Study on the flyACTChallenge · ScienceScience

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ACT Science challenge

  • 40 questions
  • 40 minutes
  • Harder than the exam
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The Science section of the ACT Science challenge — 40 questions harder than the exam. Mark your answers, then press Finish at the bottom to see how many you got right.

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Questions 1–5

Researchers studied when desert isopods seal burrow entrances with soil plugs to reduce water loss. They defined sealing onset as the first day a plug fully closed the entrance. All isopods were placed in chambers on Day 0 with entrances open. Table 1 summarizes light and cold schedules. Sealing results were withheld from three scientists, who proposed the explanations below. All other conditions were identical, and observations were made daily using the same procedure. Soil water exceeded 12% on Days 1–3 in all chambers. Table 2 shows soil water content on Days 4–9; after Day 9 water content remained below 12% in all chambers unless a question specifies rewetting.

Table 1 Light and cold schedules

TreatmentP: first day withphotoperiod ≤ 12 h(day)C: day thirdconsecutive nightat or below 8 °Cended (day)A128B125C812D85

Table 2 Soil water content (%) on Days 4–9

TreatmentDay 4 (%)Day 5 (%)Day 6 (%)Day 7 (%)Day 8 (%)Day 9 (%)A9999914B1499999C9149999D9914999

Scientist 1 Day length alone sets sealing onset. On the first day with at most 12 h of light, an irreversible internal process begins; a plug appears exactly 4 days later. Neither cold nights nor soil moisture changes this delay or initiates the process independently. Thus onset occurs on Day P + 4. Reports of differences between sites reflect inconsistent scoring of partial plugs, rather than different triggers. With standardized scoring, identical isopods with identical photoperiod schedules seal together, even when temperature and moisture schedules differ.

Scientist 2 Cool nights initiate the process. Three consecutive nights at or below 8°C are required, and a plug appears exactly 4 days after the third qualifying night ends: Day C + 4. A warmer night before the third qualifying night resets the count. Once three qualifying nights have occurred consecutively, subsequent warming cannot cancel or delay sealing onset. Day length and soil moisture have no direct effect. Their seasonal association with cooling can make either appear responsible. Consequently, shortening days without the required cool-night sequence leaves entrances open.

Scientist 3 Day length establishes the baseline onset at Day P + 4, but drought can advance it. After five consecutive days with soil water content at or below 12%, drought-induced sealing appears 2 days later. This advance requires soil to remain at or below 12% throughout those additional days. Rewetting above 12% before visible sealing cancels the drought pathway, leaving the day-length pathway intact. Rewetting after sealing appears does not reopen entrances. Without rewetting, onset is the earlier of Day P + 4 and Day D + 2, where D is the day the fifth consecutive dry day ends. Cool nights have no direct effect.

Question 1

Suppose Treatment D experiences an additional 3-day cold spell on Days 9–11, with light and moisture schedules otherwise unchanged. According to Scientist 1, sealing onset will occur on which day?

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Answer: C — Day 12

Read Scientist 1 and Table 1. Scientist 1 states day length alone sets onset at Day P plus 4 days and that neither cold nor moisture changes delay or starts process. Table 1 gives Treatment D first short day Day 8, so baseline Day 12. Added Days 9–11 cold leaves prediction unchanged. Day 9 mistakes cool-night onset, Day 11 mistakes drought onset for different treatment, and Day 16 mistakes day-length onset for different photoperiod treatments.

Question 2

In Treatment D, on which day does the fifth consecutive day with water at or below 12% end?

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Answer: D — Day 11

Count Table 2 water content under Scientist 3 definition plus both once-stated moisture sentences. Soil exceeded 12 percent Days 1–3, so counting starts Day 4. Treatment D shows Days 4–5 dry, Day 6 at 14 percent breaks, then Days 7–9 dry for three, so fifth requires Days 10–11 below 12 percent via after-Day-9 remains sentence, ending Day 11. Day 9 stops at last observed day ignoring remains sentence, Day 10 is off by one from miscounting break, and Day 8 assumes no break.

Question 3

Which set of predicted onset days agrees with Scientist 3 under the original schedules for Treatments A–D, respectively?

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Answer: C — A 16, B 11, C 12, D 12

Translate Table 1 baselines through Scientist 3 earlier-of rule after deriving D from Table 2 and checking remains requirement. Treatment A fifth dry ends Day 8 but Day 9 at 14 percent cancels to baseline 16, Treatment B fifth ends Day 9 for drought 11 beating baseline 16, Treatment C fifth ends Day 10 for drought 12 tying baseline 12, and Treatment D fifth ends Day 11 for drought 13 losing to baseline 12. First set restates day-length baseline for wrong viewpoint, second set ignores cancellation and min rule, and fourth set restates cool-night onsets for wrong viewpoint.

Question 4

In Treatment B, sealing onset is observed on Day 11. Which assessment follows from the scientists’ explanations?

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Answer: C — The result agrees with Scientist 3 and weakens Scientists 1 and 2.

Compare Day 11 to computed onsets for Treatment B: day-length 16, cool-night 9, and drought 11. Only drought matches, supporting baseline-plus-drought and contradicting other triggers. First choice cites true Scientist 1 principle but wrong day, second cites cool-night rule but predicts Day 9, and fourth confuses dry-day milestone Day 9 with onset Day 11 by omitting 2-day delay.

Question 5

A researcher claims that Treatment A alone cannot distinguish Scientist 1 from Scientist 3 under the original schedules. Is this claim correct, and why?

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Answer: A — Yes; both predict Day 16.

Check Treatment A baseline Day 16 against drought pathway with Table 2. Fifth dry ends Day 8 for drought onset Day 10, but Day 9 water at 14 percent exceeds 12 percent during required additional days and cancels drought to baseline Day 16. Thus both converge. Both-Day-12 choice mistakes cool-night onset for these viewpoints, 16-versus-10 choice assumes monotonic drying and misses reversal spike, and reversed numbers swap photoperiod treatments.

