An agricultural extension office tested the same corn fertilizer on the same farm two years in a row. Each year, workers planted 20 plots with the same corn variety on the same date. Ten plots received the fertilizer and ten plots received none. The same brand and amount of fertilizer was used both years.
Year 1 had 28 inches of rain during the growing season, which is normal for the area. Year 2 was a drought, with only 9 inches of rain. In July of each year, workers also measured soil moisture. The average soil moisture was 31% in Year 1 and 9% in Year 2.
In Year 1, the fertilized plots produced 195 bushels per acre, compared with 150 bushels per acre for the plots with no fertilizer. In Year 2, the fertilized plots produced 95 bushels per acre, compared with 92 bushels per acre for the plots with no fertilizer.
Which statement best describes how the results of the two years differ?
Hint
For each year, subtract the "no fertilizer" yield from the "fertilizer added" yield.
Step-by-step solution
Where the answer comes from: the passage states that in Year 1 "the fertilized plots produced 195 bushels per acre, compared with 150 bushels per acre," and in Year 2 "the fertilized plots produced 95 bushels per acre, compared with 92 bushels per acre."
Step by step:
Year 1 effect of fertilizer: 195 - 150 = 45 bushels per acre (that is 45 \div 150 = 0.30, a 30% increase).
Year 2 effect of fertilizer: 95 - 92 = 3 bushels per acre (only about a 3% increase).
So the fertilizer gave a large boost in the wet year and almost no boost in the drought year.
Correct answer: B. Choice A reverses the years. Choice C ignores the large Year 1 gain. Choice D is wrong because the fertilized plots still yielded slightly more (95 vs. 92) in Year 2.
Tip: Before you try to explain two findings, state each one precisely in numbers. Many "working with findings" questions start by asking you to see exactly what is different.
Question 2 of 36
Fertilizer Trials in a Wet Year and a Dry Year
An agricultural extension office tested the same corn fertilizer on the same farm two years in a row. Each year, workers planted 20 plots with the same corn variety on the same date. Ten plots received the fertilizer and ten plots received none. The same brand and amount of fertilizer was used both years.
Year 1 had 28 inches of rain during the growing season, which is normal for the area. Year 2 was a drought, with only 9 inches of rain. In July of each year, workers also measured soil moisture. The average soil moisture was 31% in Year 1 and 9% in Year 2.
In Year 1, the fertilized plots produced 195 bushels per acre, compared with 150 bushels per acre for the plots with no fertilizer. In Year 2, the fertilized plots produced 95 bushels per acre, compared with 92 bushels per acre for the plots with no fertilizer.
The same fertilizer worked well in Year 1 but barely worked in Year 2. Which explanation best accounts for both findings?
Hint
Look for the condition the passage says was very different between the two years.
Step-by-step solution
Where the answer comes from: the passage says Year 1 had "28 inches of rain" and Year 2 "was a drought, with only 9 inches of rain." Soil moisture "was 31% in Year 1 and 9% in Year 2." It also says "the same brand and amount of fertilizer was used both years."
Step by step:
The fertilizer, corn variety, number of plots and planting date stayed the same. The big difference between the years was water.
Plants need both water and nutrients. If water is in short supply, adding more nutrients cannot make the plants grow much more, because water is now the limiting factor.
This one idea explains both results: with plenty of water (Year 1), nutrients were what held the corn back, so fertilizer helped. With little water (Year 2), extra nutrients barely helped.
Correct answer: C. Choice A has no support: the passage says the fertilizer was the same. Choice B contradicts the data, since fertilized plots still did slightly better in the drought. Choice D throws away a finding instead of explaining it.
Tip: A good reconciling explanation keeps all the findings true and points to a real difference in conditions between the studies.
Question 3 of 36
Fertilizer Trials in a Wet Year and a Dry Year
An agricultural extension office tested the same corn fertilizer on the same farm two years in a row. Each year, workers planted 20 plots with the same corn variety on the same date. Ten plots received the fertilizer and ten plots received none. The same brand and amount of fertilizer was used both years.
Year 1 had 28 inches of rain during the growing season, which is normal for the area. Year 2 was a drought, with only 9 inches of rain. In July of each year, workers also measured soil moisture. The average soil moisture was 31% in Year 1 and 9% in Year 2.
In Year 1, the fertilized plots produced 195 bushels per acre, compared with 150 bushels per acre for the plots with no fertilizer. In Year 2, the fertilized plots produced 95 bushels per acre, compared with 92 bushels per acre for the plots with no fertilizer.
A researcher thinks that the fertilizer barely helped in Year 2 because the soil was too dry. Which new finding would best support this explanation?
Hint
The best evidence tests the suspected cause directly: it changes that one condition while everything else stays the same.
Step-by-step solution
Where the answer comes from: the explanation being tested is that low water (soil moisture of 9% in Year 2) kept the fertilizer from working. To support it, you need evidence where water is added back and the fertilizer effect returns.
Step by step:
Choice B describes plots in a drought year that are irrigated (watered). If the fertilizer again raised yield by about 30%, as in Year 1 (45 \div 150 = 30\%), then water really was the missing piece.
Choice A only shows the corn was smaller in Year 2. That is true for both fertilized and unfertilized plots and does not say why the fertilizer failed.
Choice C is about other farms and other practices, so it does not test the explanation.
Choice D is about temperature, a different factor; it would point to a different explanation, not support this one.
Correct answer: B.
Tip: Evidence that best supports a cause-and-effect explanation usually comes from changing the suspected cause (here, water) while keeping everything else the same.
Question 4 of 36
Soil Erosion on Bare and Grass-Covered Slopes
Students in an earth science class tested how slope and plant cover affect soil erosion. They filled six identical trays with the same garden soil. Three trays had bare soil. In the other three trays, the students grew grass for four weeks before the test.
Each tray was tilted to a slope of 10°, 20° or 30°. The students then poured 2 liters of water over the top of each tray from a watering can over one minute, as if a heavy rain were falling. They collected the soil that washed off the bottom of each tray, dried it and weighed it.
Slope
Soil lost from bare tray (g)
Soil lost from grass-covered tray (g)
10°
120
30
20°
260
45
30°
510
70
The students noticed that the grass roots formed a thick mat in the top few centimeters of soil.
Which statement best describes the students' findings?
Hint
Compare the two soil-loss columns row by row. How big is the gap at 10°, 20° and 30°?
Step-by-step solution
Where the answer comes from: the table rows "| 10° | 120 | 30 |", "| 20° | 260 | 45 |" and "| 30° | 510 | 70 |" give the soil lost (in grams) from the bare and grass-covered trays.
Step by step:
At every slope, the grass-covered tray lost less soil than the bare tray.
The gap grows with slope: 120 - 30 = 90 g, then 260 - 45 = 215 g, then 510 - 70 = 440 g.
Both columns rise with slope, but the bare column rises much faster.
Correct answer: C.
Check: the bare tray lost 4 times as much at 10° (120 \div 30) and about 7 times as much at 30° (510 \div 70 \approx 7.3), so the ratio is not a steady "about twice."
Tip: To describe a finding, state the direction (less soil lost with grass) and how the difference changes across conditions (bigger on steeper slopes).
Question 5 of 36
Hot spotTesting an Antibiotic Against Bacteria
A hospital lab tested how much of an antibiotic is needed to stop the growth of a common bacterium. Technicians added the same number of bacteria to six tubes of nutrient broth. Each tube received a different concentration of the antibiotic, measured in micrograms per milliliter (µg/mL). After 24 hours at body temperature (37 °C), they measured how cloudy each tube was. Cloudier broth means more bacteria grew. The lab counts a cloudiness reading below 0.10 as "no growth."
The readings were 0.90 at 0 µg/mL, 0.85 at 1 µg/mL, 0.60 at 2 µg/mL, 0.05 at 4 µg/mL, 0.04 at 8 µg/mL and 0.04 at 16 µg/mL.
Bacteria can change over time. Some bacteria carry a gene that lets them break down or pump out an antibiotic. Bacteria with such a gene are called resistant.
