temperature as having little effect. Finding that shaken 15°C bottles released more gas than unshaken 35°C bottles would show that 30 seconds of shaking outweighed a 20°C temperature increase, supporting that claim.

Question 3

Based on the data in Table 1, as storage temperature increased, the mean mass of gas released from unshaken bottles:

Show answer

Answer: A — increased only.

The Table 1 values rise from 0.8 g to 1.3 g to 1.9 g to 2.6 g as temperature increases.

Question 4

How do the scientists’ predictions about unshaken bottles differ?

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Answer: D — Scientist 1 predicts more gas at higher temperature, while Scientist 2 predicts little temperature effect.

Scientist 1 links warmer storage to greater release, whereas Scientist 2 expects temperature to have little effect unless shaking differs.

Question 5

Which result would Scientist 1 cite as support for that scientist’s prediction about unshaken bottles?

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Answer: C — Mean released mass from unshaken bottles increased as storage temperature increased.

Scientist 1 predicted that an unshaken warmer bottle would release more gas than an otherwise identical colder bottle, matching the increase from 0.8 g at 5°C to 2.6 g at 35°C in Table 1.

Question 6

At which storage temperature were bottles included in both unshaken and shaken treatments?

Show answer

Answer: B — 15°C

Bottles stored at 15°C were tested both without shaking and after being shaken for 30 seconds.

Questions 7–12

Researchers sampled red maple leaves in one city park to examine whether distance from a busy road was associated with particle deposits and lead concentration. They selected leaves at 0, 10, 30, and 60 m from the road. Sample sizes differed because fewer intact leaves were available near the road.

All leaves were mature red maple leaves collected on the same dry morning from a height of 1.5 to 2.0 m. At each distance, leaves came from multiple trees, and no single tree supplied more than one-fifth of that distance’s sample. The researchers recorded whether each leaf had visible black particle deposits. They then used a calibrated analyzer to determine the mean lead concentration for leaves at each distance. The same collection tools, storage bags, preparation method, and analyzer settings were used throughout. Table 1 gives the results.

Table 1

Distance (m)Leaves examinedLeaves withdepositsMean lead (mg/kg)0502518108032153010020960120126

For estimates between sampled distances, assume the local trend is linear only between the two surrounding distances.

Question 7

Do these samples justify the claim that every red maple leaf throughout the city has fewer particle deposits when it grows farther from a road?

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Answer: C — No, because the study sampled one park on one morning and did not examine every city leaf.

The data came from limited samples in one park, so a claim about every red maple leaf throughout the city exceeds the sampled population. Leaves were in fact collected at 0 m from the road.

Question 8

Table 1

Distance (m)Leaves examinedLeaves withdepositsMean lead (mg/kg)0502518108032153010020960120126

Using the stated local linear assumption, what mean lead concentration would be estimated at 20 m from the road?

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Answer: C — 12 mg/kg

Twenty meters is halfway between the surrounding sampled distances of 10 m and 30 m, whose means are 15 and 9 mg/kg. The local midpoint is 12 mg/kg; values at 0 m and 60 m are not part of this local rate.

Question 9

Table 1

Distance (m)Leaves examinedLeaves withdepositsMean lead (mg/kg)0502518108032153010020960120126

A pie chart represents the 100 leaves examined at 30 m. What percentages should its deposit and no-deposit sectors show?

Show answer

Answer: B — Deposits: 20%; no deposits: 80%

At 30 m, 20 of 100 leaves had deposits, which is 20%. The remaining 80 leaves form the 80% no-deposit sector.

Question 10

Table 1

Distance (m)Leaves examinedLeaves withdepositsMean lead (mg/kg)0502518108032153010020960120126

After normalizing by the different numbers of leaves examined, which distance had the greatest proportion of leaves with deposits?

Show answer

Answer: A — 0 m

The deposit proportions are 25/50 = 50%, 32/80 = 40%, 20/100 = 20%, and 12/120 = 10%. The greatest normalized proportion occurred at 0 m.

Question 11

Table 1

Distance (m)Leaves examinedLeaves withdepositsMean lead (mg/kg)0502518108032153010020960120126

Which conclusion is fully supported while remaining limited to the samples actually studied?

Show answer

Answer: D — Among these red maple samples, both deposit proportion and mean lead concentration were lower at greater sampled distances.

Across the four sampled distances, deposit proportion falls from 50% to 10% and mean lead falls from 18 to 6 mg/kg. Choice 3 describes those samples without asserting causation or extending the result to other species and parks.

