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Maps & landscapes: Earth Science Regents Practice

Practice Maps & landscapes for the Earth Science Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 81 · 2 pts

On a topographic map, two index (heavy) contour lines are labeled $400$ ft and $500$ ft. Four evenly spaced intermediate contour lines lie between them. What is the contour interval of this map?

Worked solution
$\text{difference} = 500 - 400 = 100 \text{ ft}$
$\text{spaces} = 4 \text{ lines} + 1 = 5$ Four lines between two index lines create five equal spaces.
$\text{contour interval} = \frac{100}{5} = 20 \text{ ft}$
Answer: $20$ ft

The contour interval is the elevation change between adjacent contour lines; divide the elevation difference between index lines by the number of spaces (not lines) between them.

Question 82 · 2 pts

A topographic map has a contour interval of $20$ m. Point P lies in the zone between a contour line labeled $160$ m and the next higher contour line, with no contour line passing through P. Which elevation is possible for point P?

Worked solution
$\text{next higher contour} = 160 + 20 = 180 \text{ m}$
$160 < \text{elevation of P} < 180 \text{ m}$ A point between lines is above the lower and below the higher contour.
$175 \text{ m lies within this range}$
Answer: $175$ m

A point located between two contour lines has an elevation greater than the lower line and less than the higher line; it cannot equal either contour value.

Question 83 · 2 pts

On a topographic map with a uniform contour interval, the contour lines on the western slope of a hill are spaced very close together, while those on the eastern slope are spaced far apart. What does this pattern indicate?

Worked solution
$\text{slope steepness} \propto \frac{1}{\text{contour spacing}}$ Same elevation change over a shorter distance means a steeper gradient.
$\text{close contours (west)} \Rightarrow \text{steeper slope}$
Answer: The western slope is steeper than the eastern slope.

Closely spaced contour lines represent a large elevation change over a short distance, indicating a steep slope; widely spaced lines indicate a gentle slope.

Question 89 · 2 pts

An observer in New York State measures the altitude of Polaris (the North Star) above the northern horizon to be $43^\circ$. What is the observer's approximate latitude?

Worked solution
$\text{altitude of Polaris} = \text{observer's latitude}$ In the Northern Hemisphere this relationship always holds.
$43^\circ \rightarrow 43^\circ\,\text{N}$ Polaris is only visible north of the equator, so the latitude is north.
Answer: $43^\circ$ N

In the Northern Hemisphere, the altitude of Polaris above the horizon is approximately equal to the observer's latitude.

Question 95 · 2 pts

A location in New York State is at $44^\circ$ N latitude and $74^\circ$ W longitude. According to the ESRT 'Generalized Landscape Regions of New York State,' in which landscape region is this location found?

Worked solution
$44^\circ\,\text{N},\ 74^\circ\,\text{W}$ Plot the point on the NYS Generalized Landscape Regions map.
$\Rightarrow \text{Adirondack Mountains}$ This falls within the northeastern NY highlands.
Answer: Adirondack Mountains

The Adirondack Mountains form the highlands of northeastern New York, centered near 44 degrees N and 74 degrees W.

Question 32 · 2 pts

On a topographic map, contour lines in one area are spaced very close together, while in another area they are spaced far apart. The area where the contour lines are closely spaced represents:

Worked solution
$Contour line spacing shows how quickly elevation changes over a horizontal distance.$
$Closely spaced contours mean a large change in elevation over a short distance, which is a steep slope.$
$Widely spaced contours would indicate a gentle slope instead.$
Answer: a steep slope

Closely spaced contour lines indicate a steep gradient (rapid elevation change over a short horizontal distance), which corresponds to a steep slope.

Question 64 · 4 pts · constructed response

On a topographic map, point A has an elevation of $850$ feet and point B has an elevation of $1{,}250$ feet. The straight-line map distance between the two points is $5$ kilometers. Calculate the gradient between point A and point B in feet per kilometer.

Model answer: The gradient between point A and point B is $80$ feet per kilometer.
Worked solution
$\Delta \text{elevation} = 1{,}250 \text{ ft} - 850 \text{ ft} = 400 \text{ ft}$
$\text{gradient} = \dfrac{\Delta \text{elevation}}{\text{distance}}$
$\text{gradient} = \dfrac{400 \text{ ft}}{5 \text{ km}} = 80 \text{ ft/km}$
Answer: The gradient between point A and point B is $80$ feet per kilometer.

