Practical Geography Practice Questions — JAMB UTME — Page 3

Question 41Question →

A farm settlement measures $6\text{ cm}$ by $8\text{ cm}$ on a topographical map drawn at a scale of $1 : 40,000$ . If the map is enlarged to a new scale of $1 : 10,000$ , what is the area of the farm settlement on the enlarged map?

768\text{ cm}^2

192\text{ cm}^2

12\text{ cm}^2

3\text{ cm}^2

Show answer & explanation

Answer: $768\text{ cm}^2$

Answer

The area of the farm settlement on the enlarged map is

768\text{ cm}^2

.

First, calculate the original area on the map:

6\text{ cm} \times 8\text{ cm} = 48\text{ cm}^2

. Next, determine the linear scale factor

n

by comparing the scale denominators:

\frac{40,000}{10,000} = 4

. Because area scale varies with the square of the linear scale, the area scale multiplier is

n^2 = 4^2 = 16

. Multiplying the original map area by

16

yields

48\text{ cm}^2 \times 16 = 768\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the original area of the farm settlement on the map.

\text{Area} = 6\text{ cm} \times 8\text{ cm} = 48\text{ cm}^2

.

The area on a rectangular grid is found by multiplying width by length.

2

Determine the linear scale factor (

n

) of the enlargement.

n = \frac{\text{Old Scale Denominator}}{\text{New Scale Denominator}} = \frac{40,000}{10,000} = 4

.

Enlarging a map decreases the scale denominator, increasing linear dimensions by a factor of

n

.

3

Calculate the area enlargement factor.

\text{Area Factor} = n^2 = 4^2 = 16

.

Area changes proportionally to the square of the linear scale factor.

4

Multiply the original map area by the area enlargement factor.

\text{Enlarged Area} = 48\text{ cm}^2 \times 16 = 768\text{ cm}^2

.

To find the new surface area on the enlarged map, multiply the initial area by

n^2

.

Key Concept

Map Area Enlargement and Reduction Ratio (

n^2

rule)

Estimated Time:1m 30s

Question 42Question →

A topographical map extract drawn at a scale of $1:50,000$ delineates a drainage basin. The total area of the basin measured on the map is $160\text{ cm}^2$ . Morphometric analysis reveals the following total stream segment lengths measured directly from the map across all stream orders:
- 1st-order stream segments: $48\text{ cm}$
- 2nd-order stream segments: $20\text{ cm}$
- 3rd-order stream segments: $12\text{ cm}$
- 4th-order stream segments: $8\text{ cm}$

Calculate the drainage density of the river basin in $\text{km/km}^2$ .

Show answer & explanation

Answer: 1.1

Answer

The drainage density of the river basin is

1.1\text{ km/km}^2

.

Drainage density (

D

) measures the ratio of total stream channel length to the total drainage basin area (

D = \frac{\sum L}{A}

). Converting map measurements using the scale

1:50,000

(

1\text{ cm} = 0.5\text{ km}

), the total ground stream length is

88\text{ cm} \times 0.5\text{ km/cm} = 44\text{ km}

, and the actual ground area is

160\text{ cm}^2 \times (0.5)^2 = 40\text{ km}^2

. Dividing ground length by area produces

1.1\text{ km/km}^2

.

Step-by-Step Solution

1

Determine linear ground scale from map scale

1\text{ cm on map} = 0.5\text{ km on ground}

Scale

1:50,000

means

1\text{ cm} = 50,000\text{ cm} = 500\text{ m} = 0.5\text{ km}

.

2

Calculate ground area of the basin

Basin Area

A = 40\text{ km}^2

Area conversion requires squaring the linear scale factor:

160\text{ cm}^2 \times (0.5\text{ km/cm})^2 = 160 \times 0.25 = 40\text{ km}^2

.

3

Calculate total actual stream length in the basin

Total Stream Length

\sum L = 44\text{ km}

Total map stream length

= 48 + 20 + 12 + 8 = 88\text{ cm}

. Ground length

= 88\text{ cm} \times 0.5\text{ km/cm} = 44\text{ km}

.

4

Compute Drainage Density (

D

)

D = 1.1\text{ km/km}^2

Drainage density is the total channel length divided by total basin area:

D = \frac{\sum L}{A} = \frac{44\text{ km}}{40\text{ km}^2} = 1.1\text{ km/km}^2

.

