ACT Science Quiz: Evaluating Trends And Making Predictions
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Evaluating Trends And Making PredictionsQuestion 1 of 20

PASSAGE IV

GEOPHYSICS: This passage is adapted from a study on the structure of Earth's interior using seismic waves.

Seismologists study the interior of the Earth by analyzing the propagation of seismic waves generated by earthquakes. There are two main types of body waves:

•P-waves (Primary waves): Compressional waves that travel through solids, liquids, and gases.

•S-waves (Secondary waves): Shear waves that travel only through solids.

The velocity of these waves depends on the density and physical state (solid or liquid) of the material they travel through. Abrupt changes in velocity indicate boundaries between Earth's layers.

Which of the following statements best describes the relationship between depth and density within the Mantle (0-2,900 km), according to Figure 2?

Question graphic
As depth increases, density decreases linearly.
As depth increases, density increases.
As depth increases, density remains constant.
As depth increases, density fluctuates randomly.
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ACT Science Quiz

ACT Science Quiz: Evaluating Trends And Making Predictions

Practice Evaluating Trends And Making Predictions 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 Trends And Making Predictions, 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.

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Question 1

PASSAGE IV

GEOPHYSICS: This passage is adapted from a study on the structure of Earth's interior using seismic waves.

Seismologists study the interior of the Earth by analyzing the propagation of seismic waves generated by earthquakes. There are two main types of body waves:

•P-waves (Primary waves): Compressional waves that travel through solids, liquids, and gases.

•S-waves (Secondary waves): Shear waves that travel only through solids.

The velocity of these waves depends on the density and physical state (solid or liquid) of the material they travel through. Abrupt changes in velocity indicate boundaries between Earth's layers.

Which of the following statements best describes the relationship between depth and density within the Mantle (0-2,900 km), according to Figure 2?

  1. As depth increases, density decreases linearly.
  2. As depth increases, density increases. (correct answer)
  3. As depth increases, density remains constant.
  4. As depth increases, density fluctuates randomly.
Explanation: This is a trend description question. Figure 2 shows that within the Mantle region (from the surface to 2,900 km), density increases from approximately 3.0 g/cm³ to 5.5 g/cm³. This is a steady upward trend. Choice B (density increases) correctly describes this relationship. Choice A (decreases) is opposite of the actual trend. Choice C (constant) would require a flat horizontal line. Choice D (fluctuates) would require up-and-down variation not present in the data. Pro tip: For relationship questions, focus on the overall direction of change within the specified range.

Question 2

PASSAGE V

BIOLOGY: This passage is adapted from a study on the metabolic rates of vertebrates. Introduction

Metabolism is the set of chemical reactions that occur in living organisms to maintain life. The metabolic rate is often measured by the amount of oxygen (O2O_2) consumed per gram of body mass per hour.

Animals can be classified based on how they regulate body temperature:

•Endotherms (e.g., mammals, birds) generate their own body heat to maintain a constant internal temperature.

•Ectotherms (e.g., reptiles, amphibians) rely on external heat sources to regulate their body temperature.

Students conducted two studies to compare the metabolic rates of a Mouse (Endotherm) and a Lizard (Ectotherm) of similar body mass.

Study 1

The students placed the mouse and the lizard in separate metabolic chambers. They varied the environmental temperature from 5°C to 35°C in 10°C increments. The animals were kept at rest. The rate of oxygen consumption (mL O2/ghrmL \ O_2 / g \cdot hr) was measured after the animals had acclimated to each temperature for 30 minutes. Findings were reported in Figure 1.

Study 2

The students investigated the effect of activity level on metabolic rate. They maintained the environmental temperature at 25°C for both animals. They measured the oxygen consumption while the animals were at rest and while they were running on a treadmill at 1.0 km/hr. Findings were reported in Table 1.

Study 3

To determine if body mass affects metabolic rate within the same group, students measured the resting metabolic rate of three different lizards at 25°C. Findings were reported in Table 2.

