Practical Geography Practice Questions — JAMB UTME — Page 5

Question 81Question →

A proposed high-voltage transmission line connecting a hydro-electric station to an urban sub-station measures $14.5\text{ cm}$ on a topographical map drawn to a scale of $1 : 50,000$ . If the map undergoes a photographic reduction such that its total surface area becomes $\frac{1}{4}$ of its original size, what is the actual ground distance, in kilometers, represented by a length of $5.8\text{ cm}$ measured on the newly reduced map?

Show answer & explanation

Answer: 5.8

Answer

The actual ground distance represented by 5.8 cm on the reduced map is 5.8 kilometers.

Reducing the surface area of a map to one-fourth reduces its linear dimensions to one-half, because linear scale factor equals the square root of the area scale factor. With linear dimensions halved, each centimeter on the new map represents twice as much ground distance as before, changing the scale from 1 : 50,000 to 1 : 100,000. At a scale of 1 : 100,000, 1 cm represents 1 km. Therefore, 5.8 cm on the reduced map corresponds to an actual ground distance of 5.8 km.

Step-by-Step Solution

1

Determine the linear scale factor from the area reduction ratio

Linear factor = \sqrt{\frac{1}{4}} = \frac{1}{2}

Map area varies as the square of the linear scale factor. Reducing area to one-fourth reduces linear dimensions by half.

2

Calculate the Representative Fraction (R.F.) of the reduced map

New R.F. = 1 : (50,000 \times 2) = 1 : 100,000

Reducing the physical dimensions of a map by half doubles the denominator of its scale ratio, making it a smaller scale map.

3

Convert the new R.F. scale to kilometers per centimeter

1\text{ cm} = 100,000\text{ cm} = 1.0\text{ km}

Dividing the scale denominator (100,000) by 100,000 converts centimeters to kilometers.

4

Compute the ground distance for a map length of 5.8 cm

5.8\text{ cm} \times 1.0\text{ km/cm} = 5.8\text{ km}

Multiplying map distance by the ground equivalence per centimeter yields actual distance.

Key Concept

Linear vs Area Scale Conversion and Ground Distance Calculation

Question 82Question →

On a topographical map, a surveyor assesses two observation posts: Post X at an elevation of $750\text{ m}$ and Post Y at an elevation of $900\text{ m}$ , separated by a straight-line distance of $4.0\text{ km}$ . An intervening ridge crest with a summit elevation of $820\text{ m}$ is located exactly $1.5\text{ km}$ along the line of sight from Post X. Assuming a direct line of sight between the two posts, which of the following statements correctly evaluates their intervisibility?

The posts are not intervisible because the elevation of the line of sight at the ridge location is

806.25\text{ m}

, which is below the ridge summit elevation of

820\text{ m}

.

The posts are intervisible because the higher station at Post Y (

900\text{ m}

) completely clears the intervening ridge elevation (

820\text{ m}

).

The posts are intervisible because the mean elevation between the two posts is

825\text{ m}

, which is greater than the intervening ridge elevation of

820\text{ m}

.

The posts are not intervisible because the gradient from Post X to the ridge (

1:21.4

) is steeper than the gradient from the ridge to Post Y (

1:31.25

), causing a convex ground depression.

Show answer & explanation

Answer: The posts are not intervisible because the elevation of the line of sight at the ridge location is $806.25\text{ m}$ , which is below the ridge summit elevation of $820\text{ m}$ .

Answer

The posts are not intervisible because the line of sight elevation at the position of the intervening ridge is

806.25\text{ m}

, which is lower than the ridge crest of

820\text{ m}

.

Intervisibility between two points on a contour map requires that no intervening landform rises above the imaginary straight ray connecting them. Calculating the line of sight height at

1.5\text{ km}

gives

750\text{ m} + \left(\frac{1.5}{4.0} \times 150\text{ m}\right) = 806.25\text{ m}

. Because the intervening ridge reaches

820\text{ m}

, it obstructs the view, making the posts non-intervisible.

Step-by-Step Solution

1

Calculate the total vertical height difference between Post X and Post Y.

Vertical Difference

= 900\text{ m} - 750\text{ m} = 150\text{ m}

.

Establishes the total elevation gain along the

4.0\text{ km}

line of sight.

2

Determine the proportional vertical rise along the line of sight at a distance of

1.5\text{ km}

from Post X.

Proportional Rise

= \frac{1.5\text{ km}}{4.0\text{ km}} \times 150\text{ m} = 0.375 \times 150\text{ m} = 56.25\text{ m}

.

Determines how much elevation the line of sight gains over the distance to the intervening ridge.

3

Calculate the absolute elevation of the line of sight above sea level at the ridge position.

