On a map, 1 inch represents 50 miles. The distance between Sommerville and Linden is 2.5 inches. What is the actual distance between these towns?
f 52.5 miles
g 20 miles
h 2.5 miles
j 125 miles
j) 125 miles
125 Miles
That is correct! The actual distance between Sommerville and Linden is 125 miles, since 1 inch on the map represents 50 miles and the distance between the towns on the map is 2.5 inches. So, we can calculate the actual distance as 2.5 inches x 50 miles per inch = 125 miles.
To find the actual distance between Sommerville and Linden, we need to use the scale on the map, which states that 1 inch represents 50 miles.
Given that the distance between Sommerville and Linden on the map is 2.5 inches, we can set up a proportion to find the actual distance.
Let x represent the actual distance between Sommerville and Linden.
The proportion can be set up as:
1 inch / 50 miles = 2.5 inches / x miles
To solve for x, we can cross-multiply and solve for x:
1 inch * x miles = 2.5 inches * 50 miles
x = 2.5 inches * 50 miles / 1 inch
Simplifying, we get:
x = 125 miles
Therefore, the actual distance between Sommerville and Linden is 125 miles.
The correct answer is j) 125 miles.