1) D) In Figure 2, observe the highest bars on the graphs for Islands B and C, since the highest bars correspond with the highest percentages. In both Island B and Island C, the highest bar corresponds with a beak of 10 mm.
2) J) The text of Study 2 indicates that small seeds are present in larger quantities during wet years. Since 1984 is the only year labeled “wet” in Figure 3, one can assume that small seeds were most abundant during this year.
3) B) G. fuliginosa finches were captured only during Study 1. While Study 1 examined two species of finches on 3 different islands, Study 2 examined only G. fortis finches on one island over a period of 10 years. Answer choice A is incorrect because G. fortis finches were captured in both studies. C and D are incorrect because both studies measured the beak depth of captured finches.
4) J) The answer to this question is not stated explicitly in the passage; instead, you must speculate on why the finches were tagged. Of the answer choices, H makes the most sense; the researches wanted to measure the beak depth of 100 members of each species, so it was important that they did not measure the beak of the same bird more than once.
5) C) According to Figure 3, 1977 was a dry year. The text of Study 2 indicates seeds tend to be larger during dry years. The text right below Figure 1 (as well as common sense) indicates that birds with larger beak depths are better able to eat large seeds, so the bird with the 9.9 mm beak would have been better able to acquire food in 1977.
6) F) Figure 2 shows that Island A contains both G. fuliginosa and G. fortis finches, while only G. fortis finches are found on Island B. According to the graphs, the G. fortis finches on Island B had beak depths ranging from 8.5 to 11.5 mm, while the G. fortis finches on Island A had beak depths ranging from 9.5 to 16.5 mm, so the researcher’s hypothesis was correct.
Passage II7) D) The bottom graph of Figure 3 shows that wet deposition of SO42– was highest in February and lowest in July, so D is the correct answer. If you chose C, you may have done so because the wet deposition of Cl– is lowest in July; however, it is highest in November, not February.
8) G) Rather than attempting to actually average the wet deposition of Cu2+ for each month, simply draw a line of best fit that accommodates the data points. This line would be a horizontal line just above 50 on the y-axis, so G is the best answer. If you answered F, your answer was very close; since this question demands a high degree of precision, you must be extremely careful when drawing your best fit line.
9) A) This statement is true based on the upper graph in Figure 2, which shows that there was more wet deposition of Cl– from November through April than there was from May through October.
10) H) The initial text of the passage defines wet deposition as the process by which ions are deposited on the Earth’s surface via precipitation. Based on this definition, if there was no precipitation for an entire year, there would also have been no wet deposition.
11) C) Figure 4 shows that the urban site experienced the most wet deposition for both ions, followed by the rural site 50 km away, with the lowest wet deposition for the rural site 100 km away.
12) F) The text of Study 1 indicates that a specific spot on a roof at the urban site was used to measure precipitation. No mention is made of this site moving, so we can assume that it stayed in the same place and was thereby held constant. G is incorrect since it is impossible to keep monthly rainfall constant. H and J are incorrect because the amount of wet deposition is a dependent variable in this study and cannot be held constant or directly controlled by the researchers.
Passage III13) B) The percent of high clouds are depicted by the dashed line in Figure 1 and measured on the graph’s lefthand y-axis. The graph shows that in January 1987, there was an average of 13.5% high clouds.
14) H) According to Table 1, every cosmic ray flux increase of 20,000 particles/m2/hr corresponds with an increase in low cloud cover of .3%. Therefore, the cosmic ray flux of 440,000 would have a low cloud cover .3 greater than the low cloud cover at a cosmic ray flux of 420,000: 29.0 + .3 = 29.3%. If you answered J, you were correct in thinking that the percent low cloud cover at a cosmic ray flux of 440,000 had to be higher than any of the percentages in the table, but you neglected to consider how much higher.
