Question

At Bank A, $600 is invested with an interest rate of 5% compounded annually. At Bank B, $500 is invested with an interest rate of 6% compounded quarterly. Which account will have a larger balance after 10 years? After 20 years?

Answer 1

The Solution:

It is given that

For Bank A:

$\begin{gathered} P=\text{ \ 600} \\ R=5\text{ \% annually} \\ T=10\text{ yrs and T=20 yrs} \end{gathered}$

For Bank B:

$\begin{gathered} \text{ P=\ 500} \\ \text{ R=6\% quarterly} \\ \text{ T=10yrs and 20 yrs} \end{gathered}$

By the formula for the compound interest, we have

$\begin{gathered} A=P(1+\frac{r}{100\alpha})^{n\alpha} \\ \text{Where} \\ \alpha=\text{ number of periods per annum} \end{gathered}$

For Bank A:

$\begin{gathered} A=P(1+\frac{r}{100\alpha})^{n\alpha} \\ \text{ In this case,} \\ \alpha=1 \\ P=\text{ \ 600} \\ r=\text{ 5\%} \\ n=10\text{ yrs, and 20 yrs} \end{gathered}$

Substituting the corresponding values, we have

$\begin{gathered} A=600(1+\frac{5}{100\times1})^{(10\times1)} \\ \\ A=600(1+0.05)^{10}=600(1.05)^{10}=977.337\approx\text{ \ 977.34} \\ \end{gathered}$

For Bank A for after 20 years:

$A=600(1.05)^{20}=1591.979\approx\text{ \ 1591.98}$

Similarly, for Bank B:

$\begin{gathered} A=P(1+\frac{r}{100\alpha})^{n\alpha} \\ \text{ In this case,} \\ \alpha=4 \\ P=\text{ \ 500} \\ r=6\text{ \%} \\ n=10\text{ yrs, and 20 yrs} \end{gathered}$

Substituting these values in the formula, we get

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