Crickets are cold-blooded, so their bodies warm and cool with the air around them. A naturalist counted how many times field crickets chirped in 15 seconds on evenings with different air temperatures. The graph shows the average number of chirps at each temperature.
According to the graph, about how many times do crickets chirp in 15 seconds when the air temperature is 70 °F?
Hint
Find 70 on the x-axis, go straight up to the line, then straight across to the y-axis.
Step-by-step solution
Read the graph directly. Find 70 °F on the x-axis (air temperature), move up to the data point, then across to the y-axis (chirps in 15 seconds). The point sits on the gridline for 30 chirps.
Step by step:
The x-axis shows temperature and the y-axis shows chirps, so start on the x-axis at 70.
Go up to the line: the point is at the height of the 30 gridline.
Check the neighbors: 65 °F gives 25 chirps and 75 °F gives 35, so 30 at 70 °F fits the pattern.
Choice D (70) repeats the temperature, which is the x-value, not the answer.
The correct answer is C, 30 chirps.
Tip: Always say the axes aloud ("temperature across, chirps up") before reading a value. It stops you from reporting the wrong axis.
Question 2 of 36
Crickets as Thermometers
Crickets are cold-blooded, so their bodies warm and cool with the air around them. A naturalist counted how many times field crickets chirped in 15 seconds on evenings with different air temperatures. The graph shows the average number of chirps at each temperature.
Based on the graph, about how many chirps in 15 seconds would you expect at an air temperature of 62 °F?
Hint
62 °F lies between two measured temperatures. Where between them does it fall?
Step-by-step solution
62 °F was not measured, so we interpolate: estimate a value between two measured points.
Step by step:
The nearest measured points are 60 °F → 20 chirps and 65 °F → 25 chirps.
Between them, the line rises 5 chirps over 5 °F, which is 1 chirp per °F.
62 °F is 2 °F above 60 °F, so add 2 chirps: 20 + 2 = 22 chirps.
Check: 22 is between 20 and 25, and closer to 20, just as 62 is closer to 60 than to 65.
The correct answer is A, about 22 chirps.
Tip: Interpolation means estimating inside the range of the data. The answer must fall between the two neighboring values. Any choice outside that range (like 25 or 62) is wrong.
Question 3 of 36
Crickets as Thermometers
Crickets are cold-blooded, so their bodies warm and cool with the air around them. A naturalist counted how many times field crickets chirped in 15 seconds on evenings with different air temperatures. The graph shows the average number of chirps at each temperature.
Which statement best describes the trend shown in the graph?
Hint
Look at how many chirps are added for each 5 °F step.
Step-by-step solution
Describe both the direction and the shape of the data.
Step by step:
Direction: as temperature rises from 55 °F to 80 °F, chirps rise from 15 to 40. The trend is upward (eliminates B).
Shape: each 5 °F step adds exactly 5 chirps (15 → 20 → 25 → 30 → 35 → 40). Equal steps make a straight line, a steady (linear) increase. It does not level off (eliminates C).
Choice D adds a cause-and-effect claim the graph cannot show. The data are about air temperature, and nothing in the study made crickets chirp on purpose.
The correct answer is A.
Tip: Equal changes in y for equal changes in x = a straight line (linear). Changes that grow larger = curving up; changes that shrink = leveling off.
Question 4 of 36
Heating a Pot of Water
A student heated 1 liter of water in a pot on a stove that stayed at the same setting the whole time. She measured the water's temperature with a thermometer every 2 minutes, starting when she turned on the burner. The experiment took place at sea level, where pure water boils at 100 °C. The water began to boil at the 10-minute reading.
According to the graph, what was the temperature of the water after 4 minutes of heating?
Hint
Start at 4 on the time axis, go straight up to the line, then straight across to the temperature axis.
Step-by-step solution
Read the value directly from the graph. The x-axis is time heated (minutes) and the y-axis is water temperature (°C). Find 4 minutes on the x-axis and move straight up to the data point. From there, move across to the y-axis: the point sits a little above the 50 °C gridline, at 52 °C.
Step by step:
Name the axes: time across, temperature up.
Locate 4 on the time axis.
Go up to the point and across: 52 °C.
The correct answer is 52 °C.
Check: The temperature rises 16 °C every 2 minutes. Starting at 20 °C, after 2 minutes it is 36 °C and after 4 minutes it is 36 + 16 = 52 °C. That matches.
Tip: Before reading any graph value, say the two axes out loud. It keeps you from reading the wrong point or the wrong axis.
Question 5 of 36
Heating a Pot of Water
A student heated 1 liter of water in a pot on a stove that stayed at the same setting the whole time. She measured the water's temperature with a thermometer every 2 minutes, starting when she turned on the burner. The experiment took place at sea level, where pure water boils at 100 °C. The water began to boil at the 10-minute reading.
The student did not take a reading at 5 minutes. Based on the graph, what was the water temperature most likely at 5 minutes?
Hint
5 minutes is halfway between two readings. What temperature is halfway between them?
Step-by-step solution
This is interpolation: estimating a value between two measured points. The graph shows 52 °C at 4 minutes and 68 °C at 6 minutes. Because the points lie on a straight line, the temperature at 5 minutes should be halfway between them.
Step by step:
Find the readings on each side of 5 minutes: 52 °C (4 min) and 68 °C (6 min).
Find the halfway value: \dfrac{52 + 68}{2} = 60 °C.
On the graph, the line crosses the 5-minute gridline exactly at the 60 °C gridline.
The correct answer is 60 °C.
Check: The water warms 8 °C per minute. From 20 °C, after 5 minutes: 20 + 5 \times 8 = 60 °C. Same answer.
Tip: For a value between two data points, use the points on both sides. Picking just the nearest point gives a value that is too high or too low.
Question 6 of 36
Drop-downTrees, Traffic and Asthma
A city health department compared six neighborhoods. For each one, it recorded the percent of land shaded by trees, the number of emergency-room (ER) visits for asthma per 10,000 residents in one year, and the distance from the center of the neighborhood to the nearest major highway. Exhaust from cars and trucks contains tiny particles and gases that can trigger asthma attacks.
