Practical Geography Practice Questions — JAMB UTME — Page 4

Question 61Question →

A geographer is constructing a statistical map using proportional circles to represent agricultural output across regions. Region A produced $100{}000\text{ metric tonnes}$ of cassava and is drawn with a circle radius of $1.5\text{ cm}$ . If Region B produced $400{}000\text{ metric tonnes}$ of cassava, what is the correct radius required for the circle representing Region B?

3.0\text{ cm}

6.0\text{ cm}

0.75\text{ cm}

4.5\text{ cm}

Show answer & explanation

Answer: $3.0\text{ cm}$

Answer

The radius required for the circle representing Region B is

3.0\text{ cm}

.

For proportional circle maps, the visual area of each symbol must be directly proportional to the magnitude of the data represented (

A \propto V

). Since the area of a circle is

A = \pi r^2

, the radius

r

is proportional to the square root of the statistical value (

r \propto \sqrt{V}

). The output ratio between Region B and Region A is

400{}000 / 100{}000 = 4

. The square root of this ratio gives a radius scale factor of

\sqrt{4} = 2

. Multiplying Region A's circle radius of

1.5\text{ cm}

by

2

gives

3.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the ratio of the quantities represented by the two regions

Data Value Ratio =

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

Determining how many times larger Region B's output is compared to Region A.

2

Apply the proportional symbol area scaling formula (

r \propto \sqrt{V}

)

Radius Scale Factor =

\sqrt{4} = 2

Because a circle's area is proportional to the statistical value (

A \propto V

), the radius must scale with the square root of the value ratio.

3

Calculate the required radius for Region B

Radius =

1.5\text{ cm} \times 2 = 3.0\text{ cm}

Multiplying the base radius of Region A by the radius scale factor.

Key Concept

Proportional Symbol Map Scaling (Square Root Radius Rule)

Question 62Question →

In practical geography, cartographers select specific statistical mapping techniques based on data characteristics, spatial continuity, and quantitative representation rules. Match each statistical mapping technique on the left with its defining cartographic principle or limitation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Choropleth Map pairs with shading predefined administrative units based on average density ranges. Isopleth Map pairs with using continuous isolines joining places of equal value. Flow Map pairs with employing bands or vectors of proportional width for directional movement volume. Dot Map pairs with using uniform point symbols representing fixed quantitative values.

Each mapping technique corresponds directly to its spatial data type and cartographic construction method: Choropleth maps shade administrative units according to calculated density intervals; Isopleth maps utilize isolines to show continuous spatial variations across continuous surfaces; Flow maps draw lines whose widths vary proportionally with trade or migration volumes; Dot maps display discrete totals by placing point symbols of fixed values across regions.

Step-by-Step Solution

1

Analyze Choropleth mapping principles

Identified that choropleth maps rely on average density ratios mapped to administrative regions with shaded fill gradients.

Administrative boundary shading is the core structural element of choropleth cartography.

2

Analyze Isopleth mapping principles

Identified that isopleths join equal values across a continuous phenomenon field using interpolated isolines.

Continuous data representation like temperature or pressure requires isoline interpolation.

3

Analyze Flow mapping principles

Identified that flow maps connect origins and destinations with proportional line widths.

Movement of goods, trade, or migrants across geographical space is measured along directional pathways.

4

Analyze Dot mapping principles

Identified that dot maps assign a unit value per point to represent absolute discrete counts.

Dot maps effectively demonstrate spatial clustering without falsely placing dots at exact coordinates.

Key Concept

Cartographic characteristics, rules, and limitations of statistical maps

Estimated Time:2m 0s

Question 63Question →

On a topographic map drawn to a scale of $1 : 50,000$ , a road ascends continuously from Point A at a contour elevation of $160\text{ m}$ to Point B at a contour elevation of $360\text{ m}$ . If the straight-line distance between Point A and Point B measured on the map is $10\text{ cm}$ , calculate the gradient of the road between these two points. Express your answer as the value of $N$ in the ratio $1 : N$ .

Show answer & explanation

Answer: 25

Answer

The value of N in the gradient ratio 1 : N is 25.

