Below are answer explanations to the full-length Science test of the released ACT practice test for the updated 2025 test. 

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Answer Explanations to the ACT 2025 Practice Science Test

Passage I

Question 1, “Based on Table 2, which of the following ratios best represents…”. The correct answer is A. “1:2”.

  1. Looking at Table 2, we can see that Behavior 2 was displayed 3 times in Habitat X and 6 times in Habitat Y, giving us a 3:6 ratio.
  2. Through division, we can simplify this 3:6 ratio to 1:2.

Question 2, “Which of the following observations for brown anoles…”. The correct answer is A. “1 only”.

  1. Looking at Table 1, we can see the average perching height for brown anoles in Habitat Y and Habitat Z is 0.6 meters.
  2. Using Table 2, we can find that the number of times Behavior 4 was displayed by brown anoles in Habitat Y was 5, but 17 in Habitat Z.
  3. Table 3 tells us the average display time for Behavior 5 by brown anoles was 49.6 seconds in Habitat Y and 33.1 seconds in Habitat Z.
  4. The only observation that was the same for brown anoles in both Habitat Y and Habitat Z was the average perching height.”

Question 3, “Based on Table 3, how many display times…”. The correct answer is D. “Cannot be determined from the given information.”.

  1. Looking at Table 3, we are given the average display time for Behavior 5 in each Habitat.
  2. However, average display time has no correlation with the number of times the behavior was displayed, meaning we have no way of finding the number of times Behavior 5 was displayed in Habitat Z.

Question 4, “Based on Figure 1, for green anoles, the difference…”. The correct answer is C. “0.7 m”.

  1. Looking at Table 1, we can determine the average perching heights for green anoles in Habitats Z and Y are 0.9 meters and 1.6 meters, respectively.
  2. Using subtraction, 1.6 meters – 0.9 meters = 0.7 meters.”

Question 5, “A student claimed that anoles are endotherms. Which …”. The correct answer is D. “reptiles and primarily absorb heat from the surrounding environment to maintain body temperature.”

  1. We are told that anoles are reptiles.
  2. Reptiles are ectotherms, meaning they really on external source for heat regulation and cannot internally maintain body temperature.

Passage II

Question 6, “Which of the following diagrams best shows the location…”. The correct answer is A. “A”.

  1. Using Student 3’s explanation, we know that “…out of the basal cavity (the area underneath an ice shelf), causing ice within the cavity to melt.”
  2. The only answer choice with the basal cavity located underneath the ice shelf is A.

Question 7, “Suppose that the air temperature along the Antarctic coastline…”. The correct answer is C. “No, because ice cannot melt at those temperatures.”

  1. Student 1 claims that the “”Antarctic ice shelves melt due to the warming of the air above the surface of the ice during the summer.”
  2. Supposing the air is at -10°C, the ice will not melt because the temperature of the air is not greater than the freezing point of the ice (0°C).

Question 8, “Based on the description of the icebergs that are calved…”. The correct answer is D. “Float; they are less dense than ocean water.”

  1. Based on the description given, icebergs come from melted ice shelves that are floating, thick sheets of ice.
  2. For a substance to float, it must be less dense than water, meaning that the floating icebergs are less dense than water.”

Question 9, “Which of Students 1 and 4, if either, implied that…”. The correct answer is C. “Both Student 1 and Student 4”.

  1. Student 1 stated that “The accumulation of many fractures in the ice over many summers gradually lead to icebergs calving from an ice shelf.”
  2. Student 4 stated that “After several winter-summer cycles, the ice shelf becomes top-heavy due to the snow and the melting from below, and calving occurs.”
  3. “…many summers…” and “…several winter-summer cycles…” both imply that the formation of icebergs takes longer than one year.

Question 10, “In regard to the season(s) involved in iceberg calving…”. The correct answer is A. “summer and winter are involved in calving, whereas Student 3 indicated that only summer is involved in calving.”

  1. Student 2 mentions “…an increase in air temperature during the summer…” and “…the air temperature lowers at the beginning of winter.”
  2. Student 3 only mentions “During the summer…”
  3. Student 3 does not mention winter, therefore Student 3 indicates only summer is involved unlike Student 2, who indicates winter and summer are involved.”

