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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?

At Bank A, $600 is invested with an interest rate of 5% compounded annually. At Bank-example-1

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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+(r)/(100\alpha))^(n\alpha) \\ \text{Where} \\ \alpha=\text{ number of periods per annum} \end{gathered}

For Bank A:


\begin{gathered} A=P(1+(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+(5)/(100*1))^((10*1)) \\ \\ 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+(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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