Click the bar that shows the lowest antibiotic concentration that stopped the bacteria from growing.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
"No growth" means a reading below the dashed 0.10 line. Find the first bar, from the left, that is below it.
Step-by-step solution
Where the answer comes from: the lab counts a reading "below 0.10" as "no growth." The passage lists "0.60 at 2 µg/mL, 0.05 at 4 µg/mL", and the graph shows the same values.
Step by step:
Bars at 0, 1 and 2 µg/mL (0.90, 0.85, 0.60) are all above the 0.10 line, so bacteria grew.
The bar at 4 µg/mL (0.05) is the first one below the line.
The bars at 8 and 16 µg/mL are also below the line, but they are higher concentrations, and the question asks for the lowest one that worked.
Correct answer: the bar at 4 µg/mL. Scientists call this the minimum concentration that stops growth.
Check: 2 µg/mL lowered the cloudiness (0.90 to 0.60) but did not reach "no growth," so it is not the answer.
Tip: On a hot-spot graph question, find the rule in the passage (here, below 0.10), then apply it point by point from left to right.
Question 6 of 36
Warm Water and Coral Bleaching
Corals are animals that get much of their food from tiny algae living inside their tissues. When the water gets too warm, many corals push the algae out and turn white. This is called bleaching. Bleached corals can recover if the water cools in time, but many die.
Marine scientists surveyed coral colonies at reef sites during one summer and recorded the highest water temperature reached at each site.
Peak summer water temperature (°C)
Coral colonies bleached (%)
28
2
29
5
30
18
31
47
32
81
Each row is the average of several sites that reached that peak temperature.
Which statement best describes the findings of the coral survey?
Hint
Find how much the bleaching percentage increases with each 1 °C step.
Step-by-step solution
Where the answer comes from: the table rows run from "| 28 | 2 |" and "| 29 | 5 |" up to "| 30 | 18 |", "| 31 | 47 |" and "| 32 | 81 |".
Bleaching stays low (2% and 5%) at 28 °C and 29 °C.
Above 29 °C, each extra degree adds a much larger amount, so bleaching rises sharply.
Correct answer: B.
Check: the steps (3, 13, 29, 34) are not equal, so a "same amount each degree" statement is wrong; and at 30 °C only 18% bleached, so "most colonies" is wrong there.
Tip: To describe a table, compute the change between rows. Unequal changes tell you the pattern speeds up or slows down.
Question 7 of 36
A Surprising Boiling Point
A high school class in Denver, Colorado (elevation about 5,280 feet) heated beakers of distilled water to find the boiling point. Before starting, every thermometer was placed in ice water and read 0.0 °C, so the thermometers were working correctly. The class textbook states that pure water boils at 100 °C.
Four groups recorded these boiling temperatures:
Group
Boiling temperature (Denver)
1
94.8 °C
2
95.1 °C
3
94.9 °C
4
95.0 °C
A partner class in Miami, Florida (at sea level) repeated the experiment with the same brand of thermometer and the same kind of distilled water. Its four groups recorded 99.9 °C, 100.1 °C, 100.0 °C and 100.0 °C.
On the day of the tests, the air pressure was about 83 kilopascals (kPa) in Denver and about 101 kPa in Miami. Air pressure is the weight of the air above pushing down on a surface. It is lower at higher elevations because there is less air above.
The Denver class expected water to boil at 100 °C, but every group measured about 95 °C. Which explanation is best supported by the information in the passage?
Hint
Check which possible causes the passage already rules out, then look at what was different between Denver and Miami.
Step-by-step solution
Where the answer comes from: the passage says every thermometer "read 0.0 °C" in ice water, the water was distilled, and the Miami class using "the same brand of thermometer and the same kind of distilled water" measured about 100 °C. The key difference: air pressure "was about 83 kilopascals (kPa) in Denver and about 101 kPa in Miami."
Step by step:
Rule out faulty thermometers: they were checked in ice water, and the same kind worked as expected in Miami.
Rule out impure water: the water was distilled in both cities. (Dissolved impurities actually raise the boiling point, so impurities would not explain a lower reading.)
Rule out a weak burner: water that is truly boiling stays at its boiling point no matter how hard you heat it.
What remains is the real difference: Denver's lower air pressure. With less air pressing down on the water's surface, water molecules escape as vapor more easily, so water boils at a lower temperature.
Correct answer: A.
Tip: When results are unexpected, first eliminate errors the passage has already ruled out, then look for a real difference in conditions.
Question 8 of 36
Drop-downA Surprising Boiling Point
A high school class in Denver, Colorado (elevation about 5,280 feet) heated beakers of distilled water to find the boiling point. Before starting, every thermometer was placed in ice water and read 0.0 °C, so the thermometers were working correctly. The class textbook states that pure water boils at 100 °C.
Four groups recorded these boiling temperatures:
Group
Boiling temperature (Denver)
1
94.8 °C
2
95.1 °C
3
94.9 °C
4
95.0 °C
A partner class in Miami, Florida (at sea level) repeated the experiment with the same brand of thermometer and the same kind of distilled water. Its four groups recorded 99.9 °C, 100.1 °C, 100.0 °C and 100.0 °C.
On the day of the tests, the air pressure was about 83 kilopascals (kPa) in Denver and about 101 kPa in Miami. Air pressure is the weight of the air above pushing down on a surface. It is lower at higher elevations because there is less air above.
Use the findings from Denver and Miami to complete the statements.
A mountain town at 10,000 feet has an air pressure of about 70 kPa. In that town, pure water would most likely boil at .
Compared with cooking at sea level, cooking pasta in boiling water in that town would take .
Choose an answer from each menu.
Hint
Line up the two cities by air pressure and boiling point, then place 70 kPa on that pattern.
Step-by-step solution
Where the answer comes from: the passage gives two data pairs. Miami: about 101 kPa, boiling at about 100 °C. Denver: about 83 kPa, boiling at about 95 °C (the four groups recorded 94.8 to 95.1 °C).
Step by step:
The pattern: lower air pressure → lower boiling point.
The mountain town's pressure (70 kPa) is even lower than Denver's (83 kPa), so its boiling point should be below 95 °C. (In real life it is about 90 °C.)
Pasta cooks because hot water transfers heat into it. Boiling water there is cooler than at sea level, so it heats the pasta more slowly and cooking takes longer.
Correct answers: b1 = below 95 °C; b2 = longer.
Tip: To apply a finding to a new situation, find the trend in the data you have, then check that the new case fits the same trend (same direction, similar conditions).
Question 9 of 36
Hot spotTemperature and an Enzyme Used in Baking
Amylase is an enzyme that breaks starch down into sugar. Bakers add amylase from a fungus to dough so yeast has more sugar to feed on. A food scientist tested how temperature affects this amylase.
She mixed equal amounts of the enzyme and a starch solution in six test tubes. Each tube was kept at a different temperature, and all tubes had the same pH (6). She measured how fast the starch was broken down (the reaction rate) in each tube.
The rate increased from 2 units per minute at 10 °C to a peak of 12 units per minute at 40 °C. Above 40 °C the rate dropped quickly, to 1 unit per minute at 60 °C. Most enzymes lose their shape (denature) when they get too hot, and a denatured enzyme stops working.
Click the data point on the graph that shows the temperature at which this amylase breaks down starch fastest.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
"Fastest" means the highest reaction rate. Look at the vertical axis.
Step-by-step solution
Where the answer comes from: the vertical axis shows reaction rate. The points are 10 °C → 2, 20 °C → 5, 30 °C → 9, 40 °C → 12, 50 °C → 6 and 60 °C → 1 units per minute. The passage also says the rate rose "to a peak of 12 units per minute at 40 °C."
Step by step:
Fastest reaction = highest point on the graph.
The highest point is at 12 units per minute, directly above 40 °C.
Points to the left (cooler) and right (hotter) are all lower.
Correct answer: the point at 40 °C (12 units per minute). This temperature is called the enzyme's optimum temperature. Below it, molecules move slowly and meet less often; above it, the enzyme starts to denature.
