ACT Past Papers with Answers

Paper 25 · ACT December 2018 · Form B05 · Science Test

The flask was capped, and the contents were shaken lor a particular time at a particular temperature.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Suppose an additional trial had been conducted at the same temperature at which Trials 1-5 were conducted. If [X] and [Y] had each been 2.00 mol/L, [Z] would most likely have been:

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on Tables l and 2, how many trials were most likely conducted at temperatures above the freezing point of water? ACT-805 39 GO ON TO THE NEXT PAGE. 40000000004 Passage 11 Experiment 2 Each of 200 large (23.3 mm wide) juvenile C. sapidus was prepared as in Experiment l. After being prepared, the C. sapidiis were equally divided into 4groups (Groups 5-8). The procedures for Groups 1-4 were repeated for Groups 5-8 (see Table 2). Juvenile Calliizectes sapidus (a species of crab) typi- cally hide among Zostera marina (a species of sea grass) to avoid detection by aquatic predators. Two experiments examined how Z. marina population density and C. sapidus bo.dy size affect the ability of juvenile C. sapidus to avoid predation in a particular coastal area. Experiment I Each of 200 small (4.5 mm wide) juvenile C. sapidus was prepared for testing as follows: One end of a 1.0 m piece of nylon fishing Iine was tied around the middle of the C. sapidus, and the knot was then securely attached to the C. sapidus with eyanoacrylale (a strong waterproof glue). The other end of the line was tied to its own metal rod. After being prepared, the 200 C. sapidus were equally divided into 4 oroups (Groups 1-4). Table 2 arina population density (shootslm ) Number of surviving Group 380 760 1,600 19 37 Each rod associated with a Group I C. sapidus was randomly placed within a 100 m plot having a Z. marina population density of O shoots/m (sand only, no plants). Twenty-four hours later, the number of surviving Group I sapidus was determined. Tables adapted from J. L. Schulman, "Habilat Complexity as a Determinant of Juvenlle Blue Crab Survival." 01996 by The Col- lege of William and Mary. The procedure for Group I was repeated for Groups2-4, except that Group2, Group3, and GTOUP 4 were placed in plots having a Z. marina popu- lation density of 3 80 shoots/m 760 shootslm and 1,600 shoots/m respectively. The results are shown in Table l. Table I marina population density (shoots/m') Nutnber of surviving sapidus 26 31 24 12

Paper 25 · ACT December 2018 · Form B05 · Science Test

Consider the temperature data in Table 2. Did the reac- tion more likely reach equilibriutn faster in Trial 6 or Trial 10 ? greater at a lower temperature. greater at a higher temperature. greater at a lower temperature. greater at a higher temperature.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Assume that substances X, Y, and Z were gases throughout each trial. The boiling points of X, Y, and Z must each be: above 40°C.

Paper 25 · ACT December 2018 · Form B05 · Science Test

According to the iesiilts of Experitnent 2, as the nzarina population density increased from O shootslm through 1,600 shoots/m , the number of surviving large juvenile C. sapidus: incieased only. increased, then deci'eased. Group 380 760 1,600 ACT-B05 40 GO ON TO THE NEXT PAGE. 40000000004

Paper 25 · ACT December 2018 · Form B05 · Science Test

Siippose that in Experiment 2 a group of C. sapidus had been placed in a Z. mai.ina plot having a popula- tion density of 562 shoots/m The number of surviving

Paper 25 · ACT December 2018 · Form B05 · Science Test

To consider whether the body size of juvenile sapidus affects their ability to avoid predation, the Tesults for which of the following 2 groups should be compared?

