A topographical map with a Representative Fraction (RF) of 1 : 100,000 undergoes map reduc… — JAMB UTME

A topographical map with a Representative Fraction (RF) of $1 : 100,000$ undergoes map reduction such that its new surface area is $\frac{1}{4}$ of the original map area. If the straight-line distance between two towns measured on the reduced map is $15\text{ cm}$ , what is the actual ground distance between the two towns in kilometers?

A
$7.5\text{ km}$
B
$15.0\text{ km}$
$30.0\text{ km}$ Answer
D
$60.0\text{ km}$

Answer

The actual ground distance between the two towns is

30.0\text{ km}

Reducing a map's area to

\frac{1}{4}

of its original size means its linear dimensions are reduced by a factor of

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

. The new scale denominator becomes

100,000 \times 2 = 200,000

, resulting in a Representative Fraction of

1 : 200,000

. Multiplying the measured map distance of

15\text{ cm}

200,000

gives

3,000,000\text{ cm}

, which equals

30.0\text{ km}

on the ground.

Step-by-Step Solution

Determine the linear scale reduction factor from the areal reduction ratio

Linear reduction factor =

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

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

Calculate the Representative Fraction (RF) of the reduced map

New scale denominator =

100,000 \times 2 = 200,000

, giving a new scale of

1 : 200,000

Reducing linear dimensions by half doubles the scale denominator of the representative fraction.

Calculate the actual ground distance using the new map distance and new scale

Ground distance =

15\text{ cm} \times 200,000 = 3,000,000\text{ cm} = 30.0\text{ km}

Converting map distance in centimeters to ground distance in kilometers by dividing

3,000,000\text{ cm}

100,000\text{ cm/km}

Key Concept

Linear versus Areal Map Scale Transformation and Ground Distance Calculation

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic