1,296 ACT Practice Questions, 3rd Edition (2013)
Science Practice Section 3
SCIENCE REASONING TEST
35 Minutes—40 Questions
DIRECTIONS: There are seven passages in the following section. Each passage is followed by several questions. After reading a passage, choose the best answer to each question and blacken the corresponding oval on your answer document. You may refer to the passages as often as necessary.
You are NOT permitted to use a calculator on this test.
Sylvatic, or jungle, Yellow Fever is caused by a virus transmitted by mosquitoes from monkeys to humans. Figure 1 shows the life cycle of the mosquitoes who carry this disease. These mosquitoes’ eggs do not hatch unless there is enough water for the next two stages of their life cycles. Yellow Fever is passed when an adult of these mosquitoes first bites a monkey that is infected with the virus and then bites a human.
A study was done on a group of ecologists who went into a jungle where the monkeys carrying the Yellow Fever virus live. These ecologists were divided into groups based on how frequently they went into the jungle. The ecologists were tested monthly for Yellow Fever. Figure 2 shows the number of new cases of Yellow Fever and the amount of rainfall in the jungle. For each group, Table 1 shows the number of ecologists in each group, number of mosquito bites, and percent of each group with Yellow Fever.
1. Based on Figure 1, what is essential in maintaining the mosquito population?
2. Based on Table 1, the average percent of ecologists affected by the yellow fever virus was closest to:
3. Suppose additional data had been gathered for Table 1 about the number of mosquito bites per month. Based on Figure 2 and Table 1, in which of the following months would you expect to have the largest total of mosquito bites per month?
4. According to Figure 2, the amount of rainfall was different for each of the following pairs of months EXCEPT:
F. May and December.
G. February and March.
H. January and October.
J. April and May.
5. Based on Table 1, as the number of trips into the jungle increased, the number of monkeys seen:
A. increased only.
B. decreased only.
C. increased, then decreased.
D. varied with no consistency.
Ethanolamines are compounds that contain both alcohol (–OH or HO–) and amine (–NH3, –RNH2, –R2NH, or –R3N) subgroups. They remove weakly acidic gases from the atmosphere of enclosed spaces such as on a submarine. An example is the use of monoethanolamine (MEA) to remove CO2 from the atmosphere as shown in Figure 1.
2 MEA(liquid) + CO2(gas) (MEA)COO–(aqueous) + (MEA)H +(aqueous) + heat
If the temperature rises sufficiently, ethanolamines will release any absorbed acidic gases back into the environment, creating a potential hazard.
Scientists studied the absorption properties of 2 ethanolamines (MEA and DEA).
At 0°C and 1 atmosphere (atm) pressure, 1 mole (6.02 × 1023 molecules) of MEA was spread at the base of a reaction vessel containing CO2 gas at a concentration of 1,000 parts per million (ppm). As the CO2 was absorbed, its ambient concentration decreased. The scrub time (time for CO2concentration to drop to at least 10 ppm) was measured. Longer scrub times indicate a slower rate of absorption. The experimental procedure was repeated at varying temperatures and for DEA, with results recorded in Table 1.
The scrub times of MEA for different acidic gases were measured using the procedures of Experiment 1 at 26°C (see Table 2). Each of the gases listed is toxic and poses a significant safety hazard if its concentration becomes elevated within an enclosed space.
6. In which of the following ways was the procedure of Experiment 2 different from that of Experiment 1? In Experiment 2:
F. temperature was varied; in Experiment 1, the temperature was held constant.
G. temperature was held constant; in Experiment 1, the temperature was varied.
H. only MEA was used; in Experiment 1, only DEA was used.
J. only DEA was used; in Experiment 1, only MEA was used.
7. In Experiment 1, during the DEA trial at 20°C, as the time progressed from 0 to 8,190 msec, the concentration of CO2 in the vessel:
A. increased from 10 ppm to 1,000 ppm.
B. increased from 1,000 ppm to 10 ppm.
C. decreased from 10 ppm to 1,000 ppm.
D. decreased from 1,000 ppm to 10 ppm.
8. If, in Experiment 1, an additional trial were done at 12°C, the scrub times (in msec) for MEA and DEA would most likely be closest to which of the following?
9. Based on the information in the passage, which of the following is a possible chemical formula for an ethanolamine?
10. A scientist claims that under the same conditions, DEA will always absorb CO2 at a faster rate than will MEA. Do the results of Experiment 1 support this claim?
