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June 2024 Earth Science Regents — Answer Key & Worked Solutions

Every question from the June 2024 Earth Science Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the June 2024 Earth Science New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

The diagrams below represent the atmosphere and inferred interior structure of the planet Uranus. Which two factors caused the inferred layered structure of this planet in our solar system?

earth question 1 figure
Worked solution
$\text{Gravity pulls denser materials toward the center}$
$\text{Density differences cause materials to settle into layers (densest core, least dense outer atmosphere)}$
Answer: gravity and density differences of materials

Planetary layering forms because gravity draws all material inward while materials of different densities settle into distinct layers, with the densest (rock/metal) sinking to the core.

Question 2 · 2 pts

Which process occurring in the Sun produces energy by converting lighter hydrogen into heavier helium?

Worked solution
$4\ ^1H \rightarrow\ ^4He + \text{energy}$
Answer: nuclear fusion

Nuclear fusion is the process in the Sun's core where hydrogen nuclei combine under extreme heat and pressure to form helium, releasing energy.

Question 3 · 2 pts

Compared to the sizes and periods of revolution of the terrestrial planets, the Jovian planets have

Worked solution
$\text{Jovian planets (Jupiter, Saturn, Uranus, Neptune) are far larger and farther from the Sun than terrestrial planets}$
$\text{Greater orbital distance} \Rightarrow \text{longer period of revolution (Kepler's third law)}$
Answer: larger sizes and longer periods of revolution

The Jovian (gas giant) planets are much larger in diameter than the terrestrial planets and, being farther from the Sun, take much longer to complete one orbit.

Question 4 · 2 pts

The photograph below shows a dust-covered, icy object that has a highly elliptical orbit in our solar system. This celestial object is most likely a

earth question 4 figure
Worked solution
$\text{Highly elliptical orbit + icy, dust-covered composition} \Rightarrow \text{comet}$
Answer: comet that leaves a trail of particles in space

Comets are icy, dusty bodies with highly elliptical orbits that develop a trail of gas and dust particles as they near the Sun.

Question 5 · 2 pts

At which location will the Sun appear to pass directly overhead for an observer?

Worked solution
$\text{The subsolar point only ranges between the Tropic of Cancer (23.5\degree N) and the Tropic of Capricorn (23.5\degree S)}$
$\text{Northern Australia lies within this band; northern Asia, Greenland, and Antarctica do not}$
Answer: northern Australia

The Sun can only appear directly overhead at locations between 23.5 degrees N and 23.5 degrees S latitude; only northern Australia falls within this tropical band.

Question 6 · 2 pts

The hydrosphere covers approximately

Worked solution
$\text{Hydrosphere (oceans, lakes, rivers, ice) covers roughly 70\% of Earth's solid surface (lithosphere)}$
Answer: 70% of Earth's lithosphere

Earth's surface, part of the lithosphere, is covered by water (the hydrosphere) over approximately 70% of its area.

Question 7 · 2 pts

The photograph below shows a one-mile diameter feature on Earth's surface. What is this feature, and how was it formed?

earth question 7 figure
Worked solution
$\text{Circular bowl-shaped depression with raised rim in flat terrain, no glacial deposits present} \Rightarrow \text{impact crater}$
Answer: crater, formed from an impact event

The circular, bowl-shaped depression with an elevated rim is characteristic of a meteorite impact crater, not a glacially formed kettle lake.

Question 8 · 2 pts

The time-lapse photograph below shows stars as they appear to move around the central star Polaris. Polaris does not appear to move in the nighttime sky because Polaris is located

earth question 8 figure
Worked solution
$\text{Polaris lies nearly along the extension of Earth's rotational axis}$
$\text{As Earth rotates on this axis, Polaris appears stationary while other stars trail in circles around it}$
Answer: above Earth's axis of rotation

Polaris appears fixed because it is positioned almost directly above Earth's axis of rotation, so Earth's spin does not change its apparent position.

Question 9 · 2 pts

The map below shows five locations, labeled A, B, C, D, and X, on Earth's surface. Solar noon is occurring at location X. At which location is the time 2:00 p.m.?

earth question 9 figure
Worked solution
$\text{Earth rotates } 15\degree \text{ of longitude per hour}$
$2:00\text{ p.m. is 2 hours after solar noon} \Rightarrow 2 \times 15\degree = 30\degree \text{ east of X}$
$\text{Locations east of X experience solar noon earlier, so it is already 2:00 p.m. there; point D is }30\degree\text{ east of X}$
Answer: D

Since Earth rotates 15 degrees of longitude per hour, a location 30 degrees east of X (where it is solar noon) will already be experiencing 2:00 p.m.; that location is D.

Question 10 · 2 pts

Approximately how many degrees per day does Earth move in its orbit around the Sun?

Worked solution
$\frac{360\degree}{365\ \text{days}} \approx 0.986\degree/\text{day} \approx 1\degree/\text{day}$
Answer: 1°/day

Earth completes one 360-degree orbit around the Sun in about 365 days, which works out to roughly 1 degree of orbital movement per day.

Question 11 · 2 pts

Which motion explains why some constellations appear in the night sky only during specific seasons?

Worked solution
$\text{As Earth revolves, the night side of Earth faces different, fixed background stars at different times of year}$
Answer: Earth revolves around the Sun.

Because Earth revolves around the Sun, different portions of the night sky (and thus different constellations) face away from the Sun and become visible at night depending on the season.

Question 12 · 2 pts

What is the relative humidity when the air temperature is 20°C and the wet-bulb temperature is 11°C?

Worked solution
$\text{Wet-bulb depression} = 20\degree C - 11\degree C = 9\degree C$
$\text{ESRT Relative Humidity table: row } 20\degree C \text{ dry-bulb, column } 9\degree C \text{ depression} \Rightarrow 30\%$
Answer: 30%

The wet-bulb depression is 20°C minus 11°C, or 9°C. On the Earth Science Reference Tables relative humidity chart, the row for a 20°C dry-bulb temperature intersects the 9°C depression column at 30%.

Question 13 · 2 pts

Earth's early atmosphere consisted of carbon dioxide, water vapor, hydrogen, and nitrogen. During this time in Earth's history, the first crust formed and was made of igneous rock. Based on this information, how did Earth's early atmosphere most likely form?

Worked solution
$\text{Early crust was igneous (volcanic) rock; volcanic outgassing releases CO}_2\text{, water vapor, hydrogen, and nitrogen}$
Answer: outgassing from volcanic eruptions

Since Earth's first crust was igneous rock, volcanic outgassing (the release of trapped gases during eruptions) is the most likely source of the early atmosphere's gases, as no plants or liquid oceans existed yet.

Question 14 · 2 pts

The subtropical jet streams are located in Earth's

Worked solution
$\text{Subtropical jet streams form at the boundary of the Hadley and Ferrel cells, high in the troposphere near latitude } 30\degree$
Answer: upper troposphere near 30° N and 30° S

The subtropical jet streams are fast-moving air currents found in the upper troposphere at approximately 30 degrees north and south latitude.

Question 15 · 2 pts

The transfer of heat energy in the oceans due to density differences is best described as

Worked solution
$\text{Density-driven circulation of a fluid transferring heat} = \text{convection}$
Answer: convection

Convection is the transfer of heat through the movement of a fluid (like ocean water) caused by density differences, typically from temperature or salinity variations.

Question 16 · 2 pts

Which list of Earth materials indicates an increasing order in the amount of energy needed to raise the temperatures of equal masses of the materials by 1°C?

Worked solution
$\text{Specific heat (ESRT Table): copper} \approx 0.39\ \frac{J}{g\cdot \degree C},\ \text{iron} \approx 0.45,\ \text{basalt} \approx 0.84$
$0.39 < 0.45 < 0.84 \Rightarrow \text{copper, iron, basalt is increasing order}$
Answer: copper, iron, basalt

Using specific heat values from the ESRT, copper has the lowest specific heat, followed by iron, then basalt, so this list correctly orders them from least to most energy needed to raise temperature.

Question 17 · 2 pts

Which color and texture of surfaces with equal areas will have the least temperature increase when heated by the Sun?

Worked solution
$\text{Light colors reflect more insolation; smooth textures reflect more than rough textures}$
$\text{Light + smooth} \Rightarrow \text{least absorption} \Rightarrow \text{least temperature increase}$
Answer: light color and smooth texture

Light-colored, smooth surfaces reflect the most incoming solar radiation and absorb the least, resulting in the smallest temperature increase.

Question 18 · 2 pts

Over the last 200 years, global warming is inferred to be primarily caused by

Worked solution
$\text{Increased CO}_2\text{ and other greenhouse gases trap more outgoing infrared radiation, enhancing the greenhouse effect}$
Answer: an increase in greenhouse gases

Scientific consensus attributes recent global warming primarily to human-driven increases in atmospheric greenhouse gas concentrations, which trap more heat; retreating glaciers and rising sea level are effects, not causes.

Question 19 · 2 pts

The graph below shows predicted air temperatures and dewpoints for a New York State location during a two-day period. What is the day and time when precipitation will most likely occur at this location?

earth question 19 figure
Worked solution
$\text{Precipitation is most likely when air temperature and dewpoint are closest (air is saturated, relative humidity} \approx 100\%)$
$\text{On the graph, the air temperature and dewpoint lines converge most closely around Monday at 8:00 p.m.}$
Answer: Monday at 8:00 p.m.

Precipitation is most likely when the air temperature and dewpoint temperature are nearly equal, indicating saturated air; on this graph that occurs around Monday at 8:00 p.m.

