Question

Megan invested $\$21000$in a savings account. If the interest rate is $5\%$, how much will be in the account in $8years$by each method of compounding?

Part (a): Compound quarterly

Part (b): Compound monthly

Part (c): Compound continuously

Short answer

Part (a): The amount after $8$years when compounded quarterly is $\$31250.74$.

Part (b): The amount after $8$years when compounded monthly is $\$31302.29$.

Part (c): The amount after $8$years when compounded continuously is$\$31328.32$.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$$\begin{gathered} P=21000, \\ r=0.05, \\ t=8, \\ n=4 \end{gathered}$$

Assume the principal, rate of interest, time and new balance to be $P,r,t,A$and when compounded $n$times in a year is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Part (a) Step 2. Substitute the values and simplify.

Substitute the values in the given formula,

$$\begin{gathered} A=21000{\left(1+\frac{0.05}{4}\right)}^{4\times 8} \\ =21000{\left(1+0.0125\right)}^{4\times 8} \\ =31250.74 \end{gathered}$$

Part (b) Step 1. Given information.

Consider the given question,

$$\begin{gathered} P=21000, \\ r=0.05, \\ t=8, \\ n=12 \end{gathered}$$

Assume the principal, rate of interest, time and new balance to be $P,r,t,A$and when compounded $n$times in a year is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Part (b) Step 2. Substitute the values and simplify.

Substitute the values in the given formula,

$$\begin{gathered} A=21000{\left(1+\frac{0.05}{12}\right)}^{12\times 8} \\ =21000{\left(\frac{12.05}{12}\right)}^{96} \\ =21000{\left(1.0125\right)}^{96} \\ =31302.29 \end{gathered}$$

Part (c) Step 1. Given information.

Consider the given question,

$$\begin{gathered} P=21000, \\ r=0.05, \\ t=8 \end{gathered}$$

When the new balance is compounded continuously $A=P{e}^{rt}$

Substitute the values,

$$\begin{gathered} A=21000{\left(e\right)}^{0.05\times 8} \\ =21000{\left(e\right)}^{0.4} \\ =31328.32 \end{gathered}$$