Question
On a map, two cities are 2.8 inches apart. The map has a scale of 1 inch to 25 miles. How far apart, in inches, would the same two cities be on a map that has a scale of 1 inch to 40 miles?
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To solve this problem, we can use the proportion:
distance on map 1 / distance in real life 1 = distance on map 2 / distance in real life 2
Let's call the distance between the two cities in real life "d".
For the first map, we know that the distance on the map is 2.8 inches and the scale is 1 inch to 25 miles. So the distance in real life is:
d1 = 2.8 inches * (1/25 miles per inch) = 0.112 miles
For the second map, we want to find the distance on the map when the scale is 1 inch to 40 miles. Let's call this distance "x". The distance in real life is still "d". So we can set up the proportion:
2.8 inches / 0.112 miles = x inches / d miles
Solving for x, we get:
x = (2.8 inches / 0.112 miles) * d miles * (1 inch / 40 miles)
x = 7 inches * d
So the distance between the two cities on the second map would be 7 times the distance between the two cities in real life.
distance on map 1 / distance in real life 1 = distance on map 2 / distance in real life 2
Let's call the distance between the two cities in real life "d".
For the first map, we know that the distance on the map is 2.8 inches and the scale is 1 inch to 25 miles. So the distance in real life is:
d1 = 2.8 inches * (1/25 miles per inch) = 0.112 miles
For the second map, we want to find the distance on the map when the scale is 1 inch to 40 miles. Let's call this distance "x". The distance in real life is still "d". So we can set up the proportion:
2.8 inches / 0.112 miles = x inches / d miles
Solving for x, we get:
x = (2.8 inches / 0.112 miles) * d miles * (1 inch / 40 miles)
x = 7 inches * d
So the distance between the two cities on the second map would be 7 times the distance between the two cities in real life.
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