Line of Sight Elevation

= 750\text{ m} + 56.25\text{ m} = 806.25\text{ m}

.

Provides the baseline height of the visual ray path at the obstacle's location.

4

Compare the line of sight elevation with the actual ground elevation of the ridge summit.

Ground Elevation (

820\text{ m}

)

>

Line of Sight Elevation (

806.25\text{ m}

). Intervisibility is obstructed.

If the terrain height is greater than the line of sight height, the view between the two points is blocked.

Key Concept

Intervisibility and Profile Line of Sight Analysis

Question 83Question →

On Map A, a straight section of a railway track connecting two transport terminals measures $18.0\text{ cm}$ . Map A is drawn to a statement scale of $2\text{ cm}$ to $1\text{ km}$ . If this identical section of railway is represented on Map B, which is drawn to a Representative Fraction (R.F.) scale of $1 : 150,000$ , what will be the length of the railway track on Map B?

24.0\text{ cm}

6.0\text{ cm}

12.0\text{ cm}

54.0\text{ cm}

Show answer & explanation

Answer: $6.0\text{ cm}$

Answer

The length of the railway track on Map B is

6.0\text{ cm}

.

The correct answer is

6.0\text{ cm}

. The statement scale of

2\text{ cm}

to

1\text{ km}

means that

1\text{ cm}

on Map A represents

0.5\text{ km}

on the ground. A measured length of

18.0\text{ cm}

therefore corresponds to an actual ground distance of

9.0\text{ km}

(

900,000\text{ cm}

). When represented on Map B with an R.F. scale of

1 : 150,000

, the new map distance is calculated by dividing

900,000\text{ cm}

by

150,000

, giving exactly

6.0\text{ cm}

.

Step-by-Step Solution

1

Determine the actual ground distance using the statement scale of Map A.

Map A scale is

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

, so

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

. Ground distance

= 18.0\text{ cm} \times 0.5\text{ km/cm} = 9.0\text{ km}

.

Before calculating distance on a second map, the actual real-world ground distance must be determined.

2

Convert the ground distance from kilometers to centimeters.

9.0\text{ km} = 9.0 \times 100,000\text{ cm} = 900,000\text{ cm}

.

Map measurements in Representative Fraction (R.F.) calculations require consistent units of measurement.

3

Calculate the map length on Map B using its R.F. scale (

1 : 150,000

).

Map length on Map B

= \frac{\text{Ground Distance}}{\text{Scale Denominator}} = \frac{900,000\text{ cm}}{150,000} = 6.0\text{ cm}

.

Dividing the ground distance in centimeters by the R.F. denominator gives the corresponding distance on the new map.

Key Concept

Multi-step conversion between statement scales, ground distances, and representative fraction (R.F.) map scales

Estimated Time:2m 0s

Question 84Question →

In topographical map interpretation, distinct contour patterns uniquely represent specific physical landforms and slope profiles. Match each contour line configuration on the left with the corresponding relief feature it describes on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Concentric closed contours decreasing inward match Crater or Depression; Widely spaced top contours becoming densely packed bottom contours match Convex Slope; Twin elevated contour loops separated by a pass match Saddle or Col; Closely packed contours merging into a single line match Vertical Cliff.

Each contour pattern corresponds directly to a unique 3D landform: decreasing inner values represent a crater/depression; wide spacing at upper elevations becoming dense at lower elevations represents a convex slope; twin enclosed summits represent a saddle/col; and coinciding merged lines represent a vertical cliff face.

Step-by-Step Solution

1

Analyze contour value progression for closed loops.

Decreasing values toward the center signify a depression or crater.

Standard hills show increasing values inward; decreasing values indicate a descent into a enclosed hollow.

2

Examine contour spacing density across elevation levels.

Wide spacing at higher elevation transitioning to dense spacing at lower elevation indicates a convex slope.

Contour spacing is inversely proportional to gradient steepness; gentle upper slopes are widely spaced while steep lower slopes are closely spaced.

3

Identify multi-peak structural contour patterns.

Twin contour summits surrounded by an outer ring contour represent a saddle or col.

A saddle is the low point between two higher relief features.

4

Interpret overlapping or coalescing contour lines.

Merged contour lines represent a vertical cliff.

A vertical surface has zero horizontal equivalent distance between different elevation contours.

Key Concept

Relief Representation and Contour Line Pattern Interpretation

Estimated Time:2m 0s

Question 85Question →

On a topographical map with a constant contour interval of $20\text{ m}$ , Point X is located on the contour line labeled $240\text{ m}$ . Point Y is located higher up the same hill, exactly three contour intervals above Point X. What is the elevation of Point Y in meters?