The vertical interval (difference in elevation) between the two points is

360\text{ m} - 160\text{ m} = 200\text{ m}

. The horizontal ground distance (Horizontal Equivalent) is calculated from the map distance of

10\text{ cm}

at a scale of

1 : 50,000

, giving

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

. Dividing the vertical interval by the horizontal equivalent gives

\frac{200\text{ m}}{5,000\text{ m}} = \frac{1}{25}

, which corresponds to a gradient ratio of

1 : 25

, so

N = 25

.

Step-by-Step Solution

1

Calculate Vertical Interval (VI)

200 m

Subtract the lower contour height from the higher contour height (360 m - 160 m).

2

Calculate Horizontal Equivalent (HE) in meters

5,000 m

Multiply map measurement by scale factor (10 cm * 50,000 = 500,000 cm = 5,000 m).

3

Compute the gradient ratio (VI / HE)

1 / 25

Divide 200 m by 5,000 m and simplify the fraction to 1 / 25.

Key Concept

Slope and Gradient Calculation from Topographical Map Contours

Question 64Question →

An environmental planning agency in Nigeria is designing a Geographic Information System (GIS) database to assess flood vulnerability in the Niger Delta. The analysis requires representing a continuous land elevation surface alongside discrete road networks. Furthermore, satellite imagery must be acquired during nighttime and heavy cloud cover conditions. Which of the following correctly identifies the appropriate GIS data model for the elevation surface and the sensor type required for imagery acquisition?

The elevation surface is best represented using a raster data model of pixel grid cells, while data acquisition requires an active microwave sensor.

The elevation surface is best represented using a vector data model of points and lines, while data acquisition requires a passive optical sensor.

The elevation surface is best represented using a raster data model based on topological polygons, while data acquisition requires a passive thermal sensor.

The elevation surface is best represented using a vector data model of pixel grid cells, while data acquisition requires an active solar sensor.

Show answer & explanation

Answer: The elevation surface is best represented using a raster data model of pixel grid cells, while data acquisition requires an active microwave sensor.

Answer

The elevation surface is best represented using a raster data model of pixel grid cells, while data acquisition requires an active microwave sensor.

The correct answer accurately pairs the raster data structure—which represents continuous surface variations via a grid of pixels—with active microwave remote sensing, which provides its own radiation source capable of penetrating atmospheric clouds and operating at night.

Step-by-Step Solution

1

Analyze spatial data representation requirements

Continuous spatial phenomena like elevation surfaces (Digital Elevation Models) are continuously varying and best stored as grid cells in a raster data model, whereas discrete features like roads use vector geometry.

Raster grid structures assign an elevation value to every pixel across a continuous space.

2

Determine remote sensing atmospheric and temporal constraints

Active sensors generate their own electromagnetic radiation (e.g., microwave RADAR signals) which can penetrate cloud cover and operate independent of daylight.

Passive sensors rely on ambient sunlight or earth radiation, which cloud cover obstructs.

3

Combine data model and sensor requirements to select correct pairing

The correct combination is raster data model paired with an active microwave sensor.

This combination fulfills both continuous spatial modeling and all-weather day/night image capture.

Key Concept

GIS Raster vs Vector Data Models and Active vs Passive Remote Sensing Sensors

Question 65Question →

Assuming an absence of surface obstructions such as trees or tall structures, two observation stations located respectively at the summit and the base of a hill possessing a convex slope are intervisible.

True

False

Show answer & explanation

Answer: False

Answer

False. Two points located at the summit and the base of a convex slope are not intervisible because the outward bulge of the slope in its upper-middle section obstructs the direct line of sight.

Intervisibility between two points on a map depends on whether the relief between them drops below the straight line of sight. On a convex slope, the upper portion has a gentle gradient (contours far apart) while the lower portion is steep (contours close together). This creates a bulging profile where the central terrain rises into the direct sightline between the summit and the valley floor, preventing mutual visibility.

Step-by-Step Solution

1

Analyze the characteristic contour arrangement and profile of a convex slope.

On a convex slope, contours are widely spaced near the crest (gentle gradient) and closely spaced near the base (steep gradient).

Understanding the physical shape of the land profile is necessary to evaluate the path of the sightline.

2

Evaluate the line of sight between the highest point (summit) and lowest point (base).

The direct line connecting the summit and base passes below the bulging middle section of the hill profile.

Intervisibility requires that no ground point along the cross-section intersects or rises above the straight line connecting the two stations.

