Question

ricky bobby wants to buy a new automobile for $23000 in 4 years. How much money must ricky’s original investment be if he makes a single deposit into an account monthly compounding and annual interest rate of 3.70% in order to reach his goal ?

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$23000\\ P=\textit{original amount deposited}\\ r=rate\to 3.70\%\to (3.70)/(100)\dotfill &0.037\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &4 \end{cases}`

`23000=P\left(1+(0.037)/(12)\right)^(12\cdot 4)\implies 23000=P\left( (12037)/(12000) \right)^(48)\implies \cfrac{23000}{~~ \left( (12037)/(12000) \right)^(48)~~}=P \\\\\\ \cfrac{23000}{~~ (12037^(48))/(12000^(48))~~}=P\implies \cfrac{23000\cdot 12000^(48)}{12037^(48)}=P\implies 19840.43\approx P`