Question 12

Which characteristic was held constant across the four distance samples?

Show answer

Answer: D — Tree species

Every sampled leaf came from a red maple, so tree species was controlled. Sample size, mean lead concentration, and deposit proportion differed by distance.

Questions 13–18

Students investigated heat transfer along solid metal rods. They used copper, aluminum, and steel rods that had equal lengths and diameters. One end of each rod was held against the same electric heater. Temperature sensors were attached 5 cm, 10 cm, and 15 cm from the heated end.

At the start of each trial, the rod and room were at 22°C. The heater was switched on, and temperatures were recorded every minute for 9 min. A shield reduced air currents around the apparatus. The same heater setting, sensor type, attachment method, and heating time were used for every metal. Three trials were completed for each condition.

Before heating, the students checked every sensor in the same water bath; all read 22°C within 0.2°C. Rods were allowed to return to room temperature between trials. The heater contacted a cleaned end of each rod, and a clamp maintained equal contact pressure.

Experiment 1 compared the 3 metals. After 6 min of heating, the students recorded the temperature at each sensor. Table 1 reports the mean of the three trials for each metal and distance.

Experiment 2 tested whether the heater, rather than room conditions alone, produced the temperature increases. The students repeated the 10 cm measurements with the heater switched off. The control rods stayed at 22°C after 6 min. Figure 1 compares those control readings with the heated 10 cm readings from Experiment 1.

The students proposed that, under their conditions, temperature at a sensor would continue to decrease as its distance from the heated end increased. They planned to test that proposal with sensors at additional distances.

Figure 1

010203040506049443422Copper, heater onAluminum, heater onSteel, heater onControl, heater offTemperature at 10cm after 6 min (°C)

Table 1

MetalTemperature at 5cm (°C)Temperature at 10cm (°C)Temperature at 15cm (°C)Copper624940Aluminum554437Steel413430

Question 13

Based on Table 1, if a sensor were added at 20 cm, which result would be most consistent with the observed direction for all 3 metals?

Show answer

Answer: C — Each 20 cm temperature would be lower than its metal’s 15 cm temperature.

For each metal, temperature decreases from 5 cm to 10 cm to 15 cm. Extending only that observed direction predicts a lower value at 20 cm, without requiring an exact numerical decrease.

Question 14

In an Experiment 1 trial, which sequence matches the procedure?

Show answer

Answer: D — Begin with the rod at 22°C; switch on the heater; record temperatures.

Each trial began with the rod at 22°C. The heater was then switched on, and the sensors recorded temperatures over the 9 min interval.

Question 15

Figure 1 separates the heater-on readings from the heater-off control at 10 cm, while Table 1 follows heater-on readings across 3 distances. Which conclusion is supported by both displays?

Show answer

Answer: D — The heater raised rod temperatures above room temperature, and the measured heating effect was smaller farther from the heated end.

The heated 10 cm readings all exceed the 22°C control, showing a heater-associated rise. Within each metal, Table 1 shows lower temperatures at greater distances from the heated end.

Question 16

Which factor was held constant rather than deliberately varied in Experiment 1?

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Answer: A — Rod diameter

The rods had equal diameters, while material and sensor distance were deliberately varied. Temperature was the measured response.

Question 17

Which additional result would directly falsify the students’ distance proposal rather than support it?

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Answer: C — An aluminum reading at 20 cm that is above its 15 cm reading of 37°C

The proposal requires a farther sensor to have a lower temperature. An aluminum value above 37°C at 20 cm would reverse the decrease from the 15 cm value and therefore falsify it.

Question 18

Based only on the adjacent aluminum measurements in Table 1, the temperature at 12.5 cm after 6 min would most likely have been:

Show answer

Answer: C — between 37°C and 44°C.

The 12.5 cm sensor would lie between the measured 10 cm and 15 cm positions. The corresponding aluminum temperatures are 44°C and 37°C, so an interpolated value should lie between them.

Questions 19–24

A laboratory prepared identical 40 g samples of a salt solution initially at 90°C. Solubility of the dissolved salt decreased as the samples were cooled from 90°C to 20°C at constant rates, and crystallization began as cooling started. The cooling rates were 2°C/min, 5°C/min, 10°C/min, and 20°C/min. After each sample reached 20°C, technicians measured the longest dimension of 50 randomly selected crystals and calculated the mean crystal size. The samples had equal composition, initial temperature, container shape, and final temperature. Figure 1 shows the means.

All 3 researchers agreed that faster cooling produced smaller crystals, but they proposed different mechanisms.

