Q50. - Algebra 1

Question

Find the final value$\mathit{\$}\mathbf{2500}$invested at an interest rate$\mathbf{2}\mathit{\%}$compounded monthly 10 year.

Step-by-Step Solution

Answer

The final amount after 10 years will be $\mathit{\$}\mathbf{3053}$.

Step 1. Write the compound interest equation.

The compound interest equation is given by:

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

Step 2. Substitute the values.

$\$2500$ is invested for 10 years with a $2\%$ interest rate compounded monthly. 

Substitute 2500 for $P$, 2 for $r$, 12 for $n$, and 10 for $t$ into the formula $A=P{(1+\frac{r}{n})}^{nt}$.

$\begin{array}{c}\text{A}=P{\left(1+\frac{r}{n}\right)}^{{}^{n·t}}\\ =2500\times {\left(1+\frac{0.02}{12}\right)}^{12\times 10}\end{array}$

Step 3. Simplify for A .

$A$Simplify the equation for .

 $\begin{array}{c}\text{A}=P{\left(1+\frac{r}{n}\right)}^{n.t}\\ =2500\times {\left(1+\frac{0.02}{12}\right)}^{12\times 10}\\ =2500\times {\left(\frac{12.02}{12}\right)}^{120}\\ =\$3053\end{array}$ 

Thus, the final amount after 10 years will be $\$3053$.