3

Determine intervisibility status.

The stations are not intervisible.

Intervening high ground caused by the convex curvature blocks the sightline.

Key Concept

Intervisibility on Convex vs. Concave Slopes

Question 66Question →

On a topographical map drawn to a scale of $1 : 25,000$ , two hilltop stations, Station A and Station B, are separated by a map distance of $16\text{ cm}$ . If Station A is situated at an elevation of $580\text{ m}$ above sea level and Station B is at an elevation of $180\text{ m}$ , what is the gradient of the slope between the two stations expressed as the value $N$ in the ratio $1 : N$ ?

Show answer & explanation

Answer: 10

Answer

The denominator

N

in the slope gradient ratio

1 : N

is

10

(representing a gradient ratio of

1 : 10

).

To calculate the gradient expressed as

1 : N

, determine the Vertical Interval (VI

= 580\text{ m} - 180\text{ m} = 400\text{ m}

) and the Horizontal Equivalent (HE

= 16\text{ cm} \times 25,000 = 400,000\text{ cm} = 4,000\text{ m}

). Dividing VI by HE gives

\frac{400\text{ m}}{4,000\text{ m}} = \frac{1}{10}

, which corresponds to a gradient of

1 : 10

, making

N = 10

.

Step-by-Step Solution

1

Determine the Vertical Interval (VI)

\text{VI} = 580\text{ m} - 180\text{ m} = 400\text{ m}

The Vertical Interval is the vertical height difference between the two specified elevations.

2

Calculate the Horizontal Equivalent (HE) on the ground

\text{HE} = 16\text{ cm} \times 25,000 = 400,000\text{ cm} = 4,000\text{ m}

Multiply the map distance by the scale denominator to find the ground distance, then divide by 100 to convert centimeters to meters.

3

Calculate the gradient ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{400\text{ m}}{4,000\text{ m}} = \frac{1}{10} = 1 : 10

Gradient is calculated as Vertical Interval divided by Horizontal Equivalent, simplified to a fraction with a numerator of 1.

Key Concept

Calculation of slope gradient using vertical interval, map distance, and representative fraction scale

Estimated Time:1m 30s

Question 67Question →

Match each statistical mapping technique on the left with the cartographic representation principle and data type it utilizes on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dot Map pairs with uniform point symbols representing discrete distributions; Choropleth Map pairs with shading intensities within administrative boundaries for average densities; Flow Map pairs with varying line thickness depicting volume of movement; Isopleth Map pairs with lines connecting points of equal value for continuous phenomena.

Each mapping technique matches its primary cartographic design function: dot maps represent absolute discrete quantities using uniform point symbols; choropleth maps represent spatial densities using shading aggregated inside administrative units; flow maps convey directional movement and volume via proportional line widths; and isopleth maps depict continuous environmental variables by linking points of equal magnitude with isolines.

Step-by-Step Solution

1

Identify the data symbol type for Dot Maps.

Dot maps use point symbols of fixed value to represent discrete counts across a region.

This establishes the relationship between Dot Map and uniform point symbols.

2

Identify the spatial boundary and shading rule for Choropleth Maps.

Choropleth maps rely on predetermined administrative boundaries filled with graded shading to show average ratios or densities.

This pairs Choropleth Map with administrative unit shading.

3

Analyze how dynamic movement data is represented cartographically.

Flow maps use proportional line widths along travel channels to quantify movements like trade or migration.

This pairs Flow Map with varying line thickness along movement paths.

4

Examine continuous spatial surface mapping using contour-like lines.

Isopleth maps interpolate continuous distributions (e.g., isotherms, isobars) by connecting equal numerical values.

This matches Isopleth Map with lines of equal value across surfaces.

Key Concept

Cartographic principles of statistical maps and quantitative spatial data representation

Question 68Question →

During a prismatic compass survey of a farmland, a student records the forward bearing of line PQ as $125^\circ$ . Assuming both stations are completely free from local attraction, what is the back bearing of line QP?

305^\circ

055^\circ

215^\circ

125^\circ

Show answer & explanation

Answer: $305^\circ$

Answer

The back bearing of line QP is

305^\circ

.