Question 20 · 2 pts

The photograph below shows a portion of the Grand Teton Mountains on a mid-July day. The inset map of the United States shows the location of the Grand Tetons. Which factor is responsible for the snow seen on these mountains in July?

earth question 20 figure
Worked solution
$\text{Temperature decreases with increasing elevation (average lapse rate} \approx 3.5\degree F\text{ per 1000 ft})$
$\text{High mountain peaks stay cold enough for snow to persist even in mid-July}$
Answer: elevation

Temperature decreases with altitude, so even in mid-July the high elevation of the Grand Teton peaks keeps them cold enough to retain snow, while the valley below is snow-free.

Question 21 · 2 pts

Which percentage of the radioactive potassium-40 in a sample will have decayed in 2.6 × 10⁹ years?

Worked solution
$\text{Half-life of potassium-40 (ESRT)} = 1.3 \times 10^9\ \text{years}$
$\frac{2.6\times10^9}{1.3\times10^9} = 2\ \text{half-lives}$
$\text{Remaining fraction} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 25\%\ \text{remaining} \Rightarrow 75\%\ \text{decayed}$
Answer: 75%

Potassium-40 has a half-life of 1.3 billion years, so 2.6 billion years represents exactly two half-lives, leaving 25% of the original undecayed and meaning 75% has decayed.

Question 22 · 2 pts

Based on Earth's geologic history, which group of organisms has existed for the shortest amount of time?

Worked solution
$\text{Using the ESRT Geologic History of New York chart: grasses and birds appeared in the Mesozoic/Cenozoic, dinosaurs in the Mesozoic}$
$\text{Humans appeared most recently (late Cenozoic, Quaternary), giving them the shortest existence span}$
Answer: humans

According to the geologic time scale, humans are the most recently evolved group listed, appearing only in the very latest part of the Cenozoic Era, so they have existed for the shortest amount of time.

Question 23 · 2 pts

Oxygen first became a major part of Earth's atmosphere from the

Worked solution
$\text{Cyanobacteria carried out photosynthesis in early oceans}$
$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$
$\text{This released free O}_2\text{, which accumulated in the atmosphere over billions of years}$
Answer: life processes of cyanobacteria in the ocean

Early Precambrian atmosphere had little free oxygen; photosynthetic cyanobacteria in the oceans released oxygen as a byproduct, gradually building up atmospheric O2 (the Great Oxidation Event).

Question 24 · 2 pts

The letter X on the map below shows the location of the present state of Florida on the early continent of North America. The region that is now Florida is inferred to have been positioned on the equator approximately

earth question 24 figure
Worked solution
$\text{Reference Tables show inferred positions of Earth's landmasses over geologic time}$
$\text{Florida's landmass plots on the equator at the } 232\text{ million years ago position (Triassic, Pangea)}$
Answer: 232 million years ago

Using the Earth Science Reference Tables chart of inferred continental positions, the plate fragment that is now Florida sat on the equator around 232 million years ago, during the time of Pangea.

Question 25 · 2 pts

The division of geologic time into eras, periods, and epochs is based upon the

Worked solution
$\text{Geologic time scale boundaries} \rightarrow \text{major changes in the fossil record}$
$\text{Mass extinctions and new life forms mark era/period/epoch boundaries}$
Answer: appearance and extinction of life forms

The geologic time scale's boundaries between eras, periods, and epochs were originally defined by major changes in the fossil record, such as mass extinctions and the appearance of new dominant life forms.

Question 26 · 2 pts

The first P-wave from an earthquake was recorded at 10:20:00 a.m. at a seismic station located 2200 km from the earthquake epicenter. At what time did the S-wave arrive?

Worked solution
$\text{At 2200 km, Reference Table travel-time graph: } t_P \approx 3\text{ min } 20\text{ s},\ t_S \approx 6\text{ min } 50\text{ s}$
$S-P\ \text{interval} \approx 6{:}50 - 3{:}20 = 3\text{ min } 30\text{ s}$
$10{:}20{:}00 + 0{:}03{:}30 = 10{:}23{:}30\ \text{a.m.}$
Answer: 10:23:30 a.m.

Using the Earthquake P-wave and S-wave Travel Time graph at a distance of 2200 km, the difference between S-wave and P-wave arrival times (the S-P interval) is about 3 minutes 30 seconds, giving an S-wave arrival of 10:23:30 a.m.

Question 27 · 2 pts

Which two hot spots are located at plate boundaries?

Worked solution
$\text{Most hot spots (Yellowstone, Canary Is., Galapagos) are intraplate, not on boundaries}$
$\text{Easter Island sits on the East Pacific Rise; Bouvet Island sits on the Mid-Atlantic Ridge}$
$\text{Both East Pacific Rise and Mid-Atlantic Ridge are divergent plate boundaries}$
Answer: Easter Island and Bouvet

Easter Island lies directly on the East Pacific Rise and Bouvet Island lies on the Mid-Atlantic Ridge, both divergent plate boundaries, unlike the other listed hot spots which occur in the interior of plates.

Question 28 · 2 pts

The block diagram below shows a portion of a meandering stream. Points A, B, C, and D are locations on the streambed. The arrows show the direction of stream flow. At which location is the greatest amount of deposition most likely occurring?

earth question 28 figure
Worked solution
$\text{Water velocity is slowest on the inside of a meander bend (point bar)}$
$\text{Slower water} \rightarrow \text{less capacity to carry sediment} \rightarrow \text{deposition}$
$\text{Point } C \text{ lies on the inside bend/point bar of the meander}$
Answer: C

On the inside of a meander bend the water moves slower, so it loses carrying capacity and drops its sediment load, building up a point bar; point C sits on this inside-bend point bar.

Question 29 · 2 pts

Instruments that record seismic waves on Mars allow scientists to infer the

Worked solution
$\text{Seismic (marsquake) waves change speed and path at internal boundaries}$
$\text{Travel times/refraction patterns reveal layers (crust, mantle, core)}$
Answer: internal structure of Mars

Just as on Earth, the way seismic waves travel, bend, and are blocked as they pass through a planet reveals the sizes and states of its internal layers, so seismometers on Mars let scientists infer its internal structure.

Question 30 · 2 pts

The block diagram below represents a landscape with two valleys. The two valleys most likely formed because the shale and limestone bedrock

earth question 30 figure
Worked solution
$\text{Differential weathering: weaker rock erodes faster, forming lowlands (valleys)}$
$\text{Shale and limestone are less resistant} \rightarrow \text{valleys}$
$\text{Igneous rock, sandstone, conglomerate are more resistant} \rightarrow \text{mountains/ridges remain}$
Answer: are less resistant to weathering than the igneous, sandstone, and conglomerate bedrock

Valleys form where weaker, less resistant rock (shale, limestone) is worn away faster than the surrounding more resistant rock types, leaving the resistant rock standing as higher mountains and ridges.

Question 31 · 2 pts

The photograph below shows rocks found in England. Which process most likely caused these rocks to become smooth and round over time?

earth question 31 figure
Worked solution
$\text{Well-rounded, smooth, similarly sized pebbles} \rightarrow \text{long transport in water}$
$\text{Tumbling/abrasion in streams or on beaches rounds and smooths clasts}$
Answer: abrasion in moving water

The rounded, smooth, well-sorted pebbles shown are typical of stream or beach cobbles that have been continually tumbled and abraded against one another by moving water.

Question 32 · 2 pts

The photograph below shows a volcanic mountain. Which map best shows the most likely complete stream drainage pattern near the summit (top) of this mountain at the time of this photograph?

earth question 32 figure
Worked solution
$\text{A conical volcanic peak slopes downward in all directions from the summit}$
$\text{Streams therefore flow outward from the summit in all directions} \rightarrow \text{radial pattern}$
Answer: Map (3)

A volcanic cone slopes away from its summit in every direction, so streams draining the peak flow outward from the central high point like spokes, producing a radial drainage pattern.

Question 33 · 2 pts

The aerial photograph below shows a landscape feature. This landscape feature is a

earth question 33 figure
Worked solution
$\text{Fan-shaped sediment deposit at a river mouth entering a standing body of water} \rightarrow \text{delta}$
$\text{Sediment dropped because stream velocity drops sharply on entering the ocean}$
Answer: delta formed by stream deposition

The fan-shaped, light-colored sediment deposit at the mouth of a stream entering the ocean is a delta, formed when the stream slows abruptly and deposits its sediment load.

Question 34 · 2 pts

What is the name and texture of a sedimentary rock composed of carbon?

Worked solution
$\text{Reference Tables Scheme for Sedimentary Rock Identification}$
$\text{Composition: carbon} \rightarrow \text{bituminous coal}$
$\text{Texture: formed from compacted plant remains} \rightarrow \text{bioclastic}$
Answer: bituminous coal with a bioclastic texture

According to the Earth Science Reference Tables, the sedimentary rock composed of carbon is bituminous coal, and because it forms from compacted organic (plant) material, its texture is classified as bioclastic.

Question 35 · 2 pts

The cross section below represents an igneous intrusion into sedimentary bedrock layers. Letters A, B, and C identify different rocks within the shaded zone of contact metamorphism. Which metamorphic rocks are most likely formed in zones A, B, and C?

earth question 35 figure
Worked solution
$\text{Contact metamorphism converts limestone} \rightarrow \text{marble}$
$\text{sandstone} \rightarrow \text{quartzite}$
$\text{conglomerate (rounded clasts, shown at C)} \rightarrow \text{metaconglomerate}$
Answer: A = marble, B = quartzite, C = metaconglomerate

In the zone of contact metamorphism surrounding the intrusion, the original sedimentary rocks recrystallize without melting: the limestone layer (A) becomes marble, the sandstone layer (B) becomes quartzite, and the conglomerate layer (C, shown with its rounded pebble texture preserved) becomes metaconglomerate.