Consider the data for the Mouse at 35°C in Figure 1. If the temperature were increased further to 45°C, which of the following predictions is most biologically likely?

  1. The metabolic rate would drop to 0.
  2. The metabolic rate would continue to decrease.
  3. The metabolic rate would increase as the mouse pants or sweats to cool down. (correct answer)
  4. The metabolic rate would become identical to the Lizard's rate.
Explanation: This is an extrapolation/prediction question requiring biological reasoning. Figure 1 shows the Mouse's metabolic rate decreases from 5°C to a minimum at 25°C, then slightly increases at 35°C (from 1.5 to 2.0). This upturn suggests the beginning of heat stress. At 45°C (well above normal), the mouse would experience severe heat stress and need to activate cooling mechanisms (panting, increased blood flow to extremities for heat dissipation). These cooling processes require energy, increasing metabolic rate. Choice C correctly predicts this biologically realistic response. Choice A (drop to 0) would mean death, not a gradual response. Choice B (continue decreasing) ignores the upturn already visible at 35°C. Choice D (identical to Lizard) is unrealistic—endotherms and ectotherms have fundamentally different metabolic strategies. Pro tip: When extrapolating trends, consider biological limits and stress responses, not just mathematical continuation.

Question 3

A materials scientist measured the thickness of a coating after each pass of a sprayer. Based on the pattern in the table, what thickness would most likely be measured after 6 passes?

  1. 21 µm
  2. 27 µm (correct answer)
  3. 24 µm
  4. 30 µm
Explanation: The data show a linear accumulation in coating thickness with each sprayer pass. The thicknesses are 4.5 µm after 1 pass, 9 µm after 2, 13.5 µm after 3, 18 µm after 4, and 22.5 µm after 5, increasing by 4.5 µm per pass. This quantifies the pattern as a constant addition of 4.5 µm each pass. To predict the thickness after 6 passes, add 4.5 µm to the 5-pass value: 22.5 + 4.5 = 27 µm. Some might assume doubling instead, leading to 45 µm, but the data confirm additive growth.

Question 4

An engineer recorded the distance a test cart traveled after different numbers of identical pushes on a smooth track. Based on the pattern in the table, what distance would most likely be traveled after 6 pushes?

  1. 18 m
  2. 21 m
  3. 24 m (correct answer)
  4. 28 m
Explanation: The data indicate a linear increase in the distance traveled by the cart with each additional push. The distances are 4 m after 1 push, 8 m after 2 pushes, 12 m after 3, 16 m after 4, and 20 m after 5, with a consistent addition of 4 m per push. This quantifies the pattern as a linear relationship where each push adds 4 m to the total distance. To predict the distance after 6 pushes, add 4 m to the 5-push distance: 20 + 4 = 24 m. One might mistakenly assume a multiplicative pattern, such as doubling, which would incorrectly predict 40 m, but the data show additive increases.

Question 5

PASSAGE II

BIOLOGY: This passage is adapted from a study on the factors affecting the rate of photosynthesis in aquatic plants.

Introduction

Photosynthesis is the process by which green plants use sunlight to synthesize nutrients from carbon dioxide (CO2CO_2) and water (H2OH_2O). The process releases oxygen (O2O_2) as a byproduct according to the following chemical equation: 6CO2+6H2O+light energyC6H12O6+6O26CO_2 + 6H_2O + \text{light energy} \rightarrow C_6H_{12}O_6 + 6O_2 Students conducted three studies to investigate how different environmental factors affect the rate of photosynthesis in Elodea, an aquatic plant. The rate was measured by counting the number of oxygen bubbles produced by a cut stem of Elodea submerged in water over a 5-minute period.

Study 1

To test the effect of light intensity, students placed a 10 cm sprig of Elodea into a test tube filled with a 0.5% sodium bicarbonate (NaHCO3NaHCO_3) solution (a source of CO2CO_2). A light source was placed at various distances from the test tube. The temperature was maintained at 25°C. The number of bubbles produced in 5 minutes was recorded.