Key Concept

Drainage Density Calculation from Topographical Maps

Question 43Question →

A wildlife sanctuary occupies an area of $48\text{ cm}^2$ on a topographical map drawn at a scale of $1 : 100,000$ . If the map is reduced to a new scale of $1 : 400,000$ , what is the area of the sanctuary on the new map in $\text{cm}^2$ ?

Show answer & explanation

Answer: 3

Answer

The area of the sanctuary on the reduced map is

3\text{ cm}^2

.

When a map scale is reduced from

1 : 100,000

to

1 : 400,000

, the linear dimension reduces by a factor of

4

(since

\frac{100,000}{400,000} = \frac{1}{4}

). Because area is two-dimensional, the area changes by the square of the linear reduction factor, which is

\left(\frac{1}{4}\right)^2 = \frac{1}{16}

. Therefore, the area on the new map is

48\text{ cm}^2 \times \frac{1}{16} = 3\text{ cm}^2

.

Step-by-Step Solution

1

Determine the linear scale factor between the original and new scales

Linear scale factor =

\frac{\text{Original Scale Denominator}}{\text{New Scale Denominator}} = \frac{100,000}{400,000} = \frac{1}{4}

When a map scale changes from

1 : 100,000

to

1 : 400,000

, every linear dimension on the map is reduced to

\frac{1}{4}

of its original length.

2

Calculate the area reduction factor

Area scale factor =

\left(\frac{1}{4}\right)^2 = \frac{1}{16}

Map area scales as the square of the linear scale factor (

n^2

).

3

Calculate the new area on the reduced map

New Area =

48\text{ cm}^2 \times \frac{1}{16} = 3\text{ cm}^2

Multiplying the original area by the area scale factor gives the area representation on the new map.

Key Concept

Map Reduction and Area Scale Factor Relationship

Question 44Question →

On a topographical map drawn to a scale of $1:20,000$ , a foot trail ascends continuously from Point A at a contour elevation of $150\text{ m}$ to Point B at a contour elevation of $300\text{ m}$ . If the straight-line map distance between Point A and Point B measures $15.0\text{ cm}$ , what is the average gradient of the slope from Point A to Point B?

1 : 10

1 : 20

1 : 200

1 : 500

Show answer & explanation

Answer: $1 : 20$

Answer

The average gradient of the slope from Point A to Point B is

1 : 20

.

To determine topographic gradient, calculate the Vertical Interval (

\text{VI}

) as

300\text{ m} - 150\text{ m} = 150\text{ m}

. Convert the map distance (

15.0\text{ cm}

) to ground distance using the map scale (

1:20,000

):

15.0\text{ cm} \times 20,000 = 300,000\text{ cm} = 3,000\text{ m}

. The ratio

\frac{\text{VI}}{\text{HE}} = \frac{150\text{ m}}{3,000\text{ m}} = \frac{1}{20}

, giving an average gradient of

1 : 20

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

\text{VI} = 300\text{ m} - 150\text{ m} = 150\text{ m}

The vertical interval represents the total vertical height gained between the starting contour line and the destination contour line.

2

Convert map distance to ground distance to determine the Horizontal Equivalent (HE)

\text{HE} = 15.0\text{ cm} \times 20,000 = 300,000\text{ cm} = 3,000\text{ m}

Using the map scale ratio of

1:20,000

, every

1\text{ cm}

on the map corresponds to

20,000\text{ cm}

(

200\text{ m}

) on the actual terrain.

3

Compute the slope gradient ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{150\text{ m}}{3,000\text{ m}} = \frac{1}{20} = 1 : 20

Gradient expresses the ratio of vertical rise to horizontal distance when both measurements are expressed in identical units.

Key Concept

Topographic Slope Gradient Calculation

Question 45Question →

A straight segment of a proposed highway measures $10\text{ cm}$ on a topographic map drawn at a scale of $1 : 50,000$ . On a newly redrawn map, the same segment of highway measures $25\text{ cm}$ . What is the Representative Fraction (R.F.) scale of the new map?

1 : 20,000

1 : 125,000

1 : 8,000

1 : 200,000

Show answer & explanation

Answer: $1 : 20,000$

Answer

The Representative Fraction (R.F.) of the new enlarged map is

1 : 20,000

.