Neighborhood
Tree cover (%)
Asthma ER visits (per 10,000 residents)
Distance to nearest highway (km)
Ash Park
10
62
0.3
Brookside
15
55
0.5
Cedar Hill
22
41
1.2
Dover
30
35
2.0
Elmwood
38
24
3.5
Fairview
45
18
4.8
Complete the sentences that describe the data.
As tree cover increases from one neighborhood to the next, the number of asthma ER visits . This means tree cover and asthma ER visits have a relationship.
Choose an answer from each menu.
Hint
Read down the tree-cover column and the asthma column together. When one goes up, what does the other do?
Step-by-step solution
Read the two columns side by side. Tree cover rises from 10% (Ash Park) to 45% (Fairview). Over the same rows, asthma ER visits fall from 62 to 55 to 41 to 35 to 24 to 18.
Step by step:
Tree cover goes up in every row.
Asthma visits go down in every row.
When one variable increases as the other decreases, the relationship is called negative (or inverse).
The correct answers are decreases and negative.
Check: A positive relationship would mean both columns climb together, like elevation and snowfall. Here they move in opposite directions.
Tip: "Negative" in a relationship does not mean bad. It only describes direction: one goes up while the other goes down. Describing a trend also does not prove that one variable causes the other.
Question 7 of 36
Fill inCharging a Phone Battery
A student tested how quickly a new charger recharges her phone. She turned the phone off, plugged it in when the battery showed 20%, and recorded the battery level every 10 minutes. The battery level is shown as a percent of a completely full battery (100%).
Time charging (minutes)
0
10
20
30
40
Battery level (%)
20
32
44
56
68
If the pattern in the table continues, what will the battery level be after 50 minutes of charging?
%
Type your answer in the box. Fractions like 3/4 are fine.
Hint
How much does the battery level go up every 10 minutes?
Step-by-step solution
Find the pattern, then extend it one step past the data.
Evidence from the table: 20% → 32% → 44% → 56% → 68%, one reading every 10 minutes.
Check: Rate = 1.2 points per minute. 20 + 1.2 \times 50 = 20 + 60 = 80%. Same answer, and 80% is still below the 100% maximum, so the prediction is reasonable.
Tip: For a steady pattern, check the change between every pair of readings, not just the first pair, before you extend it.
Question 8 of 36
Counting Tornadoes
Weather scientists compared tornado reports from five neighboring counties over 10 years (2015–2024). The counties are about the same size and have similar land. Radar records show that the five counties had about the same number of strong thunderstorms during those years. A tornado is added to the official record only if someone sees it or its damage and reports it.
County
Population density (people per square mile)
Trained storm spotters
Tornadoes reported, 2015–2024
Avery
15
4
6
Belmont
40
10
11
Corwin
90
22
19
Delling
160
41
27
Easton
300
75
34
According to the table, how many tornadoes were reported in the county that had 41 trained storm spotters?
Hint
Find 41 in the spotters column, then read across that same row.
Step-by-step solution
Read the table carefully, one row at a time. In the column Trained storm spotters, the value 41 appears in the row for Delling County. Move across that row to the last column, Tornadoes reported, 2015–2024: the value is 27.
Step by step:
Find the right column: trained storm spotters.
Find 41 in that column (Delling).
Stay in the Delling row and read the tornado column: 27.
The correct answer is 27 tornadoes.
Check: Delling's population density is 160 — a different column. Corwin (22 spotters) reported 19 tornadoes, and Easton (75 spotters) reported 34. Only Delling has 41 spotters.
Tip: Use a finger or a straight line on your erasable note board to stay on one row. Most table-reading errors come from sliding to the row above or below.
Question 9 of 36
Fill inGrowth of a Bacteria Culture
A microbiologist placed 100 bacteria cells in a flask of fresh nutrient broth kept at 37 °C. She counted the number of cells every hour for 4 hours.
Time (hours)
Number of bacteria
0
100
1
200
2
400
3
800
4
1,600
The flask holds a fixed amount of broth, and no new nutrients are added during the experiment.
If the pattern in the table continues for one more hour, how many bacteria will be in the flask at 5 hours?
bacteria
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Is the population growing by adding the same number each hour, or by multiplying by the same number?
Step-by-step solution
First find the pattern in the table, then extend it one step.
Step by step:
Compare each hour with the one before: 100 → 200 → 400 → 800 → 1,600.
The differences (100, 200, 400, 800) are not equal, so the pattern is not "add the same amount."
The ratios are equal: 200 \div 100 = 2, 400 \div 200 = 2, 1{,}600 \div 800 = 2. The population doubles every hour.
Extend one hour: 1{,}600 \times 2 = 3{,}200.
At 5 hours there would be 3,200 bacteria.
A common error is to add 800 (the last increase) to get 2,400. That treats a doubling pattern as if it were a steady, linear one.
Tip: Check both differences and ratios. Constant differences = linear growth; constant ratios = exponential growth (doubling, tripling).
Question 10 of 36
Growth of a Bacteria Culture
A microbiologist placed 100 bacteria cells in a flask of fresh nutrient broth kept at 37 °C. She counted the number of cells every hour for 4 hours.
Time (hours)
Number of bacteria
0
100
1
200
2
400
3
800
4
1,600
The flask holds a fixed amount of broth, and no new nutrients are added during the experiment.
Which statement best describes how the number of bacteria changed during the 4 hours?
Hint
Compare the increase in the first hour with the increase in the last hour.
Step-by-step solution
Test each description against the numbers in the table: "| 0 | 100 |", "| 1 | 200 |" … "| 4 | 1,600 |".
Step by step:
Increases each hour: +100, +200, +400, +800. The increase gets bigger every hour, so the population is not growing by the same amount (eliminates A and D).
Ratios each hour: ×2, ×2, ×2, ×2. The population doubles every hour (B).
Choice C says growth slowed down, but the increases are getting larger, not smaller.
D comes from taking the first increase (100) and assuming it repeats, which the later hours contradict (200 → 400 is an increase of 200).
The correct answer is B.
Tip: When a quantity doubles over equal time periods, it grows faster and faster. On a graph it curves upward more and more steeply.
Question 11 of 36
Growth of a Bacteria Culture
A microbiologist placed 100 bacteria cells in a flask of fresh nutrient broth kept at 37 °C. She counted the number of cells every hour for 4 hours.
