ACT Science Problem Cards
ACT Science Problem Task Cards • Form CJ23137
Cards 01–04 • Passage I (Biology)
Passage I • Reptiles CARD 01
Table 2: Number of Times Behavior Displayed
Behavior Habitat X Habitat Y Behavior 2 3 6
1. Based on Table 2, which of the following ratios best represents the number of times Behavior 2 was displayed in Habitat X compared to the number of times Behavior 2 was displayed in Habitat Y?
A. 1:2
B. 1:8
C. 4:5
D. 5:6
Q1 • Key: A Answer: [ ]
Passage I • Reptiles CARD 02
Brown Anoles in Habitats Y and Z
1. Perching height (Fig 1):Y = 0.6 m; Z = 0.6 m
2. Behavior 4 displays (Table 2):Y = 5; Z = 17
3. Behavior 5 display time (Tab 3):Y = 49.6 s; Z = 33.1 s
2. Which of the following observations for brown anoles was(were) the same in both Habitats Y and Z?
1. Average perching height
2. The number of times Behavior 4 was displayed
3. Average display time for Behavior 5
F. 1 only
G. 3 only
H. 1 and 2 only
J. 2 and 3 only
Q2 • Key: F Answer: [ ]
Passage I • Reptiles CARD 03
Table 3: Display Time for Behavior 5
| Species | Habitat | Avg Display Time (s) |
|---|
| Brown | Y | 49.6 |
| Brown | Z | 33.1 |
3. Based on Table 3, how many display times were measured for Behavior 5 in Habitat Z?
A. 2
B. 4
C. 12
D. Cannot be determined from the given information
Q3 • Key: D Answer: [ ]
Passage I • Reptiles CARD 04
Figure 1: Average Perching Height (m)
Habitat X (Green only) 0.9 m
vs
Habitat Y (Green & Brown) 1.6 m
4. Based on Figure 1, for green anoles, the difference in average perching height between Habitat X and Habitat Y was closest to which of the following?
F. 0.0 m
G. 0.3 m
H. 0.7 m
J. 1.0 m
Q4 • Key: H Answer: [ ]
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ACT Science Problem Task Cards
Cards 05–08 • Passage I & II
Passage I • Reptiles CARD 05
Scientific Background:
Passage I identifies anoles as "2 species of reptiles." Ectotherms primarily absorb heat from their surroundings, while endotherms generate metabolic body heat.
5. A student claimed that anoles are endotherms. Which of the following explains why this claim is incorrect? Anoles are:
A. amphibians and primarily generate heat from internal metabolic processes.
B. amphibians and primarily absorb heat from the surrounding environment.
C. reptiles and primarily generate heat from internal metabolic processes.
D. reptiles and primarily absorb heat from the surrounding environment.
Q5 • Key: D Answer: [ ]
Passage II • Ice Shelves CARD 06
Student 3 Statement:
"Antarctic ice shelves melt only from below... ocean currents circulate water into and out of the basal cavity (the area underneath an ice shelf)..."
6. Which of the following diagrams best shows the location of the basal cavity as described by Student 3?
Diagram F Basal cavity positioned on top of ice shelf
Diagram G Basal cavity located inside bedrock landmass
Diagram H Basal cavity located on surface above water
Diagram J Basal cavity located underneath ice shelf
Q6 • Key: J Answer: [ ]
Passage II • Ice Shelves CARD 07
Student 1 Mechanism:
Ice shelves melt when summer air temperature rises, causing surface ice to melt and water to pool. Note: Pure ice melting point is 0°C at 1.0 atm.
7. Suppose that the air temperature along the Antarctic coastline is never warmer than −10°C and atmospheric pressure is 1.0 atm. Does this information support Student 1's description?
A. Yes, because ice cannot melt at those temperatures.
B. Yes, because ice can melt at those temperatures.
C. No, because ice cannot melt at those temperatures.
D. No, because ice can melt at those temperatures.
Q7 • Key: C Answer: [ ]
Passage II • Ice Shelves CARD 08
Passage II Intro:
"The coastline of Antarctica consists of many ice shelves (floating 100–1,000 m thick sheets of ice)... causing them to calve large pieces known as icebergs."
8. Based on the description of the icebergs that are calved along the coastline of Antarctica, do the icebergs sink or float?
F. Sink; they are more dense than ocean water.
G. Sink; they are less dense than ocean water.
H. Float; they are more dense than ocean water.
J. Float; they are less dense than ocean water.
Q8 • Key: J Answer: [ ]
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ACT Science Problem Task Cards
Cards 09–12 • Passage II & III
Passage II • Ice Shelves CARD 09
Student 1: "...accumulation of many fractures in the ice over many summers gradually leads to icebergs calving..."
