Question

5. Find the (a) amount (b) compound interest at compounded interest that N800 would be in 2 years at 5% at the end of first year and 10% at the end of second year.

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Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$800\\ r=rate\to 5\%\to (5)/(100)\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}`

`A = 800\left(1+(0.05)/(1)\right)^(1\cdot 1) \implies A = 800( 1.05)^(1)\implies A = 840\qquad \textit{end of 1st year} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$840\\ r=rate\to 10\%\to (10)/(100)\dotfill &0.10\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}`

`A = 840\left(1+(0.10)/(1)\right)^(1\cdot 1) \implies A = 840( 1.10)^(1)\implies A = 924\qquad \textit{end of 2nd year}`

so hmm, if we want just the interest that applied to that period, well, is simply the Accumulated - Principal.

For the 1st year that'd be 840 - 800 = 40 bucks in interest only.

For the 2nd year is just 924 - 840 = 84 bucks in interest only.