Q25. - Algebra 1

Question

Money invested in a certificate of deposit $\mathbf{\left(}\mathit{C}\mathit{D}\mathbf{\right)}$ earns interest once per year. Suppose you invest 1 $\mathit{\backslash (}\mathbf{4000}$in a 2-year $\mathit{C}\mathit{D}$.

a. If the interest rate is  $\mathbf{5}\mathit{\%}$ per year, the expression  $\mathbf{4000}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.05}\mathbf{)}}^{\mathbf{2}}$ can be evaluated to find the total amount of money after two years. Explain the numbers in this expression.

b. Find the amount at the end of two years.

c. Suppose you invest $\mathit{\backslash )}\mathbf{10000}$in a $\mathit{C}\mathit{D}$ for  4 years at an annual rate of  $\mathbf{6}\mathbf{.25}\mathit{\%}$. What is the total amount of money you will have after 4 years?

Step-by-Step Solution

Answer

a. The number  $\mathbf{4000}$ is the principal or present value, $\mathbf{0}\mathbf{.05}$ is the rate of interest per annum compounded yearly expressed in decimal form, and $\mathbf{2}$ is the time taken.

b. The amount at the end of two years is  $\mathit{\$}\mathbf{4410}$.

c. The total amount of money after 4 years will be  $\mathit{\$}\mathbf{12744}\mathbf{.3}$.

Step1. Given

The value $P$ is 4000, the value $t$ is 2, and the value $r$ is 5.

The interest is earned once per year. Therefore the number of compounding in a year is 1. Therefore, the value $n$ is 1.

Step2. Write the formula to calculate the amount or future value.

The formula to calculate the amount ($A$) or future value is:

 $A=P{\left(1+\frac{{\displaystyle r}}{{\displaystyle 100n}}\right)}^{nt}$

Where $P$ is the principal or present value, $t$ is the time taken, $r$ is the rate of interest per annum in percent, and $n$ is the number of compounding in a year.

Step3. Explain the numbers in the given expression 4000 ( 1 + 0 .05 ) 2 .

$t$a.

It is given that money invested in a certificate of deposit (CD) earns interest once per year, the amount of money invested is $\$4000$ for 2 years and the interest rate is $5\%$ per year.

Therefore, the value of $P$ is 4000, the value of $t$ is 2, and the value of $r$ is  5.

The interest is earned once per year. Therefore the number of compounding in a year is 1. Therefore, the value of $n$ is 1.

Substitute these values in the expression $A=P{\left(1+\frac{{\displaystyle r}}{{\displaystyle 100n}}\right)}^{nt}$.

$\begin{array}{c}A=P{\left(1+\frac{r}{100n}\right)}^{nt}\\ =4000{\left(1+\frac{5}{100\left(1\right)}\right)}^{\left(1\right)\left(2\right)}\\ =4000{\left(1+\frac{5}{100}\right)}^2\\ =4000{\left(1+0.05\right)}^2\end{array}$

Therefore, the expression obtained and the given expression are the same and equal$4000{(1+0.05)}^2$. Therefore, the number 4000 is the principal or present value, $0.05$ is the rate of interest per annum compounded yearly expressed in decimal form, and 2 is the time taken.

b.

It is given that money invested in a certificate of deposit (CD) earns interest once per year, the amount of money invested is $\$4000$ for 2 years and the interest rate is $5\%$ per year.

Therefore, the value $P$ is 4000, the value $t$ is 2, and the value $r$ is 5.

The interest is earned once per year. Therefore the number of compounding in a year is 1. Therefore, the value $n$ is 1.

The amount ($A$) at the end of two years is given by:

$\begin{array}{c}A=P{\left(1+\frac{r}{100n}\right)}^{nt}\\ =4000{\left(1+\frac{5}{100\left(1\right)}\right)}^{\left(1\right)\left(2\right)}\\ =4000{\left(1+\frac{5}{100}\right)}^2\\ =4000{(1+0.05)}^2\\ =4000{\left(1.05\right)}^2\\ =4000\times 1.1025\\ =4410\end{array}$

Therefore, the amount at the end of two years is  $\$4410$.

 

c.

It is given that money invested in a certificate of deposit (CD) earns interest once per year, the amount of money invested is $\$10000$ for 4 years and the interest rate is $6.25\%$ per year.

Therefore, the value of $P$  is $10000$ the value of  $t$ is 4 and the value of $r$ is  $6.25$.

The interest is earned once per year. Therefore the number of compounding in a year is 1. Therefore, the value of $n$ is 1.

The amount ($A$) after 4 years is given by:

$\begin{array}{c}A=P{\left(1+\frac{r}{100n}\right)}^{nt}\\ =10,000{\left(1+\frac{6.25}{100\left(1\right)}\right)}^{\left(1\right)\left(4\right)}\\ =10000{\left(1+\frac{6.25}{100}\right)}^4\\ =10000{(1+0.0625)}^4\\ =10000{\left(1.0625\right)}^4\\ =10000\times 1.2744293212890625\\ =12744.293212890625\\ \approx 12744.3\end{array}$

Therefore, the amount after $4$ years is $\$12744.3$ .