Researcher 1

Cooling initiates the formation of crystal nuclei. Faster cooling produces more nuclei during the first minute. Because the fixed amount of dissolved material is divided among more growing crystals, each crystal remains smaller. Slower cooling produces fewer nuclei, so more material is available for each crystal.

Researcher 2

The number of nuclei is approximately the same at every cooling rate. Crystal size instead depends on growth time. During slow cooling, dissolved particles have more time to attach to existing crystals before the sample reaches 20°C. During rapid cooling, that interval is shorter, so crystals stop growing at a smaller size.

Researcher 3

Faster cooling creates steeper temperature differences between the center and edges of a sample. Those differences drive stronger convection currents. The currents break some growing crystals into fragments, and each fragment becomes another growth site. Thus, rapid cooling indirectly creates more growing sites and a smaller mean size.

In a follow-up trial, technicians counted nuclei after the first minute and mapped liquid motion with tracer beads. They could also place thin baffles in a container to reduce convection without changing its cooling rate. All measurements used the same salt solution and microscope procedure.

To check measurement reliability, a second technician independently measured 10 crystals from each sample. The two technicians’ mean estimates differed by less than 0.1 mm, and neither technician knew the assigned cooling rate while measuring them.

Figure 1

0.00.40.81.21.62.02.42.83.2251020Cooling rate (°C/min)Mean crystal size(mm)

Question 19

A second laboratory reports only that its rapidly cooled samples had smaller mean crystals than its slowly cooled samples; it did not count nuclei, measure growth time, or observe liquid motion. Relative to the researchers’ explanations, this result would:

Show answer

Answer: C — support the shared observed outcome but not distinguish among the 3 proposed mechanisms.

The result repeats the shared relationship between cooling rate and crystal size, but none of the mechanism-specific quantities was measured, so it cannot distinguish the explanations.

Question 20

According to Researcher 2, why did the sample cooled at 2°C/min have larger crystals than the sample cooled at 20°C/min?

Show answer

Answer: A — Its crystals had more time to grow before the sample reached 20°C.

Researcher 2 attributes the larger crystals during slow cooling to a longer interval for dissolved particles to attach to existing crystals.

Question 21

Researcher 1 and Researcher 2 agree that faster cooling produces smaller crystals. Which prediction for a sample cooled at 15°C/min would be consistent with both researchers and with Figure 1?

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Answer: B — Its mean crystal size would be between 0.8 mm and 1.2 mm.

Both researchers predict that faster cooling produces smaller crystals. Because 15°C/min lies between 10°C/min and 20°C/min, the corresponding mean should lie between 1.2 mm and 0.8 mm.

Question 22

At a cooling rate of 20°C/min, a baffled sample and an unbaffled sample reached 20°C in the same time and formed the same number of initial nuclei. The baffled sample had a mean crystal size of 1.4 mm, whereas the unbaffled sample had a mean of 0.8 mm. This result most directly supports:

Show answer

Answer: D — Researcher 3, because reducing convection was associated with larger crystals.

Cooling time and initial nucleus count were held the same, while reducing convection increased crystal size, as Researcher 3's fragmentation mechanism predicts.

Question 23

Which sequence gives the causal chain proposed by Researcher 3 for a rapidly cooled sample?

Show answer

Answer: A — Steeper temperature differences → stronger convection → more fragmentation → more growth sites

Researcher 3 links rapid cooling to steep temperature differences, stronger convection, fragmentation, and therefore more sites competing for material.

Question 24

Which statement accurately compares Researcher 1 with Researcher 3?

Show answer

Answer: C — Both predict that mean crystal size decreases as cooling rate increases, though they propose different mechanisms.

Both researchers predict smaller mean crystals at faster cooling rates, although their explanations for that relationship differ.

Questions 25–30

Researchers studied whether biochar, a carbon-rich material made by heating plant matter with little oxygen, can remove nitrate from water.

In Study 1, each beaker received 500 mL of water containing 40 mg/L nitrate. The researchers added 0 g, 5 g, 10 g, or 15 g of biochar to separate beakers. The mixtures were stirred for 24 hours at 20°C and then filtered. Four beakers were tested at each biochar mass. Table 1 shows the mean nitrate concentrations remaining.

Table 1

Biochar mass (g)Mean remainingnitrate (mg/L)04053110241520

In Study 2, every beaker received 10 g of biochar, but contact time varied. Remaining nitrate was measured after 1, 4, 12, and 24 hours. The starting water and stirring speed matched those used in Study 1. Table 2 shows the mean nitrate concentrations remaining.