Time (hours)
Number of bacteria
0
100
1
200
2
400
3
800
4
1,600
The flask holds a fixed amount of broth, and no new nutrients are added during the experiment.
A student uses the doubling pattern to predict that the flask will contain trillions of bacteria after 48 hours. Why is this prediction most likely incorrect?
Hint
Reread the last sentence of the passage. What will eventually run out?
Step-by-step solution
The passage's final sentence is the key: "The flask holds a fixed amount of broth, and no new nutrients are added during the experiment."
Step by step:
The data cover only 0–4 hours, a period when there was plenty of food and space.
Predicting 48 hours ahead is extrapolation far beyond the data.
As the bacteria multiply, they use up the fixed supply of nutrients and produce wastes. Growth must slow and eventually stop. Real cultures level off (and may decline) instead of doubling forever.
Choice A contradicts the table, which shows growth in every hour. Choice C is false: the experiment kept the flask at 37 °C, and nothing says the temperature changed. Choice D reverses the data, since the population grew.
The correct answer is B.
Tip: Be cautious about predictions far outside the data range. Ask what conditions might change, such as limited food, space or time.
Question 12 of 36
Ice Cream and Sunburn
A city health department compared two sets of data for each month of one year: the number of ice cream cones sold in the city (in thousands) and the number of people treated at city clinics for sunburn. Each point on the scatter plot represents one month.
A local newspaper reported the data under the headline "Is ice cream causing sunburns?"
Which conclusion is best supported by the scatter plot?
Hint
A scatter plot can show that two things change together. Can it show that one causes the other?
Step-by-step solution
Read the overall pattern. For eleven of the twelve months, points with more ice cream cones sold also have more sunburn cases: from (5, 2) at the lower left to (28, 30) at the upper right. That is a positive correlation.
Step by step:
The data show the two variables tend to rise and fall together (choice C).
A correlation alone does not prove cause and effect, so choice A goes too far. The newspaper's headline asks exactly that question, but the data cannot answer it.
Choice B describes a negative correlation, the opposite of the plot.
Choice D says "every month," but one month (22 thousand cones, 3 sunburn cases) does not follow the pattern, so the claim about every month is too absolute.
The correct answer is C.
Tip: Data like these can show that two variables are related, but only a controlled experiment can show that one causes the other. Prefer choices with careful words like "tended to."
Question 13 of 36
Ice Cream and Sunburn
A city health department compared two sets of data for each month of one year: the number of ice cream cones sold in the city (in thousands) and the number of people treated at city clinics for sunburn. Each point on the scatter plot represents one month.
A local newspaper reported the data under the headline "Is ice cream causing sunburns?"
Which factor most likely explains why ice cream sales and sunburn cases rise and fall together?
Hint
Think of something that would make people buy more ice cream AND spend more time in strong sun.
Step-by-step solution
When two variables rise and fall together but neither causes the other, a third variable may be causing both. The plot has one point per month, which hints at seasons.
Step by step:
Hot, sunny months make people want cold treats, so ice cream sales go up.
The same hot, sunny months mean more time outdoors in strong sunlight, so sunburns go up.
So sunny, warm weather (choice D) explains both trends.
Choice A (price of ice cream) might affect sales but has no link to sunburn. Choice B (number of clinics) might affect how many people are treated but not ice cream sales. Choice C (flavors) affects neither sunburn nor the seasonal pattern.
The correct answer is D.
Tip: For any correlation, ask: "Is there a hidden factor that could drive both?" Weather, age, population size and time are common hidden factors.
Question 14 of 36
Hot spotIce Cream and Sunburn
A city health department compared two sets of data for each month of one year: the number of ice cream cones sold in the city (in thousands) and the number of people treated at city clinics for sunburn. Each point on the scatter plot represents one month.
A local newspaper reported the data under the headline "Is ice cream causing sunburns?"
Click the point on the scatter plot that does not follow the overall trend of the data.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Picture a band running through most of the points. Which point is far from that band?
Step-by-step solution
An outlier is a data point that lies far from the overall pattern.
Step by step:
Find the trend: eleven points rise together, from (5, 2) up to (28, 30). For these months, the number of sunburn cases is close to the number of thousands of cones sold.
Scan for a point far from that band. The point at 22 thousand cones and only 3 sunburn cases sits far below where the trend predicts (about 20 cases).
The highest point (28, 30) and the lowest point (5, 2) are extremes, but they fit the trend, so they are not outliers.
The correct point is (22, 3). Something unusual happened that month. For example, ice cream sales could have been high for a reason unrelated to sunshine, such as a holiday event in a cold month. Scientists investigate outliers instead of ignoring them.
Tip: Outliers are judged by distance from the pattern, not by being the biggest or smallest value.
Question 15 of 36
Fill inHeating a Pot of Water
A student heated 1 liter of water in a pot on a stove that stayed at the same setting the whole time. She measured the water's temperature with a thermometer every 2 minutes, starting when she turned on the burner. The experiment took place at sea level, where pure water boils at 100 °C. The water began to boil at the 10-minute reading.
The water's temperature rose at a steady rate. At what time did the water reach 44 °C?
minutes
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First find how many degrees the water warmed each minute. Then ask how many minutes it took to warm from the start to 44 °C.
Step-by-step solution
Here you know the output (44 °C) and must work backwards to the input (time).
Evidence from the graph: 20 °C at 0 minutes, 36 °C at 2 minutes, 52 °C at 4 minutes.
Step by step:
Rate of warming: \dfrac{36 - 20}{2} = 8 °C per minute.
Warming needed: 44 - 20 = 24 °C.
Time needed: \dfrac{24}{8} = 3 minutes.
The correct answer is 3 minutes.
Check: 44 °C is exactly halfway between 36 °C (2 min) and 52 °C (4 min), so the time should be halfway between 2 and 4 minutes: 3 minutes. Both methods agree.
Tip: When a question gives a y-value and asks for x, go from the y-axis across to the line, then down — or use the rate. Do not divide 44 by 8; that forgets the 20 °C starting temperature.
Question 16 of 36
Two Digestive Enzymes and pH
Enzymes are proteins that speed up chemical reactions in living things. A biochemist measured the activity of two digestive enzymes at different pH levels. (pH tells how acidic or basic a solution is: below 7 is acidic, 7 is neutral, and above 7 is basic.)
