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ACT Science

ACT Science Quiz: Evaluating Models And Explanations

Practice Evaluating Models And Explanations in ACT Science with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Evaluating Models And Explanations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Science.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

PASSAGE VII

PHYSICS: Data Representation

Introduction

A student investigated the relationship between voltage (VVV), current (III), and resistance (RRR) in a simple direct current (DC) electrical circuit.

•Voltage (VVV) is the electrical potential difference provided by a power source, measured in volts (V).

•Current (III) is the rate of flow of electrical charge, measured in amperes (A).

•Resistance (RRR) is the opposition to the flow of charge, measured in ohms (Ω\OmegaΩ).

The student set up a circuit containing a variable voltage power supply, a resistor, and an ammeter (a device used to measure current).

Experiment 1

In the first experiment, the student used a resistor with a constant resistance of 10.0 Ω10.0 \ \Omega10.0 Ω. The student varied the voltage supplied to the circuit from 2.0 V to 10.0 V and recorded the resulting current measured by the ammeter. Results are shown in Table 1.

Experiment 2

In the second experiment, the student set the power supply to provide a constant voltage of 12.0 V12.0 \text{ V}12.0 V. The student then swapped out the resistor, testing five different resistors with varying resistance values, and recorded the resulting current for each. Results are shown in Table 2.

Based on the results of both experiments, which of the following best describes the mathematical relationships between voltage, resistance, and current?

  1. Current is directly proportional to both voltage and resistance.
  2. Current is inversely proportional to both voltage and resistance.
  3. Current is directly proportional to voltage and inversely proportional to resistance.
  4. Current is inversely proportional to voltage and directly proportional to resistance.
Explanation: The correct answer is C. Table 1 establishes the relationship between voltage and current: as voltage increases from 2.0 to 10.0 V, current increases proportionally from 0.20 to 1.00 A. When one quantity increases and the other increases in proportion, they are directly proportional. Table 2 establishes the relationship between resistance and current: as resistance increases from 2.0 to 12.0 Ω, current decreases from 6.00 to 1.00 A. When one quantity increases and the other decreases, they are inversely proportional. C correctly identifies both relationships: current is directly proportional to voltage (Table 1) and inversely proportional to resistance (Table 2). A is wrong — current is not directly proportional to resistance; Table 2 shows the opposite. B is wrong — current is not inversely proportional to voltage; Table 1 shows direct proportionality. D reverses both relationships. This question synthesizes findings from two separate tables to derive Ohm's Law (I = V/R) from the experimental data alone.

Question 2

PASSAGE II

BIOLOGY: Research Summary

Soil salinity (salt concentration) and pH can significantly affect seed germination. A botanist conducted two studies to determine how these factors influence the germination rate of Medicago sativa (alfalfa) seeds. •Note: Germination rate is the percentage of planted seeds that successfully sprout.

Study 1

The botanist prepared 5 identical planting trays. Each tray was filled with 1 kilogram (kg) of the same potting soil. The botanist adjusted the soil in each tray to have a different concentration of sodium chloride (NaCl), measured in millimoles per kilogram (mM/kg). The soil pH for all trays was kept constant at 6.5.

Fifty M. sativa seeds were planted in each tray. The trays were placed in a greenhouse with a constant temperature of 25∘C25^\circ\text{C}25∘C and watered equally every day for 14 days. On day 14, the germination rate was recorded. Results are shown in Table 1.

Study 2

The botanist prepared 5 new trays with the same potting soil. This time, the NaCl concentration in all trays was kept constant at 40 mM/kg. The botanist adjusted the soil pH in each tray to a different value, ranging from highly acidic to highly basic.

Fifty M. sativa seeds were planted in each tray. The greenhouse conditions, watering schedule, and duration were identical to those in Study 1. Results are shown in Table 2.

Suppose the botanist performs a third study using soil with an NaCl concentration of 100 mM/kg and a pH of 6.5. Based on the results of Study 1, the germination rate would most likely be:

  1. greater than 82%.
  2. between 54% and 82%.
  3. between 28% and 54%.
  4. less than 28%.
Explanation: The correct answer is C (between 28% and 54%). Study 1's data shows that at 80 mM/kg NaCl, the germination rate was 54%, and at 120 mM/kg NaCl, it was 28%. A concentration of 100 mM/kg falls between these two tested values, so the germination rate must fall between their corresponding outcomes: between 28% and 54%. F (greater than 82%) would require a lower NaCl concentration than 40 mM/kg. G (between 54% and 82%) corresponds to concentrations between 40 and 80 mM/kg — too low. H correctly identifies the range between 80 and 120 mM/kg. D (less than 28%) would require a concentration above 120 mM/kg. On interpolation questions, identify the two tested values that bracket the given value and read their corresponding outcomes.

Question 3

PASSAGE V

ASTRONOMY / EARTH SCIENCE: Data Representation

Introduction

Astronomers classify stars based on their surface temperature and their luminosity. Surface temperature is measured in Kelvin (K). Luminosity is a measure of a star's total energy output compared to the Sun (L⊙L_{\odot}L⊙​). For example, a star with a luminosity of 102L⊙10^2 L_{\odot}102L⊙​ emits 100 times more energy than the Sun. The Hertzsprung-Russell (H-R) diagram shown in Figure 1 maps stars according to these two properties.

Study

Astronomers measured the properties of four specific stars, labeled A–D, located in different regions of the H-R diagram. Results are shown in Table 1.

