Your small remodeling business has two work vehicles One is

Your small remodeling business has two work vehicles. One is a small passenger car used for job-site visits and for other general business purposes. The other is a heavy truck used to haul equipment. The car gets 25 miles per gallon (mpg). The truck gets 10 mpg. You want to improve gas mileage to save money, and you have enough money to upgrade one vehicle. The upgrade cost will be the same for both vehicles. An upgraded car will get 40 mpg; an upgraded truck will get 12.5 mpg. The cost of gasoline is $4.10 per gallon. Suppose you drive the truck 11,340 miles per year. How many miles would you have to drive the car before upgrading the car would be the better choice? Number of miles

Solution

We can begin by calculating the gallons saved by purchasing the new truck. The current and new gallon usage when driving \'x\' miles per year are:

Current truck gallons = x / 10

New truck gallons = x / 12.5

So the gallons saved by purchasing the new truck are:

Truck gallons saved = x / 10 – x / 12.5

If we let \'y\' equal the increased mileage for the car, the gallons used by the current car, the new car, and the savings by purchasing the new car are:

Current car gallons = (x + y) / 25

New car gallons = (x + y) / 40

Car gallons saved = (x + y) / 25 – (x + y) / 40

We need to set the gallon savings from the new truck purchase equal to the gallon savings from the new car purchase equal to each other, so:

x / 10 – x / 12.5 = (x + y) / 25 – (x + y) / 40

From this equation you can see again that the cost per gallon is irrelevant. Each term would be multiplied by the cost per gallon, which would cancel out since each term is multiplied by the same amount. To add and subtract fractions, we need to get the same denominator. In this case, we willchoose a denominator of 1,000 since all four of the current denominators are multiples of 1,000. Doing so, we get:

100x / 1,000 – 80x / 1,000 = 40(x + y) / 1,000 – 25(x + y) / 1,000

20x / 1,000 = 15(x + y) / 1,000

20x = 15x + 15y

5x = 15y

y = x / 3

The difference in the mileage should be 1/3 of the miles driven by the truck. So, if the truck is driven 11340 miles, the breakeven car mileage is 15,120 miles (= 11,340 + 11,340 / 3).