Questions 6–11

A researcher studied removal of dissolved copper (Cu2+) from water by sand grains coated with chitosan, a binding material. In each trial, 2.00 g of sand was added to 100 mL of copper solution, shaken for 60 min at the test temperature, and filtered. The mass of copper remaining in the filtrate was measured. The starting copper mass in 100 mL is listed as the initial mass. Each table entry is the mean of three trials; replicates differed by less than 0.05 mg. A matched blank using uncoated sand was run for each condition to account for copper lost to the container or by precipitation. Coating-attributed removal was defined as the blank remaining mass minus the coated-sand remaining mass for the matched condition.

Study 1 The researcher varied the initial solution pH while holding the test temperature at 25°C and the initial copper mass at 2.00 mg. Before mixing, sand samples were held at 25°C for 2 hr. The same acid and base additives were used at every pH, and pH was checked after mixing.

Table 1 Effect of pH on copper remaining

Solution pHCopper remainingwith coated sand(mg)Copper remainingwith uncoatedsand (mg)31.551.9041.151.9050.751.9060.501.9070.651.8081.051.60

Study 2 Coated-sand samples without copper were heated at a pretreatment temperature for 2 hr, then tested by one of two procedures. For the direct procedure, sand was immediately added to copper solution at the pretreatment temperature and shaken at that temperature. For the recovery procedure, sand was first cooled to 25°C for 2 hr and then tested at 25°C. Separate samples were used for the two procedures. Every test used pH 6 solution with 2.00 mg initial copper. Blanks with uncoated sand underwent the corresponding procedure and timing.

Table 2 Effect of heating on copper remaining

Pretreatmenttemperature (°C)Direct coatedremaining (mg)Direct blankremaining (mg)Recovery coatedremaining (mg)Recovery blankremaining (mg)150.801.900.501.90250.501.900.501.90400.301.850.551.90550.701.701.001.90701.601.601.501.90

Study 3 The researcher varied the initial copper mass at pH 6 and 25°C. Fresh sand had been held at 25°C for 2 hr. Heated sand had undergone the 55°C pretreatment and the recovery timing from Study 2. Sand mass remained 2.00 g as initial mass changed. Matched blanks for the two sand treatments gave identical remaining masses, so one blank column is listed.

Table 3 Effect of initial copper mass on copper remaining

Initial copper mass(mg)Fresh coatedremaining (mg)Heated coatedremaining (mg)Blank remaining(mg)1.000.150.400.952.000.501.001.904.001.602.403.808.004.405.207.60

Question 6

At pH 5 in Study 1, what was the coating-attributed removal?

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Answer: B — 1.15 mg

Read Table 1 pH 5 row and the introduction definition requiring blank minus coated. Blank 1.90 mg minus coated 0.75 mg gives 1.15 mg. The 0.75 mg value reports the coated remaining without subtracting the blank. The 1.25 mg value subtracts from the 2.00 mg starting mass and ignores the blank loss to container. The 1.90 mg value reports the blank remaining as if it were removal.

Question 7

Which description correctly compares the changes in coating-attributed removal for the two procedures as pretreatment temperature increased from 25°C to 40°C and then from 40°C to 55°C?

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Answer: C — The direct procedure increased and then decreased, while the recovery procedure decreased over both intervals.

Table 2 cannot be read directly because the passage defines removal as blank minus coated. Subtracting gives direct removal of 1.40, 1.55 and 1.00 mg at 25, 40 and 55 degrees, showing increase then decrease. Recovery removal is 1.40, 1.35 and 0.90 mg, showing decrease in both intervals. The both-increase option ignores recovery decline. The direct-both-increase option ignores the drop from 40 to 55. The reversed option swaps the procedures.

Question 8

If the direct procedure in Study 2 were extended to 85°C, the coating-attributed removal would most likely be:

Show answer

Answer: A — 0.00 mg, about equal to the 70°C result

Direct removal is blank minus coated, giving 1.00 mg at 55 degrees and 0.00 mg at 70 degrees where coated and blank both read 1.60 mg. The coating is therefore destroyed near 70 degrees and removal cannot become negative because coated sand then behaves like uncoated sand. The linear-continuation value assumes the prior drop continues past zero. The 55-degree value assumes a plateau ignoring the collapse. The 40-degree value uses the maximum from the wrong temperature.

Question 9

Students extend Study 3 by testing recovered, 55°C-pretreated sand at pH 5 and 4.00 mg initial copper. Each proposed blank uses uncoated sand that received the 55°C pretreatment. Which blank would allow calculation of coating-attributed removal?

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Answer: A — A blank at pH 5 tested at 25°C after cooling for 2 hr with 4.00 mg initial copper.

Coating-attributed removal requires blank minus coated under matched chemistry and timing. Study 1 shows blank remaining changes with pH, Study 3 shows it scales with initial mass, and Study 2 defines recovery as cooling to 25°C for 2 hr before testing at 25°C. Only the pH 5, 25°C recovery, 4.00 mg blank matches the new test. The pH 7 blank mismatches acidity. The immediate 55°C blank uses direct timing rather than recovery timing. The 2.00 mg blank mismatches starting mass.

Question 10

A new Study 3 trial at 2.00 mg initial copper uses 1.00 g of fresh sand plus 1.00 g of heated sand. Assume each sand contributes coating-attributed removal in direct proportion to its mass, contributions add, and blank remaining stays 1.90 mg. What total remaining mass would most likely be reported?