F. No; at all temperatures tested, the scrub time for DEA was more than that for MEA.
G. No; at all temperatures tested, the scrub time for MEA was more than that for DEA.
H. Yes; at all temperatures tested, the scrub time for DEA was more than that for MEA.
J. Yes; at all temperatures tested, the scrub time for MEA was more than that for DEA.
11. Based on the results of Experiment 2, which acidic gas had the slowest absorption by MEA at 26°C?
Taraxicum, the common dandelion, can reproduce both through spreading seeds and through vegetative reproduction. To spread its seeds, the dandelion grows seed pods shaped like globes, in which the seeds are loosely attached to a central ball; each seed grows a parachute-like tuft that lets it travel long distances on the wind (or when blown upon by humans). In vegetative reproduction, a new dandelion stalk and leaves can grow up from an existing root system. Two students discuss the spread of dandelion populations.
In Taraxicum, vegetative reproduction and seed distribution make up the only means of growing new plants. Each accounts for 50% of the growth of new dandelions.
Taraxicum grows throughout North America. In many places there is very little wind. Therefore, Taraxicum must have a non-wind-based means of spreading itself. While blowing dandelion seeds is a common pastime among humans, this human influence is very recent in evolutionary terms; it is very unlikely that Taraxicum evolved to rely on humans to distribute its seeds.
The way Taraxicum grows in a typical field shows that both vegetative reproduction and seed distribution are at work. While seeds scatter over the whole field, the dandelions tend to grow together in clumps. This suggests that individual seeds sprout the first new dandelions, which then grow several more through vegetative reproduction.
Seed distribution is the main way Taraxicum spreads itself. Without seed distribution, there are very few new dandelions. Taraxicum does use vegetative reproduction, sending new stalks from existing roots, but this is mainly to replace the above-ground plant if it has been cut or eaten. This allows the plant to survive threats in the environment but does not allow for the growth of new plants.
Plant studies show that plants which rely on vegetative reproduction to spread themselves tend to have large, complex root networks or underground root clusters. Taraxicum plants, however, each have a single large, deep taproot. This makes them very difficult to uproot, but it also means that their roots do not spread out underground, so any new plants growing from the roots would compete with each other for sunlight. Even a slight breeze or the brush of a passing animal is enough to spread dandelion seeds to a new area. Additionally, all known types of Taraxicum produce seed globes. If half the new dandelions grew from vegetative reproduction, then a seedless dandelion should not be at a competitive disadvantage and should be commonly observed in the wild.
The students proposed 3 trials using an introduced Taraxicum population in three fields in a windy area where Taraxicum can naturally thrive (see Table 1).
12. Suppose an experiment were performed in which several new Taraxicum plants were planted in a field with their roots in glass jars and with plastic bags over the flowers. Assuming that Student 1’s hypothesis is correct, the number of new dandelions in the field would most likely be what percent of the number in a control field?
13. Which of the following trials most likely provided the control group in the students’ experiment?
A. Trial 1, in which Taraxicum specimens are planted in the soil of a field with no other Taraxicum plants
B. Trial 1, in which Taraxicum specimens are planted in large glass jars, which are then buried in the soil of a field with no other Taraxicum plants
C. Trial 2, in which Taraxicum specimens are planted in the soil of a field similar to that of Trial 1
D. Trial 3, in which specimens are planted in large glass jars in a field similar to that of Trial 1
14. Student 1 states that dandelions growing in clumps “suggests that individual seeds sprout the first new dandelions, which then grow several more through vegetative reproduction.” Which of the following indicates why Student 2 believes this cannot be true? Student 2 says:
F. Taraxicum tends to grow from a root network, while vegetative reproducers grow from single roots.
G. Taraxicum tends to grow from a single root, while vegetative reproducers grow from root networks.
H. Taraxicum has seeds that are attached loosely to the stem, a fact that suggests they are not important to Taraxicum’s reproductive strategy.
J. Taraxicum has seeds that are attached loosely to the stem, but vegetative reproducers tend not to have seeds at all.
15. Student 2 would most likely agree with the statement that Taraxicum:
A. uses vegetative reproduction to compensate for windless environments.
B. improves its ability to survive by using vegetative reproduction to regenerate.
C. has evolved a dependency on humans to distribute its seeds.
D. tends to grow in clumps in fields to which it has spread itself.
16. With regard to the experiment described in the table, Students 1 and 2 would most likely agree that the increase in the Taraxicum population would be greatest in a field where:
F. neither plastic bags nor glass jars were used.
G. plastic bags were used, but not glass jars.
H. glass jars were used, but not plastic bags.
J. both plastic bags and glass jars were used.
17. Suppose Trial 3 of the experiment were performed as described. Based on Student 1’s hypothesis, the resulting population would be closest to what percentage of a control population?
