Answer:
43) 79 years and 6 months
44) 17 years and 6 months
Step-by-step explanation:
As the account increases by a constant percentage bi-monthly, we can use the compound interest formula to write a function to model the situation.
[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+\frac{r}{n}\right)^{nt}$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}[/tex]
The interest is compounded bi-monthly, which means every 2 months.
Therefore, the interest is applied 6 times per year.
Given values:
- A = $10,000
- P = $2,400
- r = 1.8% = 0.018
- n = 6 (bi-monthly)
Substitute the given values into the formula, and solve for t:
[tex]\begin{aligned}A&=P\left(1+\dfrac{r}{n}\right)^{nt}\\\\\implies 10000&=2400\left(1+\dfrac{0.018}{6}\right)^{6t}\\\\\implies 10000&=2400\left(1+0.003\right)^{6t}\\\\\dfrac{25}{6}&=\left(1.003\right)^{6t}\\\\\textsf{Take natural logs:} \quad \ln \left(\dfrac{25}{6}\right)&=\ln\left(1.003\right)^{6t}\\\\\ln \left(\dfrac{25}{6}\right)&=6t\ln\left(1.003\right)\\t&=\dfrac{\ln \left(\dfrac{25}{6}\right)}{6\ln\left(1.003\right)}\\\\t&=79.4031089...\end{aligned}[/tex]
The account balance will reach $10,000 during the 79th year (after 79 years and 4.84 months).
As the interest is applied bi-monthly (every 2 months) we need to round up to the nearest 2 month interval. Therefore, the account balance will reach $10,000 after 79 years and 6 months.
Note: After 79 years and 4 months, the account balance will be $9,987.47, and after 79 years and 6 months it will be $10,017.43.
[tex]\hrulefill[/tex]
As the account increases by a constant percentage quarterly, we can use the compound interest formula to write a function to model the situation.
[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+\frac{r}{n}\right)^{nt}$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}[/tex]
The interest is compounded quarterly, which means every 3 months.
Therefore, the interest is applied 4 times per year.
Given values:
- A = $60,000
- P = $30,000
- r = 4% = 0.04
- n = 4 (quarterly)
Substitute the given values into the formula, and solve for t:
[tex]\begin{aligned}A&=P\left(1+\dfrac{r}{n}\right)^{nt}\\\\\implies 60000&=30000\left(1+\dfrac{0.04}{4}\right)^{4t}\\\\\implies 60000&=30000\left(1+0.01\right)^{4t}\\\\2&=\left(1.01\right)^{4t}\\\\\textsf{Take natural logs:} \quad \ln \left(2\right)&=\ln\left(1.01\right)^{4t}\\\\\ln \left(2\right)&=4t\ln\left(1.01\right)\\\\t&=\dfrac{\ln \left(2\right)}{4\ln\left(1.01\right)}\\\\t&=17.4151792...\end{aligned}[/tex]
The value of the investment account will double during the 17th year (after 17 years and 4.98 months).
As the interest is applied quarterly (every 3 months) we need to round up to the nearest 3 month interval. Therefore, the investment account will double by 17 years and 6 months.
Note: After 17 years and 3 months, the account balance will be $59,606.83, and after 17 years and 6 months it will be $60,202.90.