Show answer & explanation

Answer: 300

Answer

The elevation of Point Y is

300\text{ m}

.

The contour interval indicates that consecutive contour lines differ by

20\text{ m}

in elevation. Moving up three contour line intervals from Point X (

240\text{ m}

) increases elevation by

3 \times 20\text{ m} = 60\text{ m}

. Thus, the elevation at Point Y is

240\text{ m} + 60\text{ m} = 300\text{ m}

.

Step-by-Step Solution

1

Calculate total vertical rise between the two points

Vertical rise =

3 \times 20\text{ m} = 60\text{ m}

Point Y is three contour intervals higher than Point X, and each interval represents

20\text{ m}

of vertical distance.

2

Calculate the elevation at Point Y

Elevation at Point Y =

240\text{ m} + 60\text{ m} = 300\text{ m}

Adding the total vertical rise to the baseline elevation at Point X yields the final elevation at Point Y.

Key Concept

Calculating elevation on topographical maps using contour intervals

Estimated Time:45s

Question 86Question →

A river basin covers a rectangular section measuring $4.5\text{ cm}$ by $8\text{ cm}$ on Map X, which has a representative fraction of $1 : 25,000$ . Map X is enlarged to create Map Y, where the same river basin occupies an area of $144\text{ cm}^2$ . Map Y is then reduced to construct Map Z, such that the denominator of Map Z's representative fraction is $4$ times that of Map Y. What is the area of the river basin on Map Z in $\text{cm}^2$ ?

Show answer & explanation

Answer: 9

Answer

The area of the river basin on Map Z is

9\text{ cm}^2

.

To find the area on Map Z, first calculate the basin's area on Map X:

4.5\text{ cm} \times 8\text{ cm} = 36\text{ cm}^2

. The area ratio from Map X to Map Y is

144 / 36 = 4

, which corresponds to a linear enlargement factor of

\sqrt{4} = 2

. Map Y therefore has a scale of

1 : 12,500

. Multiplying Map Y's RF denominator by

4

gives Map Z a scale of

1 : 50,000

. Comparing Map Y (

1 : 12,500

) to Map Z (

1 : 50,000

) reveals a linear reduction factor of

1/4

. The area reduction factor is

(1/4)^2 = 1/16

. Multiplying Map Y's area by

1/16

yields

144\text{ cm}^2 \times (1/16) = 9\text{ cm}^2

. Alternatively, comparing Map X (

1 : 25,000

) directly to Map Z (

1 : 50,000

) shows a linear factor of

1/2

, giving an area factor of

(1/2)^2 = 1/4

, and

36\text{ cm}^2 \times (1/4) = 9\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the initial map area on Map X.

A_X = 4.5\text{ cm} \times 8\text{ cm} = 36\text{ cm}^2

The product of the rectangular map dimensions gives the original map area.

2

Determine the linear scale factor and representative fraction of Map Y.

Area enlargement factor =

144 / 36 = 4

; Linear enlargement factor =

\sqrt{4} = 2

. Map Y scale =

1 : (25,000 / 2) = 1 : 12,500

.

Area change ratio is the square of the linear scale factor (

n^2

). Enlarging a map decreases its RF denominator by the linear factor.

3

Calculate the representative fraction of Map Z.

Map Z RF denominator =

12,500 \times 4 = 50,000

, so scale of Map Z =

1 : 50,000

.

Reducing a map increases the RF denominator proportionately.

4

Determine the final map area on Map Z.

Linear reduction factor from Y to Z =

12,500 / 50,000 = 1/4

. Area reduction factor =

(1/4)^2 = 1/16

. Final area

A_Z = 144\text{ cm}^2 \times (1/16) = 9\text{ cm}^2

.

Applying the squared linear scale reduction factor to Map Y's area yields Map Z's area.

Key Concept

Map Enlargement, Reduction, and Area-to-Linear Scale Ratio Relationships

Estimated Time:3m 0s

Question 87Question →

On a regional conservation map drawn to a Representative Fraction (R.F.) scale of $1 : 25,000$ , the distance between two forest ranger posts along a straight trail measures $16.4\text{ cm}$ . What is the actual ground distance between the two ranger posts in kilometers?

Show answer & explanation

Answer: 4.1

Answer

The actual ground distance between the two ranger posts is 4.1 kilometers.

The scale ratio 1 : 25,000 indicates that 1 centimeter on the map corresponds to 25,000 centimeters (or 0.25 kilometers) on the ground. Multiplying the measured map length of 16.4 centimeters by 0.25 kilometers per centimeter gives an actual ground distance of 4.1 kilometers.

