# Gina has a road trip and is reading a map that has a scale of 1 in to 30 miles. Her trip measures 3.5 inches on the map. What will her actual distance be for the trip?

To solve a problem like this for the distance of Gina's trip, set up a proportion with the numbers of the original scale as the numerators of the fractions, the trip inches as the denominator of the left side fraction, and the missing miles (designed by x) as the denominator of the right side fraction. Then solve for x. The best way to solve a problem like this one is to set up a proportion with the missing information designated by x and then solve for x. Let's look at a couple of examples.

If a map has a scale of 2 inches to 15 miles, and a trip measures 2.5 inches on the map, how long is the trip? We'll set up our proportion like this: 2/2.5 = 15/x. Look at what we've done here. We've set the original scale of 2 inches to 15 miles as the numerators of two fractions. Then we've put the 2.5 inches as the denominator of the first fraction. We're looking for how many miles the actual trip is, so we designate that as x and put it in the denominator of the second fraction. Now let's solve the equation. Begin by getting rid of the fractions. Multiply each side by the denominators to get 2x = (15)(2.5). Now multiply the right side of the equation for 2x = 37.5. Finally, solve for x to get 18.75 miles. That's the length of the actual trip.

Let's try one more. This time the map has a scale of 2 inches to 20 miles, and the trip measures 4.5 inches on the map. How long will the actual trip be in miles? We'll set up our proportion: 2/4.5 = 20/x. Now we'll solve for x. Getting rid of the fractions, we have 2x = (20)(4.5). Now we can multiply the right side of the equation for 2x = 90. When we solve for x, we discover that the actual trip is 45 miles long.

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