Star X is a Main Sequence star that is exactly twice as hot at its surface as Star D. Based on the mathematical trend shown on the H-R diagram, the luminosity of Star X is approximately:

  1. exactly half the luminosity of Star D.
  2. exactly twice the luminosity of Star D.
  3. between 10 and 100 times greater than the luminosity of Star D.
  4. between 10⁵ and 10⁶ times greater than the luminosity of Star D.
Explanation: The correct answer is C. Star D has a surface temperature of 6,000 K and a luminosity of 1 L☉. Twice as hot as Star D is 12,000 K. On the Main Sequence of the H-R diagram, following the diagonal band from 6,000 K (luminosity ~1) to 12,000 K, the luminosity value on the logarithmic y-axis rises to approximately 10² — meaning roughly 100 times the Sun's luminosity. Therefore the luminosity increases by a factor between 10 and 100. A and B (half or twice) dramatically underestimate the increase — the Main Sequence's trend on a log scale shows a much steeper relationship than a simple doubling. D (10⁵ to 10⁶ times greater) dramatically overestimates — these luminosities correspond to the Supergiant region, far above the Main Sequence at 12,000 K. This question tests the ability to read the nonlinear (logarithmic) trend of the Main Sequence rather than applying simple arithmetic. Pro tip: on log-scale diagrams, a visual shift of one unit on the y-axis represents a tenfold change in the actual quantity.

Question 4

PASSAGE IV

PHYSICS: Research Summary

Introduction

A group of physics students investigated the motion of a projectile. They used a toy cannon that utilizes a compressed steel spring to launch a 50-gram (g) plastic ball. The students investigated how the horizontal distance traveled by the ball (DDD) was affected by two variables:

1. The launch angle (θ\thetaθ), measured in degrees from the horizontal.

2. The spring compression distance (xxx), which is the distance the spring is pushed inward before being released.

Study 1

The students set up the cannon on a flat, level field. They kept the spring compression distance (xxx) constant at 4.0 cm4.0\text{ cm}4.0 cm for all trials. They launched the ball at five different angles (θ\thetaθ) and measured the horizontal distance (DDD) the ball traveled before hitting the ground. The results are shown in Table 1.

Study 2

Next, the students kept the launch angle (θ\thetaθ) constant at 45∘45^\circ45∘ (the angle that yielded the greatest distance in Study 1). They varied the spring compression distance (xxx) and measured the resulting horizontal distance (DDD). The results are shown in Table 2.

Based on Table 2, what is the mathematical relationship between the spring compression distance (x) and the horizontal distance (D)? As x doubles, D:

  1. stays the same.
  2. doubles (increases by a factor of 2).
  3. quadruples (increases by a factor of 4).
  4. increases by exactly 15.0 meters.
Explanation: The correct answer is C (quadruples). Check the data systematically: when x = 2.0 cm, D = 5.0 m. When x doubles to 4.0 cm, D = 20.0 m — that is 4 × 5.0 = 20.0, a factor of 4 increase. Confirm with the next doubling: when x doubles from 4.0 cm to 8.0 cm, D goes from 20.0 m to 80.0 m — that is 4 × 20.0 = 80.0, again a factor of 4. The relationship is quadratic: D is proportional to x². This makes physical sense — the energy stored in a compressed spring is proportional to x², and if energy converts directly to kinetic energy, the velocity and ultimately the distance scale with x². A (stays the same) would require constant D regardless of x. B (doubles) would give D = 10.0 at x = 4.0, but the table shows 20.0. D (increases by exactly 15.0) would give inconsistent additions. On relationship questions, always test the proposed relationship against at least two data pairs before confirming.

Question 5

PASSAGE V

ASTRONOMY / EARTH SCIENCE: Data Representation

Introduction

Astronomers classify stars based on their surface temperature and their luminosity. Surface temperature is measured in Kelvin (K). Luminosity is a measure of a star's total energy output compared to the Sun (L⊙L_{\odot}L⊙​). For example, a star with a luminosity of 102L⊙10^2 L_{\odot}102L⊙​ emits 100 times more energy than the Sun. The Hertzsprung-Russell (H-R) diagram shown in Figure 1 maps stars according to these two properties.

Study

Astronomers measured the properties of four specific stars, labeled A–D, located in different regions of the H-R diagram. Results are shown in Table 1.

According to the passage and Table 1, which of the four listed stars is most likely the Sun?

  1. Star A
  2. Star B
  3. Star C
  4. Star D
Explanation: The correct answer is D (Star D). The passage's introduction defines the luminosity scale: 'Luminosity is a measure of a star's total energy output compared to the Sun (L☉).' By definition, the Sun itself has a luminosity of exactly 1 L☉ — this is the reference point of the scale. Table 1 shows that Star D has a luminosity of 1 L☉ and a surface temperature of 6,000 K. A surface temperature of approximately 5,500–6,000 K is consistent with the Sun's known surface temperature, and a luminosity of exactly 1 L☉ is the defining property of the Sun. Star A (luminosity 10⁴) is far too bright. Star B (luminosity 10⁵) is even brighter. Star C (luminosity 10⁻³) is far too dim. This question tests whether students recognize that the unit L☉ is defined relative to the Sun, making 1 L☉ the Sun's own luminosity by definition.

Question 6

PASSAGE III

EARTH SCIENCE / BIOLOGY: Conflicting Viewpoints

Introduction

Approximately 66 million years ago, a mass extinction event occurred at the boundary between the Cretaceous and Paleogene periods (the K-Pg boundary), wiping out roughly 75% of all plant and animal species on Earth, including all non-avian dinosaurs. Geologists have discovered a distinct layer of sedimentary rock worldwide at the K-Pg boundary that contains unusually high levels of iridium, a metal rare in Earth’s crust. Two scientists present different hypotheses for the cause of the extinction and the source of the iridium.

