Question

Jaquin hopes to earn $900 in interest in 3.6 years time from $90,000 that he has available to invest. To decide if it's feasible to do this byinvesting in an account that compounds daily, he needs to determine the annual interest rate such an account would have to offer forhim to meet his goal. What would the annual rate of interest have to be? Round to two decimal places.

Answer 1

The Solution:

To find the annual interest rate that will help Jaquin to meet his goal, we shall use the formula below:

`A=P+I=P(1+(r)/(100\alpha))^(n\alpha)`

In this case,

`\begin{gathered} A=\text{amount}=90000+900=\text{ \$90900} \\ P=\text{ amount to be invested= \$90000} \\ I=\text{interest}=\text{ \$900} \\ r=\text{ interst rate per year=?} \\ n=\text{ number of years = 3.6 years} \\ \alpha=\text{ number of periods per year=365 (since it is compounded daily)} \end{gathered}`

Substituting these values in the above formula, we get

`90900=90000(1+(r)/((100*365)))^((3.6*365))`
`90900=90000(1+(r)/((36500)))^(1314)`

Dividing both sides by 90000, we get

`\begin{gathered} (90900)/(90000)=(1+(r)/((36500)))^(1314) \\ \\ 1.01=(1+(r)/(36500))^(1314) \end{gathered}`

Multiplying the power of both sides by 1/1314, we get

`\begin{gathered} 1.01^{(1)/(1314)}=(1+(r)/(36500))^{1314*(1)/(1314)} \\ \\ 1.000007573=1+(r)/(36500) \end{gathered}`

Collecting the like terms, we get

`\begin{gathered} 1.000007573-1=(r)/(36500) \\ \\ 0.000007573=(r)/(36500) \end{gathered}`

Cross multiplying, we get

`r=0.000007573*36500=0.2764145\approx0.28\text{ \%}`

Therefore, the correct answer is 0.28%