Study on the flyACTPractice testScience

ACT

ACT Science practice test

  • 40 questions
  • 40 minutes
  • Nothing is deducted for a wrong answer
  • Free

The Science section of the ACT practice test — 40 questions. Mark your answers, then press Finish at the bottom to see your score.

This is the same test the app serves as Practice Test 1. Working it here spends it: these questions will not be new when you take the test against the clock in the app.

Nothing is marked while you work, just as on the real test. The answers and the worked explanations open when you finish. What you mark is kept in this browser, so closing the tab does not lose it.

Questions 1–5

Fine dust sometimes rises above an airless body’s surface near sunrise. Two planetary physicists proposed explanations.

Dr. Chen—Electrostatic Model

Ultraviolet (UV) light removes electrons from surface grains. Similarly charged grains repel one another, and small grains lift. Temperature is not the direct cause, so increasing UV intensity at constant temperature should increase lofting. With UV blocked, warming alone should not loft many grains.

Dr. Silva—Gas-Flow Model

Small amounts of gas trapped in pores expand as the surface warms, pushing grains upward. UV has no direct role if temperature remains fixed. Increasing surface temperature should increase lofting even in darkness.

Researchers placed equal masses of dust on identical porous rocks in a vacuum chamber. In Experiment 1, surface temperature remained −20°C while UV intensity was set at 0%, 50%, or 100%. In Experiment 2, UV was blocked while surface temperature was set at −20°C, 0°C, or 20°C. Grain size, exposure time, rock porosity, and chamber pressure were held constant. Each condition was repeated 4 times. Tables 1 and 2 give the mean number of grains lofted per minute.

Table 1. Experiment 1 (surface temperature = −20°C)

UV intensity (%)Mean number ofgrains lofted perminute02501810035

Table 2. Experiment 2 (UV blocked)

Surfacetemperature (°C)Mean number ofgrains lofted perminute−2020152034

Question 1

Consider Condition A with 100% UV at −20°C and Condition B with UV blocked at 20°C. Which pair of predictions follows from the 2 models?

Show answer

Answer: B — Chen predicts more lofting in A than B, and Silva predicts more in B than A.

Condition A supplies Chen’s required UV while remaining cold; Condition B supplies Silva’s required warming while excluding UV. Thus, the models reverse which condition they expect to be more effective.

Question 2

Which choice correctly summarizes the proposed force that lifts the dust?

Show answer

Answer: A — Chen proposes repulsion between charged grains; Silva proposes pressure from expanding pore gas.

Chen’s mechanism is electrostatic repulsion after UV-driven electron loss. Silva’s mechanism is mechanical pressure from trapped gas expanding as the rock warms.

Question 3

Which set of plotted points would represent Experiment 2 with surface temperature on the horizontal axis and grains lofted per minute on the vertical axis?

Show answer

Answer: C — (−20,2),(0,15),(20,34)

Experiment 2 pairs temperatures of −20°C, 0°C, and 20°C with means of 2, 15, and 34 grains per minute, respectively. Those ordered pairs appear only in the third choice.

Question 4

How did the result at the highest UV level in Experiment 1 compare with the result at the highest temperature in Experiment 2?

Show answer

Answer: D — Experiment 1 lofted 1 more grain per minute.

At 100% UV, the Experiment 1 mean was 35 grains per minute. At 20°C, the Experiment 2 mean was 34, so Experiment 1 exceeded Experiment 2 by 1.

Question 5

Based on Table 2, do the results of Experiment 2 support Dr. Silva while weakening Dr. Chen?

Show answer

Answer: B — Yes; the results with UV blocked match Silva’s prediction rather than Chen’s.

Experiment 2 isolated warming by keeping UV blocked. Its results match Silva’s expanding-gas prediction and conflict with Chen’s claim that warming without UV should loft few grains.

Questions 6–11

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 6

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 7

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 40−24=16 mg/L. Relative to 40 mg/L, that is 16/40×100%=40%.

Question 8

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 40−31=9, 31−24=7, and 24−20=4 mg/L.

Question 9

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 10

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 11

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 12–16

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 12

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 13

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 14

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 15

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 16

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.

Questions 17–22

Researchers observed that dry field soil sometimes warmed during the 3 hr after a brief rainfall, even though the rainwater was cooler than the upper soil. They placed probes at depths of 5 cm, 15 cm, and 30 cm. Before watering, all plots had the same temperature at each depth. They then added equal amounts of 18°C water at noon to 3 types of plots: sunlit soil containing its usual microorganisms, sunlit soil sterilized to reduce microbial activity, and unsterilized soil shaded since sunrise. Figure 1 shows each mean temperature change 3 hr after watering. Positive values indicate warming.