The correct scale is

1 : 20,000

. When a map features are enlarged from

10\text{ cm}

to

25\text{ cm}

, the linear dimension increases by a factor of

2.5

. Because enlargement increases map scale (making features appear larger), the Representative Fraction denominator decreases by dividing the original denominator by

2.5

, giving

50,000 / 2.5 = 20,000

.

Step-by-Step Solution

1

Determine the linear enlargement factor

Linear Enlargement Factor =

\frac{\text{New Distance}}{\text{Original Distance}} = \frac{25\text{ cm}}{10\text{ cm}} = 2.5

Map enlargement increases map distance proportionally by a linear multiplier.

2

Calculate the new Representative Fraction (R.F.) scale denominator

\text{New Denominator} = \frac{\text{Original Denominator}}{\text{Enlargement Factor}} = \frac{50,000}{2.5} = 20,000

Enlarging a map produces a larger scale, which corresponds to a smaller scale denominator.

3

State the new R.F. scale

New R.F. scale =

1 : 20,000

The Representative Fraction expresses the scale ratio with a numerator of 1.

Key Concept

Map Enlargement Linear Scale Calculation

Question 46Question →

A statistical table shows the annual agricultural crop production for a region in Nigeria as follows:

Crop	Production (metric tonnes)
Cocoa	$540$
Oil Palm	$360$
Rubber	$180$
Groundnut	$120$

If this data is to be represented using a pie chart, what is the exact difference between the central angle allocated to Cocoa and the combined central angle allocated to Rubber and Groundnut?

72^\circ

90^\circ

144^\circ

162^\circ

Show answer & explanation

Answer: $72^\circ$

Answer

The difference between the central angle allocated to Cocoa and the combined central angle allocated to Rubber and Groundnut is

72^\circ

.

The total commodity output equals

1200

tonnes. The sector angle for Cocoa is calculated as

\frac{540}{1200} \times 360^\circ = 162^\circ

. The combined tonnage for Rubber and Groundnut is

180 + 120 = 300

tonnes, giving a combined angle of

\frac{300}{1200} \times 360^\circ = 90^\circ

. The difference between these two values is

162^\circ - 90^\circ = 72^\circ

.

Step-by-Step Solution

1

Calculate the total crop production

\text{Total} = 540 + 360 + 180 + 120 = 1200\text{ metric tonnes}

The total dataset value is required to determine the proportional weight of each commodity.

2

Calculate the central sector angle for Cocoa

\text{Angle}_{\text{Cocoa}} = \left(\frac{540}{1200}\right) \times 360^\circ = 0.45 \times 360^\circ = 162^\circ

A pie chart represents total data as a

360^\circ

circle.

3

Calculate the combined central sector angle for Rubber and Groundnut

\text{Combined Production} = 180 + 120 = 300\text{ metric tonnes}

;

\text{Angle}_{\text{Combined}} = \left(\frac{300}{1200}\right) \times 360^\circ = 0.25 \times 360^\circ = 90^\circ

Summing the metric tonnage of the two minor crops allows direct conversion to their aggregated sector angle.

4

Determine the difference between the angles

162^\circ - 90^\circ = 72^\circ

Subtracting the combined angle from the primary crop angle yields the required angular difference.

Key Concept

Calculation of Sector Angles for Pie Charts

Question 47Question →

A physical geographer constructs a cross-section across a ridge on a topographic map with a scale of $1 : 25{}000$ . On the vertical axis of the profile, $1\text{ cm}$ represents an elevation of $20\text{ metres}$ . What is the vertical exaggeration ( $VE$ ) of this cross-section profile?

Show answer & explanation

Answer: 12.5

Answer

The vertical exaggeration of the cross-section profile is 12.5.

To find vertical exaggeration (

VE

), convert both the vertical and horizontal scales to Representative Fractions (RF). The vertical scale of

1\text{ cm} = 20\text{ m}

converts to

1\text{ cm} : 2{}000\text{ cm}

(

1 : 2{}000

). The horizontal scale is given as

1 : 25{}000

. Dividing the horizontal scale denominator (

25{}000

) by the vertical scale denominator (

2{}000

) yields

12.5

.

