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How much would you need to put away at $\\var{ipa}\\%$ per annum compounded {period[0]} in order to have $\\$\\var{S}$ after $\\var{years}$ years?

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$\\$\\,$[[0]] (to the nearest cent)

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interest per period, only use for debugging, use fractions in display and calculations.

", "definition": "ipa/period[1]", "group": "Ungrouped variables", "name": "ipp"}, "P": {"templateType": "anything", "description": "

present value

", "definition": "S/(1+ippdec)^n", "group": "Ungrouped variables", "name": "P"}, "ipa": {"templateType": "anything", "description": "

interest per annum as a percentage (add the symbol afterwards)

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We are asked to find the present value using compound interest. Therefore we will use the equation

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$\\displaystyle P=\\frac{S}{(1+i)^n}$,

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where $S$ is the future value, $P$ is the present value, $i$ is the interest rate per time period and $n$ is the number of time periods.

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In our situation we have,

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$S=\\$\\var{S}$,

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$i=\\var{ipa}\\%=\\var{ipadec}$,

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$n=\\var{n}$

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and therefore

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$\\begin{align}P&=\\simplify[simplifyFractions, unitDenominator]{{S}/(1+{ipadec})^{n}}\\\\&=\\$ \\var{Prounded} \\text{        (to the nearest cent)}\\end{align}$ 

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