Show answer

Answer: B — 0.75 mg

Table 3 gives fresh removal of 1.40 mg per 2.00 g and heated removal of 0.90 mg per 2.00 g at 2.00 mg initial, with blank 1.90 mg. Per gram contributions are 0.70 and 0.45 mg, summing to 1.15 mg for one gram each. Subtracting from the unchanged blank gives 0.75 mg remaining. The fresh-only value ignores heated sand. The heated-only value ignores fresh sand. The summed-remaining value adds 0.50 and 1.00 mg without halving for mass.

Question 11

A student hypothesizes that, throughout the initial masses tested in Study 3, every doubling of initial copper mass doubles coating-attributed removal for both sand treatments. Do the results support this hypothesis?

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Answer: C — No, because doubling initial mass from 2.00 to 4.00 mg increased fresh removal from 1.40 mg to only 2.20 mg.

Table 3 removal is blank minus coated, giving fresh values 0.80, 1.40, 2.20 and 3.20 mg and heated values 0.55, 0.90, 1.40 and 2.40 mg. Doubling from 2.00 to 4.00 mg would require 2.80 mg fresh and 1.80 mg heated, but results show only 2.20 and 1.40 mg. The larger-at-larger statement is true but does not test doubling. The claimed doubled heated value does not occur. The lower-heated statement is true but does not address proportionality.

Questions 12–16

A researcher studied when dormant cysts first appear in identical cultures of a freshwater rotifer kept in growth chambers. She defined cyst onset as the first day a dormant cyst was visible in a fixed counting grid. All cultures lacked cysts on Day 0. Table 1 summarizes light, crowding, and feeding records for four dishes. All other conditions were identical, and observations were made daily using the same procedure.

Table 1 Culture records for four dishes

TreatmentL: first day withlight ≤10 hK: day thirdconsecutive daywith crowding≥400 per mL endedF: day fifthconsecutive daywith food ≤0.8 µgper individualendedA1174B1147C7114D7411

Scientist 1

Short days alone set cyst onset. On the first day with at most 10 h of light, an irreversible developmental switch begins; a visible cyst appears exactly 6 days later. Neither crowding nor food shortage changes this delay or starts the process independently. Thus onset occurs on Day L + 6, where L is the first day with light of at most 10 h. Reports of year-to-year differences reflect inconsistent sampling of dish positions rather than different triggers. With a fixed grid and standard scoring, identical cultures with identical light schedules produce cysts together even when crowding and feeding schedules differ.

Scientist 2

Crowding starts the process. Three consecutive days with density at or above 400 individuals per mL are required, and a visible cyst appears exactly 4 days after the third qualifying day ends: Day K + 4. A low-density day before the third qualifying day resets the count. Once three qualifying days have occurred consecutively, later dilution cannot cancel or delay onset. Light and food have no direct effect. Their association with crowding can make either appear responsible. Short days without the required crowding sequence will therefore leave cultures without cysts.

Scientist 3

Light sets the baseline onset at Day L + 6, but food shortage can advance it. After five consecutive days with food at or below 0.8 µg per individual, measured as total food divided by number of individuals, a food-induced cyst appears 2 days later. This advance requires food to remain at or below 0.8 µg per individual throughout those additional days and requires temperature to remain above 18°C throughout those days. Supplementing food before visible cysts appear cancels the food pathway, leaving the light pathway intact. Supplementing after cysts appear does not remove cysts. Without supplementation, onset is the earlier of Day L + 6 and Day F + 2, where F is the day the fifth low-food day ended. Crowding has no direct effect. For Table 1, light continues decreasing, crowding remains high after K, and food remains low with temperature above 18°C after F unless a question specifies a change.

Table 2 Observed cyst onset

TreatmentObserved first daywith cystA17B8C6D13

Question 12

Treatments A and C have different L days but the same F day. According to Scientist 3, how do their cyst-onset days compare?

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Answer: B — Cyst onset occurs on the same day in both.

Read Table 1 where the two runs share food Day 4 but have light Day 11 versus light Day 7, and read the third viewpoint minimum of light plus 6 versus food plus 2. The decisive step is that food plus 2 gives Day 6 for both runs, which beats light plus 6 Day 17 and Day 13, so both fall to Day 6. The earlier in the light Day 11 run reflects crowding timing. The earlier in the light Day 7 run reflects light-only timing that ignores the food advance. The neither-develops option ignores that the food pathway alone suffices.

Question 13

In Treatment B, temperature falls below 18°C on Day 8 but food and other schedules remain as in Table 1. According to Scientist 3, when will cysts appear?

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Answer: D — Day 17

Read Table 1 for the run with light Day 11 and food Day 7, giving light plus 6 Day 17 and food plus 2 Day 9, and read the third viewpoint requirement that food stay low with temperature above 18 degrees through Days 8 and 9. The decisive step is that cold on Day 8 breaks the temperature persistence, so the food pathway fails and onset falls back to Day 17. The Day 8 timing uses the crowding viewpoint. The Day 9 timing applies food plus 2 while ignoring the temperature qualifier. The Day 13 timing matches light-only timing for a different run with light Day 7.

Question 14

Which observed onset in Table 2 supports Scientist 2 while weakening both Scientist 1 and Scientist 3?

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Answer: B — Treatment B

Read Table 1 to compute light plus 6 days 17, 17, 13, 13 and crowding plus 4 days 11, 8, 15, 8 and the minimum of light plus 6 versus food plus 2 days 6, 9, 6, 13, then compare with Table 2 observed days 17, 8, 6, 13. The decisive step is that only the Day 8 observed run in the second row matches crowding timing while missing light-only Day 17 and food-advance Day 9 by one day, which still counts as a miss. The Day 17 observed run matches light-only timing. The Day 6 observed run matches food-advance timing. The Day 13 observed run in the fourth row matches both light-only and food-advance timings.

Question 15

Researchers want to modify Treatment D (L Day 7, K Day 4, F Day 11) so that Scientists 1, 2, and 3 each predict a different onset day, with other schedules retained. Which single change accomplishes this?