Step-by-Step Solution

1

Convert the vertical scale to a unitless Representative Fraction (RF)

Vertical Scale =

1\text{ cm} : 20\text{ m} = 1\text{ cm} : 2{}000\text{ cm} = 1 : 2{}000

Both horizontal and vertical scales must be expressed in the same unit (centimetres) to enable direct comparison as representative fractions.

2

State the horizontal scale as a Representative Fraction (RF)

Horizontal Scale =

1 : 25{}000

The map scale provides the horizontal ratio between map distance and ground distance.

3

Apply the vertical exaggeration formula

VE = \frac{\text{Vertical Scale (RF)}}{\text{Horizontal Scale (RF)}} = \frac{1 / 2{}000}{1 / 25{}000} = \frac{25{}000}{2{}000}

Vertical exaggeration measures how many times larger the vertical scale is relative to the horizontal scale.

4

Perform the final division

VE = 12.5

Dividing 25,000 by 2,000 yields the unitless exaggeration factor of 12.5.

Key Concept

Vertical Exaggeration of Topographic Profiles

Question 48Question →

A forest reserve covers an area of $16\text{ cm}^2$ on a topographical map with a scale of $1 : 100,000$ . If the map is enlarged such that the same forest reserve occupies an area of $64\text{ cm}^2$ on the new map, what is the scale denominator ( $N$ ) of the new map, where the scale is expressed as $1 : N$ ?

Show answer & explanation

Answer: 50000

Answer

The scale denominator of the new map is 50,000.

To find the scale denominator of the enlarged map, first determine the area multiplier by dividing the new map area by the original map area (

64\text{ cm}^2 / 16\text{ cm}^2 = 4

). Since map area scale is the square of the linear scale multiplier (

k^2 = 4

), taking the square root gives the linear factor

k = 2

. Map enlargement by a linear factor of

2

means the map detail is twice as large, so the new scale denominator is found by dividing the original denominator by

2

:

100,000 / 2 = 50,000

. The scale of the new map is

1 : 50,000

.

Step-by-Step Solution

1

Calculate the area scale multiplier

Area multiplier = 64 / 16 = 4

The area scale multiplier is the ratio of the new feature area on the map to its original area on the map.

2

Calculate the linear scale multiplier

Linear multiplier = sqrt(4) = 2

Map area varies as the square of the linear scale multiplier (k^2), so the linear multiplier k is the square root of the area multiplier.

3

Calculate the new scale denominator

New scale denominator N = 100,000 / 2 = 50,000

When a map is enlarged by a linear factor k, its representative fraction denominator is divided by k.

Key Concept

Map Enlargement and Area Scale Calculations

Estimated Time:1m 30s

Question 49Question →

Which of the following elementary surveying instruments is primarily used by geographers during fieldwork to measure vertical angles and determine the slope of a landform?

Clinometer

Ranging pole

Gunter's chain

Prismatic compass

Show answer & explanation

Answer: Clinometer

Answer

The clinometer is the instrument primarily used for measuring vertical angles and land slopes.

The clinometer is an elementary surveying instrument designed to measure angles of elevation or depression, enabling field surveyors to calculate gradients and hill slopes accurately.

Step-by-Step Solution

1

Identify the primary parameter being measured in the question.

The target parameter is vertical angles and slope gradient of a hill/landform.

Different surveying instruments measure distinct spatial parameters such as distance, bearing, alignment, or slope.

2

Evaluate candidate surveying instruments against the target parameter.

The clinometer directly measures angular elevation/depression relative to the horizontal plane to calculate slopes.

A clinometer is an optical or mechanical instrument purpose-built for measuring slope angles in physical fieldwork.

Key Concept

Functions of Elementary Surveying Instruments

Question 50Question →

Match each drainage pattern type listed on the left with its primary underlying geological control or landform characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dendritic pattern matches uniformly resistant rock strata; Trellis pattern matches alternating bands of hard and soft rocks with right-angle tributaries; Radial pattern matches streams flowing outward from a central dome; Centripetal pattern matches streams converging inward toward a central basin.

Dendritic drainage develops on uniform rock resistance. Trellis drainage forms where alternating hard and soft rocks cause tributaries to join main streams at right angles. Radial drainage flows outward from a dome peak, and centripetal drainage flows inward toward a central basin.

