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An aerodynamic three-wheeled automobile (the Dart) runs on compressed natural gas stored in two cylinders in the rear of the vehicle. The $15,000 Dart can cruise at speeds up to 80 miles per hour and it can travel 100 miles per gallon of fuel. Another two-seater automobile costs $12,000 and averages 50 miles per gallon of compressed natural gas. Assume a useful life of seven years for each car, similar salvage values, and similar maintenance costs.
a. If fuel costs $8.00 per gallon and the MARR is 10% per year, at what average annual miles driven would you be indifferent between the vehicles?
b. Over what range of annual miles driven is the Dart more economical?

Respuesta :

Chrisnando

Answer:

a) 7702.88 miles

b) If we drive more than 7702.88 miles per year, the dart would be more economical.

Step-by-step explanation:

Assuming a useful life of 7 years.

Let's calculate EUAC dart:

(15000)(A/P, 10%, 7) + (X/100mpg)(8)

= 15000 * 0.2054 + 0.08

= 3081.15 + 0.08X

EUAC For other automobile:

(12000)(A/P, 10%, 7) + (X/50mpg)(8)

= 12000 * 0.2054 + 0.16X

= 2464.92 + 0.16X

a) To solve for X, we have:

EUAC dart = EUAC other automobile

3081.15 + 0.08X = 2464.92 + 0.16X

0.08X - 0.16X = 2464.92 - 3081.15

-0.08X = - 616.23

X = 7702.88 miles

b) The range of annual miles driven is the Dart more economical.

EUAC dart < EUAC other automobile

3081.15 + 0.08X < 2464.92 + 0.16X

X > 7702.88 miles

If we drive more than 7702.88 miles per year, the dart would be more economical.

Following are the response to the given question:

Considering a service life of 7 years and computing the EUAC dart:

[tex]\to (15000)(\frac{A}{P}, 10\%, 7) + (\frac{X}{100} \ mpg)(8) \\\\\to 15000 \times 0.2054 + 0.08 X\\\\\to 3081.15 + 0.08X[/tex]

EUAC For one different vehicle:

[tex]\to (12000)(\frac{A}{P}, 10\%, 7) + ( \frac{X}{50} \ mpg)(8)\\\\\to 12000 \times 0.2054 + 0.16X\\\\\to 2464.92 + 0.16 X\\\\[/tex]

For point a)

To solve X, we have: [tex]\to \text{EUAC darts = EUAC another automobile}[/tex]

[tex]\to 3081.15 + 0.08X = 2464.92 + 0.16X\\\\\to 0.08X - 0.16X = 2464.92 - 3081.15\\\\\to -0.08X = - 616.23\\\\\to X = 7702.88\ miles[/tex]  

For point b)

The Dart is much more cost-effective in terms of annual miles driven.[tex]\to \text{EUAC dart} < \text{EUAC other automobile}\\\\\to 3081.15 + 0.08X < 2464.92 + 0.16X\\\\\to X > 7702.88 \ \ miles\\\\[/tex]

It dart would be more cost-effective if we drove more than 7702.88 miles per year.

Learn more about the three-wheeled automobile:

brainly.com/question/3929628

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