Tip: On a "rise then fall" graph, the peak is the optimum. Do not choose the last or rightmost point just because it is at the highest temperature.
Question 10 of 36
Temperature and an Enzyme Used in Baking
Amylase is an enzyme that breaks starch down into sugar. Bakers add amylase from a fungus to dough so yeast has more sugar to feed on. A food scientist tested how temperature affects this amylase.
She mixed equal amounts of the enzyme and a starch solution in six test tubes. Each tube was kept at a different temperature, and all tubes had the same pH (6). She measured how fast the starch was broken down (the reaction rate) in each tube.
The rate increased from 2 units per minute at 10 °C to a peak of 12 units per minute at 40 °C. Above 40 °C the rate dropped quickly, to 1 unit per minute at 60 °C. Most enzymes lose their shape (denature) when they get too hot, and a denatured enzyme stops working.
A bakery wants to add this amylase to dough that rests for one hour before baking. Based on the scientist's findings, at which resting temperature would the enzyme break down starch fastest?
Hint
Find where each temperature falls on the graph. Be careful with a temperature that was never tested.
Step-by-step solution
Where the answer comes from: the graph shows the rate peaking at 12 units per minute at 40 °C, with 9 at 30 °C and 6 at 50 °C. The passage says that "above 40 °C the rate dropped quickly, to 1 unit per minute at 60 °C."
Step by step:
25 °C lies between 20 °C (5) and 30 °C (9), so its rate is about 7 units per minute.
38 °C lies just below the peak at 40 °C, between 9 and 12, so its rate is about 11 units per minute — the highest of the choices.
55 °C lies between 50 °C (6) and 60 °C (1), so its rate is only about 3 to 4.
70 °C is beyond the tested range. The trend after 40 °C is sharply downward and the enzyme denatures, so it would be slower still, not faster.
Correct answer: B (38 °C).
Tip: Applying findings to a real situation often means interpolating (estimating between tested values). Be very cautious about extrapolating beyond the data, and never assume a trend reverses.
Question 11 of 36
Fill inSalt and the Freezing Point of Water
Road crews spread salt on icy roads in winter. A group of students tested why. They dissolved different amounts of table salt in 100 grams of water and slowly cooled each solution. They recorded the temperature at which each solution began to freeze.
Salt dissolved in 100 g of water
Freezing point
0 g
0.0 °C
5 g
−3.0 °C
10 g
−6.0 °C
20 g
−12.0 °C
The students concluded that adding salt lowers the freezing point of water, and that more salt lowers it further. Over the range they tested, each extra 5 grams of salt lowered the freezing point by about 3 °C.
Based on the students' results, predict the freezing point of a solution made with 15 g of salt in 100 g of water.
°C
Type your answer in the box. Fractions like 3/4 are fine.
Hint
15 g is halfway between two amounts in the table. The pattern is about 3 °C per 5 g.
Step-by-step solution
Where the answer comes from: the table shows 10 g → −6.0 °C and 20 g → −12.0 °C. The passage adds that "each extra 5 grams of salt lowered the freezing point by about 3 °C."
Step by step:
15 g is 5 g more than 10 g.
Each extra 5 g lowers the freezing point by about 3 °C: -6.0 - 3.0 = -9.0 °C.
Check with the halfway method: 15 g is halfway between 10 g and 20 g, so the freezing point is halfway between −6.0 and −12.0: \dfrac{-6.0 + (-12.0)}{2} = -9.0 °C.
Correct answer: −9 °C (answers from −9.5 to −8.5 are accepted).
Tip: Remember the negative sign. A freezing point lower than 0 °C is negative, and "lowering" it further makes the number more negative.
Question 12 of 36
Salt and the Freezing Point of Water
Road crews spread salt on icy roads in winter. A group of students tested why. They dissolved different amounts of table salt in 100 grams of water and slowly cooled each solution. They recorded the temperature at which each solution began to freeze.
Salt dissolved in 100 g of water
Freezing point
0 g
0.0 °C
5 g
−3.0 °C
10 g
−6.0 °C
20 g
−12.0 °C
The students concluded that adding salt lowers the freezing point of water, and that more salt lowers it further. Over the range they tested, each extra 5 grams of salt lowered the freezing point by about 3 °C.
A road crew expects a night temperature of −8 °C. Based on the students' data, which is the smallest amount of salt from the list that would keep 100 g of water from freezing at that temperature?
Hint
The solution stays liquid only if its freezing point is below −8 °C.
Step-by-step solution
Where the answer comes from: the table gives freezing points of −3.0 °C (5 g), −6.0 °C (10 g) and −12.0 °C (20 g). Using the pattern of about 3 °C per 5 g, 15 g gives about −9.0 °C.
Step by step:
To stay liquid at −8 °C, the solution's freezing point must be colder than −8 °C.
5 g → −3.0 °C: freezes at −8 °C. Not enough.
10 g → −6.0 °C: −6 is warmer than −8, so it still freezes. Not enough.
15 g → about −9.0 °C: −9 is colder than −8, so it stays liquid. ✓
20 g → −12.0 °C also works, but it is not the smallest amount.
Correct answer: C (15 g).
Tip: With negative temperatures, "colder" means "more negative." Compare carefully: −9 °C is colder than −8 °C, but −6 °C is warmer.
Question 13 of 36
Soil Erosion on Bare and Grass-Covered Slopes
Students in an earth science class tested how slope and plant cover affect soil erosion. They filled six identical trays with the same garden soil. Three trays had bare soil. In the other three trays, the students grew grass for four weeks before the test.
Each tray was tilted to a slope of 10°, 20° or 30°. The students then poured 2 liters of water over the top of each tray from a watering can over one minute, as if a heavy rain were falling. They collected the soil that washed off the bottom of each tray, dried it and weighed it.
Slope
Soil lost from bare tray (g)
Soil lost from grass-covered tray (g)
10°
120
30
20°
260
45
30°
510
70
The students noticed that the grass roots formed a thick mat in the top few centimeters of soil.
A town wants to reduce the soil that washes off a bare 25° hillside into a creek during heavy rain. Based only on the students' findings, which plan would most likely let the least soil wash off?
Hint
Estimate the soil loss for each plan from the table. Compare a grass-covered steep slope with a bare gentle slope.
Step-by-step solution
Where the answer comes from: the table shows soil loss for both covers at three slopes: "| 10° | 120 | 30 |", "| 20° | 260 | 45 |" and "| 30° | 510 | 70 |".
Step by step:
Plant grass, keep 25°: 25° lies between 20° and 30°, so a grass-covered tray would lose between 45 g and 70 g.
Reshape to 20°, bare: about 260 g, far more.
Reshape to 10°, bare: about 120 g. Even the gentlest bare tray lost more than the steepest grass-covered tray (70 g).
Gravel: gravel was never tested, so the findings cannot support it.
Correct answer: C.
Check: every grass-covered value (30, 45, 70 g) is lower than every bare value (120, 260, 510 g), so plant cover mattered more than slope in this test.
Tip: When you apply findings to a decision, estimate the outcome of each option from the data and stay inside what was actually tested.
Question 14 of 36
Hot spotTemperature Inside a Compost Pile
In a compost pile, microbes such as bacteria and fungi break down food scraps and yard waste. As the microbes feed and multiply, they release heat, so the center of an active pile warms up. The microbes work best when the pile is damp but not soaked.
A community garden volunteer built a new compost pile and read a long thermometer at the pile's center every two days. She tested the thermometer in ice water and in boiling water before she started, and it read correctly both times.
Her log also includes these notes:
Day 3: added a bag of grass clippings.
Day 7: a cold, heavy rain soaked the pile all night.
Day 9: turned the pile with a pitchfork to mix in air.
Click the data point that does not fit the overall pattern of the other readings.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Describe the overall shape first: does the temperature mostly rise, fall, or level off? Then look for the one reading that breaks that shape.