18. Suppose the 3 trials were performed as described. Student 2’s hypothesis about the way Taraxicum reproduces would be best supported if the number of new dandelions fit which of the following patterns?
F. The field in Trial 3 had roughly the same number of dandelions as the field in Trial 1, both of which had fewer dandelions than the field in Trial 2.
G. The field in Trial 1 had more dandelions than the field in either Trial 2 or Trial 3, while the fields in Trials 2 and 3 had roughly equal numbers of dandelions.
H. The field in Trial 2 had fewer dandelions than the field in Trial 3, which had more dandelions than the field in Trial 1.
J. The field in Trial 3 had slightly fewer dandelions than the field in Trial 1, both of which had many more dandelions than the field in Trial 2.
Three studies were conducted to analyze the content of cake mix.
Samples of 3 different cake mixes (X, Y, and Z) weighing 500.0 g were desiccated (thoroughly dried) in a 350°F oven for 36 hours and then passed through a sieve with 0.045 cm holes. Each sample was evenly spread in a trough 0.100 cm deep with a fan at one end and a secured piece of dark colored paper at the other end, downwind from the fan, as shown in Figure 1. Fan speed was slowly increased from 0 m/s until cake mix particles could be seen on the paper. This speed is called the maximum immovable speed. The maximum immovable speed was recorded for each sample.
This procedure was repeated with 35.0 g of cornmeal, a heavier ingredient, added to the cake mix prior to drying. The results are shown in Table 1.
Samples of each cake mix were prepared as in Study 1. Each sample was mixed with 500 mL of water then allowed to rest for 6 hours to ensure equal saturation of all particles in the sample. The samples were then evenly spread in a trough in an apparatus identical to that of Study 1. The fan was set to 10 m/s. Once the fan had dried the sample sufficiently, particles began to appear on the dark paper. At this point, the water content of the sample was calculated. This procedure was repeated with the 16 m/s fan speed. This set of procedures was then repeated with samples of each cake mix to which 35.0 g of cornmeal had been added. Figure 2 displays the results.
Samples of each cake mix, without any preparation, were analyzed for flour particle, sugar particle, gelatin particle, and water content. The results are shown in Table 2.
19. Based on Study 3, the water content of cake mix Z after being allowed the 6 hours to saturate with water in Study 2 was most likely:
A. greater than 10%.
B. between 6% and 10%.
C. between 1% and 6%.
D. less than 1%.
20. Which of the following statements about the maximum immovable speed in the trials without cornmeal is best supported by the results of Study 1 for the three cake mixes?
F. The maximum immovable speed for all cake mixes was 0 m/s.
G. The maximum immovable speeds for all cake mixes were roughly equal.
H. The maximum immovable speed for cake mix Y was three times the maximum immovable speed for cake mix X.
J. The maximum immovable speed for cake mix Y was half the maximum immovable speeds of cake mixes X and Z.
21. Based on the results of Study 1 for a given cake mix, the addition of cornmeal to the cake mix caused a maximum immovable speed that was approximately:
A. half as high as the maximum immovable speed when no cornmeal was added to the cake mix.
B. twice as high as the maximum immovable speed when no cornmeal was added to the cake mix.
C. between two and two and a half times as high as the immovable speed when no cornmeal was added to the cake mix.
D. between two and a half and three and a half times as high as the immovable speed when no cornmeal was added to the cake mix.
22. If equal amounts of cake mixes Y and Z were blended and then prepared as in Study 1, then tested under the conditions of Study 1 with cornmeal added, the maximum immovable speed for this sample would most likely be:
F. between 0.23 m/s and 0.25 m/s.
G. between 0.28 m/s and 0.33 m/s.
H. between 0.36 m/s and 0.38 m/s.
J. between 0.39 m/s and 0.43 m/s.
23. A food scientist hypothesized that cake mixes with higher dry particle contents of gelatin will have a higher water content than cake mixes with lower dry particle contents of gelatin. Is this hypothesis supported by Study 3?
A. Yes, because as gelatin content of cake mixes X, Y, and Z increased, water content decreased.
B. Yes, because as gelatin content of cake mixes X, Y, and Z increased, water content increased.
C. No, because as gelatin content of cake mixes X, Y, and Z increased, water content decreased.
D. No, because as gelatin content of cake mixes X, Y, and Z increased, water content increased.
24. A fourth cake mix, cake mix Q, was analyzed as in Study 3. It was determined to contain 61% flour, 28% sugar, and 10% gelatin. Based on Study 3, what range would most likely be the water content of cake mix Q?
F. Less than 3.3%
G. Between 3.3% and 6.1%
H. Between 6.1% and 10.1%
J. Greater than 10.1%
Metals differ in their relative abilities to conduct electricity. Resistance is a measurement in ohms (Ω) of how much a metal opposes electric current at a particular voltage.
