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If you need $20,000 six years from now, what is the minimum amount of money you need to deposit into a bank account that pays 3% annual interest, compounded (give your answers to the nearest cent):a) annually? $______b) monthly? $_____c) daily (assuming 365 days in a year)? $ _____

Sagot :

To solve the question, we would be making use of the compound interest formula. This is given as;

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]

A=final amount = $20,000

P=initial principal balance

r=interest rate =3%

n=number of times interest applied per time period

t=number of time periods elapsed = 6 years

Part A

If the interest is compounded annually, n =1

Therefore;

[tex]\begin{gathered} 20000=P(1+\frac{0.03}{1})^{1\times6} \\ 20000=P(1+0.03)^6 \\ (1.03)^6P=20000 \\ P=\frac{20000}{(1.03)^6} \\ P=16749.69 \end{gathered}[/tex]

Answer: The minimum amount would be $16749.69

Part B

If the interest is compounded monthly, n =12

[tex]\begin{gathered} 20000=P(1+\frac{0.03}{12})^{6\times12} \\ 20000=P(\frac{12+0.03}{12})^{72} \\ (\frac{12.03}{12})^{72}P=20000 \\ P=\frac{20000}{(\frac{12.03}{12})^{72}} \\ P=16709.16 \end{gathered}[/tex]

Answer: The minimum amount would be $16709.16

Part C

If the interest is compounded monthly, n =365

[tex]\begin{gathered} 20000=P(1+\frac{0.03}{365})^{365\times6} \\ 20000=P(\frac{365+0.03}{365})^{2190} \\ (\frac{365.03}{365})^{2190}P=20000 \\ P=\frac{20000}{(\frac{365.03}{365})^{2190}} \\ P=16705.53 \end{gathered}[/tex]

Answer: The minimum amount would be $16705.53

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