Gradient (slope) between two points on a topographic map is calculated by dividing the change in elevation by the horizontal distance separating them; a larger gradient value indicates steeper terrain.

Question 11 · 2 pts

On a topographic map, contour lines that are spaced very closely together indicate:

Worked solution
$\text{Contour lines connect points of equal elevation.}$
$\text{Closely spaced contour lines mean elevation changes rapidly over a short horizontal distance.}$
$\text{A rapid elevation change over a short distance is, by definition, a steep slope.}$
Answer: A steep slope

Contour line spacing reflects the steepness of terrain; closely spaced contours show a large change in elevation packed into a small horizontal distance, indicating a steep slope, while widely spaced contours indicate a gentle slope or flat area.

Question 47 · 2 pts

Two points on a topographic map are 5 kilometers apart along a straight line. Point X has an elevation of 850 meters, and Point Y has an elevation of 600 meters. What is the gradient (slope) between Point X and Point Y?

Worked solution
$\text{Change in elevation} = 850\text{ m} - 600\text{ m} = 250\text{ m}$
$\text{Gradient} = \dfrac{\text{change in field value}}{\text{distance}} = \dfrac{250\text{ m}}{5\text{ km}}$
$\text{Gradient} = 50\ \text{m/km}$
Answer: $50\ \text{m/km}$

Gradient is calculated by dividing the difference in elevation between two points by the horizontal distance separating them, giving 250 m divided by 5 km, or 50 m/km.

Question 72 · 4 pts · constructed response

Point A has an elevation of 620 feet and Point B has an elevation of 460 feet. The straight-line horizontal distance between Point A and Point B is 2 miles. Calculate the gradient (slope) of the land surface between Point A and Point B, in feet per mile. Show your work.

Model answer: The gradient between Point A and Point B is 80 feet per mile.
Worked solution
$\text{Change in elevation} = 620\text{ ft} - 460\text{ ft} = 160\text{ ft}$
$\text{Gradient} = \dfrac{\text{change in elevation}}{\text{horizontal distance}} = \dfrac{160\text{ ft}}{2\text{ mi}}$
$\text{Gradient} = 80\text{ ft/mi}$
Answer: The gradient between Point A and Point B is 80 feet per mile.

Gradient is calculated by dividing the difference in elevation between two points by the horizontal distance separating them, giving a rate that describes how steeply the land surface rises or falls.

Question 21 · 2 pts

Two locations are 5 kilometers apart along a straight line on a topographic map. One location has an elevation of 620 meters and the other has an elevation of 810 meters. What is the average gradient between these two locations?

Worked solution
$\text{Gradient} = \dfrac{\text{change in field value}}{\text{distance}}$ Gradient formula
$\text{change in elevation} = 810\ \text{m} - 620\ \text{m} = 190\ \text{m}$
$\text{Gradient} = \dfrac{190\ \text{m}}{5\ \text{km}} = 38\ \text{m/km}$
Answer: 38 m/km

Gradient is calculated by dividing the difference in elevation (or other field value) between two points by the horizontal distance separating them; here 190 meters divided by 5 kilometers gives 38 meters per kilometer.

Question 34 · 2 pts

On a topographic map, two adjacent contour lines have elevations of 220 meters and 240 meters. The horizontal distance between these two contour lines, measured perpendicular to them, is 2 kilometers. What is the gradient between these two points?

Worked solution
$\text{Gradient} = \dfrac{\text{change in field value}}{\text{distance}}$
$\text{Change in elevation} = 240\text{ m} - 220\text{ m} = 20\text{ m}$
$\text{Gradient} = \dfrac{20\text{ m}}{2\text{ km}} = 10\text{ m/km}$
Answer: 10 meters per kilometer

Gradient is calculated by dividing the difference in field value (here, elevation) by the horizontal distance over which that change occurs; dividing the 20-meter elevation change by the 2-kilometer distance gives a gradient of 10 meters per kilometer.