In prismatic compass surveying, the forward bearing and back bearing of any line differ by exactly

180^\circ

. When the forward bearing is less than

180^\circ

, adding

180^\circ

gives the back bearing (

125^\circ + 180^\circ = 305^\circ

).

Step-by-Step Solution

1

Identify the forward bearing and check if it is greater or less than

180^\circ

.

The forward bearing is

125^\circ

, which is less than

180^\circ

.

The rule for converting forward bearing (FB) to back bearing (BB) depends on whether FB is greater than or less than

180^\circ

.

2

Apply the bearing conversion formula:

\text{BB} = \text{FB} + 180^\circ

when

\text{FB} < 180^\circ

.

\text{BB} = 125^\circ + 180^\circ = 305^\circ

.

Back bearing represents the opposite direction (

180^\circ

difference) along the same line.

Key Concept

Forward and Back Bearing Conversion in Prismatic Compass Surveying

Estimated Time:1m 0s

Question 69Question →

In a Geographic Information System (GIS), descriptive non-spatial information such as the name, population density, and land-use category of an administrative district is classified as which type of data?

Attribute data

Spatial data

Vector data

Raster data

Show answer & explanation

Answer: Attribute data

Answer

Attribute data

Attribute data represents tabular, non-spatial characteristics such as names, counts, or categories that describe geographic features mapped within a GIS.

Step-by-Step Solution

1

Analyze the characteristic of data described in the prompt.

The parameters listed (district names, population statistics, land-use categories) are non-spatial descriptions.

GIS data is broadly divided into location geometry (spatial) and non-locational characteristics (attribute).

2

Identify the standard GIS data classification for descriptive records.

Tabular records containing names, values, or categories connected to geometric features form attribute data.

Attribute data provides answers to 'what', 'who', or 'how much' regarding spatial features.

Key Concept

GIS Fundamentals: Spatial versus Attribute Data

Question 70Question →

Match each contour line pattern listed on the left with the corresponding relief landform it represents on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: Concentric closed contours decreasing inward match with Depression or Basin; V-shaped contours pointing uphill match with River Valley or Stream Bed; Contours closely spaced above and widely spaced below match with Concave Slope; Two high closed contour sets separated by a narrow dip match with Saddle or Col.

Each contour pattern corresponds directly to a distinct landform: inward-decreasing closed contours represent a depression; uphill-pointing V-contours mark a river valley; steep-then-gentle contour density indicates a concave slope; and a low ridge between two peaks forms a saddle or col.

Step-by-Step Solution

1

Analyze contour spacing and elevation trends for closed ring patterns.

Decreasing values toward the center signify a depression rather than a hill summit.

Topographic elevation decreases toward the middle in depressions.

2

Examine V-shaped contour orientation relative to elevation values.

The apex pointing toward higher terrain indicates water flow direction down the valley line.

Valleys cut into terrain such that contour V-shapes point up-valley toward higher ground.

3

Evaluate contour density variations from higher to lower elevations.

Steep slope near the top transition to gentle slope at the bottom defines a concave profile.

Contour spacing directly correlates with slope steepness (close = steep, wide = gentle).

4

Identify relief between two high hilltops.

The low region connecting two summits represents a saddle or col.

A saddle is formed by the low area between two surrounding high points.

Key Concept

Relief Representation and Contour Patterns

Question 71Question →

On a topographic map, index contours are drawn at elevations of $400\text{ m}$ and $700\text{ m}$ . There are $5$ equal contour intervals (spaces) between these two index contours. A radio transmission mast is located on a contour line situated $3$ contour intervals higher than the $700\text{ m}$ index contour. What is the elevation, in meters, of the contour line where the radio mast is located?

Show answer & explanation

Answer: 880

Answer

The elevation of the contour line where the radio mast stands is

880\text{ m}

.

The elevation difference between the

400\text{ m}

and

700\text{ m}

index lines is

300\text{ m}

. With

5

equal spaces separating them, the contour interval is

60\text{ m}

. Adding

3

contour intervals (

180\text{ m}

) to the

700\text{ m}

index line yields an elevation of

880\text{ m}

.

Step-by-Step Solution

1

Calculate the elevation difference between the given index contours.

Vertical difference =

700\text{ m} - 400\text{ m} = 300\text{ m}

.

Index contours serve as reference lines to determine the vertical rise across intervening contour spaces.