Show answer

Answer: A — Move L from Day 7 to Day 11.

Read Table 1 for the run with light Day 7, crowding Day 4, and food Day 11, giving predictions light plus 6 Day 13, crowding plus 4 Day 8, and minimum Day 13. The decisive step is testing each proposed single change including two-day food and temperature persistence. Moving light to Day 11 gives 17, 8, and 13, all different. Moving food to Day 7 looks like 13, 8, and 9, but cold on Day 8 breaks temperature persistence so the third prediction falls back to 13. Keeping food high leaves 13, 8, 13. Moving crowding to Day 7 gives 13, 11, 13.

Question 16

Cultures A and B are both supplemented immediately after the Day 8 records are completed, and food subsequently remains above 0.8 µg per individual. According to Scientist 3, how much later will cyst onset occur in B than in A?

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Answer: C — 11 days later

Read Table 1 where both runs share light Day 11 but have food Day 4 versus food Day 7, and read the third viewpoint minimum with two-day persistence plus supplementation before visible cancels. The decisive step is checking windows Days 5 and 6 versus Days 8 and 9 against supplementation after Day 8. The first window completes and shows Day 6 before supplementation, so it stays Day 6. The second window loses Day 9 and supplementation precedes Day 9 visible onset, so it falls back to light plus 6 Day 17, giving 11 days difference. Three days ignores supplementation. Zero days wrongly cancels the already visible first run. Eight days reverses which run cancels.

Questions 17–22

Researchers studied germination timing in a desert annual. They used genetically identical seeds and defined germination onset as the first day a radicle was visible. Day 0 was sowing. Table 1 summarizes light and soil records for four growth chambers. Germination results were withheld from three scientists, who proposed the explanations below. All other growing conditions were identical, and observations were made daily using the same procedure.

Table 1 Chamber schedules for light, cold, and wetness

TreatmentL: first day withlight ≥14 h (day)K: day thirdconsecutive nightat ≤10°C ended(day)W: day fifthconsecutive daywith soil water≥20% ended (day)A1064B1049C6104D6412

Scientist 1

Day length alone sets germination onset. On the first day with at least 14 h of light, an irreversible internal process begins; visible germination appears exactly 5 days later. Neither cool nights nor soil wetness changes this delay or starts the process independently. Thus onset occurs on Day L + 5, where L is the first day with light of at least 14 h. Identical seeds exposed to identical light schedules should germinate together, even when temperature and watering schedules differ.

Scientist 2

Cool nights start the process. Three consecutive nights at or below 10°C are required, and visible germination appears exactly 6 days after the third qualifying night ends: Day K + 6, where K is the day the third consecutive qualifying night ended. A warmer night before the third qualifying night resets the count. Once three qualifying nights have occurred consecutively, later warming cannot cancel or delay onset. Day length and soil moisture have no direct effect. Without the required cool-night sequence, shortening days alone will leave seeds dormant.

Scientist 3

Day length sets the baseline onset at Day L + 5, but sustained wetness can advance it. After five consecutive days with soil water content at or above 20%, measured as water volume divided by soil volume, wet-induced germination appears 3 days later. This advance requires soil to remain at or above 20% throughout those additional days. Redrying below 20% before visible germination appears cancels the wet pathway, leaving the light pathway intact. Redrying after germination appears does not reverse it. Without redrying, onset is the earlier of Day L + 5 and Day W + 3, where W is the day the fifth consecutive wet day ended. Cool nights have no direct effect. For Table 1, light continues increasing, nights remain cool after K, and soil remains wet after W unless a question specifies a change.

Table 2 Redrying follow-up and observed onset

TreatmentDay soil first fellbelow 20% after W(day)Observed onset(day)A615B1115C97D1411

Question 17

For Treatments A and B under the original schedules, how do the predictions of Scientists 1 and 2 compare?

Show answer

Answer: B — Scientist 1 predicts the same onset day in A and B; Scientist 2 predicts earlier onset in B than in A.

Read L and K from Table 1 and the delay sentences. Scientist 1 adds 5 days to L: both L-10 cases have L 10, so both predict Day 15. Scientist 2 adds 6 days to K: the K-6 case predicts Day 12 and the K-4 case predicts Day 10, so the K-4 case is earlier. The both-earlier-in-one-case option reverses Scientist 2. The swapped-roles option assigns each scientist the other’s rule. The both-same option ignores the different K values and the reset rule.

Question 18

Researchers want to modify Treatment D so that all three scientists predict different onset days. All other environmental schedules are retained. Which single modification would accomplish this?

Show answer

Answer: A — Move L from Day 6 to Day 12.

Read the final listed Table 1 row with each onset rule. Originally the light rule and the wet-baseline rule both give Day 11 while the cold rule gives Day 10. Moving L to Day 12 makes the light rule give Day 17 and the wet-baseline rule give the earlier of Day 17 and Day 15, which is Day 15, leaving the cold rule at Day 10. Moving K alone leaves the light and wet predictions tied. Keeping soil dry leaves the light and wet-baseline rules tied at the light baseline. Moving L to Day 5 collapses all three at Day 10.

Question 19

Which treatment result in Table 2 supports Scientist 3's explanation while weakening Scientist 1's explanation?

Show answer

Answer: C — Treatment C

Read Table 1 L and W with the redrying sentence and Table 2 observed days. For the L-6 W-4 case with redrying on Day 9, light predicts Day 11 but wet predicts Day 7, and redrying on Day 9 occurs after Day 7, so the wet-baseline explanation still predicts Day 7, matching the observed Day 7 and contradicting the light-only explanation. The two L-10 cases observed on Day 15 match both the light baseline and the canceled wet pathway. The L-6 W-12 case observed on Day 11 matches both the light baseline and the canceled wet pathway.