Question 56 · 4 pts · constructed response

Points A and B lie along a straight line on Earth's surface and are 5 kilometers apart. Point A has an elevation of 620 feet, and Point B has an elevation of 380 feet. Calculate the gradient between points A and B in feet per kilometer.

Model answer: 48 feet per kilometer
Worked solution
$\text{Gradient} = \dfrac{\text{change in elevation}}{\text{distance}}$
$\text{change in elevation} = 620\ \text{ft} - 380\ \text{ft} = 240\ \text{ft}$
$\text{Gradient} = \dfrac{240\ \text{ft}}{5\ \text{km}} = 48\ \text{ft/km}$
Answer: 48 feet per kilometer

Gradient describes how steeply elevation changes over horizontal distance; it is calculated by dividing the difference in elevation between two points by the distance separating them.

Question 66 · 4 pts · constructed response

On a topographic map, Point A has an elevation of 620 meters and Point B has an elevation of 180 meters. The straight-line map distance between Point A and Point B is 5 kilometers. Calculate the gradient between Point A and Point B in meters per kilometer, using the formula: gradient = change in field value divided by distance. Show your work.

Model answer: The gradient between Point A and Point B is 88 meters per kilometer.
Worked solution
$\Delta elevation = 620\ m - 180\ m = 440\ m$
$gradient = \dfrac{440\ m}{5\ km}$
$gradient = 88\ m/km$
Answer: The gradient between Point A and Point B is 88 meters per kilometer.

Gradient measures the rate of change of a field value, here elevation, over a given distance. Dividing the elevation difference between the two points by the horizontal distance between them gives the slope's steepness in meters per kilometer.

Question 72 · 4 pts · constructed response

Point A has an elevation of 620 feet. Point B has an elevation of 200 feet. The horizontal map distance between Point A and Point B is 4 miles. Calculate the gradient between Point A and Point B in feet per mile. Show your work.

Model answer: 105 feet per mile
Worked solution
$\text{difference in elevation} = 620\text{ ft} - 200\text{ ft} = 420\text{ ft}$
$\text{gradient} = \dfrac{420\text{ ft}}{4\text{ mi}} = 105\text{ ft/mi}$
Answer: 105 feet per mile

Gradient is the change in a field value, here elevation, divided by the horizontal distance between two points, and it describes how steeply the land surface rises or falls.

Question 84 · 2 pts

Where contour lines cross Elk Creek on a topographic map, they bend into V shapes that point toward the east. The land rises toward the east and slopes downward toward the west. In which direction does Elk Creek flow?

Worked solution
$\text{Contour V's point upstream, toward higher elevation (east)}$
$\text{Water flows downhill, opposite the V's} \Rightarrow \text{toward the west}$
Answer: Toward the west

Contour lines bend upstream where they cross a stream, forming V's that point toward higher elevation; the stream flows in the opposite direction, toward lower ground.

Question 85 · 2 pts

City A lies at $75^\circ$ W longitude and City B lies at $79^\circ$ W longitude, both at $43^\circ$ N. Earth rotates $15^\circ$ of longitude each hour. What is the difference in local solar time between the two cities?

Worked solution
$\Delta\text{longitude} = 79^\circ - 75^\circ = 4^\circ$
$\frac{15^\circ}{1 \text{ hr}} = \frac{15^\circ}{60 \text{ min}} = 1^\circ \text{ per } 4 \text{ min}$ 360 degrees in 24 hours gives 15 degrees per hour, or 4 minutes per degree.
$\text{time difference} = 4^\circ \times 4 \text{ min} = 16 \text{ min}$
Answer: $16$ minutes

Earth rotates 15 degrees of longitude per hour, equal to 4 minutes of time per degree, so the local solar time difference equals the longitude difference times 4 minutes.

Question 87 · 2 pts

According to the Generalized Landscape Regions of New York State map in the Earth Science Reference Tables, which of these regions is classified as a plateau?

Worked solution
$\text{Plateau} = \text{high elevation with flat-lying (horizontal) bedrock}$
$\text{Among the choices, the Allegheny Plateau is the plateau region}$ On the ESRT map, plateaus are distinct from mountains and from lowlands/plains.
Answer: Allegheny Plateau

On the ESRT landscape regions map, plateaus such as the Allegheny Plateau are high-elevation areas with relatively flat-lying sedimentary bedrock, different from mountain regions and from low-relief lowlands and plains.