Table 2

Contact time (h)Mean remainingnitrate (mg/L)13643012262424

Researcher A proposed that nitrate attaches to binding sites on the biochar surface. Additional biochar supplies more binding sites but encounters progressively less nitrate remaining in the fixed water volume, so each additional gram of biochar should remove less nitrate than the preceding gram. Researcher B proposed that microorganisms carried on the biochar consume nitrate. According to this explanation, longer contact gives the microorganisms more time to act. Both researchers expected that adding more biochar would generally leave less nitrate in the water.

The studies did not distinguish conclusively between these mechanisms. The biochar was not sterilized, and neither the number of binding sites nor the abundance of living microorganisms was measured. As a comparison, the researchers repeated the 24-hour, 10 g treatment with washed gravel instead of biochar (Table 3). They also found that pH changed by less than 0.2 unit in every beaker.

Table 3

MaterialContact time (h)Mass (g)Mean remainingnitrate (mg/L)Biochar241024Washed gravel241039

Question 25

If Study 2 had included a contact time of 8 hours, the mean remaining nitrate concentration would most likely have been:

Show answer

Answer: B — 28 mg/L

Table 2 shows 30 mg/L remaining at 4 hours and 26 mg/L remaining at 12 hours. An 8-hour contact time lies between those times, so the remaining concentration should lie between 26 and 30 mg/L. Only 28 mg/L is in that interval.

Question 26

According to Table 1, what percent of the original nitrate concentration was removed by 10 g of biochar in Study 1?

Show answer

Answer: C — 40%

The concentration fell by 4024=1640 - 24 = 16 mg/L. Relative to 40 mg/L, that is 16/40×100%=40%16/40 \times 100\% = 40\%.

Question 27

According to Table 1, for each successive 5 g increase in biochar mass in Study 1, the additional decrease in nitrate concentration:

Show answer

Answer: D — decreased from 9 to 7 to 4 mg/L.

The successive decreases were 4031=940-31=9, 3124=731-24=7, and 2420=424-20=4 mg/L.

Question 28

Based on Studies 1 and 2, which of the following correctly matches each researcher with the study that most directly supports that researcher’s explanation?

Show answer

Answer: A — Researcher A with Study 1; Researcher B with Study 2

Study 1 shows successively smaller nitrate decreases (9, 7, then 4 mg/L) as biochar mass increased, consistent with Researcher A’s prediction that additional biochar would produce diminishing removal gains because each added gram encounters less remaining nitrate in the fixed water volume. Study 2 shows remaining nitrate decreasing as contact time increased, matching Researcher B’s time-dependent microbial prediction.

Question 29

The washed-gravel treatment in Table 3 was most likely included to determine whether:

Show answer

Answer: A — nitrate removal required a property of biochar rather than the mere presence of a solid material in the beaker

Washed gravel provided a solid surface at the same mass and contact time as the 10 g biochar treatment but left 39 mg/L nitrate, nearly the original 40 mg/L. Biochar left 24 mg/L. That contrast tests whether nitrate loss depended on biochar specifically rather than on having any solid in the water.

Question 30

Which additional experiment would best distinguish between Researcher A’s binding-site explanation and Researcher B’s microorganism explanation?

Show answer

Answer: C — Repeat the 10 g, 24-hour biochar treatment after first sterilizing the biochar.

Sterilizing the biochar would kill microorganisms while leaving surface binding sites. If remaining nitrate stayed near 24 mg/L, Researcher A’s binding-site explanation would be supported. If remaining nitrate stayed near the untreated value of 40 mg/L, Researcher B’s microorganism explanation would be supported. The other procedures would not separate the two mechanisms.

Questions 31–35

Astronomers measured the percentage of a star’s light blocked when a planet crossed the star. Table 1 lists transit depths at wavelengths of 500, 600, 700, and 800 nm during Epoch 1 and, half a stellar rotation later, during Epoch 2. Figure 1 graphs Epoch 1 transit depth versus wavelength. Measurement uncertainty was ±0.02\pm0.02 percentage point.

The Haze Model proposes that tiny particles in the planet’s atmosphere scatter shorter wavelengths, making blue-light transits deeper. It predicts that the wavelength pattern will remain stable between epochs. The Starspot Model proposes that cool, unblocked regions on the rotating star create the apparent pattern. It also predicts deeper transits at shorter wavelengths, but predicts that at least 1 depth will change by 0.10 percentage point or more after half a stellar rotation because different stellar longitudes will face Earth.

