// Numbas version: finer_feedback_settings {"name": "cash flow amount - ordinary annuity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "functions": {}, "statement": "
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", "extensions": [], "tags": [], "rulesets": {}, "parts": [{"scripts": {}, "prompt": "Suppose you borrow $\\$\\var{P}$ to buy a house. The term of the loan is $\\var{years}$ years, and you will need to make {period[0]} repayments on the loan at the end of each {period[2]}. Interest is $\\var{ipa}\\%$ per annum compounding {period[0]}. Calculate the size of your monthly repayment.
\n\n$\\$$ [[0]] (to the nearest cent)
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\n$\\displaystyle P=\\frac{C}{i}\\left(1-\\frac{1}{(1+i)^n}\\right)$
\nwhere $P$ is the present value, $C$ is the cash flow per period, $i$ is the interest rate per period, and $n$ is the number of periods.
\nIn our situation we have,
\n$P=\\var{P}$,
\n$i=\\frac{\\var{ipa}\\%}{\\var{period[1]}}=\\frac{\\var{ipadec}}{\\var{period[1]}}$,
\n$n=\\var{years}\\times \\var{period[1]}=\\var{n}$,
\nand therefore we have
\n$\\displaystyle \\var{P}=\\frac{C}{\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)}\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^\\var{n}}\\right)$
\nwhich we need to rearrange to solve for $C$.
\n\nWe want to get $C$ by itself. We start by multiplying both sides by $\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$ (to remove the division of $C$ by it on the right hand side)
\n$\\displaystyle \\var{P}\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)=C\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^\\var{n}}\\right)$
\nNow we will divide both sides by $\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^\\var{n}}\\right)$ (to remove the multiplication by it on $C$)
\n$\\displaystyle \\frac{\\var{P}\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)}{\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^\\var{n}}\\right)}=C$
\nCalculating this we find
\n$\\begin{align}C&\\approx \\var{C}\\\\&=\\$\\var{Crounded}\\quad \\text{(to the nearest cent)}\\end{align}$
", "type": "question", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}