15) A) This statement is true. Figure 3 shows that the monthly average low cloud cover, depicted by the dashed line, very closely follows the RCRF (the relative cosmic ray flux), depicted by the solid line. Figure 1 shows that the monthly average high cloud cover, depicted by the dashed line, does not closely follow the RCRF (the relative cosmic ray flux), depicted by the solid line.
16) G) Use the dashed line and the lefthand y-axis to determine the average high cloud cover (Figure 1), middle cloud cover (Figure 2), and low cloud cover (Figure 3) for January 1992. Figure 1 shows that the high cloud cover in January 1992 is just under 13.5%. Figure 2 shows that the middle cloud cover in January 1992 is about 20%. Figure 3 shows that the low cloud cover in January 1992 is about 28%. Answer choice G is the correct answer because it correctly depicts these percentages in a bar graph.
17) D) This question requires you to determine the altitudes of high clouds and low clouds using the information in the passage. The text of the passage indicates that high clouds are found at an altitude of 6.0 to 16.0 km, while low clouds are found at an altitude of 0 to 3.2 km, so you can immediately eliminate answer choices A and C, which do not include the correct altitude for high clouds. To decide between B and D, you must use the information from the question and a little bit of common sense. Since ice crystals existed in the high clouds and water droplets existed in the low clouds, the temperature of the high clouds must have been below freezing (0˚C) and the temperature of the low clouds must have been above freezing. Even if you did now know the temperature at which water freezes, you should have been able to realize that the temperature must be colder where ice existed than where liquid water existed.
Passage IV18) F) Figure 1 indicates that the solution was yellow (depicted by the small dots) at titrant volumes less than 1.00 mL, so F is the correct answer.
19) B) The word “neutral” in the question refers to a pH that is neither acidic nor basic. Neither the text of Experiment 2 nor the graph (Figure 2) discusses pH explicitly, so you must consult the initial text of the passage to figure out how Figure 2 gives you information about the pH of the solution. The last sentence of the passage’s second paragraph indicates that a pH less than 6.0 results in a yellow solution, while a pH greater than 7.0 results in a blue solution. To get this question correct, you must know that a pH of 7.0 is considered neutral. (This is one of the rare science questions – typically 1-3 per test – on which you are expected to use a bit of basic outside knowledge.) The graph in Figure 2 shows that the solution turns blue at a titrant volume of 1.00 mL, indicating that this titrant volume corresponds with a pH of 7.0 and therefore a neutral solution.
20) J) Figure 1 shows that, for titrant volumes above 1.00 mL, as titrant volume increases, conductivity increases. Therefore, as the titrant volume increases beyond 2.00 mL, one can assume that the conductivity would increase beyond 3.8 kS/cm (the conductivity at 2.00 mL). Basically, this question is just asking you to extend the line of the graph beyond the titrant volumes that are depicted on the x-axis, and the graph would simply continue with its positive slope, reaching conductivity values above 3.8 kS/cm.
21) C) Experiment 2 does not explicitly identify the titrant. It does, however, state that the experiment is the same as Experiment 1 except that it features an acetic acid solution instead of an HCl solution. Therefore, you know that acetic acid is the sample solution. If you look at your answer choices, there is only one option with a sample solution of acetic acid, answer choice C, so this is the correct answer. If you wished to make certain of the titrant, you could have found the titrant from Experiment 1, which based on the text of Experiment 2 would be the same titrant used in Experiment 2. The text of Experiment 1 states that NaOH is added to the solution, and the initial text of the entire passage states that a titrant is the solution that is slowly added to a sample solution. Therefore, NaOH was the titrant in Experiment 1, and hence in Experiment 2 as well.
22) J) The text of Experiment 1 states that the probe was used to measure conductivity. This passage never states exactly how the probe does this, so you must speculate using your scientific knowledge and/or a bit of common sense in order to answer this question correctly. If the probe is used to measure conductivity (which according to the initial text of the passage is the ability to conduct an electrical current), it makes sense that the probe would do so by transmitting a current through the solution.