Enzyme A comes from the stomach, where the contents are strongly acidic, about pH 2. Enzyme B comes from the small intestine, where the contents are slightly basic, about pH 8. All tests were run at body temperature, 37 °C. Activity is given in relative units; a higher number means the enzyme worked faster.
pH
1
2
3
4
5
6
7
8
9
Enzyme A activity
40
90
70
35
10
2
0
0
0
Enzyme B activity
0
0
5
15
35
60
85
95
70
Which statement correctly compares the two enzymes?
Hint
Find the highest number in each row and note the pH where it occurs. Then check each choice against the table, one cell at a time.
Step-by-step solution
Compare the two rows of the table. Enzyme A has its highest activity, 90 units, at pH 2. Enzyme B has its highest activity, 95 units, at pH 8.
Step by step:
Enzyme A peak: pH 2 (90). Enzyme B peak: pH 8 (95).
pH 2 is lower than pH 8, so Enzyme A is most active at a lower pH.
Test the others: at pH 7, A = 0 and B = 85, so A is not more active. At pH 9, B still has 70 units, so B has not stopped. At pH 5, A = 10 and B = 35, so neither is at its peak.
The correct answer is Enzyme A is most active at a lower pH than Enzyme B.
Check: The passage says A works in the acidic stomach and B in the slightly basic intestine — consistent with A's peak at a lower pH.
Tip: When a table has two rows, put your finger on the correct row before reading. Many wrong answers come from reading the other enzyme's number.
Question 17 of 36
Select allTwo Digestive Enzymes and pH
Enzymes are proteins that speed up chemical reactions in living things. A biochemist measured the activity of two digestive enzymes at different pH levels. (pH tells how acidic or basic a solution is: below 7 is acidic, 7 is neutral, and above 7 is basic.)
Enzyme A comes from the stomach, where the contents are strongly acidic, about pH 2. Enzyme B comes from the small intestine, where the contents are slightly basic, about pH 8. All tests were run at body temperature, 37 °C. Activity is given in relative units; a higher number means the enzyme worked faster.
pH
1
2
3
4
5
6
7
8
9
Enzyme A activity
40
90
70
35
10
2
0
0
0
Enzyme B activity
0
0
5
15
35
60
85
95
70
A food scientist needs Enzyme B to work at an activity of at least 60 units. At which pH values in the table does Enzyme B meet this requirement? Select all that apply.
Select all that apply.
Hint
Use only the Enzyme B row. "At least 60" includes 60 itself.
Step-by-step solution
This question gives the required output (activity of 60 or more) and asks for the inputs (pH values) that produce it.
Keep each value that is 60 or greater: 60 (pH 6), 85 (pH 7), 95 (pH 8), 70 (pH 9).
Reject 15 (pH 4) and 35 (pH 5) — both are below 60.
"At least" means greater than or equal to, so pH 6 (exactly 60) counts.
The correct answers are pH 6, pH 7, pH 8 and pH 9.
Check: Notice that the answers are not just the pH values above the peak. Activity drops after pH 8, but at pH 9 it is still 70, which meets the requirement.
Tip: In threshold questions, circle the words at least, more than, no more than. They decide whether the boundary value counts.
Question 18 of 36
Hot spotTemperature and Elevation
In the lower atmosphere, air generally gets colder as you climb higher. A team of earth scientists placed ten weather stations at different elevations in the Granite Range, a fictional mountain range. Each station recorded the air temperature every hour for one year. The scatter plot shows each station's average yearly temperature.
Click the point that does not fit the pattern shown by the other stations.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Picture a straight line through most of the points. Which point is far away from that line?
Step-by-step solution
Nine of the ten points fall on one straight, downward line: the higher the station, the colder it is. For example, the station at 1,500 m averages 7 °C and the station at 2,000 m averages 4 °C.
Step by step:
Identify the overall trend: temperature drops about 3 °C for every 500 m of elevation.
Predict the station at 1,750 m from its neighbors: halfway between 7 °C and 4 °C, so about 5.5 °C.
The point at 1,750 m actually shows 12 °C — about 6.5 °C warmer than expected. It sits far above the line.
The correct answer is the point at 1,750 m and 12 °C.
Check: The highest point (250 m, 14.5 °C) and the lowest point (3,000 m, −2 °C) are extreme values, but they lie right on the line, so they are not outliers.
Tip: An outlier is a point that breaks the pattern, not simply the largest or smallest value. Scientists then look for a cause (a sheltered valley, a faulty sensor, a nearby heat source) before using it.
Question 19 of 36
Temperature and Elevation
In the lower atmosphere, air generally gets colder as you climb higher. A team of earth scientists placed ten weather stations at different elevations in the Granite Range, a fictional mountain range. Each station recorded the air temperature every hour for one year. The scatter plot shows each station's average yearly temperature.
The team plans to add a station on a peak at 3,500 meters. If the pattern continues, what average yearly temperature should they expect at that station?
Hint
The highest station is at 3,000 m. How much colder should it get over 500 more meters?
Step-by-step solution
This is extrapolation: extending a trend beyond the measured data. The highest station (3,000 m) averages −2 °C. The pattern drops 6 °C per 1,000 m, which is 3 °C per 500 m.
Step by step:
Extra elevation: 3{,}500 - 3{,}000 = 500 m.
Expected drop for 500 m: \dfrac{6}{2} = 3 °C.
Expected temperature: -2 - 3 = -5 °C.
The correct answer is −5 °C.
Check: Use the whole pattern: start at 16 °C at sea level and subtract 6 °C per 1,000 m. 16 - 3.5 \times 6 = 16 - 21 = -5 °C. Both methods agree.
Tip: When you extend a trend, keep the same direction and the same rate. Going higher keeps getting colder; it does not level off or reverse unless the data or passage say so.
Question 20 of 36
Fill inCharging a Phone Battery
A student tested how quickly a new charger recharges her phone. She turned the phone off, plugged it in when the battery showed 20%, and recorded the battery level every 10 minutes. The battery level is shown as a percent of a completely full battery (100%).