Three researchers offered explanations.

Researcher 1 proposed the retained-energy model. Before watering, evaporation from dry soil removed little energy. Watering created a wet surface, but during the next hours the darkened surface absorbed more incoming solar energy instead of reflecting it. The retained energy warmed the upper soil first and then moved downward. This model predicts stronger warming in sunlit than shaded plots and does not require microorganisms.

Researcher 2 proposed the conductivity model. Sunlight stored energy near the dry surface before watering. Water filled air spaces and increased the soil’s thermal conductivity, allowing that stored energy to move downward more quickly. This model also predicts stronger warming in sunlit plots, but it predicts that downward heat transfer delays maximum warming, so the warming peak should occur progressively later at greater depths. Sterilization should have little effect.

Researcher 3 proposed the microbial model. Dormant microorganisms became active after water was added, and their metabolism released heat. Because this activity increased only gradually, the model allows a delay before maximum warming as metabolic heat production rises. This model predicts more warming in unsterilized than sterilized soil under otherwise identical conditions. It can account for warming in shaded plots, although it predicts less warming there if cooler shade slows metabolism.

Figure 1

0.00.51.01.52.02.53.03.54.05 cm15 cm30 cmProbe depthTemperature changeafter 3 hr (°C)Sunlit, unsterilizedSunlit, sterilizedShaded, unsterilized

Question 17

According to Figure 1 and the proposed models, which researcher’s mechanism is most directly supported by the warming of the shaded, unsterilized soil?

Show answer

Answer: D — Researcher 3's, because microbial metabolism provides a heat source that does not depend on strong sunlight.

Figure 1 shows warming in the shaded, unsterilized plot. This result most directly supports Researcher 3's microbial mechanism because it provides a non-solar heat source, whereas Researchers 1 and 2 predict stronger warming in sunlit plots.

Question 18

In Researcher 2's model, what is the source of the energy and what change after watering moves it downward?

Show answer

Answer: B — Earlier sunlight is the source; increased soil conductivity moves the energy.

Researcher 2 says sunlight stored energy near the dry surface, then water increased conductivity so that stored energy moved downward faster.

Question 19

Which statement describes a point of agreement among all 3 researchers and their central disagreement?

Show answer

Answer: C — They agree watered soil warmed; they disagree about whether sunlight or microbial metabolism supplied the main energy.

All 3 researchers offer explanations for the observed warming after watering, so they treat that warming as given. Researchers 1 and 2 identify sunlight as the main energy source, whereas Researcher 3 identifies microbial metabolism.

Question 20

Suppose that in a follow-up study, warming at 5 cm, 15 cm, and 30 cm all peaked 1 hr after watering. Which viewpoint would this observation challenge most directly?

Show answer

Answer: A — Researcher 2, because simultaneous peaks conflict with the predicted progressively later peaks at greater depths.

Researcher 2 predicts that the warming peak will occur progressively later at greater depths. Simultaneous peaks at all 3 depths would conflict with that specific prediction.

Question 21

Although Researchers 2 and 3 both allow a delay between watering and maximum warming, what different mechanism does each use to explain that delay?

Show answer

Answer: D — Researcher 2 assigns it to downward heat transfer; Researcher 3 assigns it to increasing microbial activity.

Researcher 2 delays maximum warming until stored heat moves downward, producing later peaks at greater depths. Researcher 3 delays it until microbial activity increases after watering.

Question 22

Which comparison from Figure 1 most directly matches a prediction made by Researcher 3 but not required by Researchers 1 and 2?

Show answer

Answer: B — At each depth, the sunlit, unsterilized soil warmed more than the sunlit, sterilized soil.

Researcher 3 predicts more warming in unsterilized than sterilized soil under otherwise identical conditions. Figure 1 shows that pattern at every depth in the sunlit plots; Researchers 1 and 2 do not require it.

Questions 23–28

Researchers investigated how filter pore diameter affects the time required to filter water. They tested papers with pore diameters of 2, 5, 10, or 20 micrometers. Because of how the papers were manufactured, papers with smaller pores were also thicker.

In Experiment 1, a researcher poured 200 mL of water at 20°C into each filter and measured the time until all the water passed through. Experiment 2 repeated the procedure with water at 40°C. For each experiment, 5 new filters of each pore diameter were tested, and the mean time was calculated. The water source, filter area, funnel, poured volume, and pouring method remained constant. Water temperature was verified before each trial, and pore-diameter order was randomized. Table 1 and Figure 1 give the results.

