Answer to Question #146137 in Algebra for De’Najah

Question

A new car is purchased for 23700 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be, to the nearest cent, after 10 years

Answer 1

First year

depreciation=(13.5/100) * 23700 = 3199.5

First year value = 23700 - 3199.5 = $20500.5

Second year

depreciation=(13.5/100) * 20500.5 = 2767.57

Second year value = 20500.5 - 2767.57 = $17732.93

Third year

depreciation=(13.5/100) * 17732.93 = 2393.95

Third year value = 17732.93 - 2393.95

= $15338.98

Fourth year

depreciation=(13.5/100) * 15338.98 = 2070.76

Fourth year value = 15338.98 - 2070.76

= $13268.22

Fifth year

depreciation=(13.5/100) * 13268.22 = 1791.21

Fifth year value = 13268.22 - 1791.21

= $11477.01

Sixth year

depreciation=(13.5/100) * 11477.01 = 1549.40

Sixth year value = 11477.01 - 1549.40

= $9927.61

Seventh year

depreciation=(13.5/100) * 9927.61 = 1340.23

Seventh year value = 9927.61 - 1340.23

= 8587.38

Eighth year

depreciation=(13.5/100) * 8587.38 = 1159.30

Eighth year value = 8587.38 - 1159.30

= $7428.08

Ninth year

depreciation=(13.5/100) * 7428.08 = 1002.79

Ninth year value = 7428.08 - 1002.79

= $6425.29

Tenth year

depreciation=(13.5/100) * 6425.29 = 867.41

Tenth year value = 6425.29 - 867.41

= $5557.88

But 1 dollar = 100 cents

5557.88 dollars = (5557.88*100) cents

= 555788 cents.