23) A) In Figure 2, a titrant volume of 0.2 mL corresponds with a yellow solution, while a titrant value of 1.8 mL corresponds with a blue solution. To determine how the color of the solution relates to pH, consult the initial text of the passage. The initial text of the passage indicates that the solution is yellow when its pH is below 6.0 and blue when its pH is above 7.0. Therefore, the blue solution produced by 1.8 mL of titrant would have a higher pH than the yellow solution produced by 0.2 mL of titrant, so the claim in the question is false. If you did not realize that the color of the solution corresponded with the pH value, the discussion of solution color in the answer choices could have provided you with a valuable hint.
Passage V24) H) This is a rare science question (typically 1-3 per test) on which you are required to use a bit of outside knowledge. The last sentence of Student 2’s second paragraph states that Algol B came into contact with the original Algol system when their orbits intersected and that due to this encounter Algol B “became part of the Algol system.” Since Student 2 does not state exactly why Algol B became part of the original system, you must use your own knowledge. Gravity is a force of attraction between bodies containing mass. It is the force that governs the motion of celestial bodies in the universe, such as stars, planets, comets, and moons, so H is the correct answer.
25) B) Student 1 states that while this process was occurring, Algol B was a post-MS star. Fact 4 states that most of the energy produced by a post-MS star is the result of the fusion of hydrogen to create helium “in a shell surrounding its center,” so B is the correct answer.
26) G) Student 2 states that Algol B was formed in a different cloud than the original Algol system (which included Algols A and C). Therefore, based on the fact provided in the question, Student 2 would claim that Algols A and C would have similar compositions and Algol B would have a different composition, so G is the correct answer.
27) C) The paragraph directly below Fact 5 states that Algol C has a solar mass of 1.7. This paragraph also defines solar mass as the mass of the Sun, so Algol C has a mass 1.7 times that of the Sun. Therefore, its mass would be 1.7 • 2.0 • 1030 = 3.4 • 1030 kg.
28) G) Fact 3 describes the fusion of hydrogen nuclei, which are made of protons, to form helium. Because this reaction involves the fusion of protons, not electrons, you can immediately eliminate H and J. G is correct over F because like charges do in fact repel, not attract each other. If you did not know that like charges repel, you still could have gotten G over F, since if it were true that like charges attracted each other, high pressure would not be needed to get like-charged protons to fuse together.
29) B) Fact 5 states that more massive stars move more quickly from pre-MS to MS to post-MS star. Since Student 1 states that Algol B had the greatest mass of the 3 stars when it was created, Student 1 would expect Algol B to move the most quickly from pre-MS to MS to post-MS star. Since Student 1 claims that the three stars were created at the same time, Student 1 would therefore claim that Algol B was the first to become an MS star.
30) H) Fact 5 states that more massive stars move more quickly from pre-MS to MS to post-MS star. Therefore, more massive stars spend less time in each of the three evolutionary stages than do less massive stars. Because Student 2 states that Algol A was more massive than Algol B, Student 2 would expect Algol A to spend less time than Algol B in each evolutionary stage (pre-MS, MS, and post-MS).
Passage VI31) C) Figure 2 shows that, as the amount of Kr increases, pressure increases. Therefore, as the amount of Kr increases beyond 10 g, one can assume that the pressure would increase beyond 390 torr (the pressure at 10 g of Kr). If you extend the line representing Kr beyond the graph to a point that would correspond with a mass of 13 grams if you were to extend the x-axis, it would correspond with a pressure between 400 and 500 torr, so C is the correct answer. If you answered J, you were correct in thinking that the pressure would be higher than the pressure at 10 g of Kr, but you neglected to consider how much higher.