A scientist performed 3 experiments using the circuit shown in Figure 1.
The metal resistor consisted of a coil of metallic wire with a known cross-sectional area and length (see Figure 2).
At the outset, the switch was open and no current flowed through the circuit. A 9-volt battery was used, and the black and red test leads of the circuit were attached to a metal resistor. When the switch was closed, electrons (negatively charged) flowed away from the negative battery terminal, through the circuit, and back to the positive battery terminal. The magnitude of current (charge per unit time) from this electron flow was measured by an ammeter, and was 1.0 × 10−3 coulombs/second for the first trial of each experiment. The resistance (R) of the metal resistor was calculated in ohms (Ω) from the resulting values for voltage (V) and current (I).
Three nickel resistor coils, each with a cross-sectional area of 7.61×× 10−10 m2 but with different lengths, were attached separately to the circuit. Results were recorded in Table 1.
Three gold resistor coils of varying cross-sectional areas were tested. Each resistor coil had a measured length of 100 m. The results were recorded in Table 2.
Three coils made of different metals were tested. Each resistor had a cross-sectional area of 2.67 × 10−10 m2 and a length of 100 m. The value ρ is related to each metal’s inherent resistivity to current flow. Results were recorded in Table 3.
25. In Experiment 2, the scientist varied which of the following aspects of the metal resistor?
A. Identity of the metal coil
B. Cross-sectional area of the coil
C. Length of the coil
D. Value ρ of the metal composing the coil
26. Assume that as ρ increases, a metal’s ability to conduct current decreases. Based on the results of Experiment 3, which of the following correctly lists gold, nickel, and tin in order of increasing ability to conduct electrons when shaped as a wire coil?
F. Gold, nickel, tin
G. Gold, tin, nickel
H. Tin, nickel, gold
J. Tin, gold, nickel
27. In the first trial of Experiments 1−3, once the resistor was attached and the switch closed, what charge returned to the positive battery terminal each second?
A. −1.0×× 10−3 coulombs
B. −2.0 ××10−3 coulombs
C. −3.0×× 10−3 coulombs
D. −4.0×× 10−3 coulombs
28. Based on the results of the 3 experiments, the resistor with which of the following values of length, cross-sectional area, and metal type will have the highest current at a given voltage?
Cross-sectional area (m2)
2.00 × −10
29. In Experiment 1, the current across the circuit increased and the resistance of the resistor decreased as the:
A. value ρ of the metal resistor increased.
B. cross-sectional area of the metal resistor decreased.
C. length of the metal resistor increased.
D. length of the metal resistor decreased.
30. When the switch is closed in the circuit described in the passage, the battery caused electrons to flow in the direction(s) shown by which of the following diagrams?
Pressure, temperature, volume, and amount of reactant are four variables that affect the rate at which a reaction in the gas phase occurs. A change in any of these variables changes the likelihood of particles running into each other and reacting: Increasing any one increases reaction rate; decreasing any one decreases reaction rate.
Pressure is measured in atmospheres, atm, where 1 atm is the sea level pressure of earth’s atmosphere. Volume is measured in liters, L. The amount of reactant is measured in moles, where 1 mole is 6.02 × 1023 molecules.
Figure 1 shows how temperature and pressure affect the gaseous reactants in an experiment. Figures 2 and 3 show how the rate of Reaction A is affected by pressure and temperature, respectively.
31. A scientist claimed that increasing temperature increases the rate at which Reaction A occurs and increasing pressure increases the rate at which Reaction A occurs. Is the scientist’s claim supported by the passage and Figures 1−3?
A. Yes; the rate at which Reaction A occurred increased as temperature increased and increased as pressure increased.
B. Yes; the rate at which Reaction A occurred increased as pressure decreased.
C. No; the rate at which Reaction A occurred increased as temperature increased, but decreased as pressure increased.
D. No; the rate at which Reaction A occurred decreased as pressure increased.
32. According to Figures 2 and 3, the reactions occur at the same rate at what pressure and temperature?
F. 20°C and 2.0 atm
G. 40°C and 1.5 atm
H. 50°C and 1.0 atm
J. 70°C and 1.5 atm
33. The amounts of reactants in Reaction A are 1 mole/L of Compound Y and 2 mole/L of Compound Z. According to the passage, the number of molecules of Compound Y is:
A. one quarter of the number of molecules of Compound Z in the reactants.
B. one half the number of molecules of Compound Z in the reactants.
C. equal to the number of molecules of Compound Z in the reactants.
D. twice the number of molecules of Compound Z in the reactants.
34. A scientist tests a new Reaction B. This reaction is conducted with the same gas phase reactants, volume, pressure, and temperature as Reaction A, but the amounts (moles) of reactants are doubled. Based only on the information in the passage and Figures 1−3, how will the rate of Reaction B compare with the rate of Reaction A?
