Question

Hunter invested $750 in an account paying an interest rate of 6\tfrac{5}{8}6 8 5 ​ % compounded continuously. London invested $750 in an account paying an interest rate of 6\tfrac{1}{2}6 2 1 ​ % compounded daily. After 18 years, how much more money would Hunter have in his account than London, to the nearest dollar?

Answer 1

Answer: $ 55

Step-by-step explanation:

When interest is compounded continuously, the final amount will be

$A=Pe^{rt}$

When interest is compounded daily, the final amount will be

$A=P(1+\dfrac{r}{365})^{365t}$

, where P= Principal , r = rate of interest , t = time

For Hunter , P= $750, r = $6\dfrac{5}{8}\%=\dfrac{53}{8}\%=\dfrac{53}{800}=0.06625$

t = 18 years

$A=750e^{0.06625(18)}=\ 2471.48$

For London , P= $750, r = $6\dfrac{1}{2}\%=\dfrac{13}{2}\%=\neq \dfrac{13}{200}=0.065$

t = 18 years

$A=750(1+\dfrac{0.065}{365})^{18(365)}=\ 2416.24$

Difference = $ 2471.48 - $ 2416.24 =$ 55.24≈$ 55

Hence, Hunter would have $ 55 more than London in his account .