Table 1

Pore diameter (μm)Experiment 1 meantime (s)Experiment 2 meantime (s)2160120511080107050204530

Figure 1

020406080100120140160180251020Pore diameter (μm)Mean filtration time(s)Experiment 1 (20°C)Experiment 2 (40°C)

Question 23

Based on Experiments 1 and 2, why do the results not establish that changing pore diameter alone caused the differences in filtration time?

Show answer

Answer: B — Paper thickness changed along with pore diameter.

Smaller-pore papers were also thicker, so pore diameter and paper thickness varied together. The water source, volume, and filter area were controlled.

Question 24

Assume that a given volume of water passes through a given filter at a constant rate. If the poured volume in Experiment 2 were increased from 200 mL to 400 mL for a 10-micrometer filter, what filtration time would be predicted?

Show answer

Answer: C — 100 s

A 10-micrometer filter passed 200 mL in 50 s at 40°C. Doubling only the controlled water volume at a constant flow rate doubles the predicted time to 100 s.

Question 25

In Experiment 1, the mean flow rate through the 20-micrometer filter was closest to which value?

Show answer

Answer: B — 4.4 mL/s

The filter passed 200 mL in 45 s. Dividing 200 mL by 45 s gives approximately 4.4 mL/s.

Question 26

On a scatterplot with pore diameter represented on a continuous numerical horizontal axis, which horizontal-axis arrangement would correctly preserve the numerical intervals among the tested pore diameters?

Show answer

Answer: C — Use distances proportional to 3, 5, and 10 micrometers between successive labels.

The successive pore-diameter differences are 3, 5, and 10 micrometers. A continuous numerical axis must use horizontal distances proportional to those unequal differences.

Question 27

Based on Table 1, for which pore diameter did increasing water temperature from 20°C to 40°C produce the greatest percentage decrease in mean filtration time?

Show answer

Answer: D — 20 micrometers

The percentage decreases are 25% for 2 micrometers, about 27% for 5, about 29% for 10, and 33% for 20. Thus the 20-micrometer filter had the greatest proportional decrease.

Question 28

What mean filtration time was recorded for the 5-micrometer paper in Experiment 2?

Show answer

Answer: C — 80 s

The 5-micrometer row lists 80 s in the Experiment 2 column.

Questions 29–34

Researchers investigated how flower characteristics affected visits by bees and hoverflies in a meadow. Each experimental array contained 12 artificial flowers, all receiving the same treatment. Flowers had identical shapes and nectar openings. A visit was counted when an insect landed and inserted its mouthparts into an opening. Repeated visits by the same insect were counted separately; therefore, visit rates did not measure numbers of individual insects.

Each study lasted 6 days. All treatments within a study were observed simultaneously for 1 h daily. Treatment locations were randomly reassigned among plots each day, and fresh flower inserts were used. Each flower initially contained 0.20 mL of nectar solution, which was restored to that volume every 5 min. Sugar concentration is expressed as grams of sugar per liter of solution. Figures report mean visits per hour to an entire array, separately for each pollinator type.

Study 1 Researchers varied petal color among blue, yellow, and white. All flowers were unscented and contained nectar at 200 g/L. An identical pad beneath each flower received a fixed volume of scent-free carrier liquid. The pads allowed subsequent scent treatments without changing flower structure. Figure 1 shows the results. For any treatment, the combined visit rate is the sum of its bee and hoverfly rates.

Figure 1

010203040BlueYellowWhitePetal colorMean visit rate(visits/h)BeesHoverflies

Study 2 Researchers tested scent presence separately on blue and yellow flowers. Scent-present pads received carrier liquid containing a floral scent; scent-absent pads received the same volume of carrier alone. Nectar remained at 200 g/L. Thus, each color had its own scent-absent comparison. Table 1 shows the results. A proportional scent effect can be calculated by dividing the rate with scent present by the rate with scent absent for the same color and pollinator.

Table 1

Petal colorScentBee visits (visits/h)Hoverfly visits(visits/h)BlueAbsent3612BluePresent5418YellowAbsent2424YellowPresent3618

Study 3 Researchers varied nectar sugar concentration in yellow, scent-present arrays, retaining the scent dose used in Study 2. Initial nectar volume and replenishment procedures remained unchanged, so increasing concentration also increased the initial sugar mass available in each flower. Table 2 shows the results. The studies measured visitation rather than sugar consumption: an insect could leave after removing only part of the available nectar.

Table 2

Sugarconcentration(g/L)Bee visits (visits/h)Hoverfly visits(visits/h)100242420036183004212400306

Question 29

In Study 1, one color had the same combined visit rate as blue but twice the hoverfly visit rate of blue. What bee visit rate was recorded for that color with scent present in Study 2?