Step-by-Step Solution

1

Determine ground distance in centimeters

16.4\text{ cm} \times 25,000 = 410,000\text{ cm}

Each unit on the map represents 25,000 identical units on the ground.

2

Convert centimeters to meters

410,000\text{ cm} \div 100 = 4,100\text{ m}

There are 100 centimeters in 1 meter.

3

Convert meters to kilometers

4,100\text{ m} \div 1,000 = 4.1\text{ km}

There are 1,000 meters in 1 kilometer.

Key Concept

Ground distance calculation using Representative Fraction (R.F.) map scale

Question 88Question →

On a topographical map drawn to a scale of $1 : 25,000$ , a road ascends continuously from Point X at an elevation of $120\text{ m}$ to Point Y at an elevation of $370\text{ m}$ . If the average gradient of the slope along this road is $1 \text{ in } 20$ , what is the distance between Point X and Point Y on the map?

20\text{ cm}

0.2\text{ cm}

5\text{ cm}

29.6\text{ cm}

Show answer & explanation

Answer: $20\text{ cm}$

Answer

The distance between Point X and Point Y on the map is

20\text{ cm}

.

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

370\text{ m} - 120\text{ m} = 250\text{ m}

. Given a gradient of

1 \text{ in } 20

, the horizontal distance on the ground is

250\text{ m} \times 20 = 5,000\text{ m}

(

500,000\text{ cm}

). Dividing this by the scale factor of

25,000

yields a map distance of

20\text{ cm}

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

\text{VI} = 370\text{ m} - 120\text{ m} = 250\text{ m}

Vertical interval is the difference in height between the two contour points.

2

Calculate the Horizontal Equivalent (HE) on the ground using the gradient formula

\text{HE} = \text{VI} \times 20 = 250\text{ m} \times 20 = 5,000\text{ m}

Gradient is given as

\text{VI} / \text{HE} = 1 / 20

, so

\text{HE} = 20 \times \text{VI}

.

3

Convert the ground distance to centimeters and calculate the map distance using the scale

1 : 25,000

\text{Ground distance} = 5,000\text{ m} = 500,000\text{ cm}

;

\text{Map distance} = \frac{500,000\text{ cm}}{25,000} = 20\text{ cm}

To find distance on the map, convert the ground measurement to centimeters and divide by the scale denominator.

Key Concept

Topographical Gradient and Scale Conversion

Estimated Time:1m 30s

Question 89Question →

A topographical map drawn to a Representative Fraction (R.F.) scale of $1 : 50,000$ is enlarged to twice its original linear dimensions. What is the scale of the new enlarged map?

1 : 100,000

1 : 25,000

1 : 12,500

1 : 200,000

Show answer & explanation

Answer: $1 : 25,000$

Answer

The scale of the new enlarged map is

1 : 25,000

.

Enlarging a map increases its scale, making the scale denominator smaller. When a map is enlarged to twice its linear size (

2\times

), the denominator of the original Representative Fraction (

50,000

) is divided by

2

, yielding a new scale of

1 : 25,000

.

Step-by-Step Solution

1

Identify the original scale denominator and the linear enlargement factor.

Original scale denominator =

50,000

; Linear factor (

k

) =

2

.

Map enlargement alters the scale denominator inversely proportional to the linear change.

2

Calculate the new scale denominator by dividing the original denominator by the linear enlargement factor.

New denominator =

\frac{50,000}{2} = 25,000

.

Enlarging a map makes features larger on paper, which corresponds to a larger scale with a smaller denominator.

3

State the new Representative Fraction (R.F.) scale.

New scale =

1 : 25,000

.

The new R.F. expresses the enlarged relationship between map distance and ground distance.

Key Concept

Linear Map Enlargement Scale Conversion

Estimated Time:45s

Question 90Question →

A proposed rural electrification line connecting a community to a regional substation measures $15.6\text{ cm}$ on a topographical map drawn to a Representative Fraction (R.F.) scale of $1 : 50,000$ . What is the actual length of the power line on the ground in kilometers?

3.12\text{ km}

7.8\text{ km}

15.6\text{ km}

78.0\text{ km}

Show answer & explanation

Answer: $7.8\text{ km}$

Answer

7.8\text{ km}

The Representative Fraction scale of

1 : 50,000

indicates that

1\text{ cm}

on the map represents

50,000\text{ cm}

(

0.5\text{ km}

) on the ground. Multiplying the map distance of

15.6\text{ cm}

by

0.5\text{ km/cm}

gives the actual ground distance of

7.8\text{ km}

.

Step-by-Step Solution

1

Calculate total ground distance in centimeters using the R.F. scale

15.6\text{ cm} \times 50,000 = 780,000\text{ cm}

An R.F. scale of

1 : 50,000

means

1\text{ unit}

on the map equals

50,000\text{ units}

on the ground.