Step-by-step solution
Where the answer comes from: the passage says microbes "release heat, so the center of an active pile warms up." The graph shows day 0: 20 °C, day 2: 35 °C, day 4: 50 °C, day 6: 58 °C, day 8: 44 °C, day 10: 60 °C and day 12: 57 °C.
Step by step:
The overall pattern: the pile heats up quickly, then levels off near 57 to 60 °C.
Day 8 breaks the pattern. The temperature drops 14 °C, then jumps back to 60 °C by day 10.
The small dip from 60 °C to 57 °C on day 12 is part of the leveling off, not a break in the pattern.
Correct answer: the point at day 8 (44 °C).
Check: if you removed the day 8 point, the line would rise and then level off smoothly, which is what an active pile should do.
Tip: To find an unexpected result, first put the overall trend into words. The point that would need its own explanation is the one that does not fit.
Question 15 of 36
Counting Stars Near a City
Volunteers in a night-sky project counted how many stars they could see inside the same small patch of sky, the constellation Orion, on clear, moonless nights. Each volunteer observed from a different site at a different distance from the center of a large city. All observers waited 20 minutes outdoors so their eyes could adjust to the dark.
Distance from city center (km)
Stars counted in Orion
2
8
10
15
25
31
50
54
80
60
120
61
The city's streetlights, signs and buildings send light up into the sky, which makes the night sky glow.
Based on the volunteers' star counts, which conclusion about distance from the city is best supported?
Hint
Look at how much the count changes between each pair of rows, especially the last three rows.
Step-by-step solution
Where the answer comes from: the counts rise from 8 stars at 2 km to 54 stars at 50 km, then "| 80 | 60 |" and "| 120 | 61 |" show only small gains.
Step by step:
From 2 km to 50 km the count climbs steeply: 8, 15, 31, 54.
From 50 km to 80 km it rises by only 6; from 80 km to 120 km, by only 1.
So the count rose with distance and then leveled off beyond about 80 km.
The data describe what observers saw. They do not show that the stars themselves changed brightness.
Correct answer: D.
Check: test the doubling claim: from 25 km to 50 km the count went from 31 to 54, not 62.
Tip: A supported conclusion describes the pattern the data actually show, including where it changes, and does not reach beyond the data range.
Question 16 of 36
Fill inStretching a Spring
A student hung slotted masses from a steel spring and measured the spring's total length with a meter stick after each mass was added.
Mass hung (g)
Spring length (cm)
0
10.0
100
12.5
200
15.0
300
17.5
400
20.0
500
22.0
600
23.0
After loads of up to 400 g were removed, the spring went back to 10.0 cm. After the 600-g load was removed, the spring stayed at 11.2 cm. A spring that is stretched too far can be permanently bent out of shape.
Based on the student's data, predict the spring's length when 250 g is hung from it.
cm
Type your answer in the box. Fractions like 3/4 are fine.
Hint
250 g is halfway between two masses in the table. How much does the length change for each 100 g in that part of the table?
Step-by-step solution
Where the answer comes from: the table gives "| 200 | 15.0 |" and "| 300 | 17.5 |". From 0 g to 400 g, each 100 g stretches the spring by 2.5 cm.
Step by step:
250 g is 50 g more than 200 g.
In this range, the stretch is 2.5 \text{ cm} \div 100 \text{ g} = 0.025 cm per gram.
Extra stretch: 50 \times 0.025 = 1.25 cm.
Length: 15.0 + 1.25 = 16.25 cm.
Correct answer: 16.25 cm (answers from 16.1 to 16.4 are accepted).
Check: 250 g is exactly halfway between 200 g and 300 g, so the length should be halfway between 15.0 and 17.5 cm: \dfrac{15.0 + 17.5}{2} = 16.25 cm.
Tip: Interpolate only inside the part of the data where the pattern is steady. Here 250 g is well inside the 0–400 g range, where every 100 g adds 2.5 cm.
Question 17 of 36
Hand Sanitizer in the Lab and in the Garage
Lab test. A company tested its hand sanitizer, which contains 70% alcohol. Volunteers first washed and dried their hands. Then technicians placed a known number of harmless test bacteria on each volunteer's palms. Each volunteer rubbed 3 mL of sanitizer over both hands until dry. Swabs showed that the sanitizer killed 99.9% of the test bacteria.
Field test. A health researcher repeated the test with auto mechanics at the end of a work shift. Their hands were covered with visible grease and dirt. The same harmless test bacteria were placed on their palms, and each mechanic rubbed in 3 mL of the same sanitizer until dry. This time, swabs showed that the sanitizer killed only about 60% of the test bacteria.
In both tests, the room temperature was 22 °C and the sanitizer came from the same batch.
Which explanation best accounts for both the lab result and the field result?
Hint
List what was the same in the two tests. Then find what was different about the hands.
Step-by-step solution
Where the answer comes from: in the lab, "Volunteers first washed and dried their hands." In the field, "Their hands were covered with visible grease and dirt." The field test used the "same harmless test bacteria" and "the sanitizer came from the same batch."
Step by step:
Same sanitizer batch, same amount (3 mL), same bacteria, same room temperature.
The one big difference is clean hands (99.9% killed) versus greasy, dirty hands (about 60% killed).
So the most reasonable explanation is that grease and dirt get in the way of the sanitizer reaching the bacteria.
Correct answer: D.
Check: the explanation keeps both results true. Saying the lab result is wrong, or that the bacteria or batch were different, contradicts the passage.
Tip: When two tests of the same product disagree, list the controlled factors first. Whatever is left over is your best explanation.
Question 18 of 36
Testing an Antibiotic Against Bacteria
A hospital lab tested how much of an antibiotic is needed to stop the growth of a common bacterium. Technicians added the same number of bacteria to six tubes of nutrient broth. Each tube received a different concentration of the antibiotic, measured in micrograms per milliliter (µg/mL). After 24 hours at body temperature (37 °C), they measured how cloudy each tube was. Cloudier broth means more bacteria grew. The lab counts a cloudiness reading below 0.10 as "no growth."
The readings were 0.90 at 0 µg/mL, 0.85 at 1 µg/mL, 0.60 at 2 µg/mL, 0.05 at 4 µg/mL, 0.04 at 8 µg/mL and 0.04 at 16 µg/mL.
Bacteria can change over time. Some bacteria carry a gene that lets them break down or pump out an antibiotic. Bacteria with such a gene are called resistant.
The lab later tested bacteria of the same species taken from a different patient. In the same test, these bacteria made the broth very cloudy (0.82) even at 16 µg/mL. Which explanation is best supported by the passage?
Hint
The first sample stopped growing at 4 µg/mL. What does the passage say can make some bacteria survive an antibiotic?
Step-by-step solution
Where the answer comes from: the first sample showed "0.04 at 16 µg/mL" (no growth). The passage also states: "Some bacteria carry a gene that lets them break down or pump out an antibiotic. Bacteria with such a gene are called resistant."
Step by step:
Same species, same test, very different result: 0.82 versus 0.04 at 16 µg/mL.
The passage gives a ready explanation: some bacteria carry a resistance gene. The second patient's bacteria most likely have it.
The idea that the antibiotic helps bacteria grow is contradicted by the first test, where 16 µg/mL stopped growth.
Nothing in the passage says the temperature changed; the test used 37 °C, and "the same test" was run.
Correct answer: B.
Check: a good explanation of an unexpected result fits both the old and new findings. Resistance explains why one sample died and the other did not.
Tip: When a result surprises you, do not throw out the earlier data. Look for a difference between the two samples that the passage gives you.
Question 19 of 36
Temperature Inside a Compost Pile
In a compost pile, microbes such as bacteria and fungi break down food scraps and yard waste. As the microbes feed and multiply, they release heat, so the center of an active pile warms up. The microbes work best when the pile is damp but not soaked.
A community garden volunteer built a new compost pile and read a long thermometer at the pile's center every two days. She tested the thermometer in ice water and in boiling water before she started, and it read correctly both times.
Her log also includes these notes:
Day 3: added a bag of grass clippings.
Day 7: a cold, heavy rain soaked the pile all night.
Day 9: turned the pile with a pitchfork to mix in air.