Time charging (minutes)
0
10
20
30
40
Battery level (%)
20
32
44
56
68
Assume the battery charged at a steady rate. After how many minutes of charging did the battery reach 50%?
minutes
Type your answer in the box. Fractions like 3/4 are fine.
Hint
50% falls between two readings in the table. Find the rate per minute first.
Step-by-step solution
You know the battery level (50%) and must find the time — work backwards.
Evidence from the table: 44% at 20 minutes and 56% at 30 minutes; the level rises 12 points every 10 minutes.
Step by step:
Rate: \dfrac{12}{10} = 1.2 percentage points per minute.
Increase needed from the start: 50 - 20 = 30 points.
Time: \dfrac{30}{1.2} = 25 minutes.
The correct answer is 25 minutes.
Check: 50% is exactly halfway between 44% (20 min) and 56% (30 min), so the time is halfway between 20 and 30 minutes: 25 minutes.
Tip: When you work backwards, subtract the starting value first. Dividing 50 by 1.2 (about 42 minutes) forgets that the battery began at 20%.
Question 21 of 36
Birds by Habitat and Season
Volunteers in a county bird count recorded how many different bird species they saw in four habitats. The same team surveyed each habitat for the same number of hours in spring (May) and again in winter (January).
Many birds that nest in the county in spring migrate south for the winter.
Which habitat had the smallest change in the number of bird species from spring to winter?
Hint
For each habitat, subtract the winter bar from the spring bar. Compare the differences, not the heights.
Step-by-step solution
The question asks about the change between seasons, so compute a difference for each habitat.
Evidence from the graph (spring − winter):
Forest: 34 - 20 = 14
Wetland: 42 - 15 = 27
Grassland: 18 - 12 = 6
Suburban: 22 - 19 = 3
Step by step:
List the four differences: 14, 27, 6, 3.
The smallest difference is 3, for the suburban habitat.
The correct answer is Suburban.
Check: Grassland has the shortest bars, but the question is about change, not size. Its change (6) is twice the suburban change (3).
Tip: In grouped bar graphs, decide first whether the question wants a value (one bar), a total, or a difference (gap between paired bars).
Question 22 of 36
Water Flea Heart Rates
Daphnia, or water fleas, are tiny pond animals with see-through bodies, so scientists can watch their hearts beat under a microscope. A Daphnia cannot control its own body temperature; its body stays at the temperature of the water around it. A biology student placed a different Daphnia in water at each of five temperatures and counted its heartbeats for one full minute. She ran three trials at each temperature.
Water temperature (°C)
Trial 1 (beats/min)
Trial 2 (beats/min)
Trial 3 (beats/min)
10
120
118
122
15
158
162
160
20
199
201
200
25
239
180
241
30
281
279
280
Which measurement in the table is most likely an error or an unusual result?
Hint
At each temperature, the three trials should be close to each other. Look for a trial that is far from the other two in its row.
Step-by-step solution
Repeated trials at the same temperature should give similar results. Compare each value with the other trials in its row and with the overall trend.
Step by step:
In most rows the three trials are within about 2 beats of each other (for example, 199, 201, 200 at 20 °C).
At 25 °C, Trials 1 and 3 are 239 and 241, but Trial 2 is 180 — about 60 beats lower.
The trend rises about 40 beats per 5 °C (160 → 200 → 240 → 280), so about 240 is expected at 25 °C. The value 180 does not fit.
The correct answer is 25 °C, Trial 2 (180 beats/min).
Check: 281 at 30 °C is the largest number in the table, but it fits both its row (279, 280) and the trend, so it is not an outlier.
Tip: An outlier is judged against its neighbors, not by being the biggest or smallest number. In a repeated-trials table, scan each row first.
Question 23 of 36
Phosphorus and Algae in Lake Marlow
Algal blooms are sudden, large growths of algae that can turn lake water green and use up oxygen that fish need. Phosphorus is a nutrient that helps algae grow. Scientists monitored Lake Marlow, a fictional lake, for 8 years. The graph shows the lake's average phosphorus level and the number of days each year with an algal bloom.
The table shows how much lawn fertilizer containing phosphorus was sold in the county around the lake. A county ban on selling lawn fertilizer that contains phosphorus took effect in the middle of Year 6.
Year
1
2
3
4
5
6
7
8
Phosphorus lawn fertilizer sold (tons)
110
125
140
160
175
60
40
35
Which statement is best supported by the graph alone?
Hint
The graph shows two lake measurements over time. Which choice describes only what those two lines show?
Step-by-step solution
Use only the two lines on the graph. Phosphorus: 20, 24, 29, 35, 41, 33, 26, 22 µg/L. Bloom days: 8, 12, 18, 25, 34, 30, 21, 15.
Step by step:
Both lines rise from Year 1 to Year 5, peak in Year 5, and fall through Year 8. Years with more phosphorus had more bloom days, so choice B is supported.
Choice A claims phosphorus caused the blooms. Phosphorus does feed algae (the passage says so), but a graph of two lines moving together cannot by itself prove that cause, so "graph alone" makes A too strong.
Choice C is false: bloom days fell after Year 5.
Choice D extrapolates to Year 10 and to zero. The decline is already slowing (drops of 8, 7, then 4 µg/L), so the graph does not support it.
The correct answer is B.
Tip: "Best supported by the data" usually means the most careful statement. A choice that describes the pattern beats a choice that claims a cause or predicts far into the future.
Question 24 of 36
Select allPhosphorus and Algae in Lake Marlow
Algal blooms are sudden, large growths of algae that can turn lake water green and use up oxygen that fish need. Phosphorus is a nutrient that helps algae grow. Scientists monitored Lake Marlow, a fictional lake, for 8 years. The graph shows the lake's average phosphorus level and the number of days each year with an algal bloom.
The table shows how much lawn fertilizer containing phosphorus was sold in the county around the lake. A county ban on selling lawn fertilizer that contains phosphorus took effect in the middle of Year 6.
Year
1
2
3
4
5
6
7
8
Phosphorus lawn fertilizer sold (tons)
110
125
140
160
175
60
40
35
Use the graph and the table. Which statements are supported by the data? Select all that apply.
Select all that apply.
Hint
Check each statement against exact numbers. Watch for words like 'only', 'zero' and 'every'.
Step-by-step solution
Check each statement against the numbers.
