ACT Past Papers with Answers

Paper 1 · ACT February 2025 · Form Z14 · Science Test

According to the results of Model B in Figure 2, at the surface, the maximum radius at which there is com- plete vaporization is closest to which of the following?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

According to the results of which model in Figure 1 is a greater area of the surface affected by the impact, and according to the results of which model in Figure 1 does the impact have a deeper effect? greater area of surface affected deeper effect

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on the results of Model B in Figure 2, the H2O that is present at a radius of 2 km and a depth of 4 km exists in what state(s)?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

According to the results of Model B in Figure 2, from a depth of 3 km to a depth of 5 km, does the maximum radius at which there is complete melting decrease or increase, and does the maximum radius at which there is 50% melting decrease or increase? complete melting 50% melting

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the results of Model A in Figure 2. Those results indicate that within an area having a radius of

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on Figure 1 and Table 1, which of the strains most likely had a defect in either Enzyme V or Enzyme W ?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on Figure 1, Enzyme U facilitates the synthesis of arginine by: citrulline. ornithine. to citrulline. to ornithine.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Strains 1−3 could not grow on GM because they could not synthesize arginine, which is a subunit of:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on Figure 1 and Table 1, which of Strains 1 and 2, if either, would have been able to grow on GM + argininosuccinate?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

A scientist hypothesized that a newly identified mutant strain of N.crassa (Strain 5) had a defect in the same enzyme as Strain 3. Strain 5 was tested according to the procedure described in the passage, and the results of the test are shown below. Are these results consis- tent with the scientist’s hypothesis? GM + ornithine. GM + citrulline and GM + arginine. GM + ornithine. GM + citrulline and GM + arginine.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

How many of the strains, if any, could convert the pre- cursor to arginine? Strain Growth on: GM GM + ornithine GM + citrulline GM + arginine 5 no no yes yes 41 GO ON TO THE NEXT PAGE. ACT-Z14 4 4 Passage III Environmental conditions, such as nighttime tempera- ture, can affect the reproduction of Utastansburiana (a species of lizard). A scientist conducted a study to examine how nighttime temperature affects egg laying by female U. stansburiana. Study Each of 20 identical tanks with 25 cm high glass walls was prepared as follows: First, sand was poured into the tank to a depth of 10 cm. Next, three 1-year-old female U.stansburiana were artificially fertilized and then placed into the tank. Finally, the top of the tank was covered with a fine wire mesh. After the tanks were prepared, they were equally divided into 4 groups (Groups Q−T). Group Q was incubated in the presence of fluorescent light for 14 hr at 29°C. Then Group Q was incubated in the dark for 10 hr at 8°C. This procedure was repeated every day for the next 39 days. During the incubation period, any eggs laid were collected, and the average clutchsize (aver- age number of eggs laid per egg-laying female), the aver- age eggmass (average mass per egg), and the average clutchtime (average time required by a female to lay a clutch of eggs) were determined for Group Q. The procedure for Group Q was then repeated for Groups R−T, except that instead of being incubated in the dark at 8°C, Group R was incubated in the dark at 18°C, Group S was incubated in the dark at 22°C, and Group T was incubated in the dark at 29°C. The results for the 4 groups are shown in the follow- ing table. Table adapted from Donald N. Clarke and Peter A. Zani, “Effects of Night-Time Warming on Temperate Ectotherm Reproduction: Potential Fitness Benefits of Climate Change for Side-Blotched Lizards.” ©2012 by The Company of Biologists Ltd.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

What percent of the height of a tank was filled with sand?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Which of the statements listed below describe(s) a function of the fine wire mesh? I. To place the U. stansburiana in a tank II. To protect the U. stansburiana from attack by prey III. To prevent the U. stansburiana from leaving a tank

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Assume that the greater an egg’s mass, the healthier its embryo. Based on the results of the study, among the 4 temperatures at which the U.stansburiana were incubated in the dark, the temperature that most likely resulted in the healthiest embryos, on average, was:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Is U.stansburiana a reptile or a mammal, and does U. stansburiana primarily maintain its body tempera- ture by generating heat from internal metabolic processes or by absorbing heat from the surrounding environment? environment processes environment processes Group Average clutch size (eggs) Average egg mass (mg) Average clutch time (days) Q 3.8 300 35.5 R 4.5 310 32.9 S 4.8 290 31.0 T 4.1 330 26.5 42 GO ON TO THE NEXT PAGE. ACT-Z14 4 4