Step-by-Step Solution

1

Identify structural control for branching patterns

Dendritic pattern pairs with uniformly resistant rock strata due to uniform erosion rates.

Tree-like dendritic networks only form where rock resistance is homogeneous.

2

Identify structural control for rectangular/parallel tributary layouts

Trellis pattern pairs with alternating bands of hard and soft rocks.

Tributaries carve valleys along softer strata and meet main streams near 90-degree angles.

3

Differentiate outward vs inward flow directions

Radial matches outward flow from elevated summits, while centripetal matches inward flow into depressed basins.

Topographic highs shed water in all directions (radial), while topographic depressions collect water (centripetal).

Key Concept

Geological Controls on Drainage Patterns

Question 51Question →

In satellite remote sensing and spatial data analysis, resolution attributes determine sensor capability and imagery utility. Match each remote sensing resolution type in Column A with its corresponding defining technical characteristic in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Spatial Resolution matches the ground dimension per pixel; Spectral Resolution matches the specific wavebands captured; Temporal Resolution matches the satellite orbital revisit frequency; Radiometric Resolution matches sensor bit-depth energy sensitivity.

Each resolution type represents a fundamental dimension of satellite imagery: Spatial Resolution determines minimum ground pixel size; Spectral Resolution determines electromagnetic waveband selection; Temporal Resolution measures repeat orbital frequency; Radiometric Resolution determines energy signal quantization bit depth.

Step-by-Step Solution

1

Identify spatial measurement properties in remote sensing.

Spatial resolution corresponds to physical pixel ground coverage dimension.

Pixel size determines spatial detail fine-scale capability.

2

Analyze electromagnetic radiation band sampling.

Spectral resolution corresponds to channel wavelength range and frequency coverage.

Spectral response distinguishes surface materials based on reflectance curves.

3

Determine time-series acquisition frequency.

Temporal resolution corresponds to satellite revisit track schedules.

Revisit intervals govern change-detection spatial monitoring capacity.

4

Evaluate sensor energy quantum quantization level.

Radiometric resolution corresponds to dynamic range bit-depth capacity.

Bit depth determines grayscale intensity level recording precision.

Key Concept

Resolution Parameters in Remote Sensing and GIS Data Acquisition

Estimated Time:2m 0s

Question 52Question →

On a topographical map drawn to a scale of $1 : 50,000$ , Point X is situated at an elevation of $150\text{ m}$ and Point Y is situated at an elevation of $400\text{ m}$ . If the straight-line distance between Point X and Point Y measured on the map is $5\text{ cm}$ , what is the slope gradient between the two points expressed as a percentage?

0.1\%

10\%

16\%

5\%

Show answer & explanation

Answer: $10\%$

Answer

The slope gradient between Point X and Point Y expressed as a percentage is

10\%

.

To find the gradient percentage, first calculate the Vertical Interval (VI):

400\text{ m} - 150\text{ m} = 250\text{ m}

. Next, calculate the Horizontal Equivalent (HE):

5\text{ cm} \times 50,000 = 250,000\text{ cm} = 2,500\text{ m}

. Expressing VI over HE gives

\frac{250\text{ m}}{2,500\text{ m}} = \frac{1}{10}

. Converting this fraction to a percentage gives

1/10 \times 100\% = 10\%

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

\text{VI} = 400\text{ m} - 150\text{ m} = 250\text{ m}

The Vertical Interval is the difference in elevation between the higher point and lower point.

2

Calculate the Horizontal Equivalent (HE) in meters

\text{HE} = 5\text{ cm} \times 50,000 = 250,000\text{ cm} = 2,500\text{ m}

Multiply the map distance by the scale denominator to find ground distance, then convert centimeters to meters.

3

Compute the slope gradient as a fraction and convert to percentage

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{250\text{ m}}{2,500\text{ m}} = 0.10 = 10\%

Gradient percentage is calculated by dividing Vertical Interval by Horizontal Equivalent in identical units and multiplying by

100\%

.

Key Concept

Slope and Gradient Calculation

Estimated Time:2m 0s

Question 53Question →

On a topographical map drawn to a scale of $1:50,000$ , a road ascends continuously from a contour elevation of $200\text{ m}$ to $450\text{ m}$ . If the distance along the road measured on the map is $5.0\text{ cm}$ , what is the average gradient of the road?