Question 36 · 2 pts

Base your answers to questions 36 and 37 on the diagram below and on your knowledge of Earth science. The diagram represents the predicted collision between the Milky Way and its sister galaxy, Andromeda. The collision is predicted to begin in about four billion years. It will take an additional two billion years to completely merge these two galaxies into one galaxy. The gravitational force pulling the galaxies together is influenced by the

earth question 36 figure
Worked solution
$F_g = G\dfrac{m_1 m_2}{d^2}$
$\text{Gravitational force depends only on the masses of the objects and the distance between them}$
Answer: mass of all the stars and the distance between the two galaxies

Newton's law of gravitation shows that gravitational attraction depends only on the masses of the objects involved and the distance separating them, not on stellar age or background radiation.

Question 37 · 2 pts

The Andromeda Galaxy has a shape that is similar to the Milky Way Galaxy. The shape of the Andromeda Galaxy is best described as

Worked solution
$\text{Milky Way is a spiral galaxy; Andromeda has a similar spiral shape}$
$\text{The universe contains an estimated hundreds of billions of galaxies}$
Answer: spiral and is one of billions of other galaxies in the universe

Andromeda, like the Milky Way, is a spiral galaxy, and it is just one among the estimated hundreds of billions of galaxies that make up the observable universe.

Question 38 · 2 pts

Base your answers to questions 38 through 40 on the diagram below and on your knowledge of Earth science. The diagram represents the water cycle. Letters A through C identify some processes. Arrows represent the movement of water. Which process is represented by letter A?

earth question 38 figure
Worked solution
$\text{Arrow A points upward from trees/vegetation into the atmosphere}$
$\text{Release of water vapor from plants} = \text{transpiration}$
Answer: transpiration

Letter A shows an upward arrow originating from the trees, representing transpiration, the release of water vapor from plant leaves into the atmosphere.

Question 39 · 2 pts

Which process and heat-energy exchange are represented by letter B?

earth question 39 figure
Worked solution
$\text{Arrow B points upward from the ocean surface} \rightarrow \text{evaporation}$
$\text{Reference Table: heat of vaporization of water} = 2260\ \text{J/g, absorbed during evaporation}$
Answer: evaporation; gain of 2260 J/g

Letter B shows water moving upward from the ocean into the atmosphere, which is evaporation; this phase change requires water to absorb (gain) 2260 J/g, the heat of vaporization listed in the Reference Tables.

Question 40 · 2 pts

Which conditions would cause runoff at location C to decrease?

earth question 40 figure
Worked solution
$\text{Runoff increases with more precipitation and with steeper slope (less time to infiltrate)}$
$\text{Runoff decreases with less precipitation and gentler slope (more infiltration time)}$
Answer: decreased precipitation and decreased slope

Runoff is reduced when there is less rainfall available to run off and when a gentler slope gives water more time to soak into the ground (infiltrate) instead of flowing overland, so both decreased precipitation and decreased slope reduce runoff at C.

Question 41 · 2 pts

Base your answers to questions 41 through 44 on the topographic map below and on your knowledge of Earth science. Lines AB and CD are reference lines on the map. Elevations are shown in feet. Letter X represents a location on Earth's surface. The map shows a Lake, a Pond, contour lines including one marked 1000 and one marked 500, points A and B near the closely spaced 500-ft contours, and points C and D connected by a straight reference line, with contour interval 100 feet and a scale bar from 0 to 2 miles. Which profile represents the most likely shape of the land between locations A and B?

earth question 41 figure
Worked solution
$\text{A and B both cross closely spaced contour lines near the shoreline}$
$\text{Elevation decreases moving from A toward B, closer to the lake}$
Answer: line declining steadily from higher elevation at A down to lower ground at B

Points A and B lie on a steadily sloping hillside where the contour lines run roughly parallel between them, with elevation decreasing from A down toward B near the lake, matching a steadily declining profile rather than a valley or hill shape.

Question 42 · 2 pts

What is the approximate gradient, in feet per mile (ft/mi), from point C to point D on the map?

earth question 42 figure
Worked solution
$\text{Gradient} = \dfrac{\text{change in elevation}}{\text{distance}}$
$\text{Elevation change from C to D} \approx 500\ \text{ft (based on contour values along line CD)}$
$\text{Map distance from C to D} \approx 3\ \text{mi (measured with the scale bar)}$
$\text{Gradient} = \dfrac{500\ \text{ft}}{3\ \text{mi}} \approx 167\ \text{ft/mi}$
Answer: 167 ft/mi

Measuring the straight-line distance between C and D with the map's scale bar (about 3 miles) and reading the elevation change between the contour lines the CD line crosses (about 500 feet) gives a gradient of roughly 167 ft/mi.

Question 43 · 2 pts

Letter X is located on which landscape feature?

earth question 43 figure
Worked solution
$\text{X sits on flat, low-lying ground beside the stream valley near the lake}$
$\text{Flat land bordering a stream that is periodically covered by floodwater} = \text{flood plain}$
Answer: flood plain

Letter X lies on the flat, low ground adjacent to the stream near where it enters the lake, the type of level land that is periodically covered by floodwaters, which defines a flood plain.

Question 44 · 2 pts

The arrows on which map best represent the direction of stream flow relative to the pond?

earth question 44 figure
Worked solution
$\text{Contour lines forming V-shapes point upstream (away from the pond)}$
$\text{Streams flow downhill, from higher to lower elevation, into the pond}$
$\text{Correct map shows tributary arrows converging into the pond}$
Answer: Map (3): arrows pointing inward toward the pond from the surrounding tributary streams

The V-shaped bends in the contour lines around the tributary streams point upstream (uphill, away from the pond), which means the streams themselves flow downhill and drain into the pond, matching the map with arrows converging inward toward the pond.

Question 45 · 2 pts

Base your answers to questions 45 through 47 on the cross sections below and on your knowledge of Earth science. The cross sections represent three plate boundaries labeled A, B, and C. 45 Compared to the density and thickness of the continental crust in diagram A, the oceanic crust is

earth question 45 figure
Worked solution
$\rho_{oceanic} \approx 3.0\ \text{g/cm}^3,\ \text{thickness} \approx 5\text{-}10\ \text{km (basalt)}$
$\rho_{continental} \approx 2.7\ \text{g/cm}^3,\ \text{thickness} \approx 30\text{-}70\ \text{km (granite)}$
Answer: more dense and thinner

Oceanic crust is basaltic, which is denser than the granitic continental crust, but oceanic crust is much thinner than continental crust.

Question 46 · 2 pts

Base your answers to questions 45 through 47 on the cross sections below and on your knowledge of Earth science. The cross sections represent three plate boundaries labeled A, B, and C. 46 Which diagram best represents the relative motion of the oceanic crust at plate boundary B?

earth question 46 figure
Worked solution
$\text{Boundary B is convergent (subduction), not divergent or transform}$ Boundary B (page 16) shows an oceanic-oceanic convergent boundary with a volcanic arc forming above a subducting oceanic plate.
$\text{Diagram 3 shows converging surface arrows with the slab subducting in the direction matching cross section B}$ Only the diagram whose diagonal slab arrow dips in the same direction as the subducting plate shown in diagram B is correct.
Answer: Diagram (3)

Plate boundary B is an oceanic-oceanic convergent (subduction) boundary, so the correct diagram must show both oceanic plates moving toward each other with one plate diving beneath the other along the same dip direction shown in the original cross section.

Question 47 · 2 pts

Base your answers to questions 45 through 47 on the cross sections below and on your knowledge of Earth science. The cross sections represent three plate boundaries labeled A, B, and C. 47 Which geologic surface feature is created at plate boundary C?

earth question 47 figure
Worked solution
$\text{Boundary C: continental crust} + \text{continental crust colliding}$
$\text{Continent-continent collision} \Rightarrow \text{folded mountains (e.g., Himalayas)}$
Answer: folded mountain

Plate boundary C shows two continental plates colliding; since neither can subduct (both are low density), the crust crumples and folds upward into folded mountains.

Question 48 · 2 pts

Base your answers to questions 48 through 50 on the passage and photograph below and on your knowledge of Earth science. The photograph shows a typical Petoskey stone. Petoskey Stones Petoskey stones are famous rock specimens found on beaches all around Lake Michigan, but especially on beaches near the city of Petoskey, Michigan. The stones are actually well-eroded pieces of coral reefs that were alive between 375 and 400 million years ago. At the time the coral animals were living and forming the reefs, the area was underwater and the coral, along with many other reef animals, thrived. Eventually, the ocean disappeared and the fossil reefs were uncovered. Advancing and retreating glaciers broke pieces off the fossil reefs, smoothed their edges, and distributed them throughout the area. Lake Michigan, along with the other Great Lakes, formed at the end of the Ice Age. Today, lake waves continue to distribute a seemingly endless supply of Petoskey stones, with their unique coral fossils, to the beaches around Lake Michigan. 48 The coral that eventually became Petoskey stones were alive during which geologic time period?

earth question 48 figure
Worked solution
$375\text{-}400\ \text{million years ago}$
$\text{Devonian} \approx 419\text{-}359\ \text{Mya}$ Per the ESRT Geologic History of New York State chart, the Devonian Period contains the 375-400 Mya range.
Answer: Devonian

The 375-400 million year age given for the coral falls within the Devonian Period on the geologic time scale.