Question 20

In Treatment B, all nights after the completed qualifying sequence on Day 4 are made warmer than 10°C. Light and soil-moisture schedules remain unchanged from the original, and germination onset is observed on Day 10. Which assessment follows from the scientists’ explanations?

Show answer

Answer: A — The result agrees with Scientist 2, because warming after the required sequence was completed cannot cancel onset.

Read the completed-sequence sentence and Table 1 K for the L-10 K-4 W-9 case. Three consecutive cold nights ending Day 4 predict Day 10, and later warming cannot cancel once the sequence is completed, so Day 10 agrees. The wet-baseline explanation predicts the earlier of Day 15 and Day 12, which is Day 12, so Day 10 disagrees. The light-only explanation starts counting from first long light day, not from cold nights. The reset option misapplies reset after completion. The wet-pathway option misorders Day 12 as before Day 10.

Question 21

Treatments A and B are both redried immediately after the environmental records for Day 8 are completed, and soil subsequently remains below 20% water content. According to Scientist 3, how much later will germination onset occur in B than in A?

Show answer

Answer: D — 8 days

Read the five-day and redrying sentences with Table 1. The L-10 W-4 case reaches five wet days ending Day 4, so wet onset Day 7 occurs before redrying after Day 8, leaving onset Day 7. The L-10 W-9 case needs wet days through Day 9 to reach the fifth-day mark, but drying after Day 8 breaks the run, so no wet onset occurs and only the light baseline Day 15 remains. The difference is Day 15 minus Day 7. The zero option cancels both. The three-day option keeps original Day 12 for the second case. The five-day option keeps original Day 12 for the second case while keeping Day 7 for the first.

Question 22

A researcher compares only Treatments A and C under the original schedules. Which sequence correctly describes the predicted comparison for Scientists 1, 2, and 3, respectively?

Show answer

Answer: C — C earlier; A earlier; same day

Read Table 1 L, K, and W with each delay rule under original schedules. The light-only rule adds 5 to L: the L-10 case gives Day 15 versus the L-6 case giving Day 11, so the L-6 case is earlier. The cold-only rule adds 6 to K: the K-6 case gives Day 12 versus the K-10 case giving Day 16, so the K-6 case is earlier. The wet-baseline rule takes the earlier of light-plus-5 and wet-plus-3: the first case gives Day 7 versus the second giving Day 7, so same day. The reversed-light option flips the first comparison. The moved-tie option moves the tie to the cold rule. The doubly-misordered option misorders both light and cold.

Questions 23–28

A researcher studied how temperature and dissolved salt affect water uptake by a soil-conditioning hydrogel. Dry granules absorb water and swell, and agronomists compare granules by grams of water absorbed per gram of dry gel rather than by swollen mass alone. All uncoated trials began with 10.0 g of dry gel in identical beakers with 200 mL of test solution, held in a water bath for 1 hour, drained briefly on mesh, and weighed immediately to give swollen mass including gel plus absorbed water. Bath temperature was held constant within 0.5 degrees, and each temperature used fresh beakers. Swollen mass was recorded to 0.1 g, and no trials were run above 50°C. Coated trials used the same procedure, timing, and solutions, except each trial began with 12.0 g of dry material containing 10.0 g of gel plus 2.0 g of inert coat that absorbs no water. The coat is intended to slow release in soil but was tested here only for its effect on initial uptake. Because salt remains in solution, any salt in absorbed water adds negligible mass.

Study 1 In Study 1 uncoated gel was tested in distilled water and in 2% salt water at four bath temperatures. Above about 40°C the polymer network contracts and squeezes out water, so uptake was expected to fall at the highest temperature. Results are shown in Table 1.

Table 1 Swollen mass of uncoated gel after 1 hour

Temperature (°C)Distilled waterswollen mass (g)Salt water swollenmass (g)2042.030.03050.034.04058.038.05046.032.0

Study 2 In Study 2 the same temperatures and distilled water were used with coated gel. Swollen mass again includes gel, absorbed water, and coat. Results are shown in Table 2. No salt-water trials were run with coated gel.

Table 2 Swollen mass of coated gel in distilled water after 1 hour

Temperature (°C)Coated gel swollenmass (g)2042.03049.04056.05045.0

Question 23

Based on Table 1, for uncoated gel in distilled water at 40°C, what is the water absorbed per gram of dry gel?

Show answer

Answer: D — 4.8 g of water per gram of dry gel

Table 1 gives 58.0 g swollen mass for uncoated distilled gel at 40°C. The text states uncoated trials began with 10.0 g dry gel, so absorbed water is 48.0 g total. Agronomists compare per gram dry gel, so dividing by 10.0 g gives 4.8 g per gram. The 48.0 response reports total water without the per-gram division. The 58.0 response reports swollen mass directly without subtracting dry mass. The 4.4 response correctly computes per-gram uptake but for coated gel at 40°C from Table 2, which is the wrong condition.

Question 24

At 30°C, how much less water per gram of dry gel did uncoated gel absorb in salt water than in distilled water?

Show answer

Answer: A — 1.6 g of water per gram of dry gel

At 30°C Table 1 gives distilled swollen mass 50.0 g and salt swollen mass 34.0 g. With 10.0 g dry gel, absorbed waters are 40.0 and 24.0 g, or 4.0 and 2.4 g per gram dry gel. The salt deficit is 1.6 g per gram. The 16.0 response reports the total-water difference without dividing by dry mass. The 2.4 response reports the salt-water per-gram value alone rather than the difference. The 4.0 response reports the distilled-water per-gram value alone rather than the difference.

Question 25

In distilled water at 30°C, how much more water per gram of dry gel did uncoated gel absorb than coated gel?