Scientist 1

The mass extinction was caused by the impact of a massive asteroid, approximately 10 kilometers in diameter. Asteroids are naturally rich in iridium. When the asteroid struck the Earth, the immense force of the collision vaporized the asteroid and a large portion of Earth's crust, ejecting a massive cloud of iridium-rich dust and debris into the atmosphere. This dust cloud enveloped the planet for months or even years, blocking out incoming sunlight. The lack of sunlight halted photosynthesis globally, causing the collapse of marine and terrestrial food webs.

Furthermore, the impact would have triggered global wildfires, acid rain, and massive tsunamis. The presence of shocked quartz (quartz crystals deformed by intense, sudden pressure) and tektites (glassy spheres formed by rapidly cooling, ejected rock) in the K-Pg boundary layer alongside the iridium firmly points to a high-velocity extraterrestrial impact as the sole trigger of the extinction.

Scientist 2

The mass extinction was not caused by a sudden impact, but rather by intense, prolonged volcanic activity. Around 66 million years ago, a massive volcanic region in what is now India, known as the Deccan Traps, experienced a series of colossal eruptions that lasted for tens of thousands of years. These eruptions released millions of cubic kilometers of lava.

While iridium is rare in Earth's surface crust, it is present in high concentrations in the deep mantle. The massive magma plumes from the Deccan Traps brought this deep-Earth iridium to the surface, where volcanic ash plumes spread it globally. The prolonged eruptions released massive quantities of sulfur dioxide (SO2SO_2SO2​) and carbon dioxide (CO2CO_2CO2​) into the atmosphere. The SO2SO_2SO2​ caused severe short-term global cooling and acid rain, while the CO2CO_2CO2​ led to long-term extreme global warming. This resulting climate instability severely stressed ecosystems over thousands of years, leading to a gradual, rather than instantaneous, mass extinction. Shocked quartz can also be formed by the explosive pressures of massive volcanic eruptions.

Both scientists would most likely agree with which of the following statements about the K-Pg boundary?

  1. The boundary layer was formed over a period of exactly ten years.
  2. The rock layer at the boundary contains an unusually high concentration of iridium.
  3. The global climate remained perfectly stable during the formation of the boundary layer.
  4. The boundary marks the exact moment a 10-kilometer asteroid struck the planet.
Explanation: The correct answer is B. The introduction — which establishes shared scientific facts before the viewpoints diverge — states: 'Geologists have discovered a distinct layer of sedimentary rock worldwide at the K-Pg boundary that contains unusually high levels of iridium.' Both scientists accept this fact; they disagree only about the source of that iridium (asteroid vs. deep mantle). A is wrong — neither scientist specifies a ten-year formation period; Scientist 2 describes processes lasting tens of thousands of years. C is directly contradicted by both scientists — Scientist 1 describes a global dust cloud causing cooling, and Scientist 2 describes SO₂ cooling followed by CO₂ warming. D represents Scientist 1's hypothesis specifically and would be rejected by Scientist 2. On agreement questions, look for claims that appear in the shared introduction or that both viewpoints independently accept.

Question 7

PASSAGE I

CHEMISTRY/PHYSICS: Data Representation

Introduction

Viscosity is a measure of a fluid's resistance to flow. Fluids with high viscosity flow slowly, while fluids with low viscosity flow quickly. The viscosity of a liquid typically decreases as its temperature increases. Students investigated the viscosity of four different synthetic motor oils (Oils W, X, Y, and Z) by measuring the time it took for a solid steel ball to drop through a 50-centimeter (cm) vertical glass cylinder filled with the oil.

Study 1

The students recorded the drop time, in seconds (s), for the steel ball in each of the four oils at 20∘C20^\circ\text{C}20∘C, 40∘C40^\circ\text{C}40∘C, 60∘C60^\circ\text{C}60∘C, and 80∘C80^\circ\text{C}80∘C. The results are shown in Figure 1.

Study 2

The students also looked up the established viscosity values, measured in millipascal-seconds (mPa·s), for each oil at 40∘C40^\circ\text{C}40∘C and 100∘C100^\circ\text{C}100∘C. The results are shown in Table 1.

Based on the relationship between drop time and viscosity described in the passage, if the students tested a new oil, Oil V, that had a viscosity of 130.0 mPa·s at 40°C, the drop time of the steel ball in Oil V at 40°C would most likely be:

  1. less than 2.5 s.
  2. between 4.5 s and 6.0 s.
  3. between 6.0 s and 8.5 s.
  4. greater than 8.5 s.
Explanation: The correct answer is D (greater than 8.5 s). The passage establishes that higher viscosity means greater resistance to flow — the ball drops more slowly — producing a longer drop time. At 40°C, Oil W has the highest viscosity of the four oils tested at 115.0 mPa·s, and it produced the longest drop time of 8.5 s. Oil V has a viscosity of 130.0 mPa·s — higher than Oil W. Therefore, the ball must take longer to fall through Oil V than through Oil W, making the drop time greater than 8.5 s. F and G describe drop times associated with lower-viscosity oils at 40°C. H describes a range between Oil Y and Oil W values — too low. Pro tip: anchor extrapolation questions by identifying the highest or lowest tested value and reasoning beyond it.

Question 8

PASSAGE III

PHYSICS: This passage presents three hypotheses regarding the discrepancy between the observed mass of galaxies and their rotational speeds.

Introduction

In the 1970s, astronomer Vera Rubin observed that stars at the edges of spiral galaxies move just as fast as stars near the center. According to Newtonian physics, stars further from the center should move slower due to the decrease in gravitational pull. This observation implies that galaxies contain far more mass than what is visible in stars and gas. This missing mass is referred to as "Dark Matter." Three scientists propose different explanations for its nature.