F. Reaction B will be slower than Reaction A because temperature will be lower.
G. Reaction B will be faster than Reaction A because temperature will be lower.
H. Reaction B will be faster than Reaction A because the concentration of reactants is greater, so the likelihood of reactant molecules colliding and reacting is greater.
J. Reaction B will be slower than Reaction A because the concentration of reactants is greater, so the likelihood of reactant molecules colliding and reacting is greater.
35. A chemist wanted to measure Reaction A at the greatest possible reaction rate. She had the ability to change either the temperature or the pressure of the gaseous reactants. Based on the data in Figures 2 and 3, which property did she most likely alter to increase the rate of Reaction A?
A. Pressure, which she decreased from 1 atm to 0.5 atm
B. Pressure, which she increased from 1 atm to 3 atm
C. Temperature, which she decreased from 50°C to 20°C
D. Temperature, which she increased from 50°C to 100°C
An experiment is set up to look at the physics of bouncing a ball, as shown in Figure 1.
When the ball is dropped, its initial velocity is 0 m/s. Velocity will increase until impact with the ground, at which point the ball’s velocity immediately drops to 0 m/s again. After impact, velocity almost immediately increases to maximum post-impact velocity, and then begins to fall again as gravity works against it, slowing it down. The ball’s velocity returns to 0 m/s when the ball is at its apex, or highest vertical point, post impact.
When a ball bounces, it deforms and becomes flatter. This is called elasticity. The more elasticity a material has, the better it is able to act like a spring and absorb force by being compressed, then use this force to “spring” back into the air. Post-impact velocity and the amount of time between velocity of 0 m/s at impact and velocity of 0 m/s at post-impact apex are affected by elasticity. Figure 2 shows the velocity of a ball versus time for balls with various elasticities and weights dropped from 1 meter height. Because gravity causes all objects to fall at the same speed regardless of weight, pre-impact velocities are identical for all balls.
36. Based on the data in Figure 2, the maximum post-impact velocity of a ball will be smallest if the elasticity of the ball is:
F. 1.5 Pa.
G. between 1 and 1.5 Pa.
H. between 0.5 and 1 Pa.
J. 0.5 Pa.
37. Based on the information in Figure 2, a ball being dropped from 1 meter height with an elasticity of 0.2 Pa and a weight of 0.5 kg would have a maximum post-impact velocity of:
A. less than 0.50 m/s.
B. 0.75 m/s.
C. 1.0 m/s.
D. greater than 1.25 m/s.
38. Consider a ball as it completes one bounce, from drop to post-impact apex. If this ball has a weight of 2 kg and an elasticity of 0.50 Pa, based on the data in Figure 2, how many times does the ball have a velocity of 1.00 m/s ?
F. One time
G. Two times
H. Three times
J. Four times
39. Based on the data in Figure 2, how does the velocity of a ball change as it goes from impact to apex?
Drop to Impact
Impact to Apex
Increases then decreases
Increases then decreases
Increases then decreases
Decreases then increases
40. A ball will deform permanently and not spring back off the ground if the velocity with which it hits the ground exceeds the ball’s elastic limit. Based on the data in Figure 2, if a ball is dropped from one meter and has a weight of 3 kg, an elasticity of 0.8 Pa, and an elastic limit of 2.75 m/s, will the ball deform permanently?
F. Yes, because the velocity with which the ball hits the ground is less than its elastic limit.
G. Yes, because the velocity with which the ball hits the ground is greater than its elastic limit.
H. No, because the velocity with which the ball hits the ground is less than its elastic limit.
J. No, because the velocity with which the ball hits the ground is greater than its elastic limit.
Answers and Explanations
SCIENCE PRACTICE 3 ANSWERS
SCIENCE PRACTICE 3 EXPLANATIONS
1. B Choice (B) is correct because it is the only choice that Figure 1 says is necessary for the mosquito life cycle to progress. If there is no water then the eggs will not hatch.
2. G To solve this problem, you need to average the five numbers in the Percent of Group Affected by Yellow Fever column. 10% + 18 % + 29% + 38% + 52% = 147%. 147% ÷ 5 = 29.4%, which is closest to 30%.