2

Determine the contour interval (the vertical distance between consecutive contour lines).

Contour Interval =

\frac{300\text{ m}}{5} = 60\text{ m}

.

Dividing the total height difference between index lines by the number of spaces gives the value of a single contour interval.

3

Calculate the target elevation at 3 contour intervals above the

700\text{ m}

index line.

Target elevation =

700\text{ m} + (3 \times 60\text{ m}) = 880\text{ m}

.

Adding three vertical intervals (

180\text{ m}

) to the

700\text{ m}

reference line gives the exact height of the target contour line.

Key Concept

Contour Interval Determination and Relief Elevation Calculation

Question 72Question →

Match each set of topographic relief measurements to its corresponding slope gradient ratio ( $1 \text{ in } N$ ).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Vertical Interval

20\text{ m}

/ Distance

200\text{ m}

matches

1 \text{ in } 10

; Vertical Interval

50\text{ m}

/ Distance

1,000\text{ m}

matches

1 \text{ in } 20

; Vertical Interval

100\text{ m}

/ Distance

500\text{ m}

matches

1 \text{ in } 5

; Vertical Interval

10\text{ m}

/ Distance

500\text{ m}

matches

1 \text{ in } 50

.

Each slope measurement set correctly matches its calculated gradient ratio by applying the standard formula

\text{Gradient} = \frac{\text{VI}}{\text{HE}}

and reducing the resulting fraction to a

1 \text{ in } N

form.

Step-by-Step Solution

1

Recall the fundamental gradient formula

Gradient =

\frac{\text{Vertical Interval (VI)}}{\text{Horizontal Equivalent (HE)}}

Gradient measures the ratio of vertical elevation change to horizontal ground distance.

2

Calculate the ratio for the first measurement set (

20\text{ m}

VI and

200\text{ m}

HE)

\frac{20}{200} = \frac{1}{10} = 1 \text{ in } 10

Dividing both numerator and denominator by 20 simplifies the fraction to 1 in 10.

3

Calculate the ratio for the second measurement set (

50\text{ m}

VI and

1,000\text{ m}

HE)

\frac{50}{1000} = \frac{1}{20} = 1 \text{ in } 20

Dividing both numerator and denominator by 50 simplifies the fraction to 1 in 20.

4

Calculate the ratio for the third measurement set (

100\text{ m}

VI and

500\text{ m}

HE)

\frac{100}{500} = \frac{1}{5} = 1 \text{ in } 5

Dividing both numerator and denominator by 100 simplifies the fraction to 1 in 5.

5

Calculate the ratio for the fourth measurement set (

10\text{ m}

VI and

500\text{ m}

HE)

\frac{10}{500} = \frac{1}{50} = 1 \text{ in } 50

Dividing both numerator and denominator by 10 simplifies the fraction to 1 in 50.

Key Concept

Slope and Gradient Calculation

Question 73Question →

A coastal lagoon covers a rectangular section measuring $12\text{ cm}$ by $15\text{ cm}$ on Map A, which is drawn to a scale of $1 : 20,000$ . If Map A is reduced to create Map B with a scale of $1 : 60,000$ , what is the area of the lagoon on Map B in square centimeters?

Show answer & explanation

Answer: 20

Answer

The area of the lagoon on Map B is

20\text{ cm}^2

.

Reducing the scale from

1 : 20,000

to

1 : 60,000

reduces all linear dimensions to

\frac{1}{3}

of their original length. Consequently, the area changes by

\left(\frac{1}{3}\right)^2 = \frac{1}{9}

. Taking the original map area of

180\text{ cm}^2

(

12\text{ cm} \times 15\text{ cm}

) and multiplying by

\frac{1}{9}

gives

20\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the surface area of the lagoon on the original map (Map A)

Area on Map A =

12\text{ cm} \times 15\text{ cm} = 180\text{ cm}^2

Determining the initial area on paper establishes the base value before scale reduction.

2

Determine the linear scale reduction ratio

Linear scale factor =

\frac{\text{Original Scale Denominator}}{\text{New Scale Denominator}} = \frac{20,000}{60,000} = \frac{1}{3}

Increasing the scale denominator from 20,000 to 60,000 means linear distances shrink to one-third of their original length.

