Question

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 $3.70 per gallon.

  

Suppose you drive the truck 12,000 miles per year. How many miles would you have to drive the car before upgrading the car would be the better choice?

Answer 1

First of all we need to calculate the gallons saved.

The current and new gallon usage when driving lets say K miles per year

Current truck = K / 10

New truck = K / 12.5

Total gallons saved = (K / 10 - K / 12.5) ---------------------equation 1

In a similar way with increase in mileage lets say L miles per year for car, we need to calculate the gallons used in both the current and the new car

Current car = (K+L) / 25

New car = (K+L) / 40

Gallons saved = (K + L / 25) - (K + L / 40) ------------------equation 2

The upgradation cost is same for both truck and car

equate equations 1 and 2,

(K / 10 - K / 12.5) = (K + L / 25) - (K + L / 40)

2.5K / 125 = 15(K+L) / 1000

2.5K * 1000 = 125 * 15 * (K+L)

2500K = 1875K + 1875L

625K = 1875L

625K / 1875 = L

K/3 = L

As given in the question lets say the truck has driven 12000 miles

We have denoted as K

K = 12000

the extra mileage for car is L = 12000/3

hence L = 4000

K+L is the miles driven by the car which is 16000

Hence the car would have to drive 16000 before upgrading the car