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

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 $1 : N$ .

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

Calculate Vertical Interval (VI)

200 m

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

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).

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

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