3

Compute the area scale conversion factor

Area scale factor =

\left(\frac{1}{3}\right)^2 = \frac{1}{9}

Map area varies as the square of the linear scale ratio.

4

Calculate the final reduced area on Map B

New area on Map B =

180\text{ cm}^2 \times \frac{1}{9} = 20\text{ cm}^2

Multiplying the original map area by the area scale factor gives the resulting map area.

Key Concept

When a map scale is reduced, linear dimensions change by the factor

k = \frac{\text{Old Denominator}}{\text{New Denominator}}

, while the map area changes by the factor

k^2

.

Question 74Question →

Match each settlement pattern observed on a topographical map extract on the left with its primary physical or socio-economic site factor on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Linear settlement along a contour line matches with spring-line site where groundwater emerges at a specific elevation; Compact nucleated settlement at a road junction matches with nodal site advantage fostering commercial activities and transport convergence; Dispersed settlement across a dissected plateau matches with topographic fragmentation forcing farmsteads to locate on scattered arable patches; Radial settlement pattern matches with ribbon development along multiple transport routes radiating from a focal hub.

Each settlement pattern correctly corresponds to its underlying geographic control: contour-aligned linear settlements reflect spring-line water availability; junction-nucleated settlements rely on nodal transport convergence; plateau-dispersed settlements are dictated by fragmented terrain; and radial settlements grow outward along arterial roads radiating from a central hub.

Step-by-Step Solution

1

Examine the linear settlement aligned along a hill slope contour line.

Recognize that settlement alignment following a constant elevation contour indicates reliance on a spring-line water source.

Spring lines emerge along permeable-impermeable rock interfaces, drawing homesteads linearly along that specific contour line.

2

Analyze the compact building cluster at the road junction.

Identify this as a nodal settlement driven by accessibility and commercial activity.

Intersections act as nodes where transportation routes meet, concentrating economic services and housing.

3

Assess the dispersed homestead arrangement on the plateau.

Associate isolated buildings with rugged relief and land fragmentation.

Uneven topography and divided farmlands prevent dense clustering, prompting families to live directly on their scattered plots.

4

Evaluate the star-shaped settlement spreading from a central town.

Link this pattern to radial expansion along main outgoing transit corridors.

Roads radiating from a central commercial hub encourage linear building growth outwards along each route.

Key Concept

Interpretation of Settlement Patterns and Site/Situation Factors on Topographical Maps

Estimated Time:1m 30s

Question 75Question →

A topographical map displays a statement scale of $2\text{ cm}$ to $1\text{ km}$ . Which of the following represents this scale as a Representative Fraction (R.F.)?

50,000 : 1

1 : 50,000

1 : 2,500,000

1 : 500

Show answer & explanation

Answer: $1 : 50,000$

Answer

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

1 : 50,000

.

The correct Representative Fraction is

1 : 50,000

. Converting

1\text{ km}

into centimeters yields

100,000\text{ cm}

. Expressing the scale as a ratio gives

2\text{ cm} : 100,000\text{ cm}

, which simplifies to

1 : 50,000

when both terms are divided by

2

.

Step-by-Step Solution

1

Convert the ground distance from kilometers to centimeters so both sides of the scale ratio share identical units.

1\text{ km} = 100,000\text{ cm}

.

Representative Fraction requires unitless comparison, so both map distance and ground distance must be in the same measurement units.

2

Express the relationship as a ratio of map distance to ground distance.

\text{Scale} = 2\text{ cm} : 100,000\text{ cm}

.

A Representative Fraction is written in the form

1 : n

, where

1

represents map distance and

n

represents ground distance.

3

Divide both terms of the ratio by the map distance value (

2

) to reduce the numerator/antecedent to

1

.

\frac{2}{2} : \frac{100,000}{2} = 1 : 50,000

.

Simplifying the antecedent to unity gives the final standard Representative Fraction format.

Key Concept

Conversion of Statement Scale to Representative Fraction (R.F.)

Estimated Time:45s

Question 76Question →

On a regional map (Map X), a straight stretch of a major highway between two towns measures $15\text{ cm}$ . Map X is drawn to a statement scale of $1\text{ cm to } 2\text{ km}$ . If Map X is reduced to produce Map Y such that Map Y has a Representative Fraction (RF) scale of $1 : 500,000$ , what is the measured length of the same highway on Map Y?

