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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/is5islmkx1ta6cqewn0eonkatffnzlnfwp.png)
For Bank B:
![\begin{gathered} \text{ P=\$500} \\ \text{ R=6\% quarterly} \\ \text{ T=10yrs and 20 yrs} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/89k84mjltnlfg3mi6q7svqs3k5hy1ny8ho.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/uyg2vrxv0jqbi0itfvrltz41gl3edbs074.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/d27rtzo8wkcokmv7513uos9w3xk396ige8.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/z7q8e7pbobpxhmomynoz5coen6y5dhxo7v.png)
For Bank A for after 20 years:
![A=600(1.05)^(20)=1591.979\approx\text{ \$1591.98}](https://img.qammunity.org/2023/formulas/mathematics/high-school/kz257wsnd5um3g7q4fffyvm7apq6cxq6yz.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5xvptwacx9u3oqi3jtcg7kr781v11j11dg.png)
Substituting these values in the formula, we get
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