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?

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sqdancefan sqdancefan · Answer 1 · 2020-04-21T22:42:48+02:00

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.

yoodyannapolis yoodyannapolis · Answer 2 · 2021-10-01T10:13:17+02:00

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

Let us represent [tex]c[/tex] and [tex]s[/tex] the cost and salvage value of Truck B.

Let [tex]x[/tex] represent the number of miles per year.

Now,

Truck A's cost for [tex]10[/tex] years is

[tex]A = c+600 +(0.06)(2.25)(10x) -(s+200)[/tex]

Truck B's cost for [tex]10[/tex] years is

[tex]B = c +(0.07)(2.25)(10x) -s[/tex]

Here we need to find the value of [tex]x[/tex] such that Truck A's cost is lower.

Therefore,

[tex]c -s +1.35x +400 < c -s +1.575x\\400 < 0.225x\\1777 7/9 < x[/tex]

So, [tex]1778[/tex] miles driven per year (or more) makes Truck A a better option.

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