Which explanation best accounts for the reading on day 8?
Hint
Check the notes for an event that happened shortly before day 8, and ask whether it would raise or lower the temperature.
Step-by-step solution
Where the answer comes from: the log says "Day 7: a cold, heavy rain soaked the pile all night." The passage also says "The microbes work best when the pile is damp but not soaked." The thermometer "read correctly both times" when it was tested.
Step by step:
The drop appears on day 8, right after the night of cold rain.
Cold water cools the pile directly, and a soaked pile slows the microbes, so they release less heat. Both lower the temperature.
By day 10, after the pile was turned and aired out, it was back to 60 °C, so the microbes were still alive.
Turning the pile happened on day 9, after the day 8 reading, so it cannot explain it.
Correct answer: C.
Check: the explanation must come before the effect in time and push the temperature in the right direction (down). Only the rain does both.
Tip: For unexpected-result questions, line up the timeline. A cause must happen before the result it explains.
Question 20 of 36
Two Walking Studies
Study 1. Researchers recruited 80 adults, ages 40 to 60, who had mildly high blood pressure. Half were assigned to a walking group and half to a no-exercise group. The walking group walked briskly for 30 minutes, 5 days a week, for 8 weeks. Average systolic blood pressure (the top number) in the walking group fell from 138 to 131 mmHg. In the no-exercise group, it went from 139 to 138 mmHg.
Study 2. A different team recruited 120 adults of the same ages, who also had mildly high blood pressure. Half walked for 30 minutes once a week for 8 weeks; the other half did not exercise. Average systolic blood pressure in the walking group went from 137 to 136 mmHg. In the no-exercise group, it stayed at 138 mmHg.
In both studies, trained nurses measured blood pressure using the same method, and people were asked not to change their diets.
A headline says the two studies "disagree about whether walking lowers blood pressure." Which statement best reconciles the two sets of findings?
Hint
List what was the same in the two studies and what was different. The difference that matches the different results is the key.
Step-by-step solution
Where the answer comes from: in Study 1, people walked "5 days a week, for 8 weeks" and pressure "fell from 138 to 131 mmHg." In Study 2, people walked "once a week for 8 weeks" and pressure "went from 137 to 136 mmHg." Study 2 used "adults of the same ages."
Step by step:
Study 1 change: 131 - 138 = -7 mmHg, versus -1 in its no-exercise group, a real difference of about 6 mmHg.
Study 2 change: 136 - 137 = -1 mmHg, versus 0 in its no-exercise group, almost no difference.
Same ages, same kind of patients, same 30-minute walks, same 8 weeks, same measuring method. The main difference is how often people walked.
Correct answer: B.
Check: this explanation keeps both findings true. It does not throw out either study.
Tip: To reconcile two findings, find the condition that differed between the studies and show how it could produce both results.
Question 21 of 36
Two Walking Studies
Study 1. Researchers recruited 80 adults, ages 40 to 60, who had mildly high blood pressure. Half were assigned to a walking group and half to a no-exercise group. The walking group walked briskly for 30 minutes, 5 days a week, for 8 weeks. Average systolic blood pressure (the top number) in the walking group fell from 138 to 131 mmHg. In the no-exercise group, it went from 139 to 138 mmHg.
Study 2. A different team recruited 120 adults of the same ages, who also had mildly high blood pressure. Half walked for 30 minutes once a week for 8 weeks; the other half did not exercise. Average systolic blood pressure in the walking group went from 137 to 136 mmHg. In the no-exercise group, it stayed at 138 mmHg.
In both studies, trained nurses measured blood pressure using the same method, and people were asked not to change their diets.
A scientist offers this explanation: The benefit of walking depends on how many days per week a person walks. Which new finding would best support this explanation?
Hint
A supporting finding should show that blood pressure changes more as the number of walking days goes up.
Step-by-step solution
Where the answer comes from: the two studies differ mainly in walking frequency: "5 days a week" in Study 1 and "once a week" in Study 2. The explanation says more days give more benefit.
Step by step:
If frequency matters, an in-between schedule should give an in-between result.
A third study at 3 days a week with a drop of about 4 mmHg falls neatly between Study 2 (about 1 mmHg) and Study 1 (about 7 mmHg). That is a dose-response pattern: more walking days, bigger drop.
Enjoying the walks says nothing about blood pressure.
A cold winter during Study 2 would offer a different explanation and so would weaken the frequency idea.
The no-exercise groups barely changing is already known from the passage; it does not test frequency.
Correct answer: D.
Check: ask whether the finding would make the explanation more likely. Only the 3-day result does.
Tip: Strong support for "more X gives more Y" is a middle amount of X that gives a middle amount of Y.
Question 22 of 36
Fill inThe Speed of Sound and Thunder
Sound travels through air at a speed that depends on the air's temperature. A physics class used published measurements of the speed of sound in dry air:
Air temperature (°C)
Speed of sound (m/s)
0
331
10
337
20
343
30
349
The speed rises by about 0.6 m/s for every 1 °C increase in temperature.
Light from a lightning flash reaches an observer almost instantly. The thunder from the same flash arrives later because sound is much slower than light. To estimate how far away lightning struck, multiply the speed of sound by the number of seconds between the flash and the thunder.
Use the pattern in the data to find the air temperature at which sound would travel at 340 m/s.
°C
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Start from the row closest to 340 m/s. How many degrees does it take to add the missing meters per second?
Step-by-step solution
Where the answer comes from: the table gives "| 10 | 337 |" and "| 20 | 343 |", and the passage says "The speed rises by about 0.6 m/s for every 1 °C increase in temperature."
Step by step:
340 m/s is between 337 m/s (10 °C) and 343 m/s (20 °C).
Correct answer: 15 °C (answers from 14.5 to 15.5 are accepted).
Check: 340 is exactly halfway between 337 and 343, so the temperature should be halfway between 10 °C and 20 °C, which is 15 °C.
Tip: Working backwards means starting from the output (speed) and finding the input (temperature). Use the same rate, just divide instead of multiply.
Question 23 of 36
Warm Water and Coral Bleaching
Corals are animals that get much of their food from tiny algae living inside their tissues. When the water gets too warm, many corals push the algae out and turn white. This is called bleaching. Bleached corals can recover if the water cools in time, but many die.
Marine scientists surveyed coral colonies at reef sites during one summer and recorded the highest water temperature reached at each site.
Peak summer water temperature (°C)
Coral colonies bleached (%)
28
2
29
5
30
18
31
47
32
81
Each row is the average of several sites that reached that peak temperature.
Forecasters expect one reef to reach a peak water temperature of 31.5 °C next summer. Based on the survey, about what percent of its coral colonies are most likely to bleach?
Hint
31.5 °C falls between two rows of the table. Which two?
Step-by-step solution
Where the answer comes from: the table gives "| 31 | 47 |" and "| 32 | 81 |".
Step by step:
31.5 °C is between 31 °C and 32 °C.
The trend rises steadily in that range, so the bleaching should fall between 47% and 81%.
A rough middle estimate is \dfrac{47 + 81}{2} = 64\%, which is inside that range.
Correct answer: C.
Check: the range 18% to 47% belongs to 30–31 °C, which is cooler than the forecast. Above 81% would need a temperature above 32 °C.
Tip: To apply a finding to a new value, locate the two data rows that bracket it. The prediction should land between their results.
Question 24 of 36
Distance From the Sun and Planet Temperatures
The table shows each planet's average distance from the sun and its average temperature. Distance is given in astronomical units (AU); 1 AU is Earth's average distance from the sun. Jupiter has no solid surface, so its temperature is measured at the cloud tops.
Planet
Distance from sun (AU)
Average temperature (°C)
Atmosphere
Mercury
0.39
167
almost none
Venus
0.72
464
very thick, mostly carbon dioxide
Earth
1.00
15
moderate, mostly nitrogen and oxygen
Mars
1.52
−65
thin, mostly carbon dioxide
Jupiter
5.20
−110
thick, mostly hydrogen and helium
Carbon dioxide is a greenhouse gas: it traps heat near a planet's surface.