Student 4: "After several winter-summer cycles, the ice shelf becomes top-heavy... and calving occurs."
9. Which of Students 1 and 4, if either, implied that the processes involved in iceberg calving will take more than one year to result in the formation of an iceberg?
A. Student 1 only
B. Student 4 only
C. Both Student 1 and Student 4
D. Neither Student 1 nor Student 4
Q9 • Key: C Answer: [ ]
Passage II • Ice Shelves CARD 10
Seasonal Descriptions:
Student 2: Summer melts surface; winter snow fills fractures causing calving.
Student 3: During summer, ocean currents melt ice from bottom; calving occurs when 50 m melted.
10. In regard to the season(s) involved in iceberg calving, how does Student 2's description differ from Student 3's description? Student 2 indicated that:
F. summer and winter are involved, whereas Student 3 indicated only summer.
G. summer and winter are involved, whereas Student 3 indicated only winter.
H. only summer is involved, whereas Student 3 indicated summer and winter.
J. only winter is involved, whereas Student 3 indicated only summer.
Q10 • Key: F Answer: [ ]
Passage II • Ice Shelves CARD 11
Student 1: Air warms above surface; surface ice melts.
Student 2: "Student 1 is correct that... leads to surface ice melting..."
Student 3: "Antarctic ice shelves melt only from below."
Student 4: "...warm air leads to melting and compaction of snow [on surface]..."
11. Which of Students 2, 3, and 4 agree(s) with Student 1 that some form of melting occurs on the ice shelf surface?
A. Student 2 only
B. Students 2 & 3 only
C. Students 2 & 4 only
D. Students 3 & 4 only
Q11 • Key: C Answer: [ ]
Passage III • Clownfish CARD 12
Table 2: Stocking Density vs. Average SGR
Density:0.5 fish/L0.50%
Density:1 fish/L0.35%
Density:2 fish/L0.25%
Density:3 fish/L0.20%
12. Which statement about the relationship between the number of A. percula per tank and average SGR is consistent with Experiment 2? On average, as the number of fish per tank increased, average SGR:
F. increased only.
G. decreased only.
H. remained the same.
J. varied with no trend.
Q12 • Key: G Answer: [ ]
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ACT Science Problem Task Cards
Cards 13–16 • Passage III (Zoology)
Passage III • Clownfish CARD 13
| Diet | Protein % | Exp 1 SGR (%/day) |
|---|
| Q | 52.5% | 0.30 (Lowest) |
| R | 48.0% | 0.40 |
| S | 41.1% | 0.50 (Highest) |
| T | 38.1% | 0.35 |
13. Which statement about percent protein and average SGR is consistent with the data? The diet that resulted in the:
A. highest average SGR also had the highest percent protein.
B. highest average SGR had the lowest percent protein.
C. lowest average SGR had the highest percent protein.
D. lowest average SGR also had the lowest percent protein.
Q13 • Key: C Answer: [ ]
Passage III • Clownfish CARD 14
Table 1 SGR Values:
Diet R: 0.40 percent/day
Diet T: 0.35 percent/day
Experiment 2 tested stocking densities while all fish were fed Diet T.
14. If Experiment 2 were repeated except all fish were fed Diet R, would average SGRs more likely have been lower or higher for each group?
F. Lower; fed Diet R had SGR 0.05% less than Diet T.
G. Lower; fed Diet R had SGR 0.10% less than Diet T.
H. Higher; fed Diet R had SGR 0.05% greater than Diet T.
J. Higher; fed Diet R had SGR 0.10% greater than Diet T.
Q14 • Key: H Answer: [ ]
Passage III • Clownfish CARD 15
Experiment 1 Feeding Protocol:
"Each group was fed the same mass of food 3 times daily."
15. Suppose that, in the experiments, 1 g of food were added to each tank at each feeding. A total of how many grams of food would have been placed into an individual tank each day?
A. 1 g
B. 3 g
C. 12 g
D. 36 g
Q15 • Key: B Answer: [ ]
Passage III • Clownfish CARD 16
Experiment 1 Setup:
"Each of 12 identical 15 L tanks received 10 L of seawater... distributed among the tanks at a stocking density of 1 fish/L."
16. How many A. percula were placed in each of the tanks in Experiment 1?
F. 1
G. 4
H. 10
J. 12
Q16 • Key: H Answer: [ ]
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ACT Science Problem Task Cards
Cards 17–20 • Passage III & IV
Passage III • Clownfish CARD 17
Experiment 1 Overview:
Investigated how different diets (Q–T) affected the growth outcome (specific growth rate; SGR) while seawater volume and density remained fixed.