Question 11, “Which of Students 2, 3, and 4 agree(s) with Student 1…”. The correct answer is C. “Students 2 and 4 only”.

  1. Student 1 stated “…summer leads to surface melting…”
  2. Student 4 stated “The compaction lowers the surface of the ice shelf… where it is melted by the ocean water.”
  3. However, Student 3 declared that “Antarctic ice shelves only melt from below.”
  4. Student 3 has the only observation that is contary to Students 1, 2, and 4.”

Passage III

Question 12, “Which of the following statements about the relationship…”. The correct answer is B. “decreased only”.

  1. Looking at Table 2, we can see the relationship between stocking density and average SGR.
  2. As stocking density increasing from 0.5 fish/Liter to 3 fish/Liter, the average SGR decreases from 0.50 percent/day to 0.20 percent/day.

Question 13, “Which of the following statements about the percent protein…”. The correct answer is C. “lowest average SGR had the highest percent protein”.

  1. Based on Table 1, the diets that had the highest and lowest average SGR were Diets S and Q, respectively.
  2. Looking at the table given in the question, Diet Q had the highest percent protein and Diet T had the lowest.
  3. The only Diet that was at the top or bottom of both tables was Diet Q, having the lowest average SGR but the highest percent protein.

Question 14, “Based on the results of Experiment 1, if Experiment 2 were repeated...”. The correct answer is C. “Higher; on average, A. percula fed Diet R had an SGR 0.05 percent/day greater than those fed Diet T.”.

  1. Looking at Table 1, Diet R had an average SGR of 0.40 percent/day while Diet T was 0.05 percent/day lower with an average SGR of 0.35 percent/day.
  2. Because Table 2 recorded Experiment 2 using only Diet T, we can assume that repeating the experiment with Diet R would increase the average SGRs for each group.

Question 15, “Suppose that, in the experiments, 1 g of food were added…”. The correct answer is B. “3 g”.

  1. The reading states that “Each group was fed the same mass of food 3 times daily.”
  2. Supposing 1 gram of food was added to each tank at each feeding, 1 gram × 3 = 3 grams, meaning each individual tank had 3 grams of food each day.

Question 16, “How many A. percula were placed in each of the tanks…”. The correct answer is C. “10”.

  1. The reading states that each tank “…received 10 L of seawater…” and that “…(the fish) were equally distributed among the tanks at a stocking density of 1 fish/L.”
  2. Using this ratio and measurement, 1 fish/L × 10 L = 10 fish per tank.

Question 17, “Which of the following was a dependent variable…”. The correct answer is B. “Specific growth rate”.

  1. A dependent variable is defined as a variable whose value depends on that of another.
  2. The only variable from the answer choices that was not controlled during the experiement in the growth rate, or average SGR.

Passage IV

Question 18, “Which of the samples in Experiment 1 was most likely…”. The correct answer is A. “Sample 1”.

  1. The hypothesis for the experiment was that heat affected the concentration on nutrients in tomatoes.
  2. Only Sample 1 can be a control because it is the element that remains unchanged by the tested variable that is heat.

Question 19, “Based on the results of Experiment 2, which incubation…”. The correct answer is D. “20 min”.

  1. Looking at the results of Experiment 2 in Figure 2, both Samples 3 and 4 sit between 5 mg/g tomato and 6 mg/g tomato.
  2. Using Table 1, the incubation times of 15 minutes and 30 minutes for Samples 3 and 4.
  3. To produce a tomato mixture with a lycopene concentration between 5 mg/g tomato and 6 mg/g tomato, we need an answer that exists between 15 minutes and 30 minutes.

Question 20, “A student claimed that heating tomatoes decreases…”. The correct answer is A. “Vitamin C only”.

  1. Looking at Figure 1, when the time in heat is increased, the concentration of Vitamin C decreased.
  2. Using Figure 2, the concentration of lycopene increases as the time in heat increases.
  3. The only concentration of nutrients that decreased when exposure to heat increases was vitamin C.

Question 21, “Assume that, in the experiments, the water bath contained…”. The correct answer is C. “No; the incubation temperature was less than the boiling point of water at 1 atm.”