A student studies the planet table. Which conclusion about distance from the sun and temperature is best supported?
Hint
Check each conclusion against every row of the table, especially Venus and Mars.
Step-by-step solution
Where the answer comes from: the table lists "| Mercury | 0.39 | 167 | almost none |", "| Venus | 0.72 | 464 | very thick, mostly carbon dioxide |" and "| Mars | 1.52 | −65 | thin, mostly carbon dioxide |", along with Earth and Jupiter.
Step by step:
From Venus outward, temperature falls with distance: 464, 15, −65, −110 °C. So farther planets are generally colder.
But Venus (0.72 AU) is hotter than Mercury (0.39 AU), even though Mercury is closer. Distance alone cannot explain that.
Mars also has a carbon dioxide atmosphere but is very cold (−65 °C), so carbon dioxide does not make every planet hot.
The drops in temperature are not equal per AU (for example, 449 °C from Venus to Earth but only 45 °C from Mars to Jupiter over a much larger distance).
Correct answer: D.
Check: a supported conclusion must fit every row, including the exceptions.
Tip: Look for the data point that breaks the general trend. The best conclusion states the trend and admits the exception.
Question 25 of 36
Two Studies of a Pesticide and Frogs
Study 1 (laboratory). Researchers raised leopard frog tadpoles in tanks of water containing different concentrations of pesticide X, measured in parts per million (ppm). Each tank held 50 tadpoles. All tanks had the same temperature, food and light. The researchers recorded the percent of tadpoles that survived to become adult frogs.
Pesticide X in tank water
Survival to adulthood
0 ppm (clean water)
82%
0.01 ppm
80%
0.05 ppm
61%
0.10 ppm
48%
Study 2 (field survey). A second team counted adult leopard frogs at 12 ponds. Six ponds were next to farms that spray pesticide X. Six ponds were in a nature preserve far from any farm. The farm ponds averaged 140 adult frogs per pond, and the preserve ponds averaged 145 adult frogs per pond. Water samples showed that the farm ponds contained an average of 0.008 ppm of pesticide X, and the preserve ponds contained none.
A news report said, "Lab and field studies disagree about pesticide X."
A news report claims the two studies "disagree." Which statement best reconciles the results of Study 1 and Study 2?
Hint
Compare the pesticide levels actually measured in the farm ponds with the levels tested in the lab.
Step-by-step solution
Where the answer comes from: Study 1 shows survival of 82% in clean water and 80% at 0.01 ppm, but only 61% at 0.05 ppm and 48% at 0.10 ppm. Study 2 says "the farm ponds contained an average of 0.008 ppm of pesticide X."
Step by step:
In the lab, harm clearly appears at 0.05 ppm and above. At 0.01 ppm, survival (80%) is nearly the same as in clean water (82%).
The farm ponds held only 0.008 ppm, which is even lower than 0.01 ppm — below the level that caused harm in the lab.
So we would expect farm ponds and preserve ponds to have similar frog numbers (140 vs. 145). Both studies can be correct at the same time.
Correct answer: D. Choice A dismisses the field data instead of explaining it. Choice B makes an unsupported claim about immunity. Choice C contradicts Study 1, which showed clear harm at 0.05 and 0.10 ppm.
Tip: Two studies that seem to disagree often tested different conditions (here, different concentrations). Look for that difference before deciding one study is wrong.
Question 26 of 36
Select allTwo Studies of a Pesticide and Frogs
Study 1 (laboratory). Researchers raised leopard frog tadpoles in tanks of water containing different concentrations of pesticide X, measured in parts per million (ppm). Each tank held 50 tadpoles. All tanks had the same temperature, food and light. The researchers recorded the percent of tadpoles that survived to become adult frogs.
Pesticide X in tank water
Survival to adulthood
0 ppm (clean water)
82%
0.01 ppm
80%
0.05 ppm
61%
0.10 ppm
48%
Study 2 (field survey). A second team counted adult leopard frogs at 12 ponds. Six ponds were next to farms that spray pesticide X. Six ponds were in a nature preserve far from any farm. The farm ponds averaged 140 adult frogs per pond, and the preserve ponds averaged 145 adult frogs per pond. Water samples showed that the farm ponds contained an average of 0.008 ppm of pesticide X, and the preserve ponds contained none.
A news report said, "Lab and field studies disagree about pesticide X."
One scientist explains the two studies this way: Pesticide X harms tadpoles only when its concentration reaches about 0.05 ppm or higher.
Select all of the new findings that would support this explanation.
Select all that apply.
Hint
A supporting finding should show harm at high concentrations, no harm at low ones, or both.
Step-by-step solution
Where the answer comes from: the explanation is a threshold: harm at about 0.05 ppm or more (Study 1: 61% survival at 0.05 ppm, 48% at 0.10 ppm), little or no harm below that (80% at 0.01 ppm vs. 82% in clean water; farm ponds at 0.008 ppm with 140 frogs vs. 145).
Step by step:
A — supports. A real pond that reached 0.09 ppm (above the threshold) saw tadpole survival drop. This shows the lab effect also happens in the field when the concentration is high enough.
B — does not support. More insects in preserve ponds is a different factor. It would offer a different explanation for any difference, not test the threshold.
C — supports. Survival at 0.008 ppm equal to clean water confirms that the low level found in farm ponds causes no measurable harm.
D — does not support. The weight of adult frogs is unrelated to the concentration threshold.
E — weakens. If 0.10 ppm had no effect in a repeat study, the claim that high concentrations cause harm would be in doubt.
Correct answers: A and C.
Tip: For "select all that support," test each choice separately: if it were true, would it make the explanation more likely? Irrelevant facts and contradicting facts are both wrong.
Question 27 of 36
Soil Erosion on Bare and Grass-Covered Slopes
Students in an earth science class tested how slope and plant cover affect soil erosion. They filled six identical trays with the same garden soil. Three trays had bare soil. In the other three trays, the students grew grass for four weeks before the test.
Each tray was tilted to a slope of 10°, 20° or 30°. The students then poured 2 liters of water over the top of each tray from a watering can over one minute, as if a heavy rain were falling. They collected the soil that washed off the bottom of each tray, dried it and weighed it.
Slope
Soil lost from bare tray (g)
Soil lost from grass-covered tray (g)
10°
120
30
20°
260
45
30°
510
70
The students noticed that the grass roots formed a thick mat in the top few centimeters of soil.
The students concluded that grass reduces erosion mainly because its roots hold the soil together. Which new finding would most weaken this conclusion?
Hint
A weakening finding shows roots present but erosion still high, or erosion low without roots.
Step-by-step solution
Where the answer comes from: the students' explanation rests on their observation that "the grass roots formed a thick mat in the top few centimeters of soil." They think the roots, not the leaves, are what hold the soil.
Step by step:
To weaken a "roots do the work" explanation, look for a case where the roots are still there but the soil still washes away.
Trays with the grass cut off at ground level still have all their roots. If they lose about as much soil as bare trays, the roots alone did not protect the soil, and the leaves (which break the force of falling water) may matter more.
Deeper roots giving less loss, or roots-only trays protecting the soil, would support the conclusion.
Sandy soil is a different variable; it says nothing about the role of roots.
Correct answer: D.
Check: ask "If this were true, would I believe the roots explanation less?" Only the cut-grass finding makes you doubt it.
Tip: Weakening evidence shows the proposed cause present without the effect (or the effect without the cause).
Question 28 of 36
Counting Stars Near a City
Volunteers in a night-sky project counted how many stars they could see inside the same small patch of sky, the constellation Orion, on clear, moonless nights. Each volunteer observed from a different site at a different distance from the center of a large city. All observers waited 20 minutes outdoors so their eyes could adjust to the dark.
Distance from city center (km)
Stars counted in Orion
2
8
10
15
25
31
50
54
80
60
120
61
The city's streetlights, signs and buildings send light up into the sky, which makes the night sky glow.
The volunteers concluded that light from the city is what reduces the number of stars people can see near the city. Which new finding would most weaken this conclusion?
