Zita’s computer depreciating at a rate of$\mathbf{3}\mathit{\%}$per year. She bought the computer for$\mathit{\$}\mathbf{1200}$.

a. Write an equation to represent this situation.

b. What will the computer’s value be after 5 years?

The compound interest equation is given by:

$A=P{\left(1+\frac{{\displaystyle r}}{{\displaystyle n}}\right)}^{nt}$, where $A$ is the amount,$P$ is the principal amount, $r$is the rate of interest, $t$is the time.

Zita bought a computer for $\$1200$which is depreciating at a rate of $3\%$per year i.e., the rate of interest must be $-3\%$.

To find the equation that represents the situation substitute 1200 for $P$, 1 for $n$, and $-3\%$for $r$into the formula $A=P{\left(1+\frac{{\displaystyle r}}{{\displaystyle n}}\right)}^{nt}$.

role="math" localid="1648203208657" $\begin{array}{c}\text{A}=P{\left(1+\frac{r}{n}\right)}^{n·t}\\ =1200\times {\left(1-\frac{0.03}{1}\right)}^{1\times t}\end{array}$

$\begin{array}{l}\text{A}=P{\left(1+\frac{r}{n}\right)}^{{}^{n·t}}\\ =1200\times {\left(1-\frac{0.03}{1}\right)}^{1\times t}\\ =1200\times {\left(\frac{1-0.03}{1}\right)}^t\\ =1200\times {\left(0.97\right)}^t\end{array}$

b.

Zita bought a computer for $\$1200$which is depreciating at a rate of $3\%$per year i.e., the rate of interest must be $-3\%$.

To find the equation that represents the situation substitute 1200 for $P$, 1 for $n$, and $-3\%$for $r$into the formula $A=P{\left(1+\frac{{\displaystyle r}}{{\displaystyle n}}\right)}^{nt}$.

$\begin{array}{c}A=P{\left(1+\frac{r}{n}\right)}^{n.t}\\ =1200\times {\left(1-\frac{0.03}{1}\right)}^{1\times t}\\ =1200\times {\left(0.97\right)}^t\end{array}$

To find the value of a computer after 5 years, substitute 5 for $t$into the equation $\text{A}=1200\times {\left(0.97\right)}^t$ and simplify.

$\begin{array}{l}\text{A}=1200\times {\left(0.97\right)}^t\\ =1200\times {\left(0.97\right)}^5\\ \approx 1030.48\end{array}$

Thus, the final value of the computer after 5 years will be $\$1030.48$.