Scientist 1

The missing mass consists of Weakly Interacting Massive Particles (WIMPs). These are subatomic particles that have mass but do not interact with electromagnetic radiation (light), making them invisible. WIMPs interact only through gravity and the weak nuclear force. They were produced in the early universe and now form a vast, spherical "halo" that surrounds every galaxy. Computer simulations of the universe’s formation match observational data only when this "cold dark matter" is included. If WIMPs exist, they should eventually be detectable by sensitive underground experiments designed to catch rare collisions between WIMPs and atomic nuclei.

Scientist 2

The missing mass is not some exotic new particle; it is simply normal ("baryonic") matter that is too dim to see. These objects are called Massive Compact Halo Objects (MACHOs). They include black holes, neutron stars, brown dwarfs (failed stars), and rogue planets. Because they emit little to no light, they have escaped detection. The gravitational pull of these objects accounts for the high rotational speeds of galaxies. Evidence for MACHOs comes from "microlensing" events, where the gravity of a massive, invisible object bends the light of a distant star, causing it to brighten temporarily.

Scientist 3

There is no missing mass. The discrepancy is caused by a flaw in our understanding of gravity itself. This theory is known as Modified Newtonian Dynamics (MOND). Newtonian laws work well on the scale of our solar system, where accelerations are high. However, on the galactic scale, where accelerations are incredibly low (\<10−10m/s2\< 10^{-10} m/s^2\<10−10m/s2), gravity behaves differently. Below this threshold, gravitational force decays more slowly with distance (1/r1/r1/r instead of 1/r21/r^21/r2). This stronger effective gravity eliminates the need for invisible halos or new particles. The rotation curves are exactly what the laws of physics predict when corrected for this scale.

A critique of Scientist 3's hypothesis is that it requires changing a fundamental law of physics. Scientist 3 would most likely respond by pointing out that:

  1. Newtonian physics has never been tested at the extremely low accelerations found in outer galaxies.
  2. WIMPs have already been detected in underground experiments.
  3. black holes and neutron stars are emitting more light than previously thought.
  4. the universe is much younger than Scientist 1 claims.
Explanation: This is a defense of argument question asking how a scientist would respond to criticism. Scientist 3's core claim is that Newtonian gravity works at solar system scales (high acceleration) but fails at galactic scales ("accelerations are incredibly low < 10⁻¹⁰ m/s²"). The natural defense against "you're changing physics" is "we've never actually tested physics at these extreme conditions." Choice A correctly predicts this defense—the untested regime justifies modification. Choice B (WIMPs detected) would support Scientist 1, not defend Scientist 3. Choice C (light emission) is irrelevant to gravity laws. Choice D (universe age) is irrelevant. Pro tip: Use a scientist's core argument to predict their defense against criticism.

Question 9

Two scientists debate the primary cause of climate change. Scientist A argues carbon emissions are the main factor, while Scientist B suggests solar activity plays a larger role. Data shows a strong correlation between carbon levels and temperature rise. Which explanation most logically accounts for the results?

  1. Neither explanation is logical.
  2. Both explanations are equally logical.
  3. Scientist B's explanation is most logical.
  4. Scientist A's explanation is most logical.
Explanation: Scientist A's explanation is most logical because it correctly identifies carbon emissions as the primary climate change driver. The data shows a strong correlation between carbon levels and temperature rise, which directly supports Scientist A's argument that carbon emissions are the main factor. Scientist B's emphasis on solar activity cannot account for the observed correlation between atmospheric carbon concentrations and global temperature increases. The evidence clearly demonstrates that carbon emissions, not solar variations, explain the documented temperature trends.

Question 10

Students test why an enzyme’s reaction rate decreases at very high substrate concentrations. Two models are proposed.

Model 1 (Saturation Only): Rate approaches a maximum as substrate increases, then stays near that maximum. Prediction: At very high substrate, rate should plateau, not decrease.

Model 2 (Substrate Inhibition): Excess substrate binds to an inhibitory site, reducing activity. Prediction: Rate should increase at first, then decrease at very high substrate.

Figure 1 shows reaction rate versus substrate concentration.

Which model is best supported by the results in Figure 1?

  1. Neither model, because the enzyme rate should be independent of substrate concentration.
  2. Model 2, because the rate decreases at very high substrate after reaching a peak at intermediate substrate.
  3. Both models, because both predict the same rate at all substrate concentrations.
  4. Model 1, because the rate increases with substrate concentration from 0 to 10 mM.
Explanation: Model 2, the Substrate Inhibition Model, is best supported by the results in Figure 1. The figure depicts reaction rate increasing from 0 to 80 units between 0–5 mM substrate, peaking at 90 units at 10 mM, and then decreasing to 60 units at 40 mM. This pattern aligns with Model 2's prediction of an initial rise followed by a decline at high substrate due to inhibitory binding, showing a non-plateauing response. Model 1 is contradicted because the rate does not plateau but instead decreases after the maximum, inconsistent with saturation-only expectations.

Question 11

Two hypotheses explain the increase in jellyfish populations. Hypothesis 1 states warmer ocean temperatures increase jellyfish numbers. Hypothesis 2 suggests overfishing of predators is the cause. Data correlates higher jellyfish numbers with warmer temperatures. Which hypothesis is more supported?

  1. Neither hypothesis is supported.
  2. Hypothesis 2 is more supported.
  3. Both hypotheses are equally supported.
  4. Hypothesis 1 is more supported.
Explanation: Hypothesis 1 is more supported because it correctly predicts that warmer ocean temperatures increase jellyfish populations. The data correlates higher jellyfish numbers with warmer temperatures, which directly aligns with Hypothesis 1's temperature-driven population growth prediction. Hypothesis 2 suggests overfishing of predators as the cause, but this explanation doesn't account for the observed temperature-population correlation patterns. The evidence clearly demonstrates that thermal conditions, not predator removal alone, drive the observed jellyfish population increases.

