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Karolina [17]
3 years ago
9

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 h

ave 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?
Mathematics
1 answer:
qwelly [4]3 years ago
5 0

Answer:

  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.

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