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the steps listed below that were performed in the study.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

For which group was the incubation temperature con- stant over a 24 hr period?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

The incubation conditions for Group R and Group T differed in which of the following ways? The incuba- tion temperature in the: Group T. Group R. Group T. Group R. 43 GO ON TO THE NEXT PAGE. ACT-Z14 4 4 Passage IV Consider a pendulum of length L that consists of a sphere of mass M suspended from one end of a string of negligible mass; the string’s other end is tied to a ring stand (see Figure 1). Figure 1 When the sphere is moved from its equilibrium (lowest) position, the string makes some initial angle, θ0, with respect to the vertical direction. If the sphere is released, the pendulum will oscillate (swing back and forth). The frequency, f, of the pendulum is the number of cycles (round trips) per second completed by the pendulum. A teacher gave her students the following table of masses and lengths for 4 such pendulums, P1−P4. She asked 2 students to predict which pendulum(s) would have the greatest f (assuming that θ0 is the same for the 4 pendulums). Student1 For a given L, as M increases, the gravitational force exerted by Earth on a sphere will increase, so sphere weight will increase. As this weight increases, gravitational acceleration will increase, so the average speed of the sphere over a cycle will increase, thereby increasing f. For a given M, as L decreases, the distance traveled by the sphere over a cycle will decrease. However, the average sphere speed will also decrease, because the length of the path traveled by the sphere over a cycle will decrease and limit gravitational acceleration. Consequently, the pendu- lum will take the same amount of time to complete a cycle, and f will remain the same. Because P3 and P4 have the same M, they will have the same f. Moreover, because both P3 and P4 have a greater mass than either P1 or P2, each will have a greater f than either P1 or P2. Student2 M will not affect f, because a sphere’s gravitational acceleration and, consequently, its average speed over a cycle are independent of the sphere’s mass. For any given M, as L increases, the distance traveled by the sphere over a cycle will increase. As a result, the pendulum will take more time to complete a cycle, so f will decrease. As P1 and P3 have the same L, they will have the same f. As both P1 and P3 are shorter than either P2 or P4, each will have a greater f than either P2 or P4. θ0 string sphere L Table 1 Pendulum M (kg) L (m) P1 1 1 P2 1 2 P3 2 1 P4 2 2 44 GO ON TO THE NEXT PAGE. ACT-Z14 4 4

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on Table 1 and Student 1’s predictions, which pendulum will have the greater gravitational accelera- tion, P1 or P3 ?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Suppose that the spheres are removed from P2 and P4, and then the spheres are simultaneously dropped from a height of 5 m above the ground. Based on Table 1 and Student 1’s predictions, the sphere more likely to strike the ground first will be the sphere removed from:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Two pendulums, P5 and P6, have the same f. How would Student 2 most likely explain this result?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Based on Table 1 and Student 2’s predictions, which pendulum will travel the greater distance over a cycle, P3 or P4 ?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Assume that if M = 1.0 kg and L = 1.0 m, fwill equal

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the statements about f, M, and L given below. I. The frequency f is affected by M but not by L. II. The frequency f is affected by L but not by M. III. The frequency f is affected by both M and L. IV. The frequency f is affected by neither M nor L. Both students would most likely disagree with which of the statements?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Which of the 2 students, if either, would be likely to agree that as sphere mass increases, frequency will decrease? 45 GO ON TO THE NEXT PAGE. ACT-Z14 4 4 Passage V Coffee beans slowly release CO2 gas after they have been roasted. Two experiments were done to compare the amounts of CO2 released by various samples of roasted coffee beans. In each trial of the 2 experiments, Steps 1−3 were performed:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the statement “As the average particle size increased, the total amount of released CO2 increased.” This statement is supported by the results of Experi- ment 2 for which of the roast colors?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

In any given trial of the experiments, if the N2 gas flowed through the apparatus at a rate of 2.0 L/hr, how many liters of N2 flowed through the apparatus?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