Question 12

PASSAGE II

BIOLOGY: Research Summary

Soil salinity (salt concentration) and pH can significantly affect seed germination. A botanist conducted two studies to determine how these factors influence the germination rate of Medicago sativa (alfalfa) seeds. •Note: Germination rate is the percentage of planted seeds that successfully sprout.

Study 1

The botanist prepared 5 identical planting trays. Each tray was filled with 1 kilogram (kg) of the same potting soil. The botanist adjusted the soil in each tray to have a different concentration of sodium chloride (NaCl), measured in millimoles per kilogram (mM/kg). The soil pH for all trays was kept constant at 6.5.

Fifty M. sativa seeds were planted in each tray. The trays were placed in a greenhouse with a constant temperature of 25∘C25^\circ\text{C}25∘C and watered equally every day for 14 days. On day 14, the germination rate was recorded. Results are shown in Table 1.

Study 2

The botanist prepared 5 new trays with the same potting soil. This time, the NaCl concentration in all trays was kept constant at 40 mM/kg. The botanist adjusted the soil pH in each tray to a different value, ranging from highly acidic to highly basic.

Fifty M. sativa seeds were planted in each tray. The greenhouse conditions, watering schedule, and duration were identical to those in Study 1. Results are shown in Table 2.

A student hypothesizes that M. sativa seeds are completely incapable of germinating in highly acidic soils (pH less than 5.0). Do the results of Study 2 support this hypothesis?

  1. Yes; Tray 6 had a pH of 4.5, and its germination rate was 0%.
  2. Yes; Tray 7 had a pH of 5.5, and its germination rate was significantly lower than Tray 8.
  3. No; Tray 6 had a pH of 4.5, and 18% of the seeds successfully germinated.
  4. No; Tray 10 had a pH of 8.5, and 44% of the seeds successfully germinated.
Explanation: The correct answer is C. The hypothesis claims M. sativa seeds are completely incapable of germinating at pH below 5.0. Tray 6 had a pH of 4.5 — which falls below 5.0 — and achieved an 18% germination rate. Because some seeds did germinate, the seeds are not completely incapable of germinating in highly acidic conditions. The hypothesis is therefore not supported. A is factually wrong — Tray 6's germination rate was 18%, not 0%. B is wrong — Tray 7 has a pH of 5.5, which does not fall below 5.0 and therefore does not test the specific condition in the hypothesis. D is wrong — Tray 10 tests basic conditions (pH 8.5), which is irrelevant to a hypothesis about acidic conditions below pH 5.0. On hypothesis-testing questions, focus precisely on whether the data satisfies the exact conditions stated in the hypothesis.

Question 13

PASSAGE IV

PHYSICS: Research Summary

Introduction

A group of physics students investigated the motion of a projectile. They used a toy cannon that utilizes a compressed steel spring to launch a 50-gram (g) plastic ball. The students investigated how the horizontal distance traveled by the ball (DDD) was affected by two variables:

1. The launch angle (θ\thetaθ), measured in degrees from the horizontal.

2. The spring compression distance (xxx), which is the distance the spring is pushed inward before being released.

Study 1

The students set up the cannon on a flat, level field. They kept the spring compression distance (xxx) constant at 4.0 cm4.0\text{ cm}4.0 cm for all trials. They launched the ball at five different angles (θ\thetaθ) and measured the horizontal distance (DDD) the ball traveled before hitting the ground. The results are shown in Table 1.

Study 2

Next, the students kept the launch angle (θ\thetaθ) constant at 45∘45^\circ45∘ (the angle that yielded the greatest distance in Study 1). They varied the spring compression distance (xxx) and measured the resulting horizontal distance (DDD). The results are shown in Table 2.

Based on the trend in Table 1, if the students had launched the ball at an angle of 90° (straight up), the horizontal distance traveled by the ball would most likely be:

  1. 0.0 m
  2. 10.0 m
  3. 20.0 m
  4. 25.0 m
Explanation: The correct answer is A (0.0 m). The data in Table 1 shows that as the angle moves away from 45° in either direction, horizontal distance decreases: at 75° the distance is 10.0 m, which is lower than at 45° (20.0 m). If the ball is launched straight up at 90°, all of the projectile's velocity is directed vertically — none is directed horizontally. The ball would travel straight up and land directly back at the launch point, traveling zero horizontal distance. The trend from 45° to 75° shows decreasing horizontal distance as the angle approaches vertical, and at exactly 90° this reaches its minimum of 0.0 m. G (10.0 m) is the value at 75°, which is a plausible wrong answer for students who don't extrapolate the trend fully. Pro tip: on extrapolation questions, follow the established trend to its logical physical endpoint.

Question 14

PASSAGE VI

BIOLOGY / GENETICS: Research Summary

Introduction

Restriction enzymes are proteins that cut DNA molecules at specific, predictable sequences of base pairs (bp). Gel electrophoresis is a technique used to separate these resulting DNA fragments by size. When an electrical current is applied to the gel, the negatively charged DNA fragments migrate toward the positive electrode. Smaller DNA fragments move through the gel much faster and travel further than larger fragments.

A researcher isolated a circular bacterial plasmid (a ring of DNA) consisting of exactly 5,000 bp. To map the plasmid, the researcher treated identical samples of the plasmid with different restriction enzymes—Enzyme 1 (E1), Enzyme 2 (E2), and Enzyme 3 (E3)—both individually and in combinations.

After allowing the enzymes to cut the DNA, the researcher ran the samples on an electrophoresis gel. A dye was added to make the DNA bands visible. The size of the fragments in each band was recorded in Table 1.