Show answer

Answer: B — 0.3 g of water per gram of dry gel

At 30°C uncoated distilled swollen mass is 50.0 g for 10.0 g dry gel, giving 40.0 g water or 4.0 g per gram. Table 2 coated swollen mass is 49.0 g for 12.0 g dry material with 10.0 g gel, giving 37.0 g water or 3.7 g per gram gel. The uncoated excess is 0.3 g per gram. The 1.0 response reports the swollen-mass difference without accounting for different dry masses. The 3.7 response reports the coated per-gram value alone. The 4.0 response reports the uncoated per-gram value alone.

Question 26

A student plots water absorbed per gram of dry gel versus temperature for the distilled-water trials in Tables 1 and 2. Which description is faithful to the data?

Show answer

Answer: A — Both per-gram series rise to 40°C then fall, with the uncoated series above the coated series at every temperature.

Per-gram uptake for uncoated distilled gel is 3.2, 4.0, 4.8, and 3.6 g per gram at 20 through 50°C, rising to 40°C then falling. For coated gel it is 3.0, 3.7, 4.4, and 3.3, also rising to 40°C then falling but lower at every temperature including 20°C. The response with both falling reverses the direction. The response with equality at 20°C is true of swollen masses, both 42.0 g, but false for per-gram uptake. The response with opposite trends and a crossing invents a coated rise-fall reversal and a crossing that the tables do not show.

Question 27

Suppose salt water reduces the total water absorbed by the same amount for coated gel as for uncoated gel at 50°C. Based on Tables 1 and 2, what swollen mass is predicted for coated gel in 2% salt water at 50°C?

Show answer

Answer: A — About 31.0 g

Table 1 at 50°C gives 46.0 g distilled and 32.0 g salt swollen mass for 10.0 g dry gel, so absorbed waters are 36.0 g and 22.0 g for a 14.0 g salt reduction. Table 2 gives 45.0 g coated swollen mass for 12.0 g dry material with 10.0 g gel, so coated distilled water is 33.0 g. Subtracting the same 14.0 g gives 19.0 g water plus 12.0 g dry material for 31.0 g. The 32.0 g response repeats uncoated salt mass from the wrong condition. The 45.0 g response repeats coated distilled mass ignoring salt. The 29.0 g response adds only 10.0 g gel omitting the 2.0 g coat.

Question 28

A student claims that the inert coat increases water absorbed per gram of dry gel. Which observation best evaluates this claim?

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Answer: A — At every temperature the coated gel absorbed less water per gram of dry gel than the uncoated gel, weakening the claim that the coat increases uptake.

The coat claim requires higher per-gram uptake with coating. Using 10.0 g gel for both series, uncoated distilled per-gram values are 3.2, 4.0, 4.8, and 3.6, while coated values are 3.0, 3.7, 4.4, and 3.3, lower at every temperature, which weakens the claim. The response with coated swollen mass greater is false since coated swollen masses are equal or lower. The response citing salt versus distilled uptake is true but tests salinity, not coating. The response citing the shared 40°C peak is true but tests temperature, not coating.

Questions 29–34

A researcher studied starch breakdown by amylase in barley extract. She used one batch of filtered extract throughout. Each reaction contained 0.5 mL of extract and 19.5 mL of buffered starch solution. Listed starch concentrations are concentrations in the final 20.0 mL mixture. The mass of maltose released during the first 5 min after mixing was measured. All masses were measured the same way. Each table entry is the mean of three trials; replicate masses differed by less than 0.2 mg. Corrected yield was defined as the reported maltose mass minus the mass from a matched blank containing water in place of extract.

Study 1 The researcher varied the final reaction pH while maintaining the reaction temperature at 30°C and the starch concentration at 10 g/L. Before mixing, extract samples were held at 30°C for 10 min. The same buffer ingredients were used at every pH, and the final pH was checked immediately after mixing. Table 1 includes a separate matched blank for each pH.

Table 1 Effect of pH on maltose yield

Reaction pHExtract maltose(mg)Blank maltose(mg)42.40.456.40.4611.40.479.40.485.40.492.40.4

Study 2 Extract samples without starch were held at a pretreatment temperature for 10 min, then tested by either of two procedures. For the direct procedure, extract was immediately mixed with starch solution at the pretreatment temperature, and the reaction remained at that temperature. For the recovery procedure, extract was transferred to 30°C for 10 min before mixing, and the reaction occurred at 30°C. Separate samples were used for the two procedures. Every reaction had pH 6 and 10 g/L starch. Blanks underwent the corresponding procedure. Table 2 reports the results.

Table 2 Effect of pretreatment temperature on maltose yield

Pretreatmenttemperature (°C)Direct extractmaltose (mg)Direct blankmaltose (mg)Recovery extractmaltose (mg)Recovery blankmaltose (mg)105.40.211.40.43011.40.411.40.44514.41.410.40.46014.94.45.40.4756.86.41.20.4

Study 3 The researcher varied starch concentration at pH 6 and 30°C. Fresh extract had been held at 30°C for 10 min. Heated extract had undergone the 60°C pretreatment and the recovery procedure from Study 2. Extract volume remained 0.5 mL as concentration changed. Matched blanks for the two extract treatments gave identical masses, so Table 3 lists one blank column.

Table 3 Effect of starch concentration on maltose yield

Starchconcentration(g/L)Fresh extractmaltose (mg)Heated extractmaltose (mg)Blank maltose(mg)2.54.42.90.458.44.40.41011.45.40.42012.96.20.4

Question 29

Consider the pretreatment temperature that produced the greatest corrected yield with the direct procedure in Study 2. At that same pretreatment temperature, the corrected yield with the recovery procedure was how much lower than the greatest corrected yield in Study 1?