32) G) A 5 L vessel has a volume in between that of the 3 L vessel in Figure 1 and the 6 L vessel in Figure 2, so 7 g of CO2 would exert a pressure between its pressure in the 3 L vessel and the 6 L vessel. Figure 1 shows that 7 g of CO2 has a pressure of about 1,000 torr in the 3 L vessel, and Figure 2 shows that 7 g of CO2 has a pressure of about 500 torr in the 6 L vessel, so its pressure in a 5 L vessel would be in between these two pressures.
33) A) As is stated in Step 5, all data was taken from gas at a temperature of 22˚C. At any mass of O2 added, the pressure in the 6 L vessel of Figure 2 is half that in the 3 L vessel of Figure 1. For instance, Figure 2 shows that 2 g of O2 has a pressure of about 200 torr in the 6 L vessel, while Figure 1 shows that 2 g of O2 has a pressure of about 400 torr in the 3L vessel. This makes sense, since one would expect that a given amount of gas would exert more pressure in a smaller vessel than in a larger one. If you answered B, you may have understood the relationship perfectly but gotten things backwards when answering the question.
34) J) Figures 1 and 2 show that O2 always exerts a greater pressure than an equal mass of CO2, so you can eliminate F and G, since they are inconsistent with the data provided. To decide between H and J, you need to use a bit of common sense or a bit of scientific knowledge. J simply makes more sense than H because it stands to reason that more molecules would exert more pressure. Also, it stands to reason that molecules with the chemical form O2 would be smaller than those with the chemical form CO2, since CO2 just adds a C to the O2. If the O2 molecules are smaller, then there would be more O2 molecules in a gram.
35) A) This is a rare science question (typically 1-3 per test) on which you are required to use a bit of outside knowledge. Specifically, you need to know that temperature and pressure are directly proportional, meaning that as temperature increases, pressure increases. If you did not know this, you may have been able to figure it out using a bit of real world common sense. As things heat up, they expand, and if gas expands within a closed container, it would exert more pressure. The temperature of the original experiment was 22˚C, so at a lower temperature of 14˚C, there would have been less pressure.
Passage VII36) G) The threshold of hearing is defined in the text as “the minimum intensity at each sound frequency that can be heard by humans.” Since the lowest frequency at which there is a threshold of hearing depicted in the figure is about 2 • 101, this represents the lowest frequency that humans can hear at any intensity level. 2 • 101 = 2 • 10 = 20. If you answered J, you found the highest frequency that humans can hear.
37) A) Higher frequencies are represented by higher values on the y-axis. A loss of high frequency hearing would mean that a person can no long hear some of the high frequencies they were once able to, so the high end of their threshold of hearing would be lowered, as is shown in answer choice A.
38) F) This question is testing you on two things, so you should treat it as two separate questions. First decide whether sound passing through air or water has a higher intensity. According to the figure, sound has a higher intensity in water than in air, since the dotted line representing water is always found at a slightly higher frequency (further to the right on the graph) than the corresponding dashed line (representing air). Now determine whether an S of 100% or 10-8% would correspond with a higher intensity. According to the figure, an S of 100% corresponds with a higher intensity since it is further to the right on the graph. Therefore, the answer is F.
39) C) The fact that the line representing the threshold of pain does not extend to a frequency of 105 Hz indicates that there is no intensity of sound at a frequency of 105 Hz that is capable of causing pain in humans. The fact that the threshold of hearing does not extend to a frequency of 105 Hz indicates that this is because humans cannot even hear sounds at such a high frequency.
40) J) The figure shows that S does not depend upon frequency, since the dashed and dotted lines show that S remains constant across a variety of frequencies. It is probably easiest to see this if you rotate the page 90˚ counterclockwise. The graph is initially difficult to read because it places frequency, the independent variable, on the y-axis, and we are used to seeing the independent variable on the x-axis. Rotating the graph in the manner described above puts the independent variable back on the x-axis, thereby making the graph easier to read. When looking at the graph this way, it is apparent that the flat horizontal dashed and dotted lines indicate that S does not depend on frequency.
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