Table 1

Wavelength (nm)Epoch 1 depth (%)Epoch 2 depth (%)5001.421.416001.341.357001.281.278001.241.25

Figure 1

1.201.251.301.351.401.451.501.421.341.281.24500600700800Transit depth (%)Wavelength (nm)

Question 31

According to Table 1, what was the mean of the 4 transit depths measured during Epoch 1?

Show answer

Answer: C — 1.32%

The Epoch 1 depths total 5.28%5.28\%, and 5.28%/4=1.32%5.28\%/4=1.32\%.

Question 32

The 2 models agree that transit depth should vary with wavelength, but they disagree about whether the pattern should:

Show answer

Answer: A — remain stable after half a stellar rotation.

Both models predict greater depths at shorter wavelengths, but only the Haze Model predicts stability between epochs.

Question 33

Considering the uncertainty and all 4 wavelength pairs in Table 1, how do the Epoch 2 results affect the models?

Show answer

Answer: D — They support the Haze Model and weaken the Starspot Model because all changes were within uncertainty and below 0.10.

Every epoch-to-epoch change was only 0.01 percentage point, within uncertainty and far below the minimum change predicted by the Starspot Model.

Question 34

Assuming a linear change between 600 nm and 700 nm during Epoch 1 in Figure 1, the transit depth at 650 nm would be:

Show answer

Answer: B — 1.31%

650 nm is halfway between 600 and 700 nm, so the interpolated depth is halfway between 1.34% and 1.28%, or 1.31%.

Question 35

According to Figure 1, as wavelength increased by equal increments during Epoch 1, transit depth:

Show answer

Answer: C — decreased by progressively smaller amounts.

The successive decreases were 0.08, 0.06, and 0.04 percentage point, so the decreases became smaller.

Questions 36–40

Scientists tested how moths corrected their headings when a white lamp appeared. In Study 1, the lamp was 60° to the right of the moths’ starting direction; in Study 2, it was 60° to the left. Separate groups experienced no lamp or a lamp at 6, 12, or 24 lux. After 4 seconds, the researchers calculated the magnitude of each group’s mean turn toward the lamp. Positive values therefore indicate a correction toward the light in either study. Each group contained 40 moths. The chamber, lamp color, release point, and airflow were unchanged. The no-lamp measurements reflect ordinary turning, so treatment effects can be found by subtracting each study’s own no-lamp value. Each moth was tested only once.

Figure 1

01020304050No lamp6 lux12 lux24 luxLamp intensityMean turn towardlamp (°)Study 1Study 2

Question 36

After each study’s no-lamp value is subtracted, how does the effect of 12-lux light in Study 1 compare with its effect in Study 2?

Show answer

Answer: B — The effects are equal at 22°.

Study 1 changes by 31° − 9° = 22°, and Study 2 changes by 36° − 14° = 22°. Subtracting each baseline makes the opposite lamp directions irrelevant to the effect size.

Question 37

A 9-lux condition is midway between the adjacent 6- and 12-lux conditions. Using straight-line interpolation over only that interval, Study 1's mean turn would be closest to:

Show answer

Answer: D — 25°.

Study 1 rises from 18° at 6 lux to 31° at 12 lux. The midpoint estimate is (18° + 31°) ÷ 2 = 24.5°, which is closest to 25°.

Question 38

If Study 2 were redrawn as points on a numeric intensity axis, which description would preserve both the values and the spacing of the tested intensities?

Show answer

Answer: A — Plot (0, 14), (6, 23), (12, 36), and (24, 48), with the last horizontal gap twice either earlier gap.

The Study 2 heights are 14, 23, 36, and 48 at 0, 6, 12, and 24 lux. On a numeric axis, the 12-to-24 gap must be twice each 6-unit gap.

Question 39

The studies began at different no-lamp values. Nevertheless, which pattern was replicated at 6 and 24 lux after each illuminated value was compared with its study’s baseline?

Show answer

Answer: C — The 6- and 24-lux effects were 9° and 34°, respectively, in both studies.

Relative to baseline, Study 1 changes by 9° at 6 lux and 34° at 24 lux; Study 2 gives the same two changes. The raw values differ by 5°, but the treatment pattern at these intensities is replicated.

Question 40

In Study 2, a mean turn of 42° lies between the observations at 12 and 24 lux. If the response changed at a constant rate only across that interval, what lamp intensity would correspond to 42°?

Show answer

Answer: A — 18 lux

From 12 to 24 lux, the turn rises from 36° to 48°. A turn of 42° is halfway through that 12° rise, so the intensity is halfway through the interval: 18 lux.

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