Step by step:
A: Fertilizer sales peak in Year 5 (175 tons), and lake phosphorus peaks in Year 5 (41 µg/L). ✓ Supported.
B: The ban started in the middle of Year 6. Lake phosphorus went 33 (Year 6) → 26 (Year 7) → 22 (Year 8): it decreased each year. ✓ Supported.
C: Bloom days fell to 30, 21 and 15, but never to zero. ✗
D: The data show phosphorus fell after the ban, but they cannot rule out other causes (for example, changes in rainfall or farming). "The only cause" goes beyond the evidence. ✗
E: Year 8 had 15 bloom days; Year 3 had 18. 15 < 18. ✓ Supported.
The correct answers are A, B and E.
Tip: When combining two sources, line them up by the shared variable (here, the year). Then reject any statement that uses absolute words (only, every, zero) the data cannot back up.
Question 25 of 36
Heating a Pot of Water
A student heated 1 liter of water in a pot on a stove that stayed at the same setting the whole time. She measured the water's temperature with a thermometer every 2 minutes, starting when she turned on the burner. The experiment took place at sea level, where pure water boils at 100 °C. The water began to boil at the 10-minute reading.
A classmate looks at the graph and predicts: "If the student keeps heating the water at the same setting, it will reach 132 °C at 14 minutes." Why is this prediction most likely wrong?
Hint
The classmate's arithmetic is fine. Think about what happens to water once it reaches 100 °C at sea level.
Step-by-step solution
The classmate extended the straight line beyond the data: 8 °C per minute for 4 more minutes gives 100 + 4 \times 8 = 132 °C. The math is correct, but the science makes the trend stop. The passage says the experiment was at sea level, "where pure water boils at 100 °C," and that "The water began to boil at the 10-minute reading."
Step by step:
A trend can only be extended while the conditions that produced it stay the same.
Once water boils, added heat energy changes liquid water into steam (a change of state). The temperature of the boiling water stays at about 100 °C.
So the line should flatten at 100 °C, not keep climbing to 132 °C.
The correct answer is the water stays at about 100 °C while it boils, because the energy goes into turning liquid water into steam.
Check: Liquid water at sea level is never at 132 °C in an open pot; steam escapes and the pot keeps boiling at 100 °C.
Tip: Before extending any trend, ask: "Is there a limit — a boiling point, a full battery, a 100% maximum, a food supply — that will stop this pattern?"
Question 26 of 36
Two Digestive Enzymes and pH
Enzymes are proteins that speed up chemical reactions in living things. A biochemist measured the activity of two digestive enzymes at different pH levels. (pH tells how acidic or basic a solution is: below 7 is acidic, 7 is neutral, and above 7 is basic.)
Enzyme A comes from the stomach, where the contents are strongly acidic, about pH 2. Enzyme B comes from the small intestine, where the contents are slightly basic, about pH 8. All tests were run at body temperature, 37 °C. Activity is given in relative units; a higher number means the enzyme worked faster.
pH
1
2
3
4
5
6
7
8
9
Enzyme A activity
40
90
70
35
10
2
0
0
0
Enzyme B activity
0
0
5
15
35
60
85
95
70
Which conclusion is best supported by the passage and the table?
Hint
Connect each enzyme's best pH from the table with the pH of the organ where the passage says it is found.
Step-by-step solution
Combine the table with the text. The passage says the stomach is "about pH 2" and the small intestine is "about pH 8." The table shows Enzyme A (stomach) peaks at pH 2 with 90 units, and Enzyme B (intestine) peaks at pH 8 with 95 units.
Step by step:
Enzyme A's best pH (2) matches the stomach's pH (2).
Enzyme B's best pH (8) matches the small intestine's pH (8).
So each enzyme works best at about the pH of the organ where it is found.
Eliminate overreaching or unsupported choices: activity does not always rise with pH (A falls after pH 2); "importance" was never measured; B at pH 2 has 0 activity.
The correct answer is each enzyme is most active at about the pH of the organ where it is found.
Check: Both data points needed for the conclusion are in the passage — nothing outside the data is assumed.
Tip: A supported conclusion uses only what the data show. Watch for words like always, more important, or predictions about conditions that contradict the table.
Question 27 of 36
Fill inTemperature and Elevation
In the lower atmosphere, air generally gets colder as you climb higher. A team of earth scientists placed ten weather stations at different elevations in the Granite Range, a fictional mountain range. Each station recorded the air temperature every hour for one year. The scatter plot shows each station's average yearly temperature.
Use the stations that fit the overall pattern. By how many degrees Celsius does the average yearly temperature drop for every 1,000 meters of elevation gained?
°C
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Choose two points on the line that are exactly 1,000 m apart, such as 1,000 m and 2,000 m.
Step-by-step solution
Find the rate of change from two points that follow the pattern.
Evidence from the scatter plot: 1,000 m → 10 °C and 2,000 m → 4 °C.
Step by step:
Change in elevation: 2{,}000 - 1{,}000 = 1{,}000 m.
Change in temperature: 10 - 4 = 6 °C (a drop).
Rate: \dfrac{6 \text{ °C}}{1{,}000 \text{ m}} = a drop of 6 °C per 1,000 m.
The correct answer is 6 °C.
Check: Try another pair: 500 m → 13 °C and 1,500 m → 7 °C. The drop is again 13 - 7 = 6 °C. Do not use the 1,750 m station; it is the outlier and would give a wrong rate.
Tip: To find a rate from a graph, pick two points that sit exactly on the trend and are easy to read. Avoid outliers — they distort the rate.
Question 28 of 36
Rain and Willow Creek
Willow Creek is a small stream fed mostly by rain. A volunteer watershed group recorded the total rainfall each month near the creek (bar graph) and the creek's average depth at a measuring post (table). Beginning in early August, farms upstream started pumping water out of the creek to irrigate their crops.
Month
Mar
Apr
May
Jun
Jul
Aug
Average creek depth (cm)
85
100
80
55
35
30
Which conclusion about August is best supported by both the graph and the table?
Hint
Compare July and August in the graph, then in the table. Do they change in the same direction? Then reread the last sentence of the passage.