Question 49 · 2 pts

Base your answers to questions 48 through 50 on the passage and photograph below and on your knowledge of Earth science. The photograph shows a typical Petoskey stone. 49 Which New York State index fossil is in the same group of organisms as the Petoskey stone coral?

earth question 49 figure
Worked solution
$\text{Lichenaria} = \text{coral}$ Lichenaria is a colonial tabulate coral, the same organism group (corals/Cnidaria) as the Petoskey stone coral.
$\text{other choices} \ne \text{coral}$ Platyceras is a snail (gastropod), Bothriolepis is an armored fish, and Cooksonia is an early land plant.
Answer: Lichenaria

Of the listed NYS index fossils, only Lichenaria is a coral, placing it in the same organism group as the coral that formed the Petoskey stone.

Question 50 · 2 pts

Base your answers to questions 48 through 50 on the passage and photograph below and on your knowledge of Earth science. The photograph shows a typical Petoskey stone, with a centimeter ruler (marked 1-6 cm) shown beside the stone for scale, labeled "(Actual size)". 50 The Petoskey stone shown in the photograph is classified as a

earth question 50 figure
Worked solution
$\text{diameter} \approx 4\text{-}5\ \text{cm}$ The photo shows the stone spanning roughly 4-5 cm against the printed centimeter scale.
$0.2\ \text{cm} < 4\text{-}5\ \text{cm} < 6.4\ \text{cm} \Rightarrow \text{pebble}$ On the ESRT particle/sediment size scale, pebble-sized particles range from about 0.2 cm to 6.4 cm.
Answer: pebble

Measured against the printed centimeter scale, the rounded stone is a few centimeters across, which falls in the pebble range on the ESRT sediment size scale (larger than sand, smaller than cobble).

Question 51 · 1 pt · constructed response

Base your answers to questions 51 through 54 on the weather map in your answer booklet and on your knowledge of Earth science. The weather map shows air pressures in millibars (mb) at locations on and around the Florida Peninsula during Hurricane Wilma. Five isobars are shown. 51 On the map in your answer booklet, draw the 1000 mb and 1004 mb isobars. Extend the isobars to the edges of the map.

earth question 51 figure
Model answer: Draw two smooth, continuous curved lines, one connecting all points where pressure equals 1000 mb and one connecting all points where pressure equals 1004 mb, nested inside the other given isobars closer to the hurricane's low-pressure center, and extend both lines fully to the edges of the map without crossing any other isobar.
Worked solution
$\text{isobars must be smooth, non-crossing curves}$ Isobars must be smooth curves, never cross each other, and must be drawn at the correct position relative to the existing isobars (lower-value isobars closer to the storm center).
$\text{each isobar reaches both map edges}$ Each isobar must reach both edges of the map, matching the pattern of the given isobars.
Answer: Draw two smooth, continuous curved lines, one connecting all points where pressure equals 1000 mb and one connecting all points where pressure equals 1004 mb, nested inside the other given isobars closer to the hurricane's low-pressure center, and extend both lines fully to the edges of the map without crossing any other isobar.

Isobars are lines of equal air pressure; for a hurricane, pressure decreases toward the center, so the 1000 mb and 1004 mb lines must be drawn inside (surrounded by) the higher-value isobars already shown, following the same smooth, roughly circular/elliptical pattern.

Question 52 · 1 pt · constructed response

Base your answers to questions 51 through 54 on the weather map in your answer booklet and on your knowledge of Earth science. The weather map shows air pressures in millibars (mb) at locations on and around the Florida Peninsula during Hurricane Wilma. Five isobars are shown. 52 Convert the air pressure of the lowest value isobar shown on the map from millibars to inches of mercury.

On the completed weather map, the lowest value isobar shown is the 1000 mb isobar (the isobars decrease inward toward the hurricane center: 1016, 1012, 1008, 1004, 1000 mb). Use the Pressure conversion scale in the Earth Science Reference Tables, which pairs millibars with inches of mercury: | Millibars (mb) | Inches of mercury (in) | | --- | --- | | 1013.2 (one atmosphere) | 29.92 | | 1000.0 | 29.53 |

Model answer: Approximately 29.53 inches of mercury (accept a range of about 29.41-29.65 in Hg)
Worked solution
$P = 1000\ \text{mb}$ After question 51, the lowest-value isobar shown on the completed map is 1000 mb, closest to the hurricane's center.
$1000\ \text{mb} \div 33.86\ \dfrac{\text{mb}}{\text{in Hg}} \approx 29.53\ \text{in Hg}$ Using the standard conversion 1 in Hg = 33.86 mb.
Answer: Approximately 29.53 inches of mercury (accept a range of about 29.41-29.65 in Hg)

The lowest isobar on the completed map is 1000 mb; converting millibars to inches of mercury using the standard conversion factor gives approximately 29.53 in Hg.

Question 53 · 1 pt · constructed response

Base your answers to questions 51 through 54 on the weather map in your answer booklet and on your knowledge of Earth science. The weather map shows air pressures in millibars (mb) at locations on and around the Florida Peninsula during Hurricane Wilma. Five isobars are shown. 53 Identify the weather instrument used to measure air pressure.

Model answer: Barometer
Worked solution
$\text{barometer}$ Air pressure is measured directly using a barometer (mercury or aneroid).
Answer: Barometer

A barometer is the standard instrument used to measure atmospheric (air) pressure.

Question 54 · 1 pt · constructed response

Base your answers to questions 51 through 54 on the weather map in your answer booklet and on your knowledge of Earth science. The weather map shows air pressures in millibars (mb) at locations on and around the Florida Peninsula during Hurricane Wilma. Five isobars are shown. 54 Write the two-letter weather map symbol for the type of air mass in which this hurricane formed.

Model answer: mT
Worked solution
$\text{hurricanes form over warm tropical ocean water}$ Hurricanes form over warm tropical ocean water, producing a warm, moist air mass.
$\text{mT} = \text{maritime tropical}$ The two-letter symbol for a maritime tropical air mass is mT.
Answer: mT

Hurricanes develop over warm tropical ocean waters, which produce a maritime tropical (mT) air mass, warm and moist.

Question 55 · 1 pt · constructed response

Base your answers to questions 55 through 57 on the passage and data table below and on your knowledge of Earth science. The data table shows garnet coefficients and rock formation temperatures. Using Garnet as a Geothermometer A geothermometer mineral is a mineral with slight differences in composition that can be used to estimate the temperature at which some types of metamorphic rock form. Garnet is a geothermometer mineral because there are several kinds of garnet that contain different amounts of iron and magnesium. The "garnet coefficient" indicates the ratio of iron to magnesium found in the garnet. When the garnet coefficient is high, there is a greater percentage of iron compared to magnesium. The garnet coefficient is an indicator of the temperature at which a metamorphic rock formed. Garnet Coefficient and Rock Formation Temperature | Garnet Coefficient | Temperature at Which Metamorphic Rock Formed (ºC) | | --- | --- | | 1.22 | 720 | | 1.34 | 680 | | 1.38 | 670 | | 1.40 | 660 | | 1.47 | 640 | | 1.63 | 590 | | 1.70 | 570 | 55 Determine the rock formation temperature for a rock sample with a garnet coefficient of 1.55.

Model answer: Approximately 615°C (accept a range of about 610-620°C)
Worked solution
$1.47 < 1.55 < 1.63$
$\text{fraction} = \dfrac{1.55-1.47}{1.63-1.47} = \dfrac{0.08}{0.16} = 0.5$
$T = 640 - 0.5\times(640-590) = 640-25=615^{\circ}\text{C}$
Answer: Approximately 615°C (accept a range of about 610-620°C)

Linear interpolation between the two nearest data points (1.47 → 640°C and 1.63 → 590°C) gives a formation temperature of about 615°C for a garnet coefficient of 1.55.

Question 56 · 1 pt · constructed response

Base your answers to questions 55 through 57 on the passage and data table below and on your knowledge of Earth science. The data table shows garnet coefficients and rock formation temperatures. 56 Describe the rock formation temperatures that are indicated when higher garnet coefficients are found.

earth question 56 figure
Model answer: Higher garnet coefficients indicate lower rock formation temperatures (there is an inverse relationship: as garnet coefficient increases, formation temperature decreases).
Worked solution
$\text{coefficient}\ 1.22\to720^{\circ}\text{C},\ \ 1.70\to570^{\circ}\text{C}$ From the table, as garnet coefficient rises from 1.22 to 1.70, formation temperature falls from 720°C to 570°C.
Answer: Higher garnet coefficients indicate lower rock formation temperatures (there is an inverse relationship: as garnet coefficient increases, formation temperature decreases).

The data table shows an inverse relationship: rock samples with higher garnet coefficients formed at lower temperatures.

Question 57 · 1 pt · constructed response

Base your answers to questions 55 through 57 on the passage and data table below and on your knowledge of Earth science. 57 Identify one foliated metamorphic rock for which the garnet coefficient might be used to determine the temperature at which the rock formed.

earth question 57 figure
Model answer: Schist (gneiss is also acceptable)
Worked solution
$\text{garnet} \in \text{schist, gneiss}$ Garnet is a common accessory mineral found in foliated metamorphic rocks such as schist and gneiss, per the ESRT scheme for metamorphic rock identification.
Answer: Schist (gneiss is also acceptable)

Garnet-bearing metamorphic rocks with visible foliation (banding/parallel mineral alignment) are typically schist or gneiss; schist is the most common garnet-bearing foliated rock.

