Consider the following​ situation, which involves two options. Determine which option is less expensive.

Question

You currently drive 375 miles per week in a car that gets 20 miles per gallon of gas. You are considering buying a new fuel-efficient car for $13 comma 000 (after trade-in on your current car) that gets 46 miles per gallon. Insurance premiums for the new and old car are $900 and $600 per year, respectively. You anticipate spending $1900 per year on repairs for the old car and having no repairs on the new car. Assume gas costs $4.00 per gallon. Over a five-year period, is it less expensive to keep your old car or buy the new car?

Answer 1

Let's look at the cost of the old car:

^gal/_{20 mi} * ^$4/_gal = ^$1/_{5 mi} (It costs $1 for every five miles you go)

375 ^mi/_wk * ^$1/_{5 mi} = ^$75/_wk (You're spending $75 a week on gas)

There are ∼52 weeks per year

^$75/_wk * ^{52 wk}/_yr * 5 years = $19,500 spent on gas

^$600/_yr * 5 years = $3,000 on insurance

^$1900/_yr * 5 years = $9,500 on repairs

 

Now for the new car:

^gal/_{46 mi} * ^$4/_gal = ^$2/_{23 mi} (It costs $2 for every 23 miles you go)

^{375 mi}/_wk * ^$2/_{23 mi} = $⁷⁵⁰/_{23 wk} (You're spending $750 every 23 weeks on gas)

^$750/_{23 wk} * ^{52 wk}/_yr * 5 years ≅ $8,478.26 spent on gas

^$900/_year * 5 years = $4,500 spent on insurance

The new car = $13,000 post trade-in

 

That's all the hard stuff. You can add the totals together to get the final tally of old vs. new.

 

Answer 2

Set up 2 equations that account for all costs with each car:

 

Old car:(375 miles)/(20 mpg)x 4 $/gal x 52 weeks x 5 years +600 insurance x 5 years + 1900 repairs x 5 years = 32000

 

New car: (375 miles)/(46 mpg) x 4 $/gal x 52 weeks x 5 years + 900 insurance x 5 years + 13000 = 25,978

 

The new car is the better value