Step-by-step solution
You need both sources. Graph: rainfall went from 20 mm in July to 30 mm in August — an increase. Table: creek depth went from 35 cm in July to 30 cm in August — a decrease. Text: "Beginning in early August, farms upstream started pumping water out of the creek."
Step by step:
More rain would normally raise the creek, yet the creek dropped.
Something else must be removing water. The passage names one: pumping for irrigation.
So the August data are consistent with pumping lowering the creek.
The correct answer is rainfall increased from July to August, but creek depth still decreased, which is consistent with water being pumped out upstream.
Check: July (20 mm), not August, had the least rain. The data cannot show exactly how many centimeters the pumping removed, because rain and pumping were acting at the same time.
Tip: In two-source questions, write down what each source says for the same time period before choosing. The right answer must agree with both.
Question 29 of 36
Charging a Phone Battery
A student tested how quickly a new charger recharges her phone. She turned the phone off, plugged it in when the battery showed 20%, and recorded the battery level every 10 minutes. The battery level is shown as a percent of a completely full battery (100%).
Time charging (minutes)
0
10
20
30
40
Battery level (%)
20
32
44
56
68
Her friend uses the table to predict: "After 2 hours (120 minutes) on this charger, the battery will show 164%." Which statement best explains why this prediction is not reasonable?
Hint
Check the friend's math first. Then reread how the battery level is measured in the passage.
Step-by-step solution
The friend extended the pattern: 20 + 1.2 \times 120 = 164%. The arithmetic is correct, but the passage defines the battery level as "a percent of a completely full battery (100%)." A battery cannot hold more than a full charge.
Step by step:
The steady pattern only describes the first 40 minutes.
At 1.2 points per minute, the battery would reach 100% after \dfrac{80}{1.2} \approx 67 minutes.
After that the level cannot keep rising; it stays at 100% (most phones also charge more slowly as they get close to full).
The correct answer is the level cannot go above 100%, so the straight-line pattern must stop once the battery is full.
Check: Any prediction above 100% contradicts the passage's own definition of the measurement.
Tip: Extrapolation is risky when it runs into a natural limit — 100% full, a boiling point, a food supply, or space in a container. Ask whether the answer is even possible.
Question 30 of 36
Hot spotBirds by Habitat and Season
Volunteers in a county bird count recorded how many different bird species they saw in four habitats. The same team surveyed each habitat for the same number of hours in spring (May) and again in winter (January).
Many birds that nest in the county in spring migrate south for the winter.
Click the winter bar for the habitat that lost the most bird species between spring and winter.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Find the habitat with the biggest gap between its spring bar and its winter bar. Then click that habitat's red (winter) bar.
Step-by-step solution
Two steps: compare the gaps, then click the right bar.
The question asks for the winter bar, which is red (right side of each pair).
Click the red wetland bar, labeled 15.
The correct answer is the wetland winter bar (15 species).
Check: The wetland spring bar (42) is the tallest bar on the graph, but it is the spring bar, not the winter one. The forest winter bar (20) is taller than the wetland winter bar, but forest lost only 14 species.
Tip: Read hot-spot directions twice. They often ask for one specific bar or point (here, the winter bar) after you identify the right group.
Question 31 of 36
Counting Tornadoes
Weather scientists compared tornado reports from five neighboring counties over 10 years (2015–2024). The counties are about the same size and have similar land. Radar records show that the five counties had about the same number of strong thunderstorms during those years. A tornado is added to the official record only if someone sees it or its damage and reports it.
County
Population density (people per square mile)
Trained storm spotters
Tornadoes reported, 2015–2024
Avery
15
4
6
Belmont
40
10
11
Corwin
90
22
19
Delling
160
41
27
Easton
300
75
34
Counties with more people reported more tornadoes. Which explanation for this pattern is best supported by the passage?
Hint
Reread the last two sentences of the passage: what do the radar records show, and how does a tornado get into the record?
Step-by-step solution
The pattern is a correlation: as population density and spotters increase (15 → 300 people per square mile; 4 → 75 spotters), reported tornadoes increase (6 → 34). But the passage gives two key facts:
"…the five counties had about the same number of strong thunderstorms."
"A tornado is added to the official record only if someone sees it or its damage and reports it."
Step by step:
If storms were about equal, the real number of tornadoes was probably similar in each county.
In places with few people, more tornadoes go unseen and unreported.
So more people and spotters lead to more reports, not more tornadoes.
The correct answer is more people and spotters mean more tornadoes are seen and reported, even though the counties probably had a similar number of tornadoes.
Check: This explanation agrees with every fact in the passage; each other choice contradicts the radar records or the rules for reporting.
Tip: Always ask how data were collected. When data depend on someone noticing an event, more observers can create a pattern that is not really there.
Question 32 of 36
Phosphorus and Algae in Lake Marlow
Algal blooms are sudden, large growths of algae that can turn lake water green and use up oxygen that fish need. Phosphorus is a nutrient that helps algae grow. Scientists monitored Lake Marlow, a fictional lake, for 8 years. The graph shows the lake's average phosphorus level and the number of days each year with an algal bloom.
The table shows how much lawn fertilizer containing phosphorus was sold in the county around the lake. A county ban on selling lawn fertilizer that contains phosphorus took effect in the middle of Year 6.
Year
1
2
3
4
5
6
7
8
Phosphorus lawn fertilizer sold (tons)
110
125
140
160
175
60
40
35
By about what percent did the number of algal bloom days decrease from Year 5 to Year 8?
Hint
Percent change = (change ÷ original value) × 100. The original value is the Year 5 value.
Step-by-step solution
Read the bloom-day values from the red line (right axis): Year 5 = 34 days, Year 8 = 15 days.
Step by step:
Change: 34 - 15 = 19 days fewer.
Divide by the original (starting) value, Year 5:
\text{percent decrease} = \frac{34 - 15}{34} \times 100 = \frac{19}{34} \times 100 \approx 55.9\%
Round: about 56%.
Traps: 19% treats the number of days as a percent. 44% is \tfrac{15}{34}, the part that remains, not the part that was lost. 127% divides by the new value (15) instead of the original.
Tip: In percent change, always divide by the value you started from. A percent decrease can never be more than 100%.