Paper 25 · ACT December 2018 · Form B05 · Science Test

Which of the nylon line and the cyanoacrylate, if either, functioned to prevent the C. sapidus from escaping. ACT-B05 41 GO ON TO THE NEXT PAGE. 40000000004 Passage 111 Experimeni 2 Student 2 placed the horn and 2 rnicrophones-m and M2-in a straight line such that the mi¢rophotLes were separated by a distance y. She then connected a timer to both microphones (see Figure 2). Two students were each asked to. experimentally determine the speed of sound in air, s,. For the attnospheric conditions in which their experiments were performed. the correct value of s. was 347.1 mlsec. Experiment I Student I placed a horn, which could emit a short puIse of sound waves, at a distance x, in m, from a tall cliff (see Figure l). horn horn cliff timer Figure 2 Figure I She set the tirner to start when the first microphone detected a sound pulse and to stop when the second micro- phone detected a sound pulse. She recorded this time inter- val, £St2, and then calculated s, using the equation He started a stopwatch when he heard the horn emit a sound pulse and stopped the stopwatch when he heard an echo return from the cliff. He recorded this time interval, ti, in sec, and then calculated s. using the equation Student I performed 3 trials at each of 3 values of x, veraging the s, results at each value of x (see Table l). Student 2 performed 3 trials at each of 3 values of y, averaging the s. results at each value of y (see Table 2). Table I Table 2 t, (sec) Average s. (mlsec) t, (sec) Average s. (mlsec) Trial (m) (mlsec) Trial (m) (mlsec) 290 270 290 10 347 350 345 200 280 50 347 12 330 320 330 13 14 15 346 347 344 500 330 75 346 340 350 340 16 17 18

Paper 25 · ACT December 2018 · Form B05 · Science Test

Which of C. sapidus or Z. marina would occupy the higher trophic level in a food chain? marina is a consumer. pnarina is a producer. sapidus is a consumer. sapidus is a producer.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Which of the following statements describes a differ- ence between the cells of C. sapidus and the cells of marina ? The cells of C. sapidus: a nucleus. a nucleus. lack a cell wall. have a cell wall.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Which of the following statements comparing an aspect of the design of the 2 experiments is accurate? 3 times as wide as the C. sapidus in Experiment l. ment I was less than the length of time needed to perform Experirnent 2. in both Experiments l and 2. 50 C. sapidus per group.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Suppose Student 2 had conducted a trial in which ￿t was 0.191 sec. The value of y in this trial would most likely have been:

Paper 25 · ACT December 2018 · Form B05 · Science Test

Which of the objects listed below were used in both Experiment l and Experiment 2 7 I. A horn II. A microphone III. A timing device

Paper 25 · ACT December 2018 · Form B05 · Science Test

Student I plotted the At data versus x and then graphed a line of best fit to the data, as shown below.

Paper 25 · ACT December 2018 · Form B05 · Science Test

In general, the farther a sound wave travels, the more it attenuates (reduces in volume), making detection more difficult. In which experiment was attenuation of greater concern? distances in Experiment I than in Experiment 2. distances in Experiment I than in Experiment 2. distances in Experiment 2 than in Experiment l. distances in Experiment 2 than in Experiment l.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on the results of both experiments, regardless of the apparatus or procedure used. on average,. as a sound pulse's travel distance increased, the recorded time interval: increased only. remained constant. varied, but with no general trend.

Paper 25 · ACT December 2018 · Form B05 · Science Test

In calculating the value of s., Student I needed to include a factor of 2 in the numerator. However, Stu- dent 2 did not incliide this factor because: in Experiment I, sound had to travel over the dis- tance x twice, whereas in Experiment 2, sound had to travel over the distance y once. in Experiment I, sound had to travel over the dis- tanoe x once, whereas in Experiment 2, sound had to travel over the distance y twice. every value of x was at least 2 times every value of y. every value of x was less than 2 times every value ofy. gap 200 400 600 800 1,000 x(rn) The gap shown near the bottom of the vertical axis suggests that even if the value of x had been O m, the value of Llti would not have been O sec, as one might expect. Based on the description of Experiment I, this gap was most likely the result of: errors due to human reaction time in starting andlor stopping Èhe stopwatch. the value of Ati. pulses. the horn being accidentally placed closer to the cliff than intended.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Overall, which student obtained the more accurate set of s. data? (Note: This question is concerned with the accuracy of a data set, NOT the precision of a data set.) spread over a larger range than are the s, values in Table 2. closer, on average, to the correct value of s. than are the values of s, in Table 2. spread over a smaller range than are the s. values in Table l. closer, on average, to the correct value of s. than are the values of s. in Table l. ACT-B05 43 GO ON TO THE NEXT PAGE. 40000000004 Passage IV Key As viewed from Earth. Jupiter's 4 largest moons-the Galilean rnoons-appear to move relative to Jupiter along an east-west line. The farther fi'om Jupiter a moon appears to be, the greater its angular separation, 8. Figure I shows these moons, positions at midnight on a certain day in June 2012 as viewed from Earth (top section) and as viewed from above Jupiter's north pole (bottom section). Europa Ganymede Callisto 30E 20E IOE Key Jupiter low Europa Ganymede ¢ Callisto 20W 30W 20 21 22 23 24 25 26 27 28 June 2012 date 8-4 noon on June 20 west (W) view from Earth east (E) 'arcseconds Figure 2 Figure 2 adapted from Sylvain Rondi, Jupiter Version 2.. Programme gratuit de calcbl d'ephémerides des satellites galiléens de Jupiter. @2009 by Sylvain Rondi. view from above Jupiter's north pole

