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 $3.45 per
gallon. Suppose you drive the truck 11,940 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: ??

Savings from the truck upgrade = 11,940 miles/(12.5 mpg – 10
mpg) * $3.45

= $16,477.20

Let the miles for car be "x". Thus savings from car = x/(40 mpg
– 25 mpg) * $3.45

= 0.23x

Now upgrading the car will be a better choice when savings from
car > savings from truck

or, 0.23x > 16,477.20

or x>71,640 miles

Thus number of miles should be = 71,641
miles