17. Which of the following was a dependent variable in Experiment 1?
A. Volume of seawater in each tank
B. Specific growth rate
C. Diet fed to the A. percula
D. Stocking density of the A. percula
Q17 • Key: B Answer: [ ]
Passage IV • Tomatoes CARD 18
Table 1: Incubation Time at 88°C
Sample 1: 0 min
Sample 2: 2 min
Sample 3: 15 min
Sample 4: 30 min
Sample 1 was immediately frozen without heat incubation.
18. Which of the samples in Experiment 1 was most likely intended as a control for the concentration of vitamin C present in the unheated tomatoes?
F. Sample 1
G. Sample 2
H. Sample 3
J. Sample 4
Q18 • Key: F Answer: [ ]
Passage IV • Tomatoes CARD 19
Figure 2: Lycopene Concentration (mg/g)
S1 (0 min): 2.0
S2 (2 min): 3.2
S3 (15 min): 5.5
S4 (30 min): 5.5
19. Based on the results of Experiment 2, which incubation time would most likely have produced a tomato mixture with a lycopene concentration between 5 mg/g and 6 mg/g?
A. 0 min
B. 0.2 min
C. 2 min
D. 20 min
Q19 • Key: D Answer: [ ]
Passage IV • Tomatoes CARD 20
Observed Nutrient Trends:
• Vitamin C (Fig 1): Decreased with heating (0.75 → 0.67 → 0.64 → 0.55 µmol/g)
• Lycopene (Fig 2): Increased with heating (2.0 → 3.2 → 5.5 → 5.5 mg/g)
20. A student claimed that heating tomatoes decreases the concentration of nutrients present. This claim is consistent with the results shown for which of vitamin C and lycopene, if either?
F. Vitamin C only
G. Lycopene only
H. Both vitamin C and lycopene
J. Neither vitamin C nor lycopene
Q20 • Key: F Answer: [ ]
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ACT Science Problem Task Cards
Cards 21–24 • Passage IV & V
Passage IV • Tomatoes CARD 21
Experimental Setup: Water bath maintained at 88°C at standard atmospheric pressure (1 atm). Note: Pure water boils at 100°C at 1 atm.
21. While the bags containing the samples were being incubated, was the water in the water bath most likely boiling?
A. Yes; incubation temperature was less than boiling point.
B. Yes; incubation temperature was greater than boiling point.
C. No; incubation temperature was less than boiling point.
D. No; incubation temperature was greater than boiling point.
Q21 • Key: C Answer: [ ]
Passage IV • Tomatoes CARD 22
Figure 1: Vitamin C Concentration (µmol/g)
S1: 0.75
S2: 0.67
S3: 0.64
S4: 0.55
Scale max on y-axis is 0.8 µmol/g.
22. In Experiment 1, how many of the samples had a vitamin C concentration of less than 1.0 µmol/g tomato?
F. 0
G. 1
H. 3
J. 4
Q22 • Key: J Answer: [ ]
Passage IV • Tomatoes CARD 23
Exp 1 Timeline: Bags incubated at 88°C, then frozen at -40°C. 2 days later, thawed, mixed with solvent, and filtered.
23. Consider the following procedures performed in Experiment 2 for Sample 2:
1. The sample was frozen.
2. The sample was incubated in the water bath.
3. The sample and solvent mixture was filtered.
These procedures were performed in what order?
A. 1, 2, 3
B. 1, 3, 2
C. 2, 1, 3
D. 2, 3, 1
Q23 • Key: C Answer: [ ]
Passage V • Molten Alloy CARD 24
Figure 1 (550°C curve):
From \(F = 10\) to \(20 \times 10^7\), \(L\) drops steadily and uniformly from ~92.2 µm to ~91.0 µm, exhibiting a straight-line slope.
24. A linear region of a graph can be approximated with a straight line. For Alloy Q at 550°C, which range of F best represents a linear region?
F. Between 0 and 10 × 10⁷
G. Between 10 × 10⁷ and 20 × 10⁷
H. Between 20 × 10⁷ and 30 × 10⁷
J. Between 30 × 10⁷ and 40 × 10⁷
Q24 • Key: G Answer: [ ]
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ACT Science Problem Task Cards
Cards 25–28 • Passage V (Materials Science)
Passage V • Molten Alloy CARD 25
Trends from F = 40 to 48 × 10⁷:
• 280°C line: Sloping downward from 97.5 µm to 95.5 µm
• 550°C line: Sloping upward from 91.5 µm to 92.5 µm
25. If these lines continue along the same trend beyond \(F = 48 \times 10^7\), at which of the following values of F would the average crystal lengths most likely be the same?