6.0\text{ cm}

37.5\text{ cm}

2.4\text{ cm}

60.0\text{ cm}

Show answer & explanation

Answer: $6.0\text{ cm}$

Answer

The measured length of the highway on Map Y is

6.0\text{ cm}

.

The ground distance between the two towns is

30\text{ km}

(

15\text{ cm} \times 2\text{ km/cm}

). On Map Y, where

1\text{ cm}

represents

500,000\text{ cm}

(

5\text{ km}

), the measured length is

\frac{30\text{ km}}{5\text{ km/cm}} = 6.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the actual ground distance using Map X.

Ground Distance

= 15\text{ cm} \times 2\text{ km/cm} = 30\text{ km}

.

Map X's scale of

1\text{ cm to } 2\text{ km}

means each centimeter on the map represents

2\text{ km}

on the ground.

2

Convert the Representative Fraction (RF) scale of Map Y to a statement scale in kilometers.

Map Y Scale

= 1 : 500,000 \Rightarrow 1\text{ cm to } \frac{500,000}{100,000}\text{ km} = 1\text{ cm to } 5\text{ km}

.

There are

100,000\text{ cm}

in

1\text{ km}

, so dividing

500,000

by

100,000

converts the scale to kilometers.

3

Calculate the distance on Map Y.

Map Y Distance

= \frac{30\text{ km}}{5\text{ km/cm}} = 6.0\text{ cm}

.

Dividing the actual ground distance by the ground equivalent per centimeter of Map Y gives the distance on Map Y.

Key Concept

Map Scale Reduction and Ground Distance Calculation

Estimated Time:1m 30s

Question 77Question →

A straight road connecting a farm settlement to a regional market town measures $8.5\text{ cm}$ on a map with a scale of $1 : 20,000$ . What is the actual ground distance between the two locations in kilometers?

Show answer & explanation

Answer: 1.7

Answer

The actual ground distance between the farm settlement and the market town is

1.7\text{ km}

.

Multiplying

8.5\text{ cm}

by the scale ratio denominator (

20,000

) gives a real-world distance of

170,000\text{ cm}

. Converting to kilometers (

170,000 \div 100,000

) yields

1.7\text{ km}

.

Step-by-Step Solution

1

Calculate the ground distance in centimeters

170,000\text{ cm}

Ground distance equals map distance multiplied by scale denominator (

8.5\text{ cm} \times 20,000

).

2

Convert centimeters into kilometers

1.7\text{ km}

Since

1\text{ km} = 100,000\text{ cm}

, divide

170,000\text{ cm}

by

100,000

.

Key Concept

Map Scale Distance Calculation

Question 78Question →

A rectangular agricultural zone measures $6\text{ cm}$ by $4\text{ cm}$ on a topographical map drawn to a Representative Fraction (R.F.) scale of $1 : 100,000$ . If the map is enlarged so that the area of the agricultural zone on the new map becomes $96\text{ cm}^2$ , what is the Representative Fraction (R.F.) scale of the enlarged map?

1 : 50,000

1 : 25,000

1 : 200,000

1 : 400,000

Show answer & explanation

Answer: $1 : 50,000$

Answer

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

1 : 50,000

.

The correct answer of

1 : 50,000

is derived by recognizing that the original map area of

24\text{ cm}^2

increases to

96\text{ cm}^2

, representing a four-fold areal enlargement. Because linear dimensions change as the square root of the areal change, the linear scale factor is

\sqrt{4} = 2

. Enlarging a map by a factor of 2 doubles linear dimensions on paper, making the scale larger by reducing the scale denominator from 100,000 to 50,000.

Step-by-Step Solution

1

Calculate the area of the agricultural zone on the original map.

Original Area

= 6\text{ cm} \times 4\text{ cm} = 24\text{ cm}^2

.

Knowing the original map area allows us to determine the area enlargement ratio.

2

Determine the area scale change ratio.

Area Ratio

= \frac{\text{New Area}}{\text{Original Area}} = \frac{96\text{ cm}^2}{24\text{ cm}^2} = 4

.

The area of the zone has been enlarged 4 times.

3

Calculate the linear scale enlargement factor.

Linear Factor

= \sqrt{\text{Area Ratio}} = \sqrt{4} = 2

.