Study 2

To test the effect of light color (wavelength), students used the same setup as in Study 1. The light source was kept at a constant distance of 10 cm. Colored filters were placed between the light and the plant to isolate specific wavelengths. Clear cellophane was used as a control.

Study 3

To test the effect of CO2CO_2 availability, students prepared five test tubes with different concentrations of sodium bicarbonate (NaHCO3NaHCO_3). A 10 cm sprig of Elodea was placed in each. The light source was kept constant at 10 cm (white light).

Based on Table 1, as the distance of the light source from the plant increases, the rate of photosynthesis:

  1. increases linearly.
  2. decreases only. (correct answer)
  3. decreases, then increases.
  4. remains constant.
Explanation: This is a trend identification question. Table 1 shows that as distance increases from 10 cm to 50 cm, the number of bubbles decreases from 45 to 5. This is a clear, consistent downward trend with no reversals or plateaus. Choice B (decreases only) is correct. Choice A (increases) is opposite of the data. Choice C (decreases then increases) would require the trend to reverse, which doesn't happen. Choice D (constant) would require the same value at all distances. Pro tip: For trend questions, look at the overall pattern from first to last data point.

Question 6

PASSAGE V

BIOLOGY: This passage is adapted from a study on the metabolic rates of vertebrates. Introduction

Metabolism is the set of chemical reactions that occur in living organisms to maintain life. The metabolic rate is often measured by the amount of oxygen (O2O_2) consumed per gram of body mass per hour.

Animals can be classified based on how they regulate body temperature:

•Endotherms (e.g., mammals, birds) generate their own body heat to maintain a constant internal temperature.

•Ectotherms (e.g., reptiles, amphibians) rely on external heat sources to regulate their body temperature.

Students conducted two studies to compare the metabolic rates of a Mouse (Endotherm) and a Lizard (Ectotherm) of similar body mass.

Study 1

The students placed the mouse and the lizard in separate metabolic chambers. They varied the environmental temperature from 5°C to 35°C in 10°C increments. The animals were kept at rest. The rate of oxygen consumption (mL O2/ghrmL \ O_2 / g \cdot hr) was measured after the animals had acclimated to each temperature for 30 minutes. Findings were reported in Figure 1.

Study 2

The students investigated the effect of activity level on metabolic rate. They maintained the environmental temperature at 25°C for both animals. They measured the oxygen consumption while the animals were at rest and while they were running on a treadmill at 1.0 km/hr. Findings were reported in Table 1.

Study 3

To determine if body mass affects metabolic rate within the same group, students measured the resting metabolic rate of three different lizards at 25°C. Findings were reported in Table 2.

Based on Table 2, what is the relationship between body mass and metabolic rate per gram for lizards?

  1. As body mass increases, metabolic rate per gram increases.
  2. As body mass increases, metabolic rate per gram decreases. (correct answer)
  3. Metabolic rate per gram is independent of body mass.
  4. Metabolic rate per gram doubles for every 50 g increase in mass.
Explanation: This is a trend identification question. Table 2 shows three lizards: as mass increases from 20 g to 50 g to 100 g, the metabolic rate per gram decreases from 0.80 to 0.50 to 0.35 mL O₂/g·hr. This is a clear inverse relationship—larger lizards have lower mass-specific metabolic rates. Choice B correctly describes this inverse trend. Choice A (increases) is opposite. Choice C (independent) would show no pattern. Choice D (doubles) is factually wrong and describes an increase, not decrease. Pro tip: This inverse relationship reflects a biological principle (Kleiber's Law)—larger animals have lower per-gram metabolic rates due to surface area to volume ratios.

Question 7

PASSAGE II

BIOLOGY: Research Summary

Introduction

Transpiration is the process by which moisture is carried through plants from roots to small pores on the underside of leaves, where it changes to vapor and is released to the atmosphere. A botanist conducted two studies to investigate how environmental factors affect the transpiration rate of Spathiphyllum (peace lily) plants.

