Question

$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. For the next 4 years. After this period the amount of money in the account would be?

Answer 1

`\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to (6)/(100)\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}`

`\bf A=20000\left(1+(0.06)/(1)\right)^(1\cdot 3)\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}`

`\bf A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to (8)/(100)\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}`

`\bf A=2382.032\left(1+(0.08)/(1)\right)^(1\cdot 4)\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762`