3. A This question asks you to look at trends in Table 1 and Figure 2 to determine amounts of mosquito bites per month. Figure 2 tells you about trends in cases of Yellow Fever. Table 1 tells you that as the number of mosquito bites increases, cases of Yellow Fever increase. Now look at Figure 2: if you know that more mosquito bites means more cases of Yellow Fever, then the month with the most mosquito bites is most likely to be the month with the most cases of Yellow Fever. April has 6 cases, June has 2 cases, August and November each have 1 case, which means that choice (A), April, is the correct answer.
4. G Choice (G) is the correct answer because February and March had the same amount of rainfall. To solve this problem you must compare the amounts of rainfall for each of the 2 months in the answer choices. Only one answer, choice (G), has the same amount for the 2 months in the pair. This question tests your ability to read a graph with 2 y-axes; be sure to check which axis you’re getting your information from. For this question you should use the one on the right, rainfall (inches.)
5. D To solve this problem, you must determine the trend of the numbers in the number of monkeys Seen column. This question is tricky because the numbers increase some and decrease some but don’t do either consistently. If the answer is not entirely right, it must be wrong, so you can eliminate choice (A) because the numbers don’t only increase, choice (B) because the numbers don’t only decrease, and choice (C) because the numbers did not increase then decrease, they partially decreased, then increased, then decreased again, then increased again. It’s intimidating to choose choice (D), varied with no consistency, but it is the only correct answer.
6. G The results in Table 1 for Experiment 1 indicate that the temperature varied for that experiment, eliminating choice (F). Table 1 also indicates that both MEA and DEA were used in Experiment 1, eliminating choices (H) and (J). The description of Experiment 2 states that all trials were done at a constant temperature of 26°C.
7. D Concentration cannot increase from 1,000 ppm to 10 ppm or decrease from 10 ppm to 1,000 ppm, eliminating choices (B) and (C). The description of Experiment 1 states that CO2 concentration started at 1,000 ppm and decreased as it was absorbed, eliminating choice (A).
8. G In Table 1, as temperature is increased, scrub times for both MEA and DEA decrease. Therefore, if an additional trial were done at 12°C, the scrub times would be expected to fall between the values listed for 10°C and 15°C for MEA and DEA. Only choice (G) has values that lie between these limits for both compounds.
9. A The first paragraph of the passage states that ethanolamines are compounds with both an alcohol and an amine subgroup. Choices (A) and (B) both have alcohol subgroups, and choices (A), (C), and (D) each have at least one amine subgroup. Only choice (A) has both.
10. J Experiment 1 compares MEA and DEA. The results in Table 1 indicate that the scrub times for DEA are always less than MEA at each temperature tested, eliminating choices (F) and (H). The passage states that longer scrub times indicate a slower rate of absorption. Therefore, DEA must have a faster rate of absorption at all temperatures.
11. B The passage states that longer scrub times indicate a slower rate of absorption. The results of Experiment 2 are shown in Table 2. The acidic gas with the longest scrub time is HCN, so it must have the slowest rate of absorption.
12. F Student 1 hypothesizes that vegetative reproduction and seed distribution together make up the only ways that Taraxicum reproduces. The experiment described would prevent vegetative reproduction (glass jars) and seed distribution (plastic bags), so if Student 1 is correct, no daughter plants should ever grow.
13. A Trials 2 and 3 both describe instances in which the experimenters have interfered with one of the ways that dandelions spread themselves. The first trial, in which Taraxicum specimens are left to grow naturally, provides the control for this experiment, meant to establish what happens without any change in the variables being investigated.
14. G Choice (G) is correct, while choice (F) states the opposite of what Student 2 claims—the actual claim was that dandelions have a single taproot, unlike vegetative reproducers. Choice (H) is wrong: the plant holds the seeds loosely not because they are not important, but because they must be distributed for the plant to spread. Choice (J) is wrong because no claim is made that vegetative reproducers never produce seeds.
15. B Student 2 says that Taraxicum does use vegetative reproduction … to replace the above-ground plant if it has been cut or lost, and that this allows Taraxicum to survive threats in the environment. Choice (A) is Student 1’s argument, which Student 2 rejects. Choice (C) is rejected by Student 1 and never mentioned by Student 2. Choice (D) is an observation made by Student 1, but Student 2 does not comment on whether it is accurate.
16. F Student 1 and Student 2 believe that dandelions spread through both seed distribution and vegetative reproduction; they disagree only on the relative importance of the two means. Thus, choice (F) is correct because neither means of reproduction is limited. Choices (G) and (H) both describe situations in which some part of the dandelion’s reproductive ability is experimentally tampered with, while choice (J) describes a situation in which both the dandelion’s reproductive strategies were negated.