Study 1

The botanist placed 5 identical Spathiphyllum plants into 5 identical environmentally controlled chambers. The relative humidity inside all chambers was kept constant at 40%, and the temperature was kept constant at 22°C. The botanist varied the light intensity—measured in micromoles of photons per square meter per second (μmol/m2/s\mu mol/m^2/s)—in each chamber. After 4 hours, the botanist measured the mass of water lost by each plant to calculate the transpiration rate in milligrams of water per square centimeter of leaf area per hour (mg/cm2/hrmg/cm^2/hr). Results are shown in Table 1.

Study 2

The botanist obtained 5 new, identical Spathiphyllum plants and placed them in the chambers. This time, the light intensity in all chambers was kept constant at 400 μmol/m2/s\mu mol/m^2/s and the temperature at 22°C. The botanist varied the relative humidity in each chamber. The transpiration rates were calculated after 4 hours. Results are shown in Table 2.

Suppose the botanist conducted a third study where a Spathiphyllum plant was exposed to a relative humidity of 50% and a light intensity of 400 μmol/m²/s. Based on Table 2, the transpiration rate for this plant would most likely be:

  1. less than 3.1 mg/cm²/hr.
  2. between 3.1 and 4.8 mg/cm²/hr. (correct answer)
  3. between 4.8 and 6.5 mg/cm²/hr.
  4. greater than 6.5 mg/cm²/hr.
Explanation: The correct answer is B. Table 2 shows that at 40% humidity the transpiration rate is 4.8 mg/cm²/hr, and at 60% humidity it is 3.1 mg/cm²/hr. Since 50% falls exactly between 40% and 60%, the transpiration rate must fall between 3.1 and 4.8 mg/cm²/hr. A is wrong — a rate less than 3.1 would require humidity higher than 60%. C is wrong — a rate between 4.8 and 6.5 would require humidity lower than 40%. D is wrong — a rate above 6.5 would require humidity below 20%. Pro tip: Interpolation questions test whether a value between two known data points produces a result between those data points. Always identify your bracketing values first.

Question 8

PASSAGE I

EARTH SCIENCE: Data Representation

Earth's atmosphere is divided into distinct layers based on how temperature changes with altitude. The boundary between each layer is called a pause (e.g., the tropopause separates the troposphere from the stratosphere). Figure 1 shows how average atmospheric temperature varies with altitude. Table 1 shows how average atmospheric pressure, measured in atmospheres (atm), changes with altitude.

A weather balloon is launched from sea level and ascends to an altitude of 25 km. Based on the provided data, during the balloon's flight, the atmospheric pressure it experiences will most likely:

  1. drop from 1.0 atm to a value between 0.05 atm and 0.01 atm. (correct answer)
  2. drop from 1.0 atm to exactly 0.01 atm.
  3. rise from 0.05 atm to 0.26 atm.
  4. remain constant at 1.0 atm until it reaches the stratosphere.
Explanation: The correct answer is A. The balloon starts at sea level (1.0 atm) and ascends to 25 km. Table 1 shows that at 20 km the pressure is 0.05 atm and at 30 km it is 0.01 atm. Since 25 km falls between these two values, the pressure at that altitude must fall between 0.05 and 0.01 atm. The balloon starts at 1.0 atm and drops to that intermediate value. B is wrong because 0.01 atm corresponds to exactly 30 km, not 25 km. C is wrong because pressure decreases as altitude increases — it cannot rise during an ascent. D is wrong because Table 1 clearly shows pressure decreasing continuously from sea level, not remaining constant. Pro tip: Interpolation questions require identifying the two data points that bracket your target value and confirming your answer falls between them.

Question 9

A hospital measured the activity of a radioactive tracer every 3 days after it was prepared. Table 1 shows the measurements.