2

Convert centimeters to kilometers

\frac{780,000\text{ cm}}{100,000\text{ cm/km}} = 7.8\text{ km}

There are

100,000\text{ centimeters}

in

1\text{ kilometer}

.

Key Concept

Map Scale Distance Calculation

Estimated Time:1m 30s

Question 91Question →

A survey map of a proposed cocoa plantation in southwestern Nigeria is drawn to a scale of $1 : 50,000$ . If the main access road across the plantation measures $12\text{ cm}$ on the map, what is the actual ground distance of the road in kilometers?

Show answer & explanation

Answer: 6

Answer

The actual ground distance of the road is

6\text{ km}

.

To determine the actual ground distance, multiply the measured map distance (

12\text{ cm}

) by the scale denominator (

50,000

) to get

600,000\text{ cm}

. Converting this value into kilometers by dividing by

100,000

gives an actual ground distance of

6\text{ km}

.

Step-by-Step Solution

1

Calculate the ground distance in centimeters

12\text{ cm} \times 50,000 = 600,000\text{ cm}

A Representative Fraction of

1 : 50,000

means

1\text{ cm}

on the map corresponds to

50,000\text{ cm}

on the ground.

2

Convert centimeters to kilometers

600,000\text{ cm} \div 100,000 = 6\text{ km}

Since

1\text{ m} = 100\text{ cm}

and

1\text{ km} = 1,000\text{ m}

,

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

.

Key Concept

Distance calculation using Representative Fraction (R.F.) map scale

Question 92Question →

Match each topographic relief profile scenario to its corresponding calculated slope gradient expressed both as a ratio ( $1 \text{ in } N$ ) and as a percentage gradient.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Trigonometric beacon to river confluence matches

1 \text{ in } 10

(

10\%

); Valley floor to plateau rim matches

1 \text{ in } 2.5

(

40\%

); Proposed railway track matches

1 \text{ in } 50

(

2\%

); Scarp face across 5 contour lines matches

1 \text{ in } 4

(

25\%

).

Each topographic scenario correctly pairs with its calculated gradient ratio and percentage by computing the exact Vertical Interval (difference in height) and Horizontal Equivalent (ground distance derived from map scale).

Step-by-Step Solution

1

Calculate the Vertical Interval (VI) for each scenario

Scenario 1:

450 - 150 = 300\text{ m}

. Scenario 2:

420 - 180 = 240\text{ m}

. Scenario 3:

220 - 100 = 120\text{ m}

. Scenario 4:

(5-1) \times 25 = 100\text{ m}

.

VI represents the difference in height between two points.

2

Calculate the Horizontal Equivalent (HE) in meters for each scenario

Scenario 1:

6\text{ cm} \times 500\text{ m/cm} = 3,000\text{ m}

. Scenario 2:

2.4\text{ cm} \times 250\text{ m/cm} = 600\text{ m}

. Scenario 3:

12\text{ cm} \times 500\text{ m/cm} = 6,000\text{ m}

. Scenario 4:

0.8\text{ cm} \times 500\text{ m/cm} = 400\text{ m}

.

Convert map distance to ground distance using the representative fraction scale.

3

Compute the slope gradient ratio (VI / HE) and express as 1 in N and percentage

Scenario 1:

300 / 3000 = 1/10 = 1 \text{ in } 10

(

10\%

). Scenario 2:

240 / 600 = 1/2.5 = 1 \text{ in } 2.5

(

40\%

). Scenario 3:

120 / 6000 = 1/50 = 1 \text{ in } 50

(

2\%

). Scenario 4:

100 / 400 = 1/4 = 1 \text{ in } 4

(

25\%

).

Gradient is calculated as VI / HE, yielding both ratio 1 in N and percentage slope (VI/HE * 100%).

Key Concept

Slope and Gradient Calculation

Question 93Question →

A topographical survey map was originally drawn to a Representative Fraction (R.F.) scale of $1 : 50,000$ . During a cartographic revision, the map was enlarged such that a forest reserve covering an area of $4.0\text{ cm}^2$ on the original map occupies $16.0\text{ cm}^2$ on the revised map. If the distance between two agricultural settlement centers measures $15.0\text{ cm}$ on the revised map, what is the actual ground distance between them in kilometers?

1.875\text{ km}

3.75\text{ km}

7.50\text{ km}

15.00\text{ km}

Show answer & explanation

Answer: $3.75\text{ km}$

Answer

The actual ground distance between the two agricultural settlement centers is

3.75\text{ km}

.