17. C Student 1 asserts that vegetative reproduction and seed distribution each make up equal parts of Taraxicum’s reproduction. Experiment 3 is meant to remove vegetative reproduction from the picture, while leaving seed distribution. If choice (D) is true, then student 2 is right; if choice (A) is true, then dandelions do not rely on seed distribution at all. If you chose Choice (B), you may have misread Student 1 to say that the two methods together account for 50% production.
18. J Student 2’s hypothesis is that seed distribution is the main reproductive strategy for Taraxicum, while some incidental vegetative reproduction is possible. For this to be true, the control field (experiment 1) would have the largest population, followed closely by the field in experiment 3 (which has no vegetative reproduction), with only very few dandelions in field 2 (which eliminated seed distribution). Choice (F) describes a scenario where vegetative reproduction alone resulted in more dandelions than seed distribution and vegetative reproduction together, which would be mysterious. Choice (H), similarly, claims that the control experiment would have fewer offspring than the experiment which eliminated vegetative reproduction. If the results in Choice (G) were obtained, then removing dandelion’s seed distribution and vegetative reproduction would have roughly equal effects, which would mean student 1 was correct.
19. A Based on the information in Table 2, you know that plain, unprepared cake mix Z has a water content of 10.1%. The question asks what the water content is after water is added, so the amount of water in the sample must have increased. Therefore, you know that the answer must be greater than 10.1%, so choice (A) is the only possible correct answer.
20. G The first thing you should do is locate the study and the data that this question asks about: Study 1 and the column in Table 1 refer to maximum immovable speed without cornmeal. Once you have done this, go through each answer to select the best one. Choice (F) cannot be right because none of the cake mixes has a maximum immovable speed of 0 m/s. Choices (H) and (J) cannot be correct because none of the cake mixes has a maximum immovable speed two or three times larger than any other. Choice (G), therefore, must be correct: even though the maximum immovable speeds aren’t exactly the same, they are approximately the same.
21. D This question asks you to compare the two columns of data for each cake mix in Table 1. Mix X is exactly 3.5 times greater, Mix Y is just under 3 times as great, and Mix Z is over 3 times greater. Among the answers, choice (D) is the best answer.
22. H Look at Study 1, the column for maximum immovable speed with cornmeal, and the speeds for cake mixes Y and Z: 0.38 m/s and 0.36 m/s, respectively. If you mix equal amounts of cake mixes Y and Z, then you would expect the maximum immovable speed for the mixture to be somewhere between the maximum immovable speeds for the individual mixes, so the answer is choice (H), between 0.36 m/s and 0.38 m/s.
23. B In short, this question asks whether increasing the amount of gelatin in a cake mix means increasing the water content. Look at Table 2 and you can see that as gelatin content increases, water content increases. Now look at your answer choices: You want a choice that says yes because the hypothesis is supported by Study 3, and you want the reason to be that as gelatin content increases, water content increases. The only correct choice is choice (B). Be sure to pick the choice that is totally right: Choices (A) and (D) are only half right!
24. H Each of the three components of cake mix Q falls between the comparable components of cake mixes Y and Z. Therefore, the water content of cake mix Q should likewise fall between that of cakes mixes Y and Z, specifically 6.1 to 10.1. Choice (H) is correct.
25. B In Experiment 2, all of the metal resistors were gold and 100 m in length, eliminating choices (A), (C), and (D). The description of the experiment and data in Table 2 indicates that cross-sectional area was varied across trials.
26. H The results of Experiment 3 shown in Table 3 indicate that as ρ increases, resistance increases and current decreases. Since gold has the lowest ρ value, it should be the best conductor and the third item in your list, eliminating choices (F), (G), and (J).
27. A According to the passage, electrons are negatively charged and return to the positive battery terminal. This explains why each answer choice is negative. The passage introduction states current is charge per unit time measured in coulombs/second, and that the magnitude of this current was 1.0 × 10−3 coulombs every second for the first trial of each experiment. Therefore, a charge of −1.0 × 10−3 coulombs returned to the positive battery terminal each second the switch was closed.
28. H The results in Table 1 indicate that current increases with decreasing length of the metal resistor, making choice (F) the least favorable by this factor. The results in Table 2 indicate that current increases with increasing cross-sectional area of the metal resistor, making choice (H) the most favorable by this factor and eliminating choice (J). The results in Table 3 indicate that gold has the highest conductance with all other factors being equal, again making choice (H) the most favorable, and eliminating choices (F) and (G).
29. D In Experiment 1, only the length of the metal resistor was varied, eliminating choices (A) and (B). The results in Table 1 indicate that as the length of the metal resistor decreased, current increased and resistance decreased, eliminating choice (C).