Hint
What should happen near the city if the city's lights were switched off? Look for the opposite.
Step-by-step solution
Where the answer comes from: the table shows "| 2 | 8 |": only 8 stars were counted at 2 km. The passage notes "The city's streetlights, signs and buildings send light up into the sky, which makes the night sky glow."
Step by step:
If city light is the cause, turning the lights off should let a nearby observer see many more stars.
A citywide power outage is a natural test. If the 2-km count stayed at about 8, removing the light did not help, so something else (such as haze or dust over the city) must be limiting the view. That weakens the conclusion.
A count of 52 during the outage would support the conclusion.
The same pattern near another city also supports it.
Haze lowering counts far from the city is a different factor; it does not show that city light is unimportant.
Correct answer: C.
Check: weakening evidence removes the suggested cause and finds that the effect does not change.
Tip: Look for a "switch-off" test: if taking away the cause leaves the effect in place, the cause is in doubt.
Question 29 of 36
Stretching a Spring
A student hung slotted masses from a steel spring and measured the spring's total length with a meter stick after each mass was added.
Mass hung (g)
Spring length (cm)
0
10.0
100
12.5
200
15.0
300
17.5
400
20.0
500
22.0
600
23.0
After loads of up to 400 g were removed, the spring went back to 10.0 cm. After the 600-g load was removed, the spring stayed at 11.2 cm. A spring that is stretched too far can be permanently bent out of shape.
Which conclusion about how the spring stretched is best supported by the student's measurements?
Hint
Find the change in length for each 100 g step. Does it stay the same all the way to 600 g?
Step-by-step solution
Where the answer comes from: the table shows the length rising 2.5 cm per 100 g up to "| 400 | 20.0 |", then "| 500 | 22.0 |" and "| 600 | 23.0 |". After the 600-g load, "the spring stayed at 11.2 cm."
Step by step:
Steps from 0 to 400 g: 10.0 → 12.5 → 15.0 → 17.5 → 20.0, each +2.5 cm.
From 400 to 500 g: +2.0 cm. From 500 to 600 g: +1.0 cm. The stretch per 100 g got smaller.
The spring did not return to 10.0 cm after 600 g, so it was permanently changed.
Doubling the mass from 100 g to 200 g changed the length from 12.5 to 15.0 cm, not to 25.0 cm.
Correct answer: B.
Check: a conclusion that uses the words "always" or "no matter how much" fails if even one row breaks the pattern, and the 500- and 600-g rows do.
Tip: Watch for where a pattern stops. A supported conclusion states the range in which it holds.
Question 30 of 36
Select allTwo Walking Studies
Study 1. Researchers recruited 80 adults, ages 40 to 60, who had mildly high blood pressure. Half were assigned to a walking group and half to a no-exercise group. The walking group walked briskly for 30 minutes, 5 days a week, for 8 weeks. Average systolic blood pressure (the top number) in the walking group fell from 138 to 131 mmHg. In the no-exercise group, it went from 139 to 138 mmHg.
Study 2. A different team recruited 120 adults of the same ages, who also had mildly high blood pressure. Half walked for 30 minutes once a week for 8 weeks; the other half did not exercise. Average systolic blood pressure in the walking group went from 137 to 136 mmHg. In the no-exercise group, it stayed at 138 mmHg.
In both studies, trained nurses measured blood pressure using the same method, and people were asked not to change their diets.
Select all of the new findings that would weaken the explanation that walking frequency is the reason the two studies got different results.
Select all that apply.
Hint
A finding weakens the explanation if it gives another difference between the studies, or shows frequent walking without the benefit.
Step-by-step solution
Where the answer comes from: Study 1 "walked briskly for 30 minutes, 5 days a week." In Study 2, "Half walked for 30 minutes once a week" (the pace is not given). Both groups "were asked not to change their diets."
Step by step:
A — weakens. If Study 2 walked slowly, pace (intensity) is a second difference between the studies and could explain the results instead of frequency.
B — weakens. If walking 5 days a week gave no drop in a repeat study, Study 1's result may not hold up, and frequency may not be the key.
C — supports. A 3-day schedule giving a middle-sized drop fits the frequency explanation.
D — does not weaken. Measuring at the same time of day makes the studies more alike, so it removes a possible difference rather than adding one.
E — weakens. Cutting salt lowers blood pressure. If only Study 1 walkers did it, diet could explain their larger drop.
Correct answers: A, B and E.
Check: each correct choice either adds a new difference between the studies or removes the effect where frequent walking happened.
Tip: Test each choice on its own. Alternative causes and failed repeats weaken; matching middle results support.
Question 31 of 36
Drop-downThe Speed of Sound and Thunder
Sound travels through air at a speed that depends on the air's temperature. A physics class used published measurements of the speed of sound in dry air:
Air temperature (°C)
Speed of sound (m/s)
0
331
10
337
20
343
30
349
The speed rises by about 0.6 m/s for every 1 °C increase in temperature.
Light from a lightning flash reaches an observer almost instantly. The thunder from the same flash arrives later because sound is much slower than light. To estimate how far away lightning struck, multiply the speed of sound by the number of seconds between the flash and the thunder.
On a 30 °C summer evening, a hiker counts 4 seconds between a lightning flash and its thunder.
The lightning struck away.
If the same 4-second delay were heard on a 0 °C day, the lightning would be .
Choose an answer from each menu.
Hint
Pick the speed for 30 °C from the table and multiply by the seconds. Then think about what a slower speed means for the same delay.
Step-by-step solution
Where the answer comes from: the table gives "| 30 | 349 |" and "| 0 | 331 |". The passage says to "multiply the speed of sound by the number of seconds between the flash and the thunder."
Step by step:
At 30 °C: 349 \text{ m/s} \times 4 \text{ s} = 1{,}396 m, which is about 1,400 m.
At 0 °C: 331 \text{ m/s} \times 4 \text{ s} = 1{,}324 m.
Sound is slower in cold air, so in the same 4 seconds it covers less distance. The lightning would be closer.
Correct answers: b1 = about 1,400 m; b2 = closer than on the 30 °C day.
Check: units: (m/s) × s = m, so multiplying is right. Dividing 349 by 4 gives about 87, which has the wrong units and is far too small for a 4-second delay.
Tip: When you apply a finding to a new case, pick the data row that matches the new conditions (here, 30 °C), then use the rule the passage gives.
Question 32 of 36
Hand Sanitizer in the Lab and in the Garage
Lab test. A company tested its hand sanitizer, which contains 70% alcohol. Volunteers first washed and dried their hands. Then technicians placed a known number of harmless test bacteria on each volunteer's palms. Each volunteer rubbed 3 mL of sanitizer over both hands until dry. Swabs showed that the sanitizer killed 99.9% of the test bacteria.
Field test. A health researcher repeated the test with auto mechanics at the end of a work shift. Their hands were covered with visible grease and dirt. The same harmless test bacteria were placed on their palms, and each mechanic rubbed in 3 mL of the same sanitizer until dry. This time, swabs showed that the sanitizer killed only about 60% of the test bacteria.
In both tests, the room temperature was 22 °C and the sanitizer came from the same batch.
Which new finding would best support the explanation that grease and dirt reduced how well the sanitizer worked in the field test?
Hint
The best support changes only the grease and dirt and shows the kill rate changing with it.
Step-by-step solution
Where the answer comes from: the sanitizer "killed 99.9% of the test bacteria" on clean hands but "killed only about 60% of the test bacteria" on greasy, dirty hands.
Step by step:
To support "grease and dirt are the cause," test the same mechanics with the grease removed.
If washing the grease off brings the kill rate back to 99.8%, then the grease and dirt, not the people or the job, made the difference.
A different sanitizer strength on clean hands says nothing about grease.
Shift length and stinging cuts are not measurements of how many bacteria were killed.
Correct answer: A.
Check: the best supporting evidence works like a controlled experiment: change only the suspected cause and watch the result change.
Tip: When asked for the "best" support, prefer a finding that isolates the proposed cause over one that is merely related to the topic.