Linear scale changes as the square root of areal scale changes.

4

Calculate the scale denominator for the enlarged map.

New Scale Denominator

= \frac{\text{Original Scale Denominator}}{\text{Linear Factor}} = \frac{100,000}{2} = 50,000

.

Enlarging a map increases its detail and linear dimensions, which reduces the scale denominator proportionally.

Key Concept

Map Enlargement and Linear vs Areal Scale Relationship

Question 79Question →

On a topographical map drawn to a Representative Fraction (RF) scale of $1 : 100,000$ , the measured distance between a railway station and a local airstrip is $7.5\text{ cm}$ . What is the actual ground distance between the two locations in kilometers?

Show answer & explanation

Answer: 7.5

Answer

The actual ground distance between the railway station and the airstrip is

7.5\text{ km}

.

Given an RF scale of

1 : 100,000

, every

1\text{ cm}

on the map corresponds to

100,000\text{ cm}

on the ground, which equals

1\text{ km}

. Multiplying the measured map length of

7.5\text{ cm}

by

1\text{ km}

per centimeter gives an actual ground distance of

7.5\text{ km}

.

Step-by-Step Solution

1

Interpret the Representative Fraction (RF) scale.

An RF scale of

1 : 100,000

means

1\text{ cm}

on the map represents

100,000\text{ cm}

on the ground.

RF scale expresses map distance to ground distance in identical units.

2

Convert the ground distance scale unit from centimeters to kilometers.

100,000\text{ cm} = \frac{100,000}{100,000}\text{ km} = 1\text{ km}

.

There are

100,000\text{ cm}

in

1\text{ km}

(

100\text{ cm/m} \times 1,000\text{ m/km}

).

3

Calculate the total actual ground distance.

\text{Ground Distance} = 7.5\text{ cm} \times 1\text{ km/cm} = 7.5\text{ km}$.

Multiplying the map length by the real-world distance represented per unit length determines the true distance.

Key Concept

Map Scale Conversion and Ground Distance Calculation

Question 80Question →

A rectangular agricultural estate measuring $5\text{ cm}$ by $12\text{ cm}$ on a topographical map represents an actual ground area of $60\text{ km}^2$ . If the map is enlarged such that the estate covers an area of $240\text{ cm}^2$ on the new map, what is the denominator of the Representative Fraction (RF) scale of the enlarged map?

Show answer & explanation

Answer: 50000

Answer

The denominator of the Representative Fraction (RF) scale of the enlarged map is 50000.

The original map area of

60\text{ cm}^2

representing

60\text{ km}^2

gives a linear scale of

1\text{ cm}

to

1\text{ km}

, which corresponds to an original RF scale of

1 : 100,000

. Enlarging the map area to

240\text{ cm}^2

increases the area by a factor of

4

. The linear enlargement factor is

\sqrt{4} = 2

. Enlarging a map by a linear factor of

2

halves the scale denominator, yielding a new Representative Fraction scale of

1 : 50,000

, with a denominator of

50,000

.

Step-by-Step Solution

1

Calculate the area of the estate on the original map.

Original Map Area =

5\text{ cm} \times 12\text{ cm} = 60\text{ cm}^2

.

Establishes the initial representation of the estate on paper.

2

Determine the original map scale from the map area and ground area.

Area scale:

60\text{ cm}^2 = 60\text{ km}^2 \implies 1\text{ cm}^2 = 1\text{ km}^2

. Linear scale:

1\text{ cm} = 1\text{ km} = 100,000\text{ cm}

. Original RF =

1 : 100,000

.

Finding the original scale denominator is necessary before applying the enlargement factor.

3

Calculate the area enlargement factor.

Area Enlargement Factor =

\frac{240\text{ cm}^2}{60\text{ cm}^2} = 4

.

Compares the new map area to the original map area.

4

Calculate the linear enlargement factor (

k

).

Linear Enlargement Factor

k = \sqrt{4} = 2

.

Linear scale changes as the square root of the area scale change.

5

Calculate the new RF scale denominator.

New RF denominator =

\frac{100,000}{2} = 50,000

.

Enlarging a map linearly by a factor of 2 makes the scale 2 times larger, dividing the denominator by 2.

Key Concept

Map Enlargement and Linear vs Area Scale Conversion

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