ACT Science Practice Test: Experimental Design and Variables
- Experimental Design and Variables
- 10 questions
- 11 minutes
This test has two original passages that each describe an experiment, followed by five questions. You will name the independent and dependent variables, find the control, describe trends and predict results that were not measured. Ask first what the scientist changed on purpose and what stayed the same.
Tap an option to mark your answer. You can change your mind, or tap the same option again to leave the question blank. The correct answers and explanations appear when you finish the test.Tap an option to answer; tap it again to leave the question blank.
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Question 1
Passage I
A student studied how the salt concentration of water affects the germination (sprouting) of radish seeds.
Five dishes were lined with paper towels, and 40 radish seeds were placed in each dish. Every day, each dish was moistened with 10 mL of a sodium chloride (NaCl) solution; each dish received a different concentration, and one dish received pure water (0 g/L). All dishes were covered and kept in the dark at 22°C. After 5 days, the student counted the seeds that had germinated. The whole procedure was then repeated with new seeds (Trial 2).
Table 1: Germinated seeds (out of 40) NaCl (g/L) Trial 1 Trial 2 Average germination (%) 0 38 37 93.8 2 36 37 91.3 4 31 29 75.0 6 20 18 47.5 8 7 9 20.0 In this study, the independent variable was the:
Correct answer: B
The student deliberately gave each dish a different NaCl concentration, so concentration is the independent variable. The number of germinated seeds is what was measured, the dependent variable; temperature and time were kept the same.
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Question 2
Passage I (see above):
Which dish served as the control in this study?
Correct answer: A
The control receives none of the factor being tested, so the pure-water dish (0 g/L) shows how the seeds germinate without salt. The dish with 8 g/L is the extreme treatment, not the control.
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Question 3
Passage I (see above):
Based on Table 1, if the student had tested a 5 g/L NaCl solution, the average germination would most likely have been:
Correct answer: C
Germination falls steadily as concentration rises; 5 g/L lies between 4 g/L (75.0%) and 6 g/L (47.5%), so the result should lie between those values.
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Question 4
Passage I (see above):
In Trial 2, how many seeds in the 6 g/L dish did NOT germinate?
Correct answer: D
In Trial 2, 18 of the 40 seeds in the 6 g/L dish germinated, so 40 − 18 = 22 did not. The value 20 is the number that did not germinate in Trial 1.
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Question 5
Passage I (see above):
The student repeated the procedure with new seeds (Trial 2) most likely in order to:
Correct answer: A
Trial 2 used the same concentrations and conditions, so its purpose was to see whether the results could be repeated. The two trials agree closely, which makes the trend more convincing. Nothing about light or temperature was changed.
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Question 6
Passage II
A pendulum is a weight (bob) that hangs from a string and swings back and forth. The period is the time for one complete swing, out and back. Students measured the time for 10 swings with a stopwatch and divided by 10 to find the period. In every trial, the bob was released from an angle of 10° from vertical.
Experiment 1: A 100 g bob was used, and the length of the string was varied.
Table 1 String length (m) Period (s) 0.25 1.00 0.50 1.42 1.00 2.01 1.50 2.46 2.00 2.84 Experiment 2: The string length was kept at 1.00 m, and bobs of different mass were used.
Table 2 Bob mass (g) Period (s) 50 2.01 100 2.00 200 2.01 400 2.00 Which variable was held constant in Experiment 2 but varied in Experiment 1?
Correct answer: D
Experiment 1 changed the string length; Experiment 2 kept it at 1.00 m. The bob mass was varied only in Experiment 2, and the release angle was 10° in both experiments, so it was never varied.
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Question 7
Passage II (see above):
According to Experiment 1, as the string length increased from 0.25 m to 2.00 m, the period:
Correct answer: B
Table 1 shows the period rising at every step, from 1.00 s to 2.84 s. It does not rise in proportion, though: doubling the length from 1.00 m to 2.00 m raises the period only from 2.01 s to 2.84 s.
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Question 8
Passage II (see above):
Based on Experiment 1, a pendulum with a 4.00 m string would most likely have a period closest to:
Correct answer: B
In Table 1, every time the length is multiplied by 4, the period doubles (0.25 m gives 1.00 s and 1.00 m gives 2.01 s). So 4.00 m, four times 1.00 m, should give about 2 × 2.01 = 4.0 s. The choice 8.0 s assumes the period is proportional to the length.
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Question 9
Passage II (see above):
Which conclusion is best supported by Experiment 2?
Correct answer: C
Doubling the mass three times, from 50 g to 400 g, left the period at 2.00-2.01 s, a difference smaller than the precision of the stopwatch timing. The release angle was never changed, so no conclusion about it is possible.
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Question 10
Passage II (see above):
The students want a pendulum with a period of 2.2 s. Based on Table 1, which string length should they try?
Correct answer: A
A period of 2.2 s lies between 2.01 s (1.00 m) and 2.46 s (1.50 m), so the length must be between 1.00 m and 1.50 m; only 1.20 m fits. The choice 2.20 m simply copies the number of seconds.
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