Question 58 · 1 pt · constructed response

Base your answers to questions 58 and 59 on the map and graph below and on your knowledge of Earth science. The map of Australia shows the locations of three cities: Brisbane, Melbourne, and Wagga Wagga. The graph shows the average monthly air temperatures in Wagga Wagga and Melbourne, in degrees Celsius (°C), with both curves peaking around January (near 20-24°C) and reaching their lowest point around July (near 9-10°C). 58 Compared to the average monthly air temperature curve for Melbourne, describe how the average monthly air temperature curve is different for a city in New York State.

earth question 58 figure
Model answer: A New York State city's temperature curve would be the reverse (inverted) of Melbourne's: it would be warmest in June, July, and August and coolest in December, January, and February, opposite of Melbourne's pattern.
Worked solution
$\text{Melbourne (S. Hemisphere): summer Dec-Feb, winter Jun-Aug}$ Melbourne is in the Southern Hemisphere.
$\text{New York (N. Hemisphere): summer Jun-Aug, winter Dec-Feb}$ New York State is in the Northern Hemisphere, exactly opposite Melbourne's seasons.
Answer: A New York State city's temperature curve would be the reverse (inverted) of Melbourne's: it would be warmest in June, July, and August and coolest in December, January, and February, opposite of Melbourne's pattern.

Because Melbourne is in the Southern Hemisphere and New York is in the Northern Hemisphere, their seasons (and therefore their monthly temperature curves) are six months out of phase with each other.

Question 59 · 1 pt · constructed response

Base your answers to questions 58 and 59 on the map and graph below and on your knowledge of Earth science. The map of Australia shows the locations of three cities: Brisbane, Melbourne, and Wagga Wagga. 59 In your answer booklet, circle warm or cool to indicate the relative temperature of the ocean current that flows off the coast of Brisbane. State the name of this current.

Model answer: Warm; East Australian Current
Worked solution
$\text{Brisbane: east coast of Australia} \to \text{East Australian Current (warm, poleward-flowing)}$ Brisbane lies on the east coast of Australia, where the East Australian Current flows poleward (southward), carrying warm water from lower latitudes.
Answer: Warm; East Australian Current

The East Australian Current is a warm current that flows south along Australia's east coast, past Brisbane.

Question 60 · 1 pt · constructed response

Base your answers to questions 60 through 62 on the cross sections below and on your knowledge of Earth science. The cross sections represent changes in the land at the Montauk Lighthouse between 1838 and 1988. The Montauk Lighthouse, built on unconsolidated sediments, is located on the easternmost tip of Long Island. The distance from the center of the lighthouse to the edge of the escarpment is indicated in each cross section: 220 feet in 1838 and 100 feet in 1988. 60 Identify one agent of erosion that removed the sediment and changed the position of the escarpment between 1838 and 1988.

earth question 60 figure
Model answer: Ocean waves (wave action)
Worked solution
$\text{waves undercut the base of the escarpment}$ The lighthouse sits atop unconsolidated sediment on a coastal bluff/escarpment directly above the ocean, which is eroded primarily by breaking waves undercutting the base of the escarpment.
Answer: Ocean waves (wave action)

Breaking ocean waves are the main agent of erosion that undercuts and removes sediment from coastal escarpments like the one at Montauk.

Question 61 · 1 pt · constructed response

Base your answers to questions 60 through 62 on the cross sections below and on your knowledge of Earth science. The distance from the center of the lighthouse to the edge of the escarpment was 220 feet in 1838 and 100 feet in 1988. 61 Calculate the rate of erosion, in feet per year, between the lighthouse and the edge of the escarpment that occurred between 1838 and 1988. Express your answer to the nearest tenth.

earth question 61 figure
Model answer: 0.8 ft/yr
Worked solution
$\Delta d = 220\ \text{ft} - 100\ \text{ft} = 120\ \text{ft}$
$\Delta t = 1988 - 1838 = 150\ \text{years}$
$\text{rate} = \dfrac{120\ \text{ft}}{150\ \text{yr}} = 0.8\ \text{ft/yr}$
Answer: 0.8 ft/yr

The escarpment retreated 120 feet closer to the lighthouse over 150 years, giving an average erosion rate of 0.8 feet per year.

Question 62 · 1 pt · constructed response

Base your answers to questions 60 through 62 on the cross sections below and on your knowledge of Earth science. 62 Describe one action that could be taken to slow down or prevent the future erosion of the land between the escarpment and the ocean.

earth question 62 figure
Model answer: Build a seawall, groin, breakwater, or place riprap/boulders at the base of the escarpment (or plant vegetation to stabilize the sediment) to reduce wave energy reaching the bluff.
Worked solution
$\text{hard structures absorb/deflect wave energy}$ Hard structures (seawalls, riprap, groins, breakwaters) absorb or deflect wave energy; vegetation/dune stabilization reduces sediment loss.
Answer: Build a seawall, groin, breakwater, or place riprap/boulders at the base of the escarpment (or plant vegetation to stabilize the sediment) to reduce wave energy reaching the bluff.

Coastal engineering structures such as seawalls or riprap, or stabilizing vegetation, reduce the wave energy that erodes the escarpment, slowing further land loss.

Question 63 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the data table below and on your knowledge of Earth science. The data table shows the Sun's altitude and the shadow length measured from the base of a five-meter-high pole on June 22 in Rochester, New York. During daylight savings time, clocks are set one hour ahead, so solar noon occurs at approximately 1:00 p.m. Data Table | Time of Day (EDT) | Altitude of Sun in Degrees (º) | Shadow Length in Meters (m) | | --- | --- | --- | | 9:00 a.m. | 35 | 7.1 | | 10:00 a.m. | 46 | 4.8 | | 11:00 a.m. | 56 | 3.3 | | 12:00 p.m. | 65 | 2.3 | | 1:00 p.m. | 70 | 1.8 | | 2:00 p.m. | 68 | 2.0 | | 3:00 p.m. | 60 | 2.9 | | 4:00 p.m. | 50 | 4.1 | | 5:00 p.m. | 40 | 6.0 | | 6:00 p.m. | 29 | 9.1 | 63 On the graph in your answer booklet, construct a line graph by plotting the altitude of the Sun for each time of day shown on the data table. Connect all ten plots with a line. The shadow-length data have already been plotted using the scale on the right.

Model answer: Plot the ten (time, altitude) points: (9am,35°), (10am,46°), (11am,56°), (12pm,65°), (1pm,70°), (2pm,68°), (3pm,60°), (4pm,50°), (5pm,40°), (6pm,29°), and connect them in order with a smooth line forming a single upside-down V (arch) shape that peaks at 1:00 p.m. (70°), mirroring the pre-plotted shadow-length curve (which is inverted, U-shaped, with its minimum at 1:00 p.m.).
Worked solution
$\text{use left altitude axis on booklet grid}$ Use the left-hand altitude scale on the answer-booklet grid to plot each of the 10 data points against time of day.
$\text{connect points chronologically}$ Connect the points in chronological order with straight line segments to form the altitude curve.
Answer: Plot the ten (time, altitude) points: (9am,35°), (10am,46°), (11am,56°), (12pm,65°), (1pm,70°), (2pm,68°), (3pm,60°), (4pm,50°), (5pm,40°), (6pm,29°), and connect them in order with a smooth line forming a single upside-down V (arch) shape that peaks at 1:00 p.m. (70°), mirroring the pre-plotted shadow-length curve (which is inverted, U-shaped, with its minimum at 1:00 p.m.).

Plotting altitude vs. time produces a bell-shaped (arch) curve peaking at solar noon (1:00 p.m., 70°), consistent with the Sun rising in altitude through the morning and declining in the afternoon.

Question 64 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the data table below and on your knowledge of Earth science. 64 Based on the data table, describe the relationship between the altitude of the Sun and the length of the shadow.

earth question 64 figure
Model answer: As the Sun's altitude increases, the shadow length decreases; as the Sun's altitude decreases, the shadow length increases (an inverse relationship).
Worked solution
$9am\!:35^{\circ}\!\to\!7.1\text{m},\ \ 1pm\!:70^{\circ}\!\to\!1.8\text{m},\ \ 6pm\!:29^{\circ}\!\to\!9.1\text{m}$ From 9am to 1pm, altitude rises while shadow length falls; from 1pm to 6pm, altitude falls while shadow length rises.
Answer: As the Sun's altitude increases, the shadow length decreases; as the Sun's altitude decreases, the shadow length increases (an inverse relationship).

The data table shows an inverse relationship between Sun altitude and shadow length: higher Sun altitude produces shorter shadows, and lower Sun altitude produces longer shadows.

Question 65 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the data table below and on your knowledge of Earth science. 65 Based on the graph, determine the shadow length of the pole, in meters, at 4:30 p.m.

earth question 65 figure
Model answer: Approximately 5.0 m (accept a range of about 4.7-5.3 m)
Worked solution
$\text{At 4:00 p.m., shadow} = 4.1\ \text{m; at 5:00 p.m., shadow} = 6.0\ \text{m}$
$\text{4:30 p.m. (midpoint)} \approx \dfrac{4.1+6.0}{2} = 5.05\ \text{m}$
Answer: Approximately 5.0 m (accept a range of about 4.7-5.3 m)

Interpolating halfway between the 4:00 p.m. (4.1 m) and 5:00 p.m. (6.0 m) shadow lengths gives an estimated shadow length of about 5.0 meters at 4:30 p.m.