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Answer: A — 1.0 mg

Read Table 2 and subtract each direct blank from its direct extract to get corrected yields; the largest is 13.0 mg at 45°C, not the raw maximum at 60°C. At 45°C the recovery corrected yield is 10.4 minus 0.4 equals 10.0 mg. Read Table 1 and subtract blanks to get the Study 1 maximum of 11.4 minus 0.4 equals 11.0 mg at pH 6. The difference is 1.0 mg. The 2.0 mg value compares the two maxima directly, the 3.0 mg value compares direct versus recovery at 45°C, and the 6.0 mg value uses the raw-maximum temperature.

Question 30

Which description correctly compares the changes in corrected yield for the two procedures as pretreatment temperature increased from 30°C to 45°C and then from 45°C to 60°C?

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Answer: C — The direct procedure showed an increase and then a decrease, while the recovery procedure showed a decrease over both intervals.

Read Table 2 and subtract the matching blank for every entry before judging direction. Direct corrected yields rise from 11.0 mg at 30°C to 13.0 mg at 45°C and then fall to 10.5 mg at 60°C. Recovery corrected yields fall from 11.0 mg to 10.0 mg and then to 5.0 mg over the same intervals. Raw direct masses rise over both intervals, which tempts the wrong trend. The both-increase-then-decrease choice misstates recovery, the both-increase choice uses raw direct masses, and the reversed choice swaps the procedures.

Question 31

Assume that fresh-extract corrected yield in Study 3 changed linearly between 5 and 10 g/L starch. Approximately what starch concentration would produce twice the corrected yield measured after the 60°C pretreatment with the recovery procedure in Study 2?

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Answer: B — 8.3 g/L

Calculate corrected yields by subtracting the 0.4 mg blank in Table 3. Fresh corrected values are 8.0 mg at 5 g/L and 11.0 mg at 10 g/L. In Study 2 the 60°C recovery corrected yield is 5.4 minus 0.4 equals 5.0 mg, so twice that target is 10.0 mg. Linear interpolation gives 5 plus 5 times 2.0 divided by 3.0 equals about 8.3 g/L. The 2.5 g/L choice matches a near-table value without doubling, the 9.0 g/L choice repeats the calculation with raw masses, and the 10.0 g/L choice selects the nearest table concentration without interpolating.

Question 32

Students extend Study 3 by testing recovered, 60°C-pretreated extract at pH 5 and 20 g/L starch. Which blank would allow them to calculate corrected yield for this new condition? Each proposed blank replaces extract with water and follows the same pretreatment and recovery timing.

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Answer: A — A blank reacted at pH 5 and 30°C with 20 g/L starch.

Study 3 reactions occur at pH 6 normally, but the new test specifies pH 5 and 20 g/L starch with recovered 60°C extract, whose reaction occurs at 30°C after cooling. The matching blank must share the reaction pH, reaction temperature, starch concentration, and timing while lacking extract. Only the pH 5, 30°C, 20 g/L blank satisfies all conditions. The pH 6 choice retains the old Study 3 pH, the 60°C choice confuses pretreatment temperature with reaction temperature, and the 10 g/L choice retains the old starch concentration.

Question 33

A new Study 3 trial at 10 g/L starch uses 0.10 mL of fresh extract plus 0.40 mL of heated extract. Assume that each extract contributes corrected yield in direct proportion to its volume, that the contributions add, and that the blank mass remains unchanged. What total maltose mass would most likely be reported?

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Answer: B — 6.6 mg

Read Table 3 at 10 g/L starch and subtract the 0.4 mg blank to obtain corrected yields of 11.0 mg for fresh extract and 5.0 mg for heated extract for 0.5 mL. Scale by volume fraction to 0.10 mL fresh as 0.2 times 11.0 equals 2.2 mg and 0.40 mL heated as 0.8 times 5.0 equals 4.0 mg. Add the contributions to 6.2 mg and add the unchanged blank to report 6.6 mg. The 6.2 mg choice omits the blank, the 7.0 mg choice scales raw masses then adds the blank again, and the 8.4 mg choice assumes equal halves.

Question 34

A student hypothesizes that, throughout the concentrations tested in Study 3, every doubling of starch concentration doubles corrected yield for both extract treatments. Do the results support this hypothesis?

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Answer: C — No, because doubling concentration from 10 to 20 g/L increased fresh-extract corrected yield from 11.0 to only 12.5 mg.

Subtract the Table 3 blank to test doubling exactly. Fresh corrected yields are 4.0, 8.0, 11.0, and 12.5 mg across 2.5, 5, 10, and 20 g/L, while heated yields are 2.5, 4.0, 5.0, and 5.8 mg. Doubling from 10 to 20 g/L raises fresh yield only from 11.0 to 12.5 mg, so the every-doubling rule fails. The doubling from 2.5 to 5 g/L is real but does not establish the throughout claim, the every-increase observation is true but does not test doubling, and the heated-below-fresh observation is true but irrelevant to the doubling claim.

Questions 35–40

A researcher studied hatching of eggs of a pond snail in laboratory chambers. Identical batches of newly laid eggs were placed in separate chambers with the same water volume, egg number, food, and light intensity. Only the schedules of daylight, cool nights, and salinity differed. Table 1 shows for Treatments A-D the first day with daylight of 11 h or less (L), the day the second consecutive night at or below 6°C ended (K), and the day the sixth consecutive day with salinity at or above 30 parts per thousand ended (S). Table 2 shows the first day hatchlings were seen and the lowest water temperature from L through L+5 and the lowest salinity from S through hatching. Observations were made daily with the same procedure.

Table 1 Timing of daylight, cool-night, and salinity thresholds

TreatmentL: first day withdaylight ≤11 h(day)K: day secondconsecutive nightat ≤6°C ended(day)S: day sixthconsecutive daywith salinity ≥30ppt ended (day)A1064B1046C6104D6410

Table 2 Observed hatching and water conditions

TreatmentFirst hatchlingsseen (day)Lowest watertemperature fromL through L+5 (°C)Lowest salinityfrom S throughhatching (ppt)A71434B92134C112135D112112

Scientist 1

Daylight alone sets hatching. On the first day with 11 h or less of light, an irreversible process begins; hatchlings appear exactly 5 days later, on Day L+5, but only if water remains above 18°C on each day from L through L+5. If water is at or below 18°C on any of those days, the daylight trigger fails and hatching is delayed beyond Day L+5. Neither cool nights nor salinity changes this delay or starts hatching by itself.