A. F = 50 × 10⁷
B. F = 60 × 10⁷
C. F = 70 × 10⁷
D. F = 80 × 10⁷
Q25 • Key: C Answer: [ ]
Passage V • Molten Alloy CARD 26
Figure 1 Values:
280°C, F=10: ~101.5 µm
280°C, F=40: ~97.5 µm
550°C, F=10: ~92.2 µm
550°C, F=40: ~91.5 µm
26. Which combination of initial temperature and F would produce the shortest average crystal length in an ingot of Alloy Q? The smallest L would be produced with a temperature of:
F. 280°C and F = 10 × 10⁷.
G. 280°C and F = 40 × 10⁷.
H. 550°C and F = 10 × 10⁷.
J. 550°C and F = 40 × 10⁷.
Q26 • Key: J Answer: [ ]
Passage V • Molten Alloy CARD 27
Table 1: Silicon (Si) = 10.8% by mass in Alloy Q.
W: 0.11 g
X: 0.14 g
Y: 2.7 g
Z: 5.4 g
Mass of Si in 50 g samples of 4 alloys.
27. Given the composition of Alloy Q, which sample is most likely Alloy Q?
A. Sample W
B. Sample X
C. Sample Y
D. Sample Z
Q27 • Key: D Answer: [ ]
Passage V • Molten Alloy CARD 28
Table 1 Composition:
Magnesium (Mg) comprises 0.22% by mass of Alloy Q.
28. Based on Table 1, if an ingot of Alloy Q had a mass of 200 g, that ingot would contain what mass of Mg?
F. 0.22 g
G. 0.44 g
H. 2.2 g
J. 4.4 g
Q28 • Key: G Answer: [ ]
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ACT Science Problem Task Cards
Cards 29–32 • Passage VI (Ecology)
Passage VI • DOC Adsorption CARD 29
Highest % Adsorbed per Clay:
• CM1 (Fig 1): Pine (~44%)
• CM2 (Fig 2): Magnolia (~42%)
• CM3 (Fig 3): Magnolia (~55%)
29. Based on the results of the studies, from which of the 5 leachates was the greatest percent of DOC adsorbed by CM1, CM2, and CM3, respectively?
A. CM1: maple; CM2: maple; CM3: rhododendron
B. CM1: oak; CM2: pine; CM3: magnolia
C. CM1: pine; CM2: magnolia; CM3: rhododendron
D. CM1: pine; CM2: magnolia; CM3: magnolia
Q29 • Key: D Answer: [ ]
Passage VI • DOC Adsorption CARD 30
Figure 3 (CM3) Adsorption Values:
Maple: 35%
Oak: 50%
Pine: 32%
Mag: 55%
Rhodo: 36%
30. Based on the results of Study 3, the percent of leachate DOC adsorbed by CM3, averaged across the 5 types of leaves, is closest to which of the following?
F. 10%
G. 20%
H. 30%
J. 40%
Q30 • Key: J Answer: [ ]
Passage VI • DOC Adsorption CARD 31
Maple Leachate Comparison:
CM2 (Figure 2): ~7%
CM3 (Figure 3): ~35%
31. Is the statement “CM2 adsorbed a greater percent of the DOC in the maple leachate than did CM3” supported by the results of Studies 2 and 3?
A. Yes; CM2 adsorbed 35%, whereas CM3 adsorbed 7%.
B. Yes; CM2 adsorbed 55%, whereas CM3 adsorbed 17%.
C. No; CM2 adsorbed 7%, whereas CM3 adsorbed 35%.
D. No; CM2 adsorbed 17%, whereas CM3 adsorbed 55%.
Q31 • Key: C Answer: [ ]
Passage VI • DOC Adsorption CARD 32
Oak Leachate Adsorption:
CM1: ~22%
CM2: ~23%
CM3: ~50% (not > 50%)
32. Based on the results of the studies, which of the 3 clay minerals, if any, reduced the DOC in the oak leachate by more than 50%?
F. CM1 only
G. CM2 only
H. CM1 and CM3 only
J. None of the 3 clay minerals
Q32 • Key: J Answer: [ ]
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ACT Science Problem Task Cards
Cards 33–36 • Passage VI & VII
Passage VI • DOC Adsorption CARD 33
Passage Definition: DOC is adsorbed by "clay mineral particles that are suspended in the water." In a solution, solutes are dissolved.