Enlarging a map's area by a factor of

4

(

16.0\text{ cm}^2 / 4.0\text{ cm}^2

) increases its linear scale by a factor of

\sqrt{4} = 2

. Consequently, the original scale denominator of

50,000

is divided by

2

, yielding a revised map scale of

1 : 25,000

(

1\text{ cm} = 0.25\text{ km}

). A measured distance of

15.0\text{ cm}

on the revised map represents

15.0 \times 0.25\text{ km} = 3.75\text{ km}

on the ground.

Step-by-Step Solution

1

Calculate the area enlargement ratio

\text{Area Ratio} = \frac{16.0\text{ cm}^2}{4.0\text{ cm}^2} = 4

Determines how many times larger the surface area has become on the revised map.

2

Determine the linear scale enlargement factor

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

Linear scale factor is the square root of the area scale factor.

3

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

\text{New R.F. Denominator} = \frac{50,000}{2} = 25,000 \implies 1 : 25,000

Enlarging a map increases detail and decreases the scale denominator proportionally by the linear factor.

4

Compute ground distance from map distance

\text{Ground Distance} = 15.0\text{ cm} \times 25,000 = 375,000\text{ cm} = 3.75\text{ km}

Multiply map distance by the scale denominator and convert centimeters to kilometers (

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

).

Key Concept

Relationship between map area enlargement ratio and linear scale conversion for ground distance calculation.

Question 94Question →

On a topographical map with a scale of $1 : 50,000$ , Point X lies on a hilltop at an elevation of $420\text{ m}$ , while Point Y sits near a stream bed at an elevation of $170\text{ m}$ . If the measured distance between Point X and Point Y on the map is $12.5\text{ cm}$ , calculate the slope gradient between the two points. Express your answer as the value of $N$ in the standard gradient ratio $1 : N$ (or $1 \text{ in } N$ ). What is the value of $N$ ?

Show answer & explanation

Answer: 25

Answer

The denominator N in the gradient ratio 1 : N is 25.

The vertical interval (V.I.) between the hilltop at

420\text{ m}

and the stream at

170\text{ m}

is

250\text{ m}

. The horizontal equivalent (H.E.) on the ground is calculated by multiplying the map distance of

12.5\text{ cm}

by the map scale factor of

50,000

, which gives

625,000\text{ cm}

(

6,250\text{ m}

). Dividing V.I. by H.E. yields

250 / 6,250 = 1 / 25

. Thus, the gradient ratio is

1 : 25

, making

N = 25

.

Step-by-Step Solution

1

Calculate Vertical Interval (V.I.)

V.I. = 420 m - 170 m = 250 m

Vertical interval is the difference in height between the highest and lowest points.

2

Convert map distance to ground distance to determine Horizontal Equivalent (H.E.) in meters

H.E. = 12.5 cm × 50,000 = 625,000 cm = 6,250 m

Horizontal equivalent must be in the same units as the vertical interval before computing ratio.

3

Divide V.I. by H.E. to express the gradient as a ratio 1 : N

Gradient = 250 / 6,250 = 1 / 25

Simplifying the fraction V.I. / H.E. yields 1 / 25, giving N = 25.

Key Concept

Slope and Gradient Calculation

Question 95Question →

A county boundary encloses an area of $72\text{ cm}^2$ on Map A, which is drawn at a scale of $1 : 20,000$ . Map A is reduced to produce Map B, which has a Representative Fraction of $1 : 60,000$ . What is the area of the county boundary on Map B?

8\text{ cm}^2

24\text{ cm}^2

216\text{ cm}^2

648\text{ cm}^2

Show answer & explanation

Answer: $8\text{ cm}^2$

Answer

The area of the county boundary on Map B is

8\text{ cm}^2

.

The linear reduction ratio from a scale of

1 : 20,000

to

1 : 60,000

is

\frac{20,000}{60,000} = \frac{1}{3}

. Because area is a two-dimensional quantity, the area scale factor is the square of the linear factor, which is

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

. Applying this area scale factor to the original map area of

72\text{ cm}^2

gives

72\text{ cm}^2 \times \frac{1}{9} = 8\text{ cm}^2

.

Step-by-Step Solution

1

Determine the linear scale reduction factor (

k

) between Map A and Map B.

k = \frac{\text{Scale Denominator of Map A}}{\text{Scale Denominator of Map B}} = \frac{20,000}{60,000} = \frac{1}{3}

Map scale reduction decreases linear dimensions proportionally to the ratio of the original scale denominator to the new scale denominator.

2

Calculate the area scale change factor (

k^2

).

\text{Area Scale Factor} = k^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}

Surface area changes according to the square of the linear scale factor.

3

Calculate the new area on Map B using the original map area.