  1. Looking at Table 1, the incubation time may have varied, but the temperature was consistently at 88°C.
  2. Water has a boiling point of 100°C, so the water was not boiling during the experiments.

Question 22, “In Experiment 1, how many of the samples had a vitamin C…”. The correct answer is D.“4”.

  1. Using Table 1, the highest concentration of vitamin C found in a sample was 0.75 μmol/g tomato in Sample 1.
  2. Since the highest concentration of vitamin C found during the experiment is less than 1.0 μmol/g, all 4 samples have a concentration of vitamin C less than 1.0 μmol/g.

Question 23, “These procedures were performed in what order?”. The correct answer is C. “2, 1, 3”.

  1. Notice that for Experiment 2, “Experiment 1 was repeated…”
  2. In the reading, it states that “Samples 2-4 were each incubated in a water bath…”
  3. “Then, 2 days later… the sample was thawed, and then 100 g of the sample was placed in a beaker containing 200 mL of Solvent A.”
  4. Using the quotes above, in Experiment 2, Sample 2 was first incubated in the water bath, then frozen, and finally filtered with the solvent mixture.

Passage V

Question 24, “A linear region of a graph is a range of data that…”. The correct answer is D. “Between 30 × 10⁷ and 40 × 10⁷”.

  1. A linear region means there will have a straight line (y = mx + b) for the domain given.
  2. Looking at Figure 1, the only domain out of the choices given that resembles of linear equation in between 30 × 10⁷ and 40 × 10⁷.

Question 25, “Consider the 2 trends shown for Alloy Q initially…”. The correct answer is B. “F = 60 × 10⁷”.

  1. Looking at Figure 1, it is evident that the two graphs have a similar slope value, although the slope of the 280°C initial temperature graph is negative.
  2. This slope can be estimated at ± 1 L per F = 10 × 10⁷.
  3. At F = 48 × 10⁷, the two graphs are approximately 3 L apart.
  4. Using the estimates above, the two graphs will intersect at roughly F = 63 × 10⁷, which is closest to F = 60 × 10⁷.”

Question 26, “Based on Figure 1, which of the following combinations…”. The correct answer is D. “550°C and F = 40 × 10⁷”.

  1. Both answers with 280°C must be incorrect as the 280°C initial temperature graph produces crystals with greater average lengths when compared to the initial temperature of 550°C graph.
  2. Looking at Figure 1, 550°C and F = 10 × 10⁷ produces crystals with the average length of 92 L, which is greater than 550°C and F = 40 × 10⁷’s average length of 91.5 L.

Question 27, “Given the composition of Alloy Q, which sample is most likely Alloy Q?”. The correct answer is D. “Sample Z”.

  1. It is stated in Table 1 that Alloy Q is 10.8% silicon by mass.
  2. 10.8% of 50 g is 0.108 × 50 = 5.4 g.
  3. Sample Z is the only sample that contains 5.4 g of silicon.”

Question 28, “Based on Table 1, if an ingot of Alloy Q had a mass…”. The correct answer is B. “0.44 g”.

  1. Looking at Table 1, Alloy Q consists of 0.22% magnesium.
  2. 0.22% of 200 g is 0.0022 × 200 = 0.44 g.

Passage VI

Question 29, “Based on the results of the studies, from which of the 5 leachates…”. The correct answer is D. “CM1: pine, CM2: magnolia, CM3: magnolia”.

  1. Using Figure 1, the leachate that had the highest percent of DOC adsorbed by CM1 was pine at roughly 45%.
  2. Repeating the step above using Figures 2 and 3, the leachate that had the highest percent of DOC adsorbed by CM2 and CM3 was magnolia at roughly 32% and 55% respectively.

Question 30, “Based on the results of Study 3, the percent of leachate DOC…”. The correct answer is D. “40%”.

  1. Using Figure 3, the percents of DOC adsorbed by CM3 for all 5 leachates were roughly 35%, 50%, 30%, 55%, and 35%.
  2. The average of these percents is (35 + 50 + 30 + 55 + 35) / 5 = 205 / 5 = 41.
  3. 41% is closest to 40% out of the answer choices.