33. Is a mixture of any one of the leachates and any one of the clay minerals properly considered a solution?
A. Yes, because the clay mineral particles are dissolved.
B. Yes, because the clay mineral particles are not dissolved.
C. No, because the clay mineral particles are dissolved.
D. No, because the clay mineral particles are not dissolved in the leachate.
Q33 • Key: D Answer: [ ]
Passage VI • DOC Adsorption CARD 34
Experimental Procedure: "The mixture was then placed in the dark for 10 weeks at 4°C while leaching occurred."
34. In lake water, DOC is broken down into simpler compounds by visible light. What action was taken in the studies to prevent this process from occurring?
F. Each mixture of leaves and filtered lake water was placed in the dark.
G. Each mixture of filtrate and clay mineral was placed in the dark.
H. Each mixture of leaves and lake water was filtered.
J. Each mixture of leachate and clay mineral was filtered.
Q34 • Key: F Answer: [ ]
Passage VII • Physics CARD 35
Figure 1 Ammeter: 19.8 mA Table 2 Intensity:
100%: 40.0 mA
50%: 19.8 mA
25%: 9.8 mA
12.5%: 4.8 mA
35. Consider the current shown on the ammeter in Figure 1 (19.8 mA). Based on Experiment 2, when this current was measured, what was the relative intensity of the light?
A. 100%
B. 50%
C. 25%
D. 12.5%
Q35 • Key: B Answer: [ ]
Passage VII • Physics CARD 36
Exp 1: Filters varied light frequency and color (Red, Yellow, Green, Blue, Violet).
Exp 2: Performed "without a filter" (constant light color); light intensity varied.
36. What aspect of the experimental setup was held constant in Experiment 2 but not in Experiment 1?
F. Color of light
G. Light source
H. Type of metal plate
J. Distance between metal plate and electrode
Q36 • Key: F Answer: [ ]
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ACT Science Problem Task Cards
Cards 37–40 • Passage VII (Physics)
Passage VII • Physics CARD 37
Figure 1 Apparatus: Ejected particles are electrons. Arrows show electrons traveling from the illuminated metal plate across to the electrode.
37. Based on Figure 1, are the particles ejected from the metal plate moving toward or away from the electrode, and are those particles positively or negatively charged?
A. Toward; positively charged
B. Toward; negatively charged
C. Away from; positively charged
D. Away from; negatively charged
Q37 • Key: B Answer: [ ]
Passage VII • Physics CARD 38
Photoelectric Equation: \(K = E - M \implies M = E - K\)
Green: E=2.31, K=0.11
Blue: E=2.60, K=0.40
Violet: E=3.10, K=0.90
38. Based on the equation in the passage and the results of Experiment 1, what was the value of M for the metal plate used in the setup?
F. 2.14 eV
G. 2.20 eV
H. 2.31 eV
J. 2.42 eV
Q38 • Key: G Answer: [ ]
Passage VII • Physics CARD 39
Table 1 Frequency vs. Ejection:
• Yellow (\(5.2 \times 10^{14}\) Hz): K = N.A. (no electrons ejected)
• Green (\(5.6 \times 10^{14}\) Hz): K = 0.11 eV (electrons ejected!)
39. The cutoff frequency is the lowest frequency of light at which electrons are ejected. The cutoff frequency for the metal plate was:
A. less than 4.4 × 10¹⁴ Hz.
B. between 4.4 × 10¹⁴ Hz and 5.2 × 10¹⁴ Hz.
C. between 5.2 × 10¹⁴ Hz and 5.6 × 10¹⁴ Hz.
D. greater than 5.6 × 10¹⁴ Hz.
Q39 • Key: C Answer: [ ]
Passage VII • Physics CARD 40
Formula: \(E = hf \implies h = \frac{E}{f}\)
Table 1 Data for Green Light: Frequency \(f = 5.6 \times 10^{14}\) Hz, Energy \(E = 2.31\) eV, \(K = 0.11\) eV
40. Based on the data for green light in Table 1, which expression could be used to determine the value of h?
F. \(\frac{5.6 \times 10^{14}\text{ Hz}}{0.11\text{ eV}}\)
G. \(\frac{0.11\text{ eV}}{5.6 \times 10^{14}\text{ Hz}}\)
H. \(\frac{5.6 \times 10^{14}\text{ Hz}}{2.31\text{ eV}}\)
J. \(\frac{2.31\text{ eV}}{5.6 \times 10^{14}\text{ Hz}}\)
Q40 • Key: J Answer: [ ]
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