Show answer

Answer: C — 36 visits/h

Blue and yellow each had 48 combined visits/h in Figure 1, while yellow had 24 hoverfly visits/h compared with blue’s 12. The scented yellow treatment in Table 1 had 36 bee visits/h; 24 uses the unscented yellow treatment, and 54 uses scented blue.

Question 30

How did adding scent affect the ratio of bee visits to hoverfly visits for each color in Study 2?

Show answer

Answer: D — The ratio remained unchanged for blue and doubled for yellow.

For blue, the ratio was 36/12 = 3 without scent and 54/18 = 3 with scent. For yellow, it increased from 24/24 = 1 to 36/18 = 2. Although bee rates increased by 50% for both colors, the different hoverfly responses prevent the ratios from changing alike.

Question 31

Researchers want to determine whether the proportional scent effect on hoverfly visits to yellow flowers differs between 100 and 400 g/L. Without assuming a relationship between concentration and visit rate, which two additional treatments would provide the missing comparisons for Study 3?

Show answer

Answer: D — Scent-absent yellow flowers at 100 and 400 g/L.

Table 2 supplies scent-present yellow rates of 24 and 6 hoverfly visits/h at 100 and 400 g/L. Calculating the proportional scent effect at each concentration requires its matching scent-absent yellow rate. Blue flowers change color as well as scent, while measurements at 200 and 300 g/L cannot establish the missing rates without an additional assumption.

Question 32

A follow-up experiment will compare yellow, scented flowers at 100 and 400 g/L. Each flower must initially contain the same sugar mass as a flower at 200 g/L in Study 3. Which initial nectar volumes should be used at 100 and 400 g/L, respectively?

Show answer

Answer: C — 0.40 mL and 0.10 mL

The reference flower contained 200 g/L × 0.00020 L = 0.040 g of sugar. Providing that mass requires 0.40 mL at 100 g/L and 0.10 mL at 400 g/L. Keeping both volumes at 0.20 mL holds volume constant rather than sugar mass, while 0.80 mL and 0.20 mL provide equal masses that are twice the required mass.

Question 33

Suppose that for yellow flowers, each pollinator’s proportional scent effect measured in Study 2 also applies at 100 g/L. Using Study 3 to predict visit rates for an unscented yellow array at 100 g/L, approximately what percentage of its combined visits would be made by bees?

Show answer

Answer: A — 33%

For yellow flowers in Table 1, the scent-present/scent-absent factors are 36/24 = 1.5 for bees and 18/24 = 0.75 for hoverflies. Dividing the scented rates of 24 and 24 visits/h at 100 g/L by those factors predicts unscented rates of 16 and 32 visits/h. Bees would therefore make 16/48, or about 33%, of visits; 50% leaves scent present, while 67% results from applying the factors in the opposite direction.

Question 34

A researcher hypothesized that every tested treatment change that increases the combined visit rate also increases the percentage of visits made by bees. Do the results support this hypothesis?

Show answer

Answer: B — No, because adding scent to blue flowers increased the combined rate while leaving the bee percentage unchanged.

Adding scent to blue flowers increased the combined rate from 48 to 72 visits/h, but the bee percentage remained 36/48 = 54/72 = 75%, providing a counterexample. The yellow scent comparison and the 100-to-200 g/L comparison agree with the hypothesis but cannot establish its universal claim. The 300-to-400 g/L comparison starts with a decrease in combined rate, so it does not test the stated condition.

Questions 35–40

Trees of one maple species appear to begin changing color earlier in some years than in others. Researchers investigated whether day length, cool nights, or late-summer drought explains this variation. They used genetically identical saplings and defined color onset as the first day that yellow pigment was visible in a fixed proportion of marked leaves. All saplings were green on Day 0. Table 1 summarizes environmental records for four chamber treatments. Color-onset results were withheld from three scientists, who proposed the explanations below. All other growing conditions were identical, and observations were made daily using the same procedure.

Table 1

TreatmentP: first day withday length ≤12 hC: day thirdconsecutive nightat ≤8°C endedD: day fifthconsecutive daywith soil watercontent ≤12%endedA1285B1258C8125D8512

Scientist 1

Day length alone sets color onset. On the first day with at most 12 h of light, an irreversible internal process begins; visible color appears exactly 4 days later. Neither cool nights nor drought changes this delay or initiates the process independently. Thus, onset occurs on Day P + 4. Reports of differences between years reflect inconsistent sampling of canopy positions, rather than different environmental triggers. With marked leaves and standardized scoring, identical saplings exposed to identical day-length schedules should begin changing color together, even when their temperature and watering schedules differ.