30. J The passage states that when the circuit was closed, electrons flowed away from the negative battery terminal, through the circuit, and back to the positive battery terminal. This describes flow in one direction only, eliminating choices (F) and (G). In Figure 1, the negative battery terminal is on the left and the positive battery terminal is on the right. The only way that electrons could flow from negative to positive while passing through the circuit is to go counter-clockwise, eliminating choice (H).
31. A You can see in Figure 2 that increasing pressure increases the rate of reaction, and you can see in Figure 3 that increasing temperature increases the rate of reaction. The scientist’s claim is true based on these two things so the answer is choice (A). Be careful of choices like choice (B), which are only half right. Choice (B) says that the scientist is right, but gives an incorrect reason.
32. H To solve this problem, compare the reaction rates for each answer choice, which shows that choice (H) is the only correct answer. Be sure to look at the y-axis in problems like these to be sure both charts are measuring the same thing.
33. B This is a deceptive question—you don’t need to convert mole/L back to number of molecules, and you don’t need to use 6.02 × 1023 at all! Because the units for the amount of Compound Y and Compound Z are the same, you can just divide the amount of Compound Y by the amount of Compound Z, [1 mole/L]/[2 mole/L] to get the correct answer, choice (B).
34. H You can eliminate choices (F) and (G) right away because they talk about changing temperature, and nothing in the passage or Figures 1−3 tells you that changing the amount of reactants will change temperature. You know from the passage that the rate of a gaseous reaction is changed by changing temperature, pressure, volume, or amount of reactants. Figure 1 shows that increasing temperature and pressure increases the likelihood of particles running into each other. Figure 2 shows that increasing pressure increases the rate of reaction and Figure 3 shows that increasing temperature increases the rate of reaction. Putting all of this together, you know that increasing the likelihood of a collision increases the rate of reaction. This means that increasing the concentration of reactants increases the rate of reaction, so choice (H) is correct. For the same reason, choice (J) must be incorrect.
35. D The best way to solve this problem is to look at each answer choice and see how the reaction rate changes. Looking at choice (A) and Figure 2, decreasing pressure would cause a decrease in reaction rate so it cannot be the correct answer. Choice (C) and Figure 3 show the same, so that cannot be correct either. Choice (B), increasing pressure from 1 atm to 3 atm, would increase the reaction rate to between 125% and 150% of the reaction rate at 50°C and 1 atm. Looking at Choice (D) and Figure 3, you can see that increasing temperature from 50°C to 100°C will increase the reaction rate to 350% of the reaction rate at 50°C and 1 atm. Since choice (D) represents the biggest increase in reaction rate, it is the correct answer.
36. J Looking at Figure 2, you can see that the maximum velocity post-impact, the second peaks on the graphs, decreases as elasticity decreases. The answer to this question must therefore be the smallest elasticity, or 0.1 Pa., choice (J).
37. A Looking only at the graph in Figure 2 that shows balls with an elasticity of 0.2 Pa., you can see that as weight increases, maximum post-impact velocity increases. This question asks about a ball with a weight of 0.5 kg. The smallest ball in this graph has a weight of 1 kg and a maximum post-impact velocity between 0.50 and 0.75 m/s, so you know your 0.5 kg ball must have a smaller maximum post-impact velocity. Choice (A) is the only choice that satisfies that requirement.
38. J To solve this problem, you should first locate the graph and the line on the graph that matches the ball in the problem. Then, the easiest way to solve this is to draw a line from 1.00 m/s on the y-axis all the way across the graph and count how many times it hits the line that represents the 2.0 kg ball: 4 times, choice (J).
39. C You know from the passage and Figure 2 that drop is the very beginning when time is 0 s and velocity is 0 m/s the first time, impact is when velocity is 0 m/s the second time, and apex is when velocity is 0 m/s the third time. Looking at the graphs in Figure 2, you can see that velocity increases after drop then decreases to impact, then increases after impact, and then finally decreases to the apex, so choice (C) is the only correct answer. This question could be tricky if you forgot that apex refers to vertical height, not maximum velocity. Be careful of partial answers like choices (A), (B), or (D); if the answer says only, be sure that it’s really the only thing that happens.
40. H Look at Figure 2, and find the graph and the line on the 0.8 Pa graph, which matches the 3.0 kg ball. The passage tells you that at impact, the ball has a high velocity, then almost immediately slows to 0 m/s, then almost immediately increases to a high velocity again, so you know to look at the maximum velocity prior to impact. This velocity is about 2.25 m/s, which is less than the elastic limit, 2.75 m/s, of the ball in the question, so the only correct answer is choice (H). Note: Be sure to read the entire answer in questions like this. There are two answers that correctly say No, but only one that gives the correct reason.