Scientist 2

Cool nights start hatching. Two consecutive nights at or below 6°C are required, and hatchlings appear exactly 5 days after the second night ends, on Day K+5. A warmer night before the second night resets the count. Once two qualifying nights have occurred consecutively, later warming cannot cancel or delay hatching. Daylight and salinity have no direct effect. Their seasonal association with cooling can make either appear responsible. Shorter days without the required cool-night sequence leave eggs unhatched beyond Day K+5.

Scientist 3

Daylight sets a baseline of Day L+5, but salinity can advance hatching. After six consecutive days with salinity at or above 30 parts per thousand, salinity-induced hatching appears 3 days later, on Day S+3. This advance requires salinity to remain at or above 30 parts per thousand through those 3 days. Dilution below 30 parts per thousand before hatchlings appear cancels the salinity pathway, leaving the daylight baseline intact. Without dilution, hatching is the earlier of Day L+5 and Day S+3. Cool nights have no direct effect. Daylight keeps decreasing, nights stay cool after K, and salinity stays high after S unless a question states a change.

Question 35

Based on Tables 1 and 2, the hatching observed in Treatment C agrees with the prediction of which scientist(s), when qualifiers are applied?

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Answer: A — Scientist 1 only

Read Table 1 Treatment C timing L6 K10 S4 with delays L+5 K+5 S+3, then Table 2 observed Day 11 with warm water and high salinity so qualifiers pass. Scientist 1 predicts Day 11 which matches observed, Scientist 2 predicts Day 15 which is four days late, Scientist 3 predicts Day 7 which is four days early. The second option misreads K timing as matching, the third ignores that high salinity should have advanced hatching, the fourth wrongly adds Scientist 3 whose early prediction fails.

Question 36

According to Scientist 1, if water remains above 18°C, on what day will hatchlings appear in Treatment B?

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Answer: C — 15

Scientist 1 sets hatching at Day L+5 when water stays above 18 degrees, and Table 1 gives Treatment B L10 so predicted Day 15 under the stated warm assumption. Table 2 temperature is not needed because the stem stipulates warm water. The 9-day option uses K+5 or S+3 from the wrong pathway, the 11-day option uses L+5 for Treatments C or D with L6, the 7-day option uses S+3 for Treatments A or C.

Question 37

For Treatments A and C, how do the predictions of Scientists 1 and 2 compare, assuming water stays above 18°C and salinity stays high?

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Answer: C — Scientist 1 predicts earlier hatching in C; Scientist 2 predicts earlier hatching in A.

Table 1 gives Treatment A L10 and Treatment C L6 so Scientist 1 L+5 predicts Day 15 versus Day 11 with Treatment C earlier, while Treatment A K6 and Treatment C K10 gives Scientist 2 K+5 predicting Day 11 versus Day 15 with Treatment A earlier under stated assumptions that remove Table 2 cold and dilution effects. The first option extends Scientist 1 order to Scientist 2, the second extends Scientist 2 order to Scientist 1, the fourth swaps both orders by misreading L and K columns.

Question 38

In Treatment A, the lowest water temperature from L through L+5 was 14°C and hatchlings appeared on Day 7. What does this result imply about Scientist 1's explanation?

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Answer: A — Weakens it, because hatchlings appeared well before Day L+5 though no other pathway is allowed to cause early hatching.

Scientist 1 allows only Day L+5 when warm and delay beyond L+5 when cold, explicitly denying that cool nights or salinity can start early hatching, with Table 1 Treatment A L10 giving L+5 Day 15. Table 2 shows cold 14 degrees plus Day 7 hatching eight days before L+5, which contradicts exclusivity despite qualifier failure. The second option misreads the qualifier as excusing early hatching, the third misstates K timing and direction since Day 7 precedes K+5 Day 11, the fourth reverses chronology because Day 7 is before not after Day 11.

Question 39

Treatment C is diluted below 30 parts per thousand on Day 5 and remains below thereafter. According to Scientist 3, on what day will hatchlings appear?

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Answer: C — 11

Scientist 3 predicts the earlier of L+5 and S+3 without dilution, with Table 1 Treatment C L6 giving L+5 Day 11 and S4 giving S+3 Day 7 so undiluted prediction is Day 7. Dilution on Day 5 occurs before Day 7 and cancels the salinity pathway while leaving daylight baseline intact, so prediction reverts to Day 11. The 7-day option ignores dilution, the 9-day option uses S+3 for Treatment B, the 15-day option uses K+5 for Treatment C from the wrong scientist.

Question 40

Under the original schedules in Tables 1 and 2, how do the predicted hatching days for Treatments B and D compare for Scientists 1, 2, and 3, respectively?

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Answer: A — Treatment D earlier; same day; Treatment B earlier.

Table 1 gives Treatment B L10 and Treatment D L6 so Scientist 1 L+5 predicts Day 15 versus Day 11 with Treatment D earlier, Treatment B K4 and Treatment D K4 gives Scientist 2 K+5 predicting Day 9 versus Day 9 same day, and Scientist 3 earlier-of gives Treatment B Day 9 from S6+3 versus Treatment D Day 11 from L+5 because Treatment D S10+3 Day 13 exceeds L+5 and Table 2 dilution confirms salinity does not advance Treatment D. The second option rotates correct relations forward, the third rotates them differently using wrong scientist order, the fourth swaps Scientist 2 and Scientist 3 relations by confusing K and S timings.

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