Truck A costs $600 more that Truck B. However, it burns 0.06 gallon/mile compared to 0.07 gallon/mile for truck B. Both trucks have a useful life of 10 years. Salvage value of Truck B is $200 less than Truck A's. The cost of fuel is $2.25/gallon. If the value of $ remains the same in 10 years, what is the number of miles driven per year (X) beyond which Truck A is a better option than Truck B?

1,778

Step-by-step explanation:

Let C and S represent the cost and salvage value of Truck B, respectively. Let X represent the number of miles per year that answers the question.

Truck A's cost for 10 years is ...

A = C+600 + (0.06) (2.25) (10X) - (S+200)

Truck B's cost for 10 years is ...

B = C + (0.07) (2.25) (10X) - S

We want to find X such that Truck A's cost is lower, so ...

C - S + 1.35X + 400 < C - S + 1.575X

400 < 0.225X

1777 7/9 < X

1778 miles driven per year (or more) makes Truck A a better option.