Based on the results in Table 1, where is the cut site for Enzyme 1 (E1) located relative to the cut sites for Enzyme 2 (E2)?

  1. E1 cuts the 2,000 bp fragment produced by E2 in half.
  2. E1 cuts the 3,000 bp fragment produced by E2 into two unequal pieces (2,500 bp and 500 bp).
  3. E1 cuts the uncut plasmid exactly 500 bp away from the E3 cut site.
  4. E1 cuts at the exact same location as E2, producing identical fragments.
Explanation: The correct answer is B (E1 cuts the 3,000 bp E2 fragment into 2,500 bp and 500 bp). This requires systematic reasoning across lanes. E2 alone (Lane 3) produces two fragments: 3,000 bp and 2,000 bp. When E1 is added to E2 (Lane 5), the results are: 2,500 bp, 2,000 bp, and 500 bp. Comparing these: the 2,000 bp fragment appears unchanged in both Lane 3 and Lane 5 — E1 did not cut it. The 3,000 bp fragment from Lane 3 disappears in Lane 5 and is replaced by 2,500 bp and 500 bp. Confirming: 2,500 + 500 = 3,000 ✓. Therefore, E1 cut the 3,000 bp fragment into two unequal pieces. A is wrong — the 2,000 bp fragment is unchanged between Lane 3 and Lane 5. C incorrectly references E3 — the E3 data is not needed to answer this question. D is wrong — if E1 cut at the same location as E2, the fragment pattern would be identical to E2 alone, which it is not. Pro tip: on restriction mapping questions, compare the combined enzyme lane to the individual enzyme lanes to identify exactly which fragment was cut.

Question 15

PASSAGE VI

BIOLOGY / GENETICS: Research Summary

Introduction

Restriction enzymes are proteins that cut DNA molecules at specific, predictable sequences of base pairs (bp). Gel electrophoresis is a technique used to separate these resulting DNA fragments by size. When an electrical current is applied to the gel, the negatively charged DNA fragments migrate toward the positive electrode. Smaller DNA fragments move through the gel much faster and travel further than larger fragments.

A researcher isolated a circular bacterial plasmid (a ring of DNA) consisting of exactly 5,000 bp. To map the plasmid, the researcher treated identical samples of the plasmid with different restriction enzymes—Enzyme 1 (E1), Enzyme 2 (E2), and Enzyme 3 (E3)—both individually and in combinations.

After allowing the enzymes to cut the DNA, the researcher ran the samples on an electrophoresis gel. A dye was added to make the DNA bands visible. The size of the fragments in each band was recorded in Table 1.

A chemical agent is introduced that induces a mutation at the exact DNA sequence where Enzyme 1 (E1) normally binds, preventing E1 from cutting the DNA. If this mutated plasmid is treated with the E1 + E2 combination, what fragment sizes will be produced?

  1. 5,000 bp only
  2. 3,000 bp and 2,000 bp only
  3. 4,000 bp and 1,000 bp only
  4. 2,500 bp, 2,000 bp, and 500 bp
Explanation: The correct answer is B (3,000 bp and 2,000 bp only). If the mutation destroys E1's binding site, E1 can no longer cut the DNA — it becomes functionally absent from the reaction. Therefore, treating the mutated plasmid with 'E1 + E2' produces the same result as treating it with E2 alone. Lane 3 shows that E2 alone produces 3,000 bp and 2,000 bp fragments. These are the only fragments produced when E2 is the only functional enzyme. A (5,000 bp only) would result if neither enzyme could cut — but E2 is unaffected by the mutation. C (4,000 bp and 1,000 bp) are the fragments produced by E3 alone — E3 is not present in this reaction. D (2,500 bp, 2,000 bp, and 500 bp) is the result of the normal E1 + E2 treatment from Lane 5 — but E1 cannot cut in the mutated plasmid. This question tests the logical principle that a non-functional enzyme is equivalent to an absent enzyme.

Question 16

PASSAGE V

ASTRONOMY / EARTH SCIENCE: Data Representation

Introduction

Astronomers classify stars based on their surface temperature and their luminosity. Surface temperature is measured in Kelvin (K). Luminosity is a measure of a star's total energy output compared to the Sun (L⊙L_{\odot}L⊙​). For example, a star with a luminosity of 102L⊙10^2 L_{\odot}102L⊙​ emits 100 times more energy than the Sun. The Hertzsprung-Russell (H-R) diagram shown in Figure 1 maps stars according to these two properties.

Study

Astronomers measured the properties of four specific stars, labeled A–D, located in different regions of the H-R diagram. Results are shown in Table 1.

Star B and Star D both emit radiation into space. However, Star B is significantly cooler than Star D at its surface, yet it emits 100,000 times more total energy. Based on the groups shown in Figure 1, which of the following best explains this phenomenon?

  1. Star B is a White Dwarf, which generates energy much more efficiently than a Main Sequence star.
  2. Star B is a Supergiant, meaning its massive surface area allows it to emit vastly more total energy despite being cooler.
  3. Star B is located on the extreme lower-right of the Main Sequence, where stars burn nuclear fuel more rapidly.
  4. Star D has a logarithmic energy output, which suppresses its luminosity compared to Star B.
Explanation: The correct answer is B. Table 1 shows Star B has a surface temperature of 3,000 K (cool) and a luminosity of 10⁵ L☉ (extremely bright). Star D has a temperature of 6,000 K (warmer) but only 1 L☉ (much less luminous). The apparent paradox — cooler surface yet vastly more total energy output — is resolved by size. A Supergiant is enormously larger than a Main Sequence star. Although its surface is cooler, the sheer area of surface radiating energy makes the total output vastly greater. Figure 1 places Stars with low temperature and very high luminosity in the Supergiant region (upper-right), where Star B's coordinates (3,000 K, 10⁵) fall. A is wrong — White Dwarfs are in the lower-left (hot but dim), not the upper-right, and Star B is dim by White Dwarf standards. C is wrong — lower-right Main Sequence stars are cool and dim, the opposite of Star B. D invents a 'logarithmic suppression' concept that does not appear in the passage.