Question 66 · 1 pt · constructed response

Base your answers to questions 66 and 67 on the cross section below and on your knowledge of Earth science. Letters G and H identify some rock layers. The wavy line WX indicates an unconformity, and line YZ indicates a fault. Index fossils are represented in some rock layers. Rock layers have not been overturned. 66 Identify the most likely geologic time period during which unconformity WX was forming.

earth question 66 figure
Model answer: Silurian Period (the time period represented by the missing rock record between the index fossil found at the top of layer G, just below the unconformity, and the index fossil found in the rock layer directly above WX)
Worked solution
$\text{unconformity} = \text{gap in rock record (erosion/non-deposition)}$ An unconformity represents a gap in the rock record caused by an interval of erosion and/or non-deposition.
$\text{compare index fossil ages above and below WX using ESRT chart}$ The missing time period is identified by comparing the age of the index fossil in the youngest layer below WX (top of layer G) to the age of the index fossil in the oldest layer above WX.
Answer: Silurian Period (the time period represented by the missing rock record between the index fossil found at the top of layer G, just below the unconformity, and the index fossil found in the rock layer directly above WX)

The unconformity WX formed during the time period(s) missing from the rock record, that is, the interval between the top of layer G (below WX) and the layer directly above WX, as determined by comparing the index fossils in those layers on the ESRT geologic time scale.

Question 67 · 1 pt · constructed response

Base your answers to questions 66 and 67 on the cross section below and on your knowledge of Earth science. Letters G and H identify some rock layers. The wavy line WX indicates an unconformity, and line YZ indicates a fault. Index fossils are represented in some rock layers. Rock layers have not been overturned. In your answer booklet, circle the term that best describes the relative age of fault YZ compared to layers G and H. Describe the evidence that infers this relative age.

earth question 67 figure
Model answer: Fault YZ is younger than layers G and H. Evidence: the fault line YZ cuts through (offsets) both rock layers G and H, and a fault can only cut through rock layers that already existed, so the fault must have formed after layers G and H were deposited.
Worked solution
$\text{Principle of cross-cutting relationships: any feature that cuts across rock layers (a fault, intrusion, etc.) must be younger than the layers it cuts.}$
$\text{Fault } YZ \text{ is shown cutting through both layer } G \text{ and layer } H.}$
$\Rightarrow \text{Fault } YZ \text{ is younger than layers } G \text{ and } H.$
Answer: Fault YZ is younger than layers G and H. Evidence: the fault line YZ cuts through (offsets) both rock layers G and H, and a fault can only cut through rock layers that already existed, so the fault must have formed after layers G and H were deposited.

By the law of cross-cutting relationships, any fault or intrusion must be younger than all the rock layers it cuts through, so since YZ cuts both G and H, the fault formed after those layers were already in place.

Question 68 · 1 pt · constructed response

Base your answers to questions 68 through 71 on the data table and passage below, and on your knowledge of Earth science. The data table shows the location and characteristics of seven stars, numbered 1 through 7, found in the constellation Cygnus. Location and Characteristics of Seven Stars in Cygnus | Star Number | Star Name | Celestial Longitude (hours) | Celestial Latitude (degrees) | Temperature (K) | Luminosity | | --- | --- | --- | --- | --- | --- | | 1 | Deneb | 20.7 | 45 | 8500 | 197,000 | | 2 | Sadr | 20.3 | 40 | 5800 | 60,000 | | 3 | Delta Cygni | 19.8 | 45 | 9800 | 180 | | 4 | Epsilon Cygni | 20.8 | 34 | 4800 | 60 | | 5 | Eta Cygni | 20 | 35 | 4840 | 54 | | 6 | Alberio | 19.5 | 28 | 4400 | 950 | | 7 | Tabby's Star | 20.1 | 44 | 6200 | 1.5 | The Mystery of Tabby's Star Located in the constellation of Cygnus, there is a mysterious star named Tabby's Star that has a constant luminosity, but has a brightness that appears to change in a cyclic manner. A team of astronomers has developed a plausible explanation for this change. A dust cloud with a variable thickness has an approximate 700-day period of revolution around Tabby's Star. Why dust? Astronomers observed that a portion of the electromagnetic spectrum having wavelengths just shorter than visible light is blocked out, causing a decrease in apparent brightness. While astronomers don't know the origin of the dust cloud around Tabby's Star, dust unevenly spaced in its orbit around the star would explain this cyclic dimming of Tabby's Star. On the grid in your answer booklet, plot the positions of stars 1 through 6. Record the star number of each star beside the plot. In order to create the correct shape of Cygnus, connect the plots with two lines: the first line connects the plots numbered 1-2-5-6; the second line connects the plots numbered 4-2-3. This position of Tabby's Star, number 7, has been indicated with a ⊕.

Model answer: Each of stars 1 through 6 is plotted on the celestial longitude/latitude grid at its coordinates from the data table (e.g., Star 1/Deneb at longitude 20.7 h, latitude 45°; Star 2/Sadr at 20.3 h, 40°; etc.), each point labeled with its star number, and lines are drawn connecting points 1-2-5-6 and 4-2-3 to form the outline of Cygnus (the Northern Cross).
Worked solution
$\text{Plot each star at (celestial longitude, celestial latitude): }(20.7,45),(20.3,40),(19.8,45),(20.8,34),(20,35),(19.5,28)$
$\text{Label each point with its star number (1-6).}$
$\text{Draw a line connecting points in order } 1\to2\to5\to6$
$\text{Draw a second line connecting points in order } 4\to2\to3$
Answer: Each of stars 1 through 6 is plotted on the celestial longitude/latitude grid at its coordinates from the data table (e.g., Star 1/Deneb at longitude 20.7 h, latitude 45°; Star 2/Sadr at 20.3 h, 40°; etc.), each point labeled with its star number, and lines are drawn connecting points 1-2-5-6 and 4-2-3 to form the outline of Cygnus (the Northern Cross).

This is a plotting task on a celestial coordinate grid provided in the separate answer booklet, using the longitude/latitude values from the data table for each star.

Question 69 · 1 pt · constructed response

Identify the planet in our solar system that has a period of revolution most similar to the dust cloud orbiting Tabby's Star.

Model answer: Mars
Worked solution
$\text{Dust cloud period of revolution} \approx 700 \text{ days}$
$\text{Mars' period of revolution} \approx 687 \text{ Earth days}$
$687 \text{ days is closest among the planets to } 700 \text{ days}$
Answer: Mars

Mars orbits the Sun in about 687 Earth days, which is closer to the dust cloud's 700-day period than any other planet's orbital period.

Question 70 · 1 pt · constructed response

State the name of the shortwave electromagnetic energy that is mostly blocked out by the dust cloud orbiting Tabby's Star.

Model answer: Ultraviolet radiation (UV)
Worked solution
$\text{Wavelengths just shorter than visible light on the EM spectrum} = \text{ultraviolet}$
Answer: Ultraviolet radiation (UV)

On the electromagnetic spectrum, the region with wavelengths just shorter than visible light is the ultraviolet band, so this is the shortwave energy being blocked by the dust.

Question 71 · 1 pt · constructed response

Complete the table in your answer booklet by identifying the color and classification for two of the stars in the constellation Cygnus based on temperature and luminosity shown in the table. Color and classification for Deneb have been completed as an example.

Answer table to complete (Deneb is the completed example): | Star | Color | Classification | | --- | --- | --- | | Deneb (8500 K, luminosity 197,000) | White | Supergiant | | (choose a star from the data table) | ? | ? | | (choose a star from the data table) | ? | ? | Use the Characteristics of Stars chart in the Earth Science Reference Tables. Its horizontal axis is surface temperature in kelvins (hotter at the left), with star colors labeled across the top from hottest to coolest: Blue, Blue White, White, Yellow, Orange, Red. The vertical axis is luminosity (relative to the Sun = 1), from about 10^-6 at the bottom to about 10^6 at the top. Star groups on the chart: Supergiants (very high luminosity, top of chart), Giants (high luminosity, cooler side), Main Sequence (diagonal band from hot bright to cool dim; the Sun, 5800 K and luminosity 1, is a Yellow main sequence star), and White Dwarfs (hot but very dim, lower left). Classify each star by locating its temperature and luminosity on this chart: for example, stars near 4400 to 4840 K read as Orange in color; a star at 6200 K with luminosity 1.5 falls on the Main Sequence; a star at 4400 K with luminosity 950 falls in the Giants region.

Model answer: Example (Deneb, given): temperature 8500 K, luminosity 197,000 -> color: white, classification: supergiant. For two additional stars, using the Earth Science Reference Tables Characteristics of Stars chart (temperature vs. luminosity): Sadr (5800 K, 60,000) -> color: yellow, classification: supergiant; Delta Cygni (9800 K, 180) -> color: blue-white, classification: main sequence (main sequence/dwarf).
Worked solution
$\text{Locate each star's temperature (x-axis) and luminosity (y-axis) on the Characteristics of Stars chart from the Earth Science Reference Tables.}$
$\text{Temperature determines color: hotter} \Rightarrow \text{blue/blue-white; cooler} \Rightarrow \text{red/orange.}$
$\text{Position relative to the main sequence band determines classification: on the band = main sequence; far above (very luminous for its temperature) = giant/supergiant; below = white dwarf.}$
$\text{Sadr: } T=5800K \text{ (yellow)}, L=60{,}000 \text{ (far above main sequence)} \Rightarrow \text{supergiant}$
$\text{Delta Cygni: } T=9800K \text{ (blue-white)}, L=180 \text{ (near main sequence for that temperature)} \Rightarrow \text{main sequence}$
Answer: Example (Deneb, given): temperature 8500 K, luminosity 197,000 -> color: white, classification: supergiant. For two additional stars, using the Earth Science Reference Tables Characteristics of Stars chart (temperature vs. luminosity): Sadr (5800 K, 60,000) -> color: yellow, classification: supergiant; Delta Cygni (9800 K, 180) -> color: blue-white, classification: main sequence (main sequence/dwarf).