Question 31, “Is the statement ‘CM2 adsorbed a greater percent…”. The correct answer is C. “No; CM2 adsorbed 7% of the leachate DOC, whereas CM3 adsorbed 35%.”

  1. Looking at Figure 2, CM2 adsorbed roughly 7% of the DOC in the maple leachate.
  2. Looking at Figure 3, CM3 adsorbed roughly 35% of the DOC in the maple leachate.
  3. The statement is not true as CM3 adsorbed roughly 28% more of the DOC in the maple leachate than CM2.

Question 32, “Based on the results of the studies, which…”. The correct answer is D. “None of the 3 clay minerals”.

  1. Looking at Figure 1, CM1 adsorbed roughly 30% of the DOC in the oak leachate.
  2. Repeating the step above using Figures 2 and 3, CM2 and CM3 adsorbed roughly 8% and 50% of the DOC in the oak leachate respectively.
  3. None of 30%, 8%, and 50% are greater than 50%.

Question 33, “Is a mixture of any one of the leachates and any one…”. The correct answer is D. “No, because the clay mineral particles are not dissolved in the leachate.”

  1. The definition given for adsorbed is to become adhered to the surface of.
  2. To be considered a solution, the solute mist be completely dissolved in the solvent to create a homogenous solution.
  3. In none of the figures given are the clay mineral particles completely dissolved.

Question 34, “In lake water, DOC is broken down into simpler compounds…”. The correct answer is A. “Each mixture of leaves and filtered lake water was placed in the dark.”

  1. To avoid electromagnetic energy in the visible wavelength, there would need to be a lack of visible light.
  2. It is stated that “A 100 g sample of leaves was mixed with a 1 L volumes of the filtered lake water… then placed in the dark…”

Passage VII

Question 35, “Consider the current shown on the ammeter in Figure 1…”. The correct answer is B. “50%”.

  1. Looking at Figure 1, the ammeter reads 19.8 mA.
  2. Using Table 2, the relative intensity that matches the reading on the ammeter from Figure 1 is 50%.

Question 36, “What aspect of the experimental setup was held constant…”. The correct answer is A. “Color of light”.

  1. Experiment 1 was performed with multiple filters that changed the color of the light.
  2. Experiment 2 had “…the same setup as in Experiment 1 without a filter…”
  3. No filters used in Experiment 2 means the color of light became a constant.

Question 37, “Based on Figure 1, are the particles ejected from the metal plate…”. The correct answer is B. “Toward; negatively charged”.

  1. Looking at Figure 1, there are arrows from the ejected electrons that signify they are moving toward the electrode.
  2. Electrons are always negatively charged particles.

Question 38, “Based on the equation in the passage and the results…”. The correct answer is B. “2.20 eV”.

  1. It is given that K = E – M, which can be rearranged to be M = E – K.
  2. Looking at Table 1, the E and K values for the green filter experiment are 2.31 eV and 0.11 eV respectively.
  3. Plugging into the rearranged formula, M = 2.31 – 0.11 = 2.20 eV.
  4. Note: the same process can be done by using the E and K values from either the blue or violet experiments, and the answer will be the same.”

Question 39, “The cutoff frequency for a particular metal is the lowest frequency…”. The correct answer is C. “between 5.2 × 10¹⁴ Hz and 5.6 × 10¹⁴ Hz.”.

  1. Looking at Table 1, electrons from the metal plate were not ejected with the use of yellow light, but were when green light was used instead.
  2. The yellow light frequency is 5.2 × 10¹⁴, and the green light frequency is 5.6 × 10¹⁴ Hz.
  3. The cutoff frequency for the metal plate exists between those two frequencies: 5.2 × 10¹⁴ Hz and 5.6 × 10¹⁴ Hz.

Question 40, “The relationship between E and the frequency of light is given…”. The correct answer is D. “2.31eV / (5.6 × 10¹⁴ Hz)”.

  1. Looking at Table 1, the energy of green light is 2.31 eV and the frequency is 5.6 × 10¹⁴ Hz.
  2. Plug these values into the given formula and solve for h.
  3. 2.31 eV = h × (5.6 × 10¹⁴ Hz).
  4. h = 2.31eV / (5.6 × 10¹⁴ Hz).

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