Paper 25 · ACT December 2018 · Form B05 · Science Test

The number of orbits per week is a rneasiire of orbilal freque71cy. Based on Figure 2, which Galilean moon has the highest orbital frequency. Note: The figure is not drawn to scale. In the top section, only e for Ganymede is shown. Figure I

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on Figure 2, Callisto completes l orbit around Jupiter in approximately how many days? Figure 2 shows O for each moon from midnight on June 20, 2012, to midnight on June 28, 2012. For each moon, I wave cycle represents l orbit around Jupiter. (The time to complete l orbit is the time between any 2 consecu- tive peaks or any 2 consecutive troughs.) ACT-B05 44 GO ON TO THE NEXT PAGE. 0000000004

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on Figure 2, as viewed from Earth at midnight on June 27, which Galilean moon appeared farthest west of Jupiter?

Paper 25 · ACT December 2018 · Form B05 · Science Test

According to Figure 2. at noon on June 23, 8 for Ganymede was approximately: 14 arcsec west. ACT-B05 45 GO ON TO THE NEXT PAGE. 40000000004 Passage V

Paper 25 · ACT December 2018 · Form B05 · Science Test

According to Figure 2, at which of the following titnes in 2012 was the angular separation for Callisto closest to 20 arcsec east?

Paper 25 · ACT December 2018 · Form B05 · Science Test

Thebe is one of Jupiter's smaller moons. Thebe com- pletes l orbit of Jupiter in approximately 0.7 days. Based on Figure 2, regardless of direction. the maxi- mum value of 8 for Thebe is most likely:

Paper 25 · ACT December 2018 · Form B05 · Science Test

Both scientists would be likely to agree that modern birds: evolved from crocodiles. evolved from nonavian dinosaurs. Two scientists discussed the evolution of 3 animal groups". crocodiles. nondvian dinosaurs (all dinosaurs except modern birds), and modern birds. The scientists characterized the animals in each group by their resting melabolic rate, RMR (the rate at which an animal uses caloric energy when at rest), and according to whether they were poikilothermic (had body temperatures that varied significantly) or homeothermic (had body temperatures that remained relatively constant).

Paper 25 · ACT December 2018 · Form B05 · Science Test

Would Scientist 2 more likely expect to find bones that have growth rings in a crocodile or in a modern bird? Indicate a high RMR. indicate a low RMR. rings indicate a high RMR. rings indicate a low RMR. Scientist I Crocodiles and nonavian dinosaurs evolved from a common ancestor that had a low RMR and was poikilother- mic. Crocodiles retained these traits. However, nonavian dinosaurs had a high RMR and were homeothermic. Modern birds inherited these traits when the group first evolved from nonavian dinosaurs. and both traits have been retained by all modern birds. The bones of nonavian dinosaurs had a porous struc- ture, no growth rings, and extensive vascularization (chan- nels containing blood vessels). Extensive vascularization, which is also seen in modern birds, is evidence of an effi- cient circulatory system and a high RMR. Efficient circula- tion distributes heat evenly throughout the body. Thus, nonavian dinosaurs were homeotherrnic and could have survived in cold climates.

Paper 25 · ACT December 2018 · Form B05 · Science Test

In which of the 3 animal groups would Scientist I most likely expect to find animals having bones with exten- sive vascularization? Scientist 2 Crocodiles and nonavian dinosaurs evolved from a common ancestor that had a low RMR and was poikilother- mic. Crocodiles retained these traits. Nonavian dinosaurs also had a low RMR and were poikilothermic. In contrast, modern birds have had a high RMR ever since the group first evolved from nonavian dinosaurs. Modern birds only became homeothertnic after nonavian dinosaurs went extinct 65 tnillion years ago.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Consider a crocodile and a modern bird, each having the same body tnass. Based on Scientist 2's discussion, which animal would more likely produce CO2 at a faster rate: the crocodile or the modern bird? have a high RMR and that crocodiIes have a low RMR. have a low RMR and that crocodiles have a high RMR. have a high RMR and that crocodiles have a low RMR. have a low RMR and that crocodiles have a high RMR. The bones of nonavian dinosaurs were dense and had growth rings. Each of these features indicates a low RMR, and neither is seen in the bones of modern birds. Nonavian dinosaur bones also lacked extensive vascularization, sug- gesting an inefficient circulatory system that would not have been able to distribute heat evenly throughout the body. Thus, nonavian dinosaurs were poikilothermic and could not have survived in cold climates. ACf-B05 46 GO ON TO THE NEXT PAGE. 40000000004