Color is determined from a star's surface temperature and classification (main sequence, giant, supergiant, dwarf) is determined by plotting temperature against luminosity on the H-R type chart in the Earth Science Reference Tables.

Question 72 · 1 pt · constructed response

Base your answers to questions 72 through 74 on the weather map below and on your knowledge of Earth science. The weather map shows the high (H) and low (L) pressure centers and fronts across the United States on a January morning. The darker gray shaded areas show radar images of precipitation. A partial station model represents weather conditions in southern Texas. Two different types of fronts are represented along the line from X to Y. Identify the two fronts represented along line XY.

earth question 72 figure
Model answer: A cold front and a warm front (the front symbols change type along the line from X to Y, going from a cold front near X to a warm front near Y).
Worked solution
$\text{Line } XY \text{ trails from the low-pressure center and shows two distinct symbol types.}$
$\text{Solid triangles pointing in the direction of movement} = \text{cold front}$
$\text{Solid semicircles pointing in the direction of movement} = \text{warm front}$
$\Rightarrow \text{the segment nearer } X \text{ is a cold front and the segment nearer } Y \text{ is a warm front (both trailing off the same low-pressure system).}$
Answer: A cold front and a warm front (the front symbols change type along the line from X to Y, going from a cold front near X to a warm front near Y).

A single low-pressure system typically has a cold front trailing on one side and a warm front on the other; the two different symbol types along line XY represent these two different front types.

Question 73 · 1 pt · constructed response

Identify the compass direction toward which the low-pressure system located just south of New York State will move if it follows a normal storm track.

Model answer: Northeast
Worked solution
$\text{In the middle latitudes of the United States, the prevailing winds aloft (westerlies) steer surface storm systems.}$
$\text{Normal storm tracks in the U.S. move from southwest to northeast.}$
$\Rightarrow \text{the L just south/east of New York will move toward the northeast (out over the North Atlantic).}$
Answer: Northeast

Storms in the United States are steered by the prevailing westerlies and typically follow storm tracks moving from the southwest toward the northeast.

Question 74 · 1 pt · constructed response

Complete the table in your answer booklet by filling in the values for the four weather variables that are represented by the station model. The partial station model over southern Texas shows the number 46 to the upper left of the station circle, 37 to the lower left of the station circle, and 217 to the upper right of the station circle, with an unfilled (open) station circle and no wind barb shown.

earth question 74 figure
Model answer: Temperature = 46°F; Dew point = 37°F; Barometric pressure = 1021.7 mb (from the coded value 217); Sky cover (cloud cover) = clear (0%, shown by the open/unfilled station circle), with calm winds (no wind barb shown).
Worked solution
$\text{Upper-left number on a station model} = \text{temperature (°F)} = 46°F$
$\text{Lower-left number} = \text{dew point (°F)} = 37°F$
$\text{Upper-right three-digit code} = \text{pressure in tenths of millibars, last three digits of the value + implied leading 9 or 10}$
$217 \Rightarrow 1021.7 \text{ mb (since } 900 < \text{value} < 1050\text{, prepend a 10 and insert a decimal before the last digit)}$
$\text{The unfilled circle indicates 0% cloud cover (clear sky); the absence of a wind barb indicates calm wind.}$
Answer: Temperature = 46°F; Dew point = 37°F; Barometric pressure = 1021.7 mb (from the coded value 217); Sky cover (cloud cover) = clear (0%, shown by the open/unfilled station circle), with calm winds (no wind barb shown).

Station model conventions: upper-left is temperature, lower-left is dew point, upper-right coded number is sea-level pressure in tenths of a millibar (missing leading digits are inferred), and the shading/fill of the station circle indicates percent cloud cover.

Question 75 · 1 pt · constructed response

Base your answers to questions 75 and 76 on the maps and data table below and on your knowledge of Earth science. Map 1 shows the area where a total lunar eclipse was visible on November 8, 2022. Map 2 shows the area where a total solar eclipse was visible on April 20, 2023. The data table shows the date, type, and duration of totality for each total eclipse that has occurred or will occur at a specific location during the years 2022 through 2025. Total Eclipses 2022-2025 | Date | Type of Eclipse | Duration of Totality | | --- | --- | --- | | May 16, 2022 | lunar | 1 hour 25 minutes | | November 8, 2022 | lunar | 1 hour 25 minutes | | April 20, 2023 | solar | 1 minute 16 seconds | | April 8, 2024 | solar | 4 minutes 28 seconds | | March 14, 2025 | lunar | 1 hour 5 minutes | | September 7, 2025 | lunar | 1 hour 22 minutes | Determine whether each of the eclipses shown on the maps were visible or not visible to an observer in Hawaii. In your answer booklet, circle 'visible' or 'not visible' on the line next to the date of each eclipse.

earth question 75 figure
Model answer: May 16, 2022 (lunar): visible. November 8, 2022 (lunar): visible (Map 1 shows Hawaii within the region where the total lunar eclipse was visible). April 20, 2023 (solar): not visible (Map 2 shows the path of totality over the western Pacific/Indonesia/Australia region, not over Hawaii). April 8, 2024 (solar): not visible (path of totality crossed North America, not Hawaii). March 14, 2025 (lunar): visible. September 7, 2025 (lunar): not visible (this eclipse's visibility region was centered on Asia, Africa, Europe, and Australia, occurring during Hawaiian daytime).
Worked solution
$\text{Total lunar eclipses are visible from the entire night-side hemisphere of Earth where the Moon is above the horizon during the event.}$
$\text{Total solar eclipses are visible only along a narrow path of totality that sweeps across part of Earth's surface.}$
$\text{Map 1 (Nov 8, 2022 lunar): the hatched visibility region includes Hawaii} \Rightarrow \text{visible.}$
$\text{Map 2 (Apr 20, 2023 solar): the crossed-hatch path of totality does not pass over Hawaii} \Rightarrow \text{not visible.}$
$\text{Apply the same night-side/day-side and path-of-totality reasoning to the remaining dates using known eclipse geometry.}$
Answer: May 16, 2022 (lunar): visible. November 8, 2022 (lunar): visible (Map 1 shows Hawaii within the region where the total lunar eclipse was visible). April 20, 2023 (solar): not visible (Map 2 shows the path of totality over the western Pacific/Indonesia/Australia region, not over Hawaii). April 8, 2024 (solar): not visible (path of totality crossed North America, not Hawaii). March 14, 2025 (lunar): visible. September 7, 2025 (lunar): not visible (this eclipse's visibility region was centered on Asia, Africa, Europe, and Australia, occurring during Hawaiian daytime).

Lunar eclipses are visible from wherever the Moon is above the horizon (roughly half of Earth at once), while solar eclipse totality is confined to a narrow path, so each eclipse's visibility from Hawaii depends on whether Hawaii fell within the relevant shaded region or path shown on the maps (or, by extension, in the actual eclipse geometry for dates not mapped).

Question 76 · 1 pt · constructed response

On the diagram in your answer booklet, place an X on the Moon's orbit to indicate the position of the Moon on September 7, 2025.

The answer booklet diagram shows Earth at the center, viewed from above the North Pole, with the Moon's orbit drawn as a circle around Earth. Arrows labeled "Sunlight" enter the diagram from one side, indicating the direction of the Sun. On September 7, 2025, a total lunar eclipse occurred, so the Moon was in its full moon position: place the X on the Moon's orbit at the point directly opposite the Sun, on the far side of Earth from the incoming sunlight (in line with Earth's shadow).

Model answer: The X should be placed at the point on the Moon's orbit where the Moon is on the opposite side of Earth from the Sun (the full moon/lunar eclipse position), since September 7, 2025 was a lunar eclipse.
Worked solution
$\text{A lunar eclipse occurs only at full moon, when the Moon is on the side of Earth directly opposite the Sun.}$
$\text{September 7, 2025 is listed as a lunar eclipse in the data table.}$
$\Rightarrow \text{Mark the X on the Moon's orbit diametrically opposite the Sun's position, in Earth's shadow.}$
Answer: The X should be placed at the point on the Moon's orbit where the Moon is on the opposite side of Earth from the Sun (the full moon/lunar eclipse position), since September 7, 2025 was a lunar eclipse.

Since a total lunar eclipse can only happen at the full moon phase, when Earth is directly between the Sun and Moon, the Moon's position on that date must be plotted on the far side of Earth's orbit from the Sun.

Question 77 · 1 pt · constructed response

Base your answers to questions 77 and 78 on the diagram below and on your knowledge of Earth science. The diagram represents a North polar view of Earth as viewed from space on September 23. The curved arrows indicate the direction of Earth's rotation. State the number of degrees Earth's axis is tilted to a line perpendicular to the plane of its orbit.

earth question 77 figure
Model answer: 23.5°
Worked solution
$\text{Earth's axis is tilted } 23.5^\circ \text{ from a line perpendicular to the plane of Earth's orbit around the Sun.}$
Answer: 23.5°

This is a standard Earth science fact: Earth's rotational axis is tilted 23.5 degrees from the perpendicular to its orbital plane, which causes the seasons.