\text{New Area} = 72\text{ cm}^2 \times \frac{1}{9} = 8\text{ cm}^2

Multiplying the initial map area by the area scale change factor yields the reduced map area.

Key Concept

Relationship between linear scale change and area scale change in map reduction

Estimated Time:1m 30s

Question 96Question →

A wildlife sanctuary covers an area of $36\text{ cm}^2$ on Map X, which is drawn to a scale of $1 : 40,000$ . If the map is reduced to a scale of $1 : 120,000$ , what is the area of the sanctuary on the new map?

12\text{ cm}^2

4\text{ cm}^2

108\text{ cm}^2

324\text{ cm}^2

Show answer & explanation

Answer: $4\text{ cm}^2$

Answer

The area of the sanctuary on the new map is

4\text{ cm}^2

.

When a map is reduced from a scale of

1 : 40,000

to

1 : 120,000

, the linear dimensions become

\frac{40,000}{120,000} = \frac{1}{3}

of the original size. Because area is proportional to the square of linear dimensions, the area factor is

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

. Reducing

36\text{ cm}^2

by a factor of 9 gives

4\text{ cm}^2

.

Step-by-Step Solution

1

Determine the linear scale factor of reduction

Linear scale factor

k = \frac{\text{Old Scale Denominator}}{\text{New Scale Denominator}} = \frac{40,000}{120,000} = \frac{1}{3}

Going from

1 : 40,000

to

1 : 120,000

reduces all linear map dimensions to one-third of their original length.

2

Calculate the area scale factor

Area scale factor

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

Surface area changes proportionally to the square of the linear scale factor.

3

Calculate the new map area

New Area = Original Area

\times k^2 = 36\text{ cm}^2 \times \frac{1}{9} = 4\text{ cm}^2

Multiplying the original map area by the area scale factor gives the area on the reduced map.

Key Concept

Map Reduction and Area Scale Relationship

Estimated Time:1m 30s

Question 97Question →

On a topographical map with a scale of $1 : 25,000$ and a contour interval of $20\text{ m}$ , a river flows down a valley. Contour line $V_1$ , whose V-shape apex points upstream toward higher elevation, crosses the river channel at an elevation of $360\text{ m}$ . Further downstream, the river crosses contour line $V_2$ , which is separated from $V_1$ by $4$ contour intervals. If the measured map distance along the stream between these two crossing points is $8\text{ cm}$ , what is the average gradient of the river channel between $V_1$ and $V_2$ ?

1 in 25

1 in 2,500

1 in 20

1 in 2.5

Show answer & explanation

Answer: 1 in 25

Answer

The average gradient of the river channel between points V1 and V2 is 1 in 25.

The vertical interval (VI) represents the height difference between the two points along the stream, which is equal to

4 \text{ intervals} \times 20\text{ m} = 80\text{ m}

. The horizontal equivalent (HE) is calculated by multiplying the map distance (

8\text{ cm}

) by the scale factor (

25,000

), yielding

200,000\text{ cm}

or

2,000\text{ m}

. Dividing VI by HE yields

\frac{80}{2000} = \frac{1}{25}

, expressed as a ratio of 1 in 25.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI) between contour crossings

Vertical Interval (VI) =

4 \text{ intervals} \times 20\text{ m} = 80\text{ m}

Since the stream flows downstream, elevation drops by 4 contour intervals from the initial elevation of 360m.

2

Convert the map distance to actual ground distance (Horizontal Equivalent, HE)

Ground distance (HE) =

8\text{ cm} \times 25,000 = 200,000\text{ cm} = 2,000\text{ m}

Map scale 1 : 25,000 means 1 cm on the map equals 25,000 cm (or 250 m) on the ground.

3

Calculate the gradient using the ratio formula VI / HE

Gradient =

\frac{80\text{ m}}{2,000\text{ m}} = \frac{1}{25}

or

1 \text{ in } 25

Both Vertical Interval and Horizontal Equivalent must be expressed in the same unit (meters) to compute the gradient ratio.

Key Concept

Topographic Gradient and Relief Interpretation

Question 98Question →

Match each contour line pattern description on the left with the correct relief landform or slope type it represents on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

V-shaped contours pointing uphill match with River valley; V-shaped contours pointing downhill match with Spur; contours closely spaced near the top match with Concave slope; contours widely spaced near the top match with Convex slope.

V-shaped contours with their apex pointing uphill reflect a river valley, while those pointing downhill depict a spur. Slope profiles are determined by contour density: closely spaced lines near the summit combined with widely spaced lines at the base form a concave slope, whereas widely spaced lines near the summit combined with closely spaced lines at the base form a convex slope.