If the pattern in Table 1 continues, what will the activity be 15 days after the tracer was prepared?

  1. 0 kBq
  2. 63 kBq
  3. 125 kBq (correct answer)
  4. 250 kBq
Explanation: The activity in Table 1 halves every 3 days: 4000, 2000, 1000, 500 kBq. Continuing that pattern gives 250 kBq at 12 days and 125 kBq at 15 days. Halving never reaches zero, so 0 kBq cannot be right, and 250 kBq is the 12-day figure.

Question 10

The power output of a rooftop solar array was recorded every 2 hours on a clear day. Sunset that day was shortly after 18:00. Figure 1 shows the results.

Based on Figure 1, what will the output be closest to at 18:00?

  1. 0.3 kW (correct answer)
  2. 2.0 kW
  3. 4.0 kW
  4. 5.4 kW
Explanation: Output in Figure 1 peaks at 5.0 kW at noon and then falls, reaching 4.1 kW at 14:00 and 2.0 kW at 16:00, a drop that is getting steeper. With sunset shortly after 18:00, output at 18:00 will be near zero. Options at or above 2.0 kW would require the afternoon decline to stop.

Question 11

A sealed rigid flask of gas was warmed in stages and its pressure was recorded at four temperatures. The flask did not leak. Figure 1 shows the results.

Based on Figure 1, what pressure would the sample reach at 400 K?

  1. 160 kPa (correct answer)
  2. 170 kPa
  3. 180 kPa
  4. 200 kPa
Explanation: Pressure in Figure 1 rises 20 kPa for every 50 K, a constant rate. Continuing from 140 kPa at 350 K gives 160 kPa at 400 K. The other options overshoot the steady 20 kPa step.

Question 12

A seed crystal was suspended in a saturated solution that was kept saturated throughout, and the crystal was weighed every 2 days. Table 1 shows the results.

Based on Table 1, what would the crystal have weighed on day 5?

  1. 3.2 g
  2. 3.6 g
  3. 3.8 g
  4. 4.0 g (correct answer)
Explanation: The crystal in Table 1 gains 1.6 g every 2 days, or 0.8 g per day. Day 5 falls halfway between day 4 (3.2 g) and day 6 (4.8 g), so the mass is halfway between them: 4.0 g.

Question 13

A tide gauge recorded the clock time of four consecutive high tides at one harbour. Table 1 shows the times; the third and fourth tides occurred on the day after the first and second.

Based on the pattern in Table 1, at about what time will the next high tide occur?

  1. 02:25
  2. 02:50 (correct answer)
  3. 03:15
  4. 14:50
Explanation: Each high tide in Table 1 comes 12 hours 25 minutes after the one before it: 01:10 to 13:35, 13:35 to 02:00, and 02:00 to 14:25. Adding another 12 hours 25 minutes to 14:25 gives 02:50 the following morning.

Question 14

Fish of one species were placed one at a time in a swim tunnel and the fastest speed each could hold for 20 minutes was recorded. Five water temperatures were tested, with the same oxygen level in every trial. Figure 1 shows the mean result at each temperature.

Based on Figure 1, what would the sustained swimming speed most likely be at 35 °C?

  1. 4 cm/s (correct answer)
  2. 20 cm/s
  3. 34 cm/s
  4. 40 cm/s
Explanation: Speed in Figure 1 climbs to 32 cm/s at 25 °C and then drops sharply to 18 cm/s at 30 °C, so the trend has already reversed before the last reading. Continuing that decline puts the speed at 35 °C well below 18 cm/s: repeating the 14 cm/s drop seen from 25 °C to 30 °C gives 18 − 14 = 4 cm/s. That is also the only option below the 30 °C reading, and the options above 32 cm/s would require the early rise to resume.

Question 15

A rechargeable battery was charged and discharged repeatedly, and its capacity was measured every 200 cycles as a percentage of its capacity when new. The manufacturer replaces a battery once its capacity falls to 80%. Table 1 shows the measurements.