In Experiment 1, for a roasting time of 240 sec, as roasting temperature increased, the amount of released CO2:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Suppose that a trial had been performed in Experi- ment 2 with a sample of light-roast beans ground to an average particle size of 0.420 mm. The total amount of CO2 released by the sample would most likely have been:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Suppose that, in the trial of Experiment 1 involving the medium-S sample, the inlet port had NOT been sealed after the sample was poured into the apparatus. Would the total amount of CO2 measured by the detector during this trial more likely have been greater than

Paper 1 · ACT February 2025 · Form Z14 · Science Test

A scientist predicted that, on average, all the coffee beans roasted at a temperature greater than 235°C would release more than 1.50 mg of CO2/g. Do the results of Experiment 1 support this prediction? released more than 1.50 mg of CO2/g. released more than 1.50 mg of CO2/g. released less than 1.50 mg of CO2/g. released less than 1.50 mg of CO2/g.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Upon the completion of Experiment 1, what was the minimum number of hours that N2 gas had flowed through the apparatus? 47 GO ON TO THE NEXT PAGE. ACT-Z14 4 4 Passage VI Carotenoids (CTs) are colored compounds that are found in many foods. Solutions of CTs can absorb ultra- violet (UV) light, which has wavelengths ranging from 200 nanometers (nm) to 400 nm, and visible light, which has wavelengths ranging from 400 nm to 700 nm. (1 nm = 10−9 m.) The amount of light absorbed (the absorbance, A) by a CT solution varies with wavelength, λ. The wavelength that results in the maximum Ais known as λmax. When a CT solution absorbs visible light of a certain color at λmax, the solution appears to be a different color. On a color wheel (Figure 1), the color of the solution is across from the color of light absorbed. Figure 1 For example, if a CT solution has a λmax of 642 nm, then the solution absorbs red light and appears to be green. Study Students determined the λmax of each of 6 CTs in solu- tion. For each CT, Steps 1−5 were performed:

Paper 1 · ACT February 2025 · Form Z14 · Science Test

The solution containing which of the following CTs absorbed green light at its λmax ?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

What percent of the volume of the CT solution pre- pared in Step 1 was placed into a vial in Step 2 ? orange 600–640 nm yellow 560–600 nm green 480–560 nm blue 450–480 nm violet 400–450 nm red 640–700 nm light source detector beam of light solution in vial Table 1 CT λmax (nm) Phytoene 286 Auroxanthin 399 Beta-carotene 451 Lycopene 468 Astaxanthin 488 Spirilloxanthin 492 48 GO ON TO THE NEXT PAGE. ACT-Z14 4 4

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Which of the following graphs best shows how the absorbance of the astaxanthin sample varied with wavelength?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the following results for a CT solution of torulene. At its λmax, does the torulene solution absorb UV light or visible light? 400 nm. 700 nm. 400 nm. 700 nm.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Consider the vial that contained each CT solution when in the spectrophotometer. Based on the descrip- tion of the study and Figure 2, is it more likely that the students chose to use this vial because it was com- posed of a plastic that transmitted all or almost all light between 200 nm and 700 nm or because it was com- posed of a plastic that reflected all or almost all light between 200 nm and 700 nm ? solution contained in the vial. from passing into the solution contained in the vial. solution contained in the vial. from passing into the solution contained in the vial.

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Violerythrin is a CT that appears violet when in solu- tion. Based on Figure 1, the λmax of a violerythrin solution could be which of the following values?

Paper 1 · ACT February 2025 · Form Z14 · Science Test

Each time a sample’s A was determined at a specific λ in Step 4, the sample is said to have been scanned. Based on the description of the study, the total number of scans performed in the study was approximately: wavelength (nm) absorbance 300 250 350 400 450 500 550 0 absorbance 0 absorbance 0 absorbance 0 absorbance wavelength (nm) 250 350 450 550 wavelength (nm) 250 350 450 550 wavelength (nm) 250 350 450 550 wavelength (nm) 250 350 450 550 END OF TEST 4 STOP! DO NOT RETURN TO ANY OTHER TEST. 49 ACT-Z14 4 4 50