Question 33 of 36
Trees, Traffic and Asthma
A city health department compared six neighborhoods. For each one, it recorded the percent of land shaded by trees, the number of emergency-room (ER) visits for asthma per 10,000 residents in one year, and the distance from the center of the neighborhood to the nearest major highway. Exhaust from cars and trucks contains tiny particles and gases that can trigger asthma attacks.
Neighborhood
Tree cover (%)
Asthma ER visits (per 10,000 residents)
Distance to nearest highway (km)
Ash Park
10
62
0.3
Brookside
15
55
0.5
Cedar Hill
22
41
1.2
Dover
30
35
2.0
Elmwood
38
24
3.5
Fairview
45
18
4.8
A newspaper headline says, "Trees Prevent Asthma." Based on the passage and the table, which factor could also explain why neighborhoods with more trees have fewer asthma ER visits?
Hint
Look at the column the headline ignores. How does it change from Ash Park to Fairview, and what does the passage say about it?
Step-by-step solution
A correlation (two things changing together) does not prove that one causes the other, because a third factor might be involved. The table has a third column: distance to the nearest highway. It rises from 0.3 km in Ash Park to 4.8 km in Fairview — the same rows where tree cover rises and asthma visits fall. The passage also states that traffic exhaust "can trigger asthma attacks."
Step by step:
Neighborhoods with the most trees are also the farthest from a highway.
Living farther from heavy traffic could reduce asthma attacks by itself.
So distance from traffic is a possible confounding variable; the data cannot show how much of the difference, if any, comes from the trees.
The correct answer is neighborhoods with more trees are also farther from a major highway, so residents may breathe less traffic exhaust.
Check: The other choices either use data the table does not contain, predict the opposite pattern, or describe the method rather than a cause.
Tip: When a question asks what else could explain a pattern, look for a variable that changes along with both of the others.
Question 34 of 36
Select allRain and Willow Creek
Willow Creek is a small stream fed mostly by rain. A volunteer watershed group recorded the total rainfall each month near the creek (bar graph) and the creek's average depth at a measuring post (table). Beginning in early August, farms upstream started pumping water out of the creek to irrigate their crops.
Month
Mar
Apr
May
Jun
Jul
Aug
Average creek depth (cm)
85
100
80
55
35
30
Using the rainfall graph together with the creek-depth table, decide which statements the data support. Select all that apply.
Select all that apply.
Hint
Test each statement month by month. A single month that breaks an "every month" statement makes it false.
Step-by-step solution
Check each statement against the numbers. Rainfall (graph): Mar 90, Apr 110, May 70, Jun 40, Jul 20, Aug 30 mm. Depth (table): Mar 85, Apr 100, May 80, Jun 55, Jul 35, Aug 30 cm.
Step by step:
Most rainfall = April (110 mm); deepest creek = April (100 cm). Supported.
Depth from April to August: 100 → 80 → 55 → 35 → 30, falling every month. Supported.
Rainfall from April to August: 110 → 70 → 40 → 20 → 30. It rose in August. Not supported.
"Only factor" — the passage describes pumping in August, and the data cannot rule out other causes. Not supported.
June vs. March: rainfall 40 < 90 mm and depth 55 < 85 cm. Supported.
The correct answers are the month with the most rainfall also had the deepest creek; creek depth decreased every month from April through August; and June had less rainfall and a shallower creek than March.
Check: Each supported statement can be confirmed with specific numbers from both sources.
Tip: Absolute words — every, only, always — are the first things to test. One exception is enough to reject the statement.
Question 35 of 36
Birds by Habitat and Season
Volunteers in a county bird count recorded how many different bird species they saw in four habitats. The same team surveyed each habitat for the same number of hours in spring (May) and again in winter (January).
Many birds that nest in the county in spring migrate south for the winter.
By about what percent did the number of bird species seen in the wetland decrease from spring to winter?
Hint
Percent change = (change ÷ starting value) × 100. Which season is the starting value?
Step-by-step solution
Percent decrease compares the change with the starting value.
Evidence from the graph: Wetland spring = 42 species; wetland winter = 15 species.
Step by step:
Change: 42 - 15 = 27 species.
Starting value (spring): 42.
Percent decrease: \dfrac{27}{42} \times 100 \approx 64.3\%, so about 64%.
The correct answer is about 64%.
Check: 64% of 42 is about 27, and 42 - 27 = 15, the winter count. ✔
Tip: Always divide by the value you started from. Dividing by the ending value (27 ÷ 15 = 180%) or reporting the remaining part (15 ÷ 42 ≈ 36%) are the two most common percent-change mistakes.
Question 36 of 36
Water Flea Heart Rates
Daphnia, or water fleas, are tiny pond animals with see-through bodies, so scientists can watch their hearts beat under a microscope. A Daphnia cannot control its own body temperature; its body stays at the temperature of the water around it. A biology student placed a different Daphnia in water at each of five temperatures and counted its heartbeats for one full minute. She ran three trials at each temperature.
Water temperature (°C)
Trial 1 (beats/min)
Trial 2 (beats/min)
Trial 3 (beats/min)
10
120
118
122
15
158
162
160
20
199
201
200
25
239
180
241
30
281
279
280
Using only the trials that agree with each other, what heart rate would you expect for a Daphnia in water at 27.5 °C (halfway between 25 °C and 30 °C)?
Hint
Find a typical value at 25 °C without the unusual trial, and a typical value at 30 °C. Then go halfway.
Step-by-step solution
This combines interpolation with handling an outlier.
Evidence from the table: At 25 °C the trials that agree are 239 and 241 (Trial 2, 180, is the unusual result). At 30 °C the trials are 281, 279 and 280.
Step by step:
Typical value at 25 °C: \dfrac{239 + 241}{2} = 240 beats/min.
Typical value at 30 °C: \dfrac{281 + 279 + 280}{3} = 280 beats/min.
Halfway: \dfrac{240 + 280}{2} = 260 beats/min.
The correct answer is about 260 beats per minute.
Check: The trend rises about 8 beats per °C (for example, 200 at 20 °C to 280 at 30 °C). From 240 at 25 °C, 2.5 °C more adds 2.5 \times 8 = 20: 260. ✔
Tip: If you average in an outlier, every value you calculate from it is off. Here, the mean at 25 °C with the outlier is (239 + 180 + 241) \div 3 \approx 220, which would give a wrong estimate of about 250.