Question 17

PASSAGE I

CHEMISTRY/PHYSICS: Data Representation

Introduction

Viscosity is a measure of a fluid's resistance to flow. Fluids with high viscosity flow slowly, while fluids with low viscosity flow quickly. The viscosity of a liquid typically decreases as its temperature increases. Students investigated the viscosity of four different synthetic motor oils (Oils W, X, Y, and Z) by measuring the time it took for a solid steel ball to drop through a 50-centimeter (cm) vertical glass cylinder filled with the oil.

Study 1

The students recorded the drop time, in seconds (s), for the steel ball in each of the four oils at 20∘C20^\circ\text{C}20∘C, 40∘C40^\circ\text{C}40∘C, 60∘C60^\circ\text{C}60∘C, and 80∘C80^\circ\text{C}80∘C. The results are shown in Figure 1.

Study 2

The students also looked up the established viscosity values, measured in millipascal-seconds (mPa·s), for each oil at 40∘C40^\circ\text{C}40∘C and 100∘C100^\circ\text{C}100∘C. The results are shown in Table 1.

Is the statement 'At any given temperature, the oil with the highest viscosity will have the shortest drop time' supported by the data in Figure 1 and Table 1?

  1. Yes; Oil W has the highest viscosity and the shortest drop times.
  2. Yes; Oil Z has the highest viscosity and the shortest drop times.
  3. No; Oil W has the highest viscosity but the longest drop times.
  4. No; Oil Z has the highest viscosity but the longest drop times.
Explanation: The correct answer is C. This question requires cross-referencing Figure 1 (drop times) with Table 1 (viscosity values). Table 1 shows Oil W has the highest viscosity at 40°C (115.0 mPa·s), which is higher than Oils X, Y, and Z. Figure 1 shows that Oil W consistently has the longest drop times across all temperatures — for example, 12.0 s at 20°C compared to Oil Z's 4.0 s. Therefore, the oil with the highest viscosity has the longest drop time, not the shortest. The statement reverses the actual relationship: high viscosity = more resistance to flow = slower ball = longer drop time. A is wrong because while it correctly identifies Oil W as highest viscosity, it incorrectly states it has the shortest drop time. B and D both misidentify Oil Z as having the highest viscosity when in fact Oil Z has the lowest.

Question 18

PASSAGE VII

PHYSICS: Data Representation

Introduction

A student investigated the relationship between voltage (VVV), current (III), and resistance (RRR) in a simple direct current (DC) electrical circuit.

•Voltage (VVV) is the electrical potential difference provided by a power source, measured in volts (V).

•Current (III) is the rate of flow of electrical charge, measured in amperes (A).

•Resistance (RRR) is the opposition to the flow of charge, measured in ohms (Ω\OmegaΩ).

The student set up a circuit containing a variable voltage power supply, a resistor, and an ammeter (a device used to measure current).

Experiment 1

In the first experiment, the student used a resistor with a constant resistance of 10.0 Ω10.0 \ \Omega10.0 Ω. The student varied the voltage supplied to the circuit from 2.0 V to 10.0 V and recorded the resulting current measured by the ammeter. Results are shown in Table 1.

Experiment 2

In the second experiment, the student set the power supply to provide a constant voltage of 12.0 V12.0 \text{ V}12.0 V. The student then swapped out the resistor, testing five different resistors with varying resistance values, and recorded the resulting current for each. Results are shown in Table 2.

Based on the trend shown in Table 1, if the student had set the voltage to 14.0 V using the same 10.0 Ω resistor, the current measured by the ammeter would most likely have been:

  1. 1.20 A
  2. 1.40 A
  3. 1.60 A
  4. 14.00 A
Explanation: The correct answer is B (1.40 A). Table 1 reveals a perfect linear relationship: Current = Voltage ÷ 10. Checking: 2.0 V ÷ 10 = 0.20 A ✓; 4.0 V ÷ 10 = 0.40 A ✓; 10.0 V ÷ 10 = 1.00 A ✓. Extending this to 14.0 V: 14.0 ÷ 10 = 1.40 A. A (1.20 A) is the correct answer for 12.0 V — one step below the target. C (1.60 A) is the answer for 16.0 V — one step above. D (14.00 A) would only be correct if the resistance were 1.0 Ω, not 10.0 Ω — it represents a mathematical error of using voltage as current directly. On linear extrapolation questions, identify the mathematical rule from existing data pairs and apply it to the new value.

Question 19

PASSAGE III

BIOLOGY / PALEONTOLOGY: Conflicting Viewpoints

Introduction

For over 135 million years, dinosaurs dominated Earth’s terrestrial ecosystems. A major debate in paleontology surrounds dinosaur metabolism. Were they ectotherms (cold-blooded animals, like modern reptiles, that rely on environmental heat), endotherms (warm-blooded animals, like modern mammals and birds, that generate their own internal heat), or mesotherms (animals with an intermediate strategy)? Three scientists present their viewpoints.