Step-by-Step Solution

1

Determine the landform associated with V-shaped contour orientation

A 'V' shape pointing toward higher ground indicates a valley cut into the terrain by water flow, whereas a 'V' shape pointing toward lower ground represents a protruding ridge section (spur).

Water flows downhill, creating re-entrants that cut into higher land, while spurs project outward into lower land.

2

Analyze contour spacing relative to slope profile steepness

Tight spacing indicates a steep slope, while wide spacing indicates a gentle slope. Steep top with gentle base yields a concave slope; gentle top with steep base yields a convex slope.

Contour line density directly corresponds to gradient changes across relief features.

Key Concept

Relief Representation using Contour Line Patterns

Question 99Question →

Four survey transects were evaluated on a topographical map drawn to a scale of $1 : 50,000$ . Match each transect scenario on the left with its corresponding calculated slope gradient ratio on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Transect P matches

1 \text{ in } 25

; Transect Q matches

1 \text{ in } 4

; Transect R matches

1 \text{ in } 50

; Transect S matches

1 \text{ in } 10

.

Each scenario is correctly matched by converting map distance to real-world ground distance (HE) using the

1 : 50,000

scale multiplier (

1\text{ cm} = 500\text{ m}

) and then computing rise over run (VI / HE).

Step-by-Step Solution

1

Convert map distances to Horizontal Equivalent (HE) in meters for each transect using the map scale ratio (

1 : 50,000

, where

1\text{ cm} = 500\text{ m}

).

Transect P HE:

5\text{ cm} \times 500\text{ m/cm} = 2,500\text{ m}

; Transect Q HE:

2\text{ cm} \times 500\text{ m/cm} = 1,000\text{ m}

; Transect R HE:

6\text{ cm} \times 500\text{ m/cm} = 3,000\text{ m}

; Transect S HE:

3\text{ cm} \times 500\text{ m/cm} = 1,500\text{ m}

.

Gradient calculation requires both Vertical Interval (VI) and Horizontal Equivalent (HE) to be in identical linear measurement units.

2

Calculate the slope gradient for each transect using the formula

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

and simplify to ratio form (

1 \text{ in } N

).

Transect P:

\frac{100\text{ m}}{2,500\text{ m}} = \frac{1}{25} = 1 \text{ in } 25

; Transect Q:

\frac{250\text{ m}}{1,000\text{ m}} = \frac{1}{4} = 1 \text{ in } 4

; Transect R:

\frac{60\text{ m}}{3,000\text{ m}} = \frac{1}{50} = 1 \text{ in } 50

; Transect S:

\frac{150\text{ m}}{1,500\text{ m}} = \frac{1}{10} = 1 \text{ in } 10

.

Expressing

\frac{\text{VI}}{\text{HE}}

as a unit fraction yields the standard ratio representation used in topographic map reading.

Key Concept

Slope gradient is the ratio of vertical elevation change (Vertical Interval) to ground horizontal distance (Horizontal Equivalent), expressed as a fraction or ratio

1 \text{ in } N

.

Question 100Question →

A forest reserve covers an area of $18\text{ cm}^2$ on a topographical map drawn to a scale of $1 : 150,000$ . If the map is enlarged so that a statement scale of $1\text{ cm}$ to $0.5\text{ km}$ is used for the new map, what is the area of the forest reserve on the enlarged map in $\text{cm}^2$ ?

Show answer & explanation

Answer: 162

Answer

The area of the forest reserve on the enlarged map is

162\text{ cm}^2

.

The original scale

1 : 150,000

is enlarged to

1 : 50,000

(

1\text{ cm}

to

0.5\text{ km}

), giving a linear enlargement factor of

n = 150,000 / 50,000 = 3

. Since area changes by the square of the linear scale multiplier (

n^2 = 3^2 = 9

), the new area on the map is

18\text{ cm}^2 \times 9 = 162\text{ cm}^2

.

Step-by-Step Solution

1

Convert the new statement scale to a Representative Fraction (R.F.)

New scale R.F. is

1 : 50,000

Both scales must be in the same format to compare denominators directly (

0.5\text{ km} = 50,000\text{ cm}

).

2

Calculate the linear enlargement factor (

n

)

n = \frac{150,000}{50,000} = 3

The linear enlargement factor is found by dividing the original scale denominator by the new scale denominator.

3

Calculate the area enlargement factor (

n^2

)

Area enlargement factor

= 3^2 = 9

Area changes by the square of the linear scale factor.

4

Calculate the area on the enlarged map

18\text{ cm}^2 \times 9 = 162\text{ cm}^2

Multiplying the original area on the map by the area enlargement factor yields the new map area.

Key Concept

Relationship between linear scale factor and area scale factor in map enlargement

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