Based on Table 1, after about how many cycles will the battery reach the 80% replacement point?

  1. 900 cycles
  2. 1,000 cycles (correct answer)
  3. 1,200 cycles
  4. 1,600 cycles
Explanation: Capacity in Table 1 falls 4 percentage points every 200 cycles. From 84% at 800 cycles, one more step of 200 cycles brings it to 80%, so the battery reaches the replacement point at about 1,000 cycles.

Question 16

Divers surveyed reef sites during four summers and recorded the percentage of coral colonies that had bleached, along with how far the sea surface temperature exceeded its long-term average. Figure 1 shows the results. No more than 100% of the colonies at a site can bleach.

If the pattern in Figure 1 continues, what percentage of colonies would be expected to bleach at an anomaly of 2.5 °C?

  1. 75%
  2. 85%
  3. 100% (correct answer)
  4. 130%
Explanation: The increases in Figure 1 are getting larger: 13, 22, and 30 percentage points. Adding another step of roughly 38 points to 70% gives about 108%, which is impossible because a site cannot have more than 100% of its colonies bleached. The prediction is therefore essentially total bleaching, 100%. Choosing 130% extends the arithmetic past what the quantity allows.

Question 17

A monitoring station recorded the atmospheric CO₂ concentration each May and each September for three years. Concentrations are always higher in May than in the following September because plants take up CO₂ over the northern summer. Figure 1 shows the readings.

Based on Figure 1, what reading is expected in May 2025?

  1. 415 ppm
  2. 418 ppm
  3. 424 ppm
  4. 427 ppm (correct answer)
Explanation: The May readings in Figure 1 rise by 3 ppm each year: 418 in 2022, 421 in 2023, and 424 in 2024. May 2025 should therefore be about 427 ppm. Comparing a May reading with a September reading instead would suggest the concentration is flat or falling, which is why 418 ppm is tempting.

Question 18

A reservoir was measured on the first day of each month during a drought. Water cannot be drawn from it once the level falls below 22.0 m. Table 1 shows the readings.

If the level keeps falling at the same rate, on the first day of which month will the recorded level first be below 22.0 m?

  1. June
  2. July (correct answer)
  3. August
  4. September
Explanation: The level in Table 1 drops 3.5 m between readings. Continuing from 28.0 m on 1 May gives 24.5 m on 1 June and 21.0 m on 1 July, so the July reading is the first one below 22.0 m. The June reading is still 2.5 m above the limit.

Question 19

A ball was dropped onto a hard floor and the height it reached after each bounce was measured. Table 1 shows the first four bounces.

Based on the pattern in Table 1, what height will the ball reach after the fifth bounce?

  1. 7.8 cm
  2. 10.8 cm
  3. 13.0 cm (correct answer)
  4. 18.0 cm
Explanation: Each bounce in Table 1 reaches 0.6 times the previous height (60 ÷ 100, 36 ÷ 60, and 21.6 ÷ 36 all equal 0.60). The fifth bounce therefore reaches 21.6 × 0.6 = 13.0 cm. Applying the ratio twice by mistake gives 7.8 cm, which is the height after the sixth bounce rather than the fifth.

Question 20

A light trap was run on the same schedule in every season for two years, and the mean number of moths caught per night was recorded. Figure 1 shows the counts.

Based on Figure 1, what is the most reasonable prediction for spring 2025?

  1. About 25 moths per night
  2. About 95 moths per night
  3. About 140 moths per night (correct answer)
  4. About 285 moths per night
Explanation: The counts in Figure 1 repeat a seasonal cycle that peaks in summer and bottoms out in winter, and each season is slightly higher than the same season a year earlier (spring rose from 120 to 130). Spring 2025 should therefore be a little above 130, so about 140. The other options come from reading the wrong season: about 25 is a winter count, about 95 is a fall count, and about 285 is where the summer peak is heading.