On a topographical map drawn to a scale of 1:50,000, a road ascends continuously from a co… — JAMB UTME

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?

$1\text{ in }10$ Answer
B
$1\text{ in }50$
C
$1\text{ in }100$
D
$1\text{ in }5.5$

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

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.

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.

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

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