1\text{ in }10

1\text{ in }50

1\text{ in }100

1\text{ in }5.5

Show answer & explanation

Answer: $1\text{ in }10$

Answer

The average gradient of the road is

1\text{ in }10

.

To find the gradient, first compute the Vertical Interval (V.I.) as the difference between the two contour elevations:

450\text{ m} - 200\text{ m} = 250\text{ m}

. Next, determine the Horizontal Equivalent (H.E.) on the ground by converting the map distance using the scale:

5.0\text{ cm} \times 50,000 = 250,000\text{ cm} = 2,500\text{ m}

. Dividing the V.I. by the H.E. gives

\frac{250}{2,500} = \frac{1}{10}

, which expresses an average gradient of

1\text{ in }10

.

Step-by-Step Solution

1

Calculate the Vertical Interval (V.I.)

\text{V.I.} = 450\text{ m} - 200\text{ m} = 250\text{ m}

The Vertical Interval is the difference in height between the upper and lower contour elevations.

2

Calculate the Horizontal Equivalent (H.E.) in ground distance

\text{H.E.} = 5.0\text{ cm} \times 50,000 = 250,000\text{ cm} = 2,500\text{ m}

The map scale of

1:50,000

means

1\text{ cm}

on the map represents

50,000\text{ cm}

(

500\text{ m}

) on the ground.

3

Calculate the Gradient ratio

\text{Gradient} = \frac{\text{V.I.}}{\text{H.E.}} = \frac{250\text{ m}}{2,500\text{ m}} = \frac{1}{10} = 1\text{ in }10

Topographic gradient is defined as the ratio of Vertical Interval to Horizontal Equivalent expressed in the same unit.

Key Concept

Calculating Topographic Gradient from Contour Lines and Map Scale

Estimated Time:1m 30s

Question 54Question →

Match each type of statistical map or graphical representation on the left with its most appropriate geographical application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Choropleth map matches displaying population density using shaded administrative areas; Dot map matches showing absolute spatial distribution of population using point symbols; Flow map matches illustrating volume and direction of cargo or passenger movements; Pie chart matches showing percentage component breakdown of a total statistical value.

Each mapping technique is specifically designed for a particular data format: Choropleth maps represent area-based densities, Dot maps present absolute point distributions, Flow maps display movement volumes, and Pie charts represent percentage proportions of a total.

Step-by-Step Solution

1

Analyze Choropleth map characteristics

Matches displaying population density using shaded administrative areas

Choropleth maps use graded shading over administrative units to convey continuous density or ratio data.

2

Analyze Dot map characteristics

Matches showing absolute spatial distribution of population using point symbols

Dot maps represent raw numerical quantities distributed spatially using point symbols.

3

Analyze Flow map characteristics

Matches illustrating volume and direction of cargo or passenger movements

Flow diagrams display dynamic movements along paths with width scaled to quantity.

4

Analyze Pie chart characteristics

Matches showing percentage component breakdown of a total statistical value

Pie charts show proportional sub-divisions of 360 degrees corresponding to 100% of a dataset.

Key Concept

Selection and Functions of Statistical Maps and Diagrams

Question 55Question →

On a topographical survey map drawn to a scale of $1:40,000$ , Point X lies at an elevation of $250\text{ m}$ on a hillside, and Point Y lies further uphill along a straight spur at an elevation of $650\text{ m}$ . If the distance between Point X and Point Y measured on the map is $5\text{ cm}$ , the gradient of the slope between the two points can be expressed in the ratio form $1 : x$ . What is the numerical value of $x$ ?

Show answer & explanation

Answer: 5

Answer

The numerical value of

x

is 5.

The difference in elevation (Vertical Interval) between the two points is

400\text{ m}

. Converting the

5\text{ cm}

map measurement to ground distance using the

1:40,000

scale gives

2,000\text{ m}

(Horizontal Equivalent). Dividing the Vertical Interval by the Horizontal Equivalent yields

\frac{400}{2000} = \frac{1}{5}

, meaning the gradient expressed as

1 : x

has

x = 5

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

400 m

The vertical interval represents the elevation difference between Point Y and Point X:

650\text{ m} - 250\text{ m} = 400\text{ m}

.

