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A new car is purchased for $20000$ dollars. The value of the car depreciates at $12.5\%$ per year. What will the value of the car be, to the nearest cent, after $11$ years?

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

20000

dollars. The value of the car depreciates at

12.5\%

per year. What will the value of the car be, to the nearest cent, after

11

years?

Identify initial value and rate: Identify the initial value of the car and the annual depreciation rate. $\newline$ The initial value of the car is $\$20,000$ and it depreciates at a rate of $12.5\%$ per year.
Convert rate to decimal: Convert the annual depreciation rate into a decimal. $\newline$ To convert a percentage to a decimal, divide by $100$ . $\newline$ $12.5\% = \frac{12.5}{100} = 0.125$
Determine depreciation multiplier: Determine the depreciation multiplier. $\newline$ The value of the car decreases by $12.5\%$ each year, so it retains $100\% - 12.5\% = 87.5\%$ of its value each year. $\newline$ Convert $87.5\%$ to a decimal to get the multiplier: $\frac{87.5}{100} = 0.875$
Calculate value after $11$ years: Calculate the value of the car after $11$ years using the formula for exponential decay. $\newline$ The formula for exponential decay is $V = P(1 - r)^t$ , where $V$ is the final value, $P$ is the initial value, $r$ is the rate of depreciation, and $t$ is the time in years. $\newline$ In this case, $V = 20000 \times 0.875^{11}$
Perform the calculation: Perform the calculation. $\newline$ $V = 20000 \times 0.875^{11}$ $\newline$ $V \approx 20000 \times 0.2756$ (using a calculator for $0.875^{11}$ ) $\newline$ $V \approx 5512.00$

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