Paper 25 · ACT December 2018 · Form B05 · Science Test

Consider a rock formation containing many bones and nesting sites from nonavian dinosaurs as well as indi- cators of a stable but cold climate. Suppose it were dis- covered that both the nonavian dinosaur fossils and climate indicators have been present together through- out the entire 50-million-year history of the formation. This discovery would support the viewpoint(s) of which of the scientists. if either?

Paper 25 · ACT December 2018 · Form B05 · Science Test

Scientist I characterized the animals in which of the 3 groups as poikilothermic? ACT-B05 47 GO ON TO THE NEXT PAGE. 40000000004 Passage Vl Key Trial 5, 30°C Trial 6, 40°C Trial 7. 50°C Biosorption is a process that uses certain biological materials (bioinoss) to remove metaI ions from a solution. Three experimeiits were done to study the removal of the chromium ion Cr from aqueous solutions using a biomass composed of crushed shells of the fruit Litchi chinensis. In each trial in each experiment, the following steps were performed: 100 l. In a flask, 1.0 g. of fresh biomass was added to l 00 ML of an aqueous solution having a particular pH and Cr concentration. 60 50 40 30

Paper 25 · ACT December 2018 · Form B05 · Science Test

Suppose it.were confirmed that the common ancestor of nonavian dinosaurs and crocodiles had a low RMR. This finding would be consistent with the viewpoint(s) of which of the scientists, if either?

Paper 25 · ACT December 2018 · Form B05 · Science Test

According to the results of Experiment 2, as the tem- perature was increased from 30°C to 50°C from ti'ial to trial, the time needed to remove all the Cr ' from the solution: increased only. lllcreased and then decreased.

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on Figure I, if a satnple from Trial 3 had been analyzed at 95 min, the percentage of Cr in the solution would most Iikely have been: 6+ remaining

Paper 25 · ACT December 2018 · Form B05 · Science Test

According to the results of ExperiTnent I, as the pH was increased frorn trial to trial, the time needed to remove 90% of the Cr ' from the solution: increased only. increased and then decE'eased.

Paper 25 · ACT December 2018 · Form B05 · Science Test

In how many trials in Experiments 1-3 did the percent- age of Cr ' remaining in solution reach Olo in less than Ihr? 200 mg x 5(Yo 200 mg 1,900 ML x510 200 mg x30% IOOMLX 200 mg 1,000 n]L x 3010 END OF TEST 4 STOP! DO NOT RETURN TO ANY OTHER TEST. ACT.B05 49 Scoring Keys for Form B05 Use the scoring key for each test to score your answer document for the multiple-choice tests. Mark a "1" in the blank for each question you answered correctly. Add up the numbers in each reporting category and enter the total number correct for each reporting category in the blanks provided. Also enter the total number correct for each test in the blanks provided. The total number correct for each test is the sum of the number correct in each reporting category. Test 1: English-scoring Key Reporting Category, pow Reporting Category. pow KLA CSE 'Reporting Categories POW - Production of Writing KLA - Knowledge of Language CSE Conventions of Standard English Key KLA CSE Key

Paper 25 · ACT December 2018 · Form B05 · Science Test

Biosoiption using L. chiiiensis involves the one-time transfer of 3 electrons from the biomass to each chromium ion. Based on inforrnation in the passage,. the resulting charge on each chromium ion after it gains 3 electrons is:

Paper 25 · ACT December 2018 · Form B05 · Science Test

How many of the trials performed in Experiment I were done at an acidic pH ?

Paper 25 · ACT December 2018 · Form B05 · Science Test

Based on Figure 3, wbich of the following expressions best approximates the mass of Cr ' remaining in the 200 mg/L solution at 5 min ?