Question 33 of 36
Drag and dropDistance From the Sun and Planet Temperatures
The table shows each planet's average distance from the sun and its average temperature. Distance is given in astronomical units (AU); 1 AU is Earth's average distance from the sun. Jupiter has no solid surface, so its temperature is measured at the cloud tops.
Planet
Distance from sun (AU)
Average temperature (°C)
Atmosphere
Mercury
0.39
167
almost none
Venus
0.72
464
very thick, mostly carbon dioxide
Earth
1.00
15
moderate, mostly nitrogen and oxygen
Mars
1.52
−65
thin, mostly carbon dioxide
Jupiter
5.20
−110
thick, mostly hydrogen and helium
Carbon dioxide is a greenhouse gas: it traps heat near a planet's surface.
Drag each statement into the correct column.
Supported by the data
Not supported by the data
Tap a tile, then tap a box to place it.
Hint
A statement is supported only if the table (or the sentence about carbon dioxide) backs it up for every planet it mentions.
Step-by-step solution
Where the answer comes from: the table rows (for example "| Mercury | 0.39 | 167 | almost none |", "| Mars | 1.52 | −65 | thin, mostly carbon dioxide |" and "| Jupiter | 5.20 | −110 | thick, mostly hydrogen and helium |") and the sentence "Carbon dioxide is a greenhouse gas: it traps heat near a planet's surface."
Supported:
Mercury, the closest planet, is not the hottest. Venus (464 °C) is hotter than Mercury (167 °C).
The farthest planet listed is the coldest. Jupiter (5.20 AU) has the lowest temperature, −110 °C.
A thick carbon dioxide atmosphere may help explain Venus's heat. Venus has a very thick CO₂ atmosphere, and CO₂ traps heat. The word "may" keeps the claim modest.
Not supported:
4. Mars is cold because it has no carbon dioxide. Mars's atmosphere is mostly carbon dioxide.
5. Closer planets always have thicker atmospheres. Mercury, the closest, has almost none.
6. Earth is the only planet with water. The table gives no information about water.
Check: for each statement, point to the exact row that backs it up. If you cannot, it is not supported.
Tip: "Not supported" includes statements the data contradict and statements the data never address.
Question 34 of 36
Two Studies of a Pesticide and Frogs
Study 1 (laboratory). Researchers raised leopard frog tadpoles in tanks of water containing different concentrations of pesticide X, measured in parts per million (ppm). Each tank held 50 tadpoles. All tanks had the same temperature, food and light. The researchers recorded the percent of tadpoles that survived to become adult frogs.
Pesticide X in tank water
Survival to adulthood
0 ppm (clean water)
82%
0.01 ppm
80%
0.05 ppm
61%
0.10 ppm
48%
Study 2 (field survey). A second team counted adult leopard frogs at 12 ponds. Six ponds were next to farms that spray pesticide X. Six ponds were in a nature preserve far from any farm. The farm ponds averaged 140 adult frogs per pond, and the preserve ponds averaged 145 adult frogs per pond. Water samples showed that the farm ponds contained an average of 0.008 ppm of pesticide X, and the preserve ponds contained none.
A news report said, "Lab and field studies disagree about pesticide X."
A new farm will spray pesticide X near a pond that now has no pesticide. Computer models predict that runoff will raise pesticide X in the pond to about 0.10 ppm each spring. Based on both studies, which prediction is best supported?
Hint
Which study tested conditions closest to 0.10 ppm? Then use its numbers, without claiming more than they show.
Step-by-step solution
Where the answer comes from: Study 1 found 82% survival in clean water and 48% at 0.10 ppm. Study 2's farm ponds had only 0.008 ppm, far below 0.10 ppm.
Step by step:
The new pond's predicted level (0.10 ppm) matches a concentration tested in Study 1, so Study 1 is the relevant finding.
Survival would fall from about 82% to about 48%: 82 - 48 = 34 percentage points, or 34 \div 82 \approx 0.41, about 41% fewer tadpoles surviving.
Study 2 cannot be used to predict "no change," because its ponds had about one-twelfth of that concentration (0.10 \div 0.008 = 12.5).
With fewer young frogs surviving each spring, the adult population would likely shrink over time.
Correct answer: A. Choice B misapplies Study 2. Choice C overreaches: 48% survival is not zero, and nothing supports extinction "within one year." Choice D has no support at all.
Tip: When you apply findings to a new case, use the study whose conditions match the new case, and pick the prediction that stays closest to what the data actually show.
Question 35 of 36
Select allHand Sanitizer in the Lab and in the Garage
Lab test. A company tested its hand sanitizer, which contains 70% alcohol. Volunteers first washed and dried their hands. Then technicians placed a known number of harmless test bacteria on each volunteer's palms. Each volunteer rubbed 3 mL of sanitizer over both hands until dry. Swabs showed that the sanitizer killed 99.9% of the test bacteria.
Field test. A health researcher repeated the test with auto mechanics at the end of a work shift. Their hands were covered with visible grease and dirt. The same harmless test bacteria were placed on their palms, and each mechanic rubbed in 3 mL of the same sanitizer until dry. This time, swabs showed that the sanitizer killed only about 60% of the test bacteria.
In both tests, the room temperature was 22 °C and the sanitizer came from the same batch.
Select all of the findings that would support the idea that grease and dirt, not the sanitizer itself, caused the lower kill rate in the field test.
Select all that apply.
Hint
Supporting findings show the sanitizer working when hands are clean, or failing when grease is added. Ignore findings about other factors.
Step-by-step solution
Where the answer comes from: the lab volunteers' hands were clean ("Volunteers first washed and dried their hands."), while the mechanics' "hands were covered with visible grease and dirt." The question asks which findings point to the grease and dirt as the cause.
Step by step:
A — supports. Clean-handed office workers in the field setting, with the same batch, got 99.9%. The sanitizer works outside the lab when hands are clean.
B — supports. Adding grease to clean lab hands dropped the kill rate to about 62%, close to the field result. Grease alone reproduces the problem.
C — does not support. Warmer hands would be a different possible cause, which competes with the grease explanation.
D — does not support. A label claim is not new evidence; the company's own claim does not test grease.
E — supports. Mechanics with clean, gloved hands got 99.7%, so the job itself was not the problem; the grease was.
Correct answers: A, B and E.
Check: every correct choice changes only how clean the hands are and shows the kill rate moving with it.
Tip: In select-all questions, judge each choice separately. Alternative causes and claims without data are traps.
Question 36 of 36
Warm Water and Coral Bleaching
Corals are animals that get much of their food from tiny algae living inside their tissues. When the water gets too warm, many corals push the algae out and turn white. This is called bleaching. Bleached corals can recover if the water cools in time, but many die.
Marine scientists surveyed coral colonies at reef sites during one summer and recorded the highest water temperature reached at each site.
Peak summer water temperature (°C)
Coral colonies bleached (%)
28
2
29
5
30
18
31
47
32
81
Each row is the average of several sites that reached that peak temperature.
One scientist claims that warm water by itself determines how much coral bleaches. Which new finding would most weaken this claim?
Hint
To weaken "temperature by itself," find a case where the temperature is the same but the bleaching is very different.
Step-by-step solution
Where the answer comes from: the survey "recorded the highest water temperature reached at each site" and found, for example, "| 32 | 81 |": 81% of colonies bleached at 32 °C.
Step by step:
If temperature alone decided bleaching, colonies at the same temperature should bleach about equally.
Shaded colonies at 32 °C bleaching far less than unshaded colonies at the same 32 °C shows that something besides temperature (here, sunlight) also matters. That directly weakens "by itself."
92% bleaching at 33 °C continues the temperature trend and supports the role of temperature.
Bleaching that starts a few days after the peak temperature fits temperature as a cause.
Aquarium corals bleaching at 32 °C also supports temperature as a cause.
Correct answer: A.
Check: none of the wrong choices shows a difference in bleaching when temperature is held the same.
Tip: Claims with "alone," "only" or "by itself" are weakened by any case where the factor stays the same but the result changes.