Scientist 1

Dinosaurs were endotherms. Like modern birds and mammals, they possessed high metabolic rates that allowed them to sustain rapid movement and thrive in a variety of climates. Fossilized bone cross-sections from many dinosaur species reveal dense networks of blood vessels called Haversian canals. In modern animals, this specific highly vascularized bone tissue is only found in endotherms, as it is required to support rapid, continuous, year-round growth. Furthermore, numerous dinosaur fossils have been discovered in paleolandscapes that were located near the Earth's poles during the Cretaceous period. These environments experienced months of freezing temperatures and complete darkness. An ectotherm would not survive these conditions, but an internally heated endotherm could.

Scientist 2

Dinosaurs were ectotherms. The "continuous growth" model proposed by Scientist 1 is flawed. A closer microscopic analysis of dinosaur bones reveals Lines of Arrested Growth (LAGs). These are rings within the bone, similar to tree rings, which form when an animal's growth temporarily stops due to a lack of resources or seasonal drops in temperature. LAGs are a hallmark of modern ectotherms like crocodiles and turtles.

The primary reason dinosaurs could remain active is a phenomenon called gigantothermy. Because many dinosaurs were massive, they had a very low surface-area-to-volume ratio. This allowed them to absorb heat from the sun and retain it for long periods, essentially keeping their massive bodies warm without the incredibly high energy cost of burning food to generate internal heat.

Scientist 3

Both Scientist 1 and Scientist 2 represent extreme ends of a spectrum. Dinosaurs were mesotherms. They could raise their body temperature through the heat generated by their massive muscles during activity, but they could not strictly regulate their temperature at a constant internal set point like modern endotherms.

By analyzing the spacing of LAGs alongside the estimated adult mass of various dinosaurs, we can calculate their specific growth rates. Dinosaur growth rates fall exactly in the middle—faster than modern reptiles (ectotherms) but significantly slower than modern mammals (endotherms). Additionally, modern endotherms possess complex nasal structures called respiratory turbinates, which prevent excessive water loss caused by the high breathing rates needed to sustain a warm-blooded metabolism. Dinosaur skulls completely lack respiratory turbinates, proving they could not have sustained the metabolic rates of true endotherms.

Suppose paleontologists discover the fossilized remains of a very small dinosaur species (roughly the size of a chicken) that lived exclusively in the freezing, dark climates of the Cretaceous polar regions. This discovery would most heavily weaken the hypothesis of:

  1. Scientist 1, because small animals require fewer Haversian canals.
  2. Scientist 2, because an animal that small could not rely on gigantothermy to survive freezing temperatures.
  3. Scientist 3, because mesotherms must grow to massive sizes to form Lines of Arrested Growth.
  4. Scientists 1 and 3, because chickens are modern endotherms.
Explanation: The correct answer is B. Scientist 2's entire mechanism for maintaining warmth is gigantothermy — the idea that massive bodies have a low surface-area-to-volume ratio, allowing them to absorb and retain environmental heat. A chicken-sized dinosaur would have a very high surface-area-to-volume ratio and would lose body heat almost immediately in freezing polar temperatures. If this tiny dinosaur survived in that environment, it could not have been relying on gigantothermy, directly undermining Scientist 2's central mechanism. A is wrong — Scientist 1's endothermy argument does not depend on animal size; small endotherms (like birds and mice) are well-documented. C is wrong — Scientist 3 never claims LAGs only form in massive dinosaurs. D is wrong because the chicken comparison is irrelevant — modern chicken endothermy does not undermine Scientist 1's argument. Pro tip: 'Weaken the hypothesis' questions require identifying the specific mechanism each scientist relies on and finding which mechanism is most directly contradicted by the new evidence.

Question 20

PASSAGE VII

PHYSICS: Research Summary

Introduction

When an object falls through a fluid (like air), it experiences a downward gravitational force (FgF_gFg​) and an upward air resistance, or drag force (FdF_dFd​). As the object's falling speed increases, FdF_dFd​ also increases. Eventually, FdF_dFd​ becomes exactly equal to FgF_gFg​. At this point, the net force on the object is zero, and it stops accelerating, falling at a constant maximum speed known as terminal velocity (vtv_tvt​). Students investigated terminal velocity by dropping standard paper coffee filters from a height of 5 meters.

Study 1

The students nested (stacked) different numbers of coffee filters (NNN) together. Nesting the filters increased the total mass (mmm) of the falling object without significantly changing its cross-sectional area. They dropped the nested filters and used a motion sensor to record the terminal velocity (vtv_tvt​) in meters per second (m/s). Findings are shown in Table 1.

Study 2

The students investigated how cross-sectional area (AAA) affects terminal velocity. They built 4 small parachutes of different cross-sectional areas. They attached a constant 10.0-gram mass to each parachute and dropped them from 5 meters, recording the vtv_tvt​. Findings are shown in Table 2.

A mathematical analysis of Study 1 reveals that the terminal velocity (vt) is directly proportional to the square root of the number of filters (√N). Note that for N=1, √1 = 1.0, and the measured vt = 1.0 m/s. Based on this relationship, if the students dropped a stack of N = 9 filters, the expected terminal velocity would be:

  1. 2.5 m/s
  2. 3.0 m/s
  3. 4.0 m/s
  4. 4.5 m/s
Explanation: The correct answer is B. The relationship states vt ∝ √N, with the anchor point N=1 giving vt = 1.0 m/s. This means vt = 1.0 × √N. For N=9: vt = 1.0 × √9 = 1.0 × 3.0 = 3.0 m/s. This can be verified with the existing data: for N=4, √4 = 2.0, and the table shows vt = 2.0 m/s ✓. A (2.5 m/s) would correspond to N ≈ 6.25. C (4.0 m/s) would correspond to N=16. D (4.5 m/s) would correspond to N ≈ 20. Pro tip: When given a proportionality relationship with an anchor data point, use that anchor to set the constant of proportionality. Here, vt = 1.0 × √N is fully defined by the N=1 case.