2

Calculate the Horizontal Equivalent (HE)

2,000 m

At a map scale of

1:40,000

,

1\text{ cm}

on the map corresponds to

40,000\text{ cm} = 400\text{ m}

on the ground. A map measurement of

5\text{ cm}

equals

5 \times 400\text{ m} = 2,000\text{ m}

.

3

Determine the Gradient Ratio

1 : 5 (x = 5)

Gradient is determined by dividing Vertical Interval by Horizontal Equivalent:

\frac{400\text{ m}}{2,000\text{ m}} = \frac{1}{5}

. Writing this ratio as

1 : x

yields

x = 5

.

Key Concept

Slope and Gradient Calculation from Topographical Contour Maps

Question 56Question →

Two adjacent drainage basins, Basin A and Basin B, experience identical climatic conditions and precipitation levels. A geomorphic survey records the following morphometric data:

Drainage Basin	Total Channel Length ( $L$ )	Total Basin Area ( $A$ )
Basin A	$180\text{ km}$	$60\text{ km}^2$
Basin B	$50\text{ km}$	$50\text{ km}^2$

Based on the drainage density of each basin, which of the following statements accurately compares their hydrological and geological characteristics?

Basin A has a higher drainage density (

3.0\text{ km/km}^2

), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

Basin A has a lower drainage density (

0.33\text{ km/km}^2

) than Basin B (

1.0\text{ km/km}^2

), indicating greater surface permeability and enhanced groundwater infiltration in Basin A.

Basin A's drainage density of

3.0\text{ km/km}^2

is primarily produced by lower-course floodplain depositional features such as oxbow lakes and meander loops that lengthen channels.

Basin B has a higher drainage density than Basin A because V-shaped contour lines across its main channels point downstream towards the basin outlet.

Show answer & explanation

Answer: Basin A has a higher drainage density ( $3.0\text{ km/km}^2$ ), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

Answer

Basin A has a higher drainage density (

3.0\text{ km/km}^2

), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

The option identifying Basin A as having a higher drainage density (

3.0\text{ km/km}^2

) is correct because

D_d

is calculated by dividing total stream length (

180\text{ km}

) by total basin area (

60\text{ km}^2

). High values of drainage density reflect dense channel development, typical of impermeable rock strata where water cannot easily infiltrate, resulting in high surface runoff velocity.

Step-by-Step Solution

1

Calculate the drainage density (

D_d

) for Basin A using the formula

D_d = \frac{L}{A}

.

D_d(\text{Basin A}) = \frac{180\text{ km}}{60\text{ km}^2} = 3.0\text{ km/km}^2

.

Drainage density measures the total channel length per unit basin area.

2

Calculate the drainage density (

D_d

) for Basin B.

D_d(\text{Basin B}) = \frac{50\text{ km}}{50\text{ km}^2} = 1.0\text{ km/km}^2

.

Provides the baseline comparison value for Basin B.

3

Interpret the geomorphic and hydrological implications of high vs. low drainage density.

Basin A (

3.0\text{ km/km}^2 > 1.0\text{ km/km}^2

) has a higher stream network concentration, which signifies impermeable bedrock/clay soils, reduced infiltration, and rapid surface runoff leading to intense flood peaks.

High drainage density correlates with surface impermeability, sparse vegetation, and efficient surface runoff networks.

Key Concept

Drainage Density (

D_d

) and Basin Hydrology

Estimated Time:2m 0s

Question 57Question →

Match each fundamental GIS data model or remote sensing concept on the left with its correct defining characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Raster Data Model corresponds to the continuous grid of square pixels or cells; Vector Data Model corresponds to discrete points, lines, and polygon boundaries; Passive Remote Sensing corresponds to detecting naturally occurring radiation; Active Remote Sensing corresponds to emitting its own energy signal and measuring the reflected response.

Each concept is matched correctly according to standard GIS and remote sensing definitions: Raster uses pixel grids; Vector uses points/lines/polygons; Passive sensing detects natural sunlight/thermal radiation; Active sensing transmits and records its own artificial energy pulse.

Step-by-Step Solution

1

Identify the core structural difference between spatial data models.

