A teacher decides to purchase a new car and considers two options. Option one is the new Zoomba for $60,000 with an expected depreciation of 2% per month. Option two is the new Starfish for $40,000 with an expected depreciation of $200 per month. The teacher plans to keep the car for two years and chooses the option that is expected to retain the most value. Which option did the teacher choose (exponential or linear), and what was the value of the chosen option after two years

The teacher chose option 2, the Starfish and it was valued at $35,200 after two months.

Step-by-step explanation:

Step 1; The Zoomba is priced at $60,000 and has an expected depreciation of 2% per month. So the annual depreciation is 12 months * 2% per month = 24% per year. The teacher plans to keep either car for two years do total depreciation is 2 years * 24% depreciation per year = 48 % in two years.

Value of car after two years = Buying price - Depreciation amount

= $60,000 - 48% of $60,000 = $60,000 - 0.48*$60,000 = $31,200

The car will be valued at $31,200 after two years. To determine how much value it has retained we divide the value in two years to the current value i. e $31,200 divided by $60,000 which equals 0.52 (52% of initial price)

Step 2; The Starfish is priced at $40,00 and will depreciate $200 every month. For a year it will depreciate 12 months * $200 per month = $2,400 a year. So for a total of two years a depreciaition of $4,800.

Value of car after two years = Buying price - Depreciation amount

= $40,000 - $4,800 = $35,200

The car will be valued at $35,200 after two years. To see the retained value we divide the depreciated value by the current value i. e. $35,200 divided by $40,000 which equals 0.88 (88% of the initial price)

Step 3; Of the two options, the Starfish retains more of its value over two years. So the teacher chose the Starfish over the Zoomba.