Question 78 · 1 pt · constructed response

Explain why a Foucault pendulum located on the Arctic Circle appears to change direction as it swings.

Model answer: The plane in which the pendulum swings stays fixed in space, but Earth rotates on its axis underneath the pendulum, so from the point of view of an observer standing on the rotating Earth, the pendulum's swing direction appears to gradually change (rotate) over time.
Worked solution
$\text{A Foucault pendulum's plane of swing remains fixed relative to the stars (inertia).}$
$\text{Earth rotates on its axis beneath the pendulum.}$
$\Rightarrow \text{relative to the ground (which is rotating), the pendulum's swing direction appears to slowly change, providing evidence of Earth's rotation.}$
Answer: The plane in which the pendulum swings stays fixed in space, but Earth rotates on its axis underneath the pendulum, so from the point of view of an observer standing on the rotating Earth, the pendulum's swing direction appears to gradually change (rotate) over time.

A Foucault pendulum demonstrates Earth's rotation: the pendulum itself keeps swinging in the same fixed plane in space, while the ground beneath it turns, making the swing direction appear to shift relative to the surface.

Question 79 · 1 pt · constructed response

Base your answers to questions 79 through 81 on the block diagram below and on your knowledge of Earth science. The diagram represents the magnetic orientation of the seafloor bedrock on both sides of the Southeast Indian Ridge. The age of the igneous bedrock and its distance, in kilometers, from the ridge center are shown. Determine the age and the magnetic orientation of the seafloor bedrock located 80 kilometers south of the ridge center.

earth question 79 figure
Model answer: The bedrock 80 km south of the ridge center is approximately 2 million years old, and it has reversed polarity.
Worked solution
$\text{Assume roughly uniform seafloor spreading rate: the age scale (0 to 4 million years) corresponds proportionally to the distance scale (0 to 150 km) on each side of the ridge.}$
$\text{Rate} \approx \dfrac{150 \text{ km}}{4 \text{ million years}} \approx 37.5 \text{ km per million years}$
$\text{Age at } 80\text{ km} \approx \dfrac{80}{37.5} \approx 2.1 \text{ million years} \approx 2 \text{ million years}$
$\text{Reading the polarity band at that position on the diagram shows it falls within a dark gray (reversed polarity) band.}$
Answer: The bedrock 80 km south of the ridge center is approximately 2 million years old, and it has reversed polarity.

Using the roughly linear relationship between distance from the ridge and bedrock age shown on the two aligned scales, 80 km from the ridge corresponds to about 2 million years old bedrock, which the diagram shows lies within a reversed-polarity band.

Question 80 · 1 pt · constructed response

Identify the two tectonic plates that are diverging at the Southeast Indian Ridge.

Model answer: The Australian (Indo-Australian) Plate and the Antarctic Plate.
Worked solution
$\text{The Southeast Indian Ridge is a divergent (spreading) boundary located in the Indian Ocean south of Australia.}$
$\text{It separates the Australian Plate (to the north) from the Antarctic Plate (to the south).}$
Answer: The Australian (Indo-Australian) Plate and the Antarctic Plate.

The Southeast Indian Ridge is the mid-ocean spreading boundary between the Australian Plate and the Antarctic Plate, where new oceanic crust forms as the two plates move apart.

Question 81 · 1 pt · constructed response

Explain, in terms of cooling rate, why surface oceanic crust forming at this ocean ridge is most likely basalt instead of gabbro.

The diagram represents the magnetic orientation of the seafloor bedrock on both sides of the Southeast Indian Ridge. The age of the igneous bedrock and its distance, in kilometers, from the ridge center are shown.

Model answer: Because the lava erupting at the ridge is in direct contact with cold ocean water at Earth's surface, it cools very quickly, which does not give crystals enough time to grow large; this produces fine-grained (extrusive) basalt rather than the coarse-grained (intrusive) gabbro that forms when magma cools slowly underground.
Worked solution
$\text{Surface lava at a mid-ocean ridge is exposed directly to cold seawater} \Rightarrow \text{rapid cooling.}$
$\text{Rapid cooling} \Rightarrow \text{little time for mineral crystals to grow} \Rightarrow \text{fine-grained texture} = \text{basalt.}$
$\text{Gabbro has the same composition as basalt but forms from slow cooling at depth, producing large, coarse-grained crystals.}$
Answer: Because the lava erupting at the ridge is in direct contact with cold ocean water at Earth's surface, it cools very quickly, which does not give crystals enough time to grow large; this produces fine-grained (extrusive) basalt rather than the coarse-grained (intrusive) gabbro that forms when magma cools slowly underground.

Basalt and gabbro have the same (mafic) chemical composition but differ in cooling rate and resulting texture; rapid surface cooling at the ridge produces basalt's fine grains, while gabbro requires slow, deep cooling to form large crystals.

Question 82 · 1 pt · constructed response

Base your answers to questions 82 through 85 on the passage and photograph below and on your knowledge of Earth science. The photograph shows a Herkimer diamond located in a cavity in dolostone. Herkimer Diamonds Herkimer 'diamonds' are actually quartz crystals that are double terminated (pointed at both ends) and found within exposed outcrops of dolostone around the Mohawk River Valley. The dolostone was formed about 500 million years ago in a shallow sea that was filled with sediment from the eroded ancient Adirondack Mountains to the north. While buried, cavities were formed in the dolostone by the slightly acidic ocean water. As sediment buried the rock, crystals grew in the cavities, slowly creating quartz crystals of exceptional clarity. The quartz crystals in the cavities are believed to have formed during the Carboniferous Period. Source: "Herkimer Diamonds" Geology.com Identify the chemical composition of the quartz that makes up a Herkimer diamond.

earth question 82 figure
Model answer: $SiO_2$ (silicon dioxide)
Worked solution
$\text{Quartz is composed of silicon dioxide: } SiO_2$
Answer: $SiO_2$ (silicon dioxide)

Quartz, the mineral making up Herkimer diamonds, has the chemical formula SiO2 (silicon dioxide), as listed in the Earth Science Reference Tables mineral chart.

Question 83 · 1 pt · constructed response

Identify one physical characteristic of a Herkimer diamond, other than its color, that allows it to be used in jewelry.

Model answer: Its hardness (quartz has a Mohs hardness of 7, making it durable and resistant to scratching).
Worked solution
$\text{Quartz hardness} = 7 \text{ on the Mohs scale (per the Earth Science Reference Tables).}$
$\text{A hardness of 7 resists scratching by most other materials, including dust and sand}, \Rightarrow \text{ suitable for jewelry that will be worn and handled.}$
Answer: Its hardness (quartz has a Mohs hardness of 7, making it durable and resistant to scratching).

Besides color, gem-quality minerals need durability to be worn as jewelry; quartz's hardness of 7 (or alternatively its glassy luster/transparency/clarity) makes it resistant to scratching, which is why Herkimer diamonds are usable in jewelry.

Question 84 · 1 pt · constructed response

Identify the orogeny that was responsible for the metamorphism of bedrock now exposed in the Adirondack Mountains.

Model answer: The Grenville orogeny
Worked solution
$\text{The bedrock core of the Adirondack Mountains was metamorphosed during the Grenville orogeny (a Precambrian mountain-building event).}$
Answer: The Grenville orogeny

The Adirondack Mountains' metamorphic bedrock (gneiss, etc.) resulted from the Grenville orogeny, a major Precambrian episode of mountain building and metamorphism, as identified in the Earth Science Reference Tables geologic history chart.

Question 85 · 1 pt · constructed response

In your answer booklet, circle the term that describes the relative age of the Herkimer diamonds compared to the relative age of the dolostone. Explain how the evidence cited in the reading passage supports your answer. Use the terms Herkimer diamonds and dolostone in your explanation.

Model answer: The Herkimer diamonds are younger than the dolostone. Evidence: the passage states that the dolostone had to form first (about 500 million years ago) and that cavities were later formed within the dolostone by acidic water; only after these cavities existed did quartz crystals grow inside them to form the Herkimer diamonds (during the Carboniferous Period). Since the Herkimer diamonds grew inside cavities that already existed within the dolostone, the dolostone must have formed before the Herkimer diamonds.
Worked solution
$\text{Passage: dolostone formed about 500 million years ago.}$
$\text{Cavities later formed in the dolostone from acidic ocean water.}$
$\text{Quartz crystals (Herkimer diamonds) grew within those cavities afterward, during the Carboniferous Period.}$
$\Rightarrow \text{Herkimer diamonds are younger than the dolostone that contains them.}$
Answer: The Herkimer diamonds are younger than the dolostone. Evidence: the passage states that the dolostone had to form first (about 500 million years ago) and that cavities were later formed within the dolostone by acidic water; only after these cavities existed did quartz crystals grow inside them to form the Herkimer diamonds (during the Carboniferous Period). Since the Herkimer diamonds grew inside cavities that already existed within the dolostone, the dolostone must have formed before the Herkimer diamonds.

Using the principle that a mineral filling a cavity in a rock must be younger than the rock that contains the cavity, the passage's sequence (dolostone forms, then cavities form, then Herkimer diamond quartz crystals grow inside the cavities) shows the Herkimer diamonds are younger than the dolostone.

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