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In the year 2006, a person bought a new car for
$24000. For each consecutive year after that, the value of the car depreciated by
9%. How much would the car be worth in the year 2008, to the nearest hundred dollars?
Answer:

In the year $2006$ , a person bought a new car for $\$ 24000$ . For each consecutive year after that, the value of the car depreciated by $9 \%$ . How much would the car be worth in the year $2008$ , to the nearest hundred dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. In the year

2006

, a person bought a new car for

\$ 24000

. For each consecutive year after that, the value of the car depreciated by

9 \%

. How much would the car be worth in the year

2008

, to the nearest hundred dollars?

\newline

Answer:

Determine Initial Value and Rate: Determine the initial value of the car and the annual depreciation rate. $\newline$ The initial value of the car, $P$ , is $\$24000$ , and the annual depreciation rate, $r$ , is $9\%$ or $0.09$ in decimal form.
Calculate Value After $1$ Year: Calculate the value of the car after one year ( $2007$ ). $\newline$ The value of the car after one year is given by $P \times (1 - r)$ . $\newline$ So, after one year, the value is $\$24000 \times (1 - 0.09) = \$24000 \times 0.91$ .
Value After $1$ Year Calculation: Perform the calculation for the value after one year. $\newline$ $\$24000 \times 0.91 = \$21840$ .
Value After $2$ Years: Calculate the value of the car after the second year ( $2008$ ). $\newline$ The value after the second year is given by the value after the first year multiplied by $(1 - r)$ again. $\newline$ So, after the second year, the value is $\$21840 \times (1 - 0.09) = \$21840 \times 0.91$ .
Value After $2$ Years Calculation: Perform the calculation for the value after the second year. $\newline$ $\$21840 \times 0.91 = \$19874.4$ .
Round Value to Nearest Hundred: Round the value to the nearest hundred dollars. $\newline$ The value of the car in $2008$ , rounded to the nearest hundred dollars, is approximately $\$19900$ .

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