Question

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 three 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 three years?

Answer 1

Answer:

She chose Option 2 which is a linear option, because it offers a smaller lose    

in value compared to option 1 which is an exponential option.

The final value for option 2=$32,800

Step-by-step explanation:

Option 1

New Zoomba for 60000 with a depreciation rat of 2%per month for 3 years

Exponential equation;

y=a(1-r)^t

where;

y=future value

a=initial value=60000

r=depreciation rate=2% per month

t=time interval=12×3=36 months

Replacing;

y=60000(1-2/100)^36

y=60000(0.98)^36=28,992.79

The value after 3 years=$28,992.79

Initial value-Final value=(60000-28992.79)=$31007.21

Percentage of initial value lost=((Final value-Initial Value)/(Initial Value))×100

(31007.21/60000)×100=51.68%

Option 2

New starfish for $40,000 with a depreciation of $200 per month for 3 years

Linear equation;

y=a-bt

where;

y=Future value

a=Initial value=$40,000

b=the depreciation amount per time interval=$200 per month for 3 years

t=time interval=(3×12)=36 months

Replacing;

y=40000-(200×36)

y=32,800

Final value=y=$32,800

Initial value-Final value=(40000-32800)=$7200

Percentage of initial value lost=((Final value-Initial Value)/(Initial Value))×100

(7200/40000)×100=18%

Option 1(51.68%)>Option 2(18%) therefor Option 1 loses value at a faster rate than Option 2

She chose Option 2 which is a linear option, because it offers a smaller lose    

in value compared to option 1 which is an exponential option