Raster uses continuous pixel grids, whereas vector uses coordinate-based points, lines, and polygons.

Raster data stores attributes cell-by-cell across space, while vector data outlines discrete geographical entities explicitly.

2

Distinguish between remote sensing systems based on their energy source.

Passive sensing measures existing natural radiation (sunlight/heat), while active sensing supplies its own illumination signal (radar pulses).

The distinction hinges entirely on whether the sensor generates energy or passively records available environmental radiation.

Key Concept

GIS Data Structure Fundamentals and Remote Sensing Energy Sources

Question 58Question →

On a topographical map, a series of V-shaped contour lines point towards lower elevation values as they extend across the landscape. Which of the following relief features is represented by this contour pattern?

Spur

River valley

Col

Escarpment

Show answer & explanation

Answer: Spur

Answer

Spur

V-shaped contour lines whose apexes point towards lower elevation represent a spur, which is a tongue of elevated land projecting downwards from a main hill or ridge. Conversely, when V-shapes point upstream towards higher elevation, they indicate a river valley.

Step-by-Step Solution

1

Analyze the contour line geometry described in the prompt

The contour lines form V-shapes pointing towards decreasing (lower) elevation.

Understanding the orientation of V-shaped contours relative to elevation is key to identifying linear relief features.

2

Distinguish between a spur and a valley using contour apex orientation

Apexes pointing upstream (higher elevation) represent valleys, whereas apexes pointing downstream/down-slope (lower elevation) represent spurs.

A spur is a salient tongue of high ground sloping downwards between two valleys.

Key Concept

Contour pattern recognition for spurs and valleys

Question 59Question →

On a topographical map, a hillside shows contour lines that are densely packed near the hilltop at higher elevations and gradually become widely spaced towards the foot of the hill at lower elevations. Which type of relief slope does this contour pattern represent?

Concave slope

Convex slope

Uniform slope

Terraced slope

Show answer & explanation

Answer: Concave slope

Answer

Concave slope

The correct answer is a concave slope because contour lines that are closely spaced near the crest and widely spaced towards the base represent a steep upper slope transitioning into a gentle lower slope, which forms an inward-curving profile.

Step-by-Step Solution

1

Analyze the contour line spacing at higher elevations.

Densely packed contours indicate a steep gradient near the top of the hill.

When contour lines are close together, elevation changes rapidly over a short horizontal distance.

2

Analyze the contour line spacing at lower elevations.

Widely spaced contours indicate a gentle gradient near the base of the hill.

When contour lines are far apart, elevation changes gradually over a longer horizontal distance.

3

Combine the gradient observations to identify the slope profile.

A slope that curves inward—steep near the summit and gentle near the base—is defined as a concave slope.

This physical shape corresponds to a concave profile.

Key Concept

Contour Spacing and Slope Profiles

Question 60Question →

Match each relief feature or slope type with the contour line pattern that uniquely describes its representation on a topographical map.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Concave Slope pairs with closely spaced contours at higher elevations and widely spaced contours at lower elevations; Convex Slope pairs with widely spaced contours at higher elevations and closely spaced contours at lower elevations; Col (Saddle) pairs with a low pass between two peaks bounded by contour lines curving away on opposing sides; Escarpment (Cliff) pairs with contour lines merging along a steep vertical face.

The paired features match their exact cartographic representations: concave slopes are steeper at high elevations; convex slopes are steeper at low elevations; cols are pass features between summits; and escarpments display merged contours due to vertical cliffs.

Step-by-Step Solution

1

Analyze the gradient changes of concave and convex slope profiles.

A concave slope grows steeper with altitude (contours close together near the top), while a convex slope flattens out near the top (contours widely spaced near the top).

Contour spacing directly corresponds to terrain gradient; closer contours indicate steeper slopes.

2

Identify the structural contour representation of a col or saddle.

A col appears as a low point between two adjacent high areas where contours loop away from the center gap.

The neck between two peaks creates a characteristic hourglass-shaped contour arrangement.

3

Examine how extreme vertical drops are shown on topographic maps.

A cliff or escarpment has very small or zero horizontal distance between elevation changes, making contour lines run together or merge.

Vertical relief over negligible horizontal equivalent results in overlapping contour lines.

Key Concept

Contour Line Patterns and Relief Feature Identification

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