Question 91 · 2 pts

A topographic map has a scale of $1\!:\!24{,}000$. On the map, two benchmarks are $5.0$ cm apart. What is the actual ground distance between the benchmarks?

Worked solution
$5.0\,\text{cm} \times 24{,}000 = 120{,}000\,\text{cm}$ Multiply the map distance by the scale factor to get true distance.
$120{,}000\,\text{cm} \div 100{,}000\,\text{cm/km} = 1.2\,\text{km}$ There are 100,000 cm in 1 km.
Answer: $1.2$ km (1,200 m)

On a fractional-scale map, true distance equals the map distance times the scale factor, after converting to the desired units.

Question 92 · 2 pts

According to the ESRT graph 'Relationship of Transported Particle Size to Water Velocity,' a stream flowing at a velocity of $100$ cm/s can transport particles up to what maximum size?

Worked solution
$v = 100\,\text{cm/s}$ Locate this velocity on the ESRT graph and read across to the curve.
$\text{max particle diameter} \approx 1\,\text{cm}$ A diameter near 1 cm falls within the pebble range of 0.2 to 6.4 cm.
$\Rightarrow \text{largest particle} = \text{pebble}$ Cobbles need about 190 cm/s, so they cannot be moved at 100 cm/s.
Answer: Pebbles

The faster a stream flows, the larger the particles it can carry; at about 100 cm/s the largest transportable particles are pebbles (diameter near 1 cm).

Question 93 · 2 pts

On a topographic map with a contour interval of $10$ ft, a hachured (depression) contour line is labeled $500$ ft. Point $X$ lies inside this depression but is not enclosed by any other contour line. What is the possible elevation of point $X$?

Worked solution
$\text{contour interval} = 10\,\text{ft}$ Hachures point downslope into the depression.
$490\,\text{ft} < \text{elevation of } X < 500\,\text{ft}$ Land inside the 500 ft depression contour is one interval lower, but no inner contour is crossed.
Answer: Between 490 ft and 500 ft

Hachured contour lines mark closed depressions, where elevation decreases toward the center, so a point just inside a 500 ft depression contour lies between 490 and 500 ft.

Question 94 · 2 pts

On a topographic map, point $P$ is located between the $300$ ft and $320$ ft contour lines. $P$ is about one-quarter of the distance from the $300$ ft line toward the $320$ ft line. What is the best estimate of the elevation at point $P$?

Worked solution
$\text{interval} = 320\,\text{ft} - 300\,\text{ft} = 20\,\text{ft}$
$300 + \tfrac{1}{4}(20) = 305\,\text{ft}$ P is one-quarter of the way up from the 300 ft line.
Answer: 305 ft

Interpolation estimates an elevation between two contour lines based on the fractional distance across the contour interval.

Question 11 · 2 pts

On a topographic map, a region where the contour lines are spaced very close together indicates

Worked solution
$Contour lines connect points of equal elevation, and each line represents a fixed change in elevation (the contour interval).$
$When contour lines are spaced close together, a large elevation change occurs over a short horizontal distance.$
$A large elevation change over a short distance corresponds to a steep slope.$
Answer: a steep slope

Because each contour line represents a set change in elevation, closely spaced contour lines mean elevation changes rapidly over a short horizontal distance, which is the definition of a steep slope.

Question 19 · 2 pts

On a topographic map, contour lines in one section are drawn very close together, while in another section they are spaced far apart. The closely spaced contour lines most likely represent:

Worked solution
$Contour lines connect points of equal elevation on a topographic map.$
$The spacing between contour lines shows how quickly elevation changes over a given horizontal distance.$
$Closely spaced contour lines mean elevation changes rapidly over a short distance, indicating a steep slope, while widely spaced lines indicate a gentle slope.$
Answer: a steep slope

Closely spaced contour lines indicate a large change in elevation over a small horizontal distance, which corresponds to a steep slope; widely spaced lines instead indicate a gentle, gradual slope.

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