Examkin

Question

Difficulty: MediumScale Drawings and Map Measurements

A search and rescue grid map uses a scale of 23 inch=10 miles\frac{2}{3}\text{ inch} = 10\text{ miles}. If a distress beacon is located 60 miles60\text{ miles} away from the rescue station, what is the distance between these two points on the map in inches?

  1. 44Answer
  2. B
    66
  3. C
    99
  4. D
    4040

Answer

The distance on the map is 44 inches.
To find the map distance, first determine the scale multiplier by dividing the total actual distance by the reference distance in the scale (60÷10=660 \div 10 = 6). Next, multiply this factor by the scale measurement on the map (6×23=123=46 \times \frac{2}{3} = \frac{12}{3} = 4). Thus, the distance on the map is exactly 44 inches.

Step-by-Step Solution

1
Determine how many 1010-mile units fit into the total actual distance of 6060 miles.
60÷10=660 \div 10 = 6 units.
This establishes the scale multiplier, telling us how many times we need to apply the map's specified scale distance.
2
Multiply the scale multiplier by the map's fractional measurement.
6×23=123=46 \times \frac{2}{3} = \frac{12}{3} = 4 inches.
Because each 1010-mile unit represents 23\frac{2}{3} of an inch on the map, multiplying them gives the total map distance.

Key Concept

Scale Drawings and Map Measurements
Rate this question

Topics

Scale Drawings and Map Measurements

More questions from this topic