Scientist 2

Cool nights initiate the process. Three consecutive nights at or below 8°C are required, and visible color appears exactly 4 days after the third qualifying night ends: Day C + 4. A warmer night before the third qualifying night resets the count. Once three qualifying nights have occurred consecutively, subsequent warming cannot cancel or delay color onset. Day length and soil moisture have no direct effect. Their seasonal association with cooling can make either appear responsible. Consequently, shortening days without the required cool-night sequence will leave these saplings green.

Scientist 3

Day length establishes the baseline onset at Day P + 4, but drought can advance it. After five consecutive days with soil water content at or below 12%, measured as water volume divided by soil volume, drought-induced color appears 2 days later. This advance requires the soil to remain at or below 12% throughout those additional days. Rewatering before visible color appears cancels the drought pathway, leaving the day-length pathway intact. Rewatering after color appears does not restore green coloration. Without rewatering, onset is the earlier of Day P + 4 and Day D + 2. Cool nights have no direct effect. For Table 1, day length continues decreasing, nights remain cool after C, and soil remains dry after D unless a question specifies a change.

Question 35

For Treatments A and B, how do the predictions of Scientists 2 and 3 compare?

Show answer

Answer: B — Scientist 2 predicts earlier onset in B; Scientist 3 predicts earlier onset in A.

Scientist 2 predicts Days 12 and 9 for A and B because their C values are 8 and 5. Scientist 3 predicts Days 7 and 10 because D + 2 precedes the shared baseline of Day 16. The identical P values imply identical onset only for Scientist 1.

Question 36

Researchers want to modify Treatment D so that all three scientists predict different onset days. Which single modification would accomplish this, with the other environmental schedules retained?

Show answer

Answer: A — Move P from Day 8 to Day 12.

Moving P to Day 12 gives Scientist 1 an onset of Day 16, Scientist 2 an onset of Day 9, and Scientist 3 an onset of Day 14 from D = 12. Moving C to Day 8 makes all three predict Day 12; preventing drought leaves Scientists 1 and 3 at Day 12. Moving P to Day 5 makes all three predict Day 9.

Question 37

Treatment C is modified so that its third consecutive qualifying cool night ends on Day 8 instead of Day 12. Its light and soil-moisture schedules remain unchanged. Which observed onset would agree with Scientists 1 and 2 while weakening Scientist 3?

Show answer

Answer: C — Day 12

Scientist 1 predicts 8 + 4 = Day 12, and Scientist 2 now also predicts 8 + 4 = Day 12. Scientist 3 still predicts the earlier drought onset, 5 + 2 = Day 7. Day 16 uses the original cool-night schedule, while Day 9 uses another treatment’s C value.

Question 38

In Treatment B, all nights following the completed qualifying sequence on Day 5 are made warmer than 8°C. The light and soil-moisture schedules remain unchanged, and color onset occurs on Day 9. Which assessment follows from the scientists’ explanations?

Show answer

Answer: A — The result agrees with Scientist 2 because warming occurred after the required sequence was completed.

Scientist 2 predicts Day 5 + 4 = Day 9 and explicitly states that later warming cannot cancel the process. Scientist 3 predicts Day 10 from D = 8, and Scientist 1 predicts Day 16 from P = 12. Resetting the cool-night count applies only before the qualifying sequence is completed.

Question 39

Treatments A and B are both rewatered immediately after the environmental records for Day 8 are completed, and their soil subsequently remains above 12% water content. According to Scientist 3, how much later will color onset occur in B than in A?

Show answer

Answer: D — 9 days

In A, drought-induced color has already appeared on Day 5 + 2 = Day 7, so rewatering cannot reverse it. In B, rewatering precedes the predicted drought onset on Day 10 and cancels that pathway, leaving onset at Day 12 + 4 = Day 16. The difference is 16 − 7 = 9 days; 3 days would apply without rewatering.

Question 40

A researcher compares only Treatments A and C under their original schedules. Which sequence correctly describes the predicted comparison for Scientists 1, 2, and 3, respectively?

Show answer

Answer: C — C earlier; A earlier; same day

Scientist 1 predicts Days 16 and 12 for A and C, so C begins earlier. Scientist 2 predicts Days 12 and 16, reversing that order, while Scientist 3 predicts Day 7 for both because D = 5 in both treatments. Equal drought timing therefore creates agreement only within Scientist 3's comparison, despite the opposing light and temperature schedules.

In the app the clock runs, you can pause and pick up where you left off, every question is explained afterwards, and your score is estimated on the test’s own scale. Extended time (1.5× or 2×) is there if you have an accommodation.

Take the full practice test in the app

The rest of the test

The full ACT practice test · Everything on the ACT