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24 votes
24 votes
The mortgage was a personal, amortized loan for $91,000, at an interest rate of 3.75%, with monthly payments for a term of 30 years. Find Ahmad's monthly payment

User Anit Kumar
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1 Answer

12 votes
12 votes


~~~~~~~~~~~~ \textit{Amortized Loan Value} \\\\ pymt=P\left[ \cfrac{(r)/(n)}{1-\left( 1+ (r)/(n)\right)^(-nt)} \right]\implies pymt=P\left[ \cfrac{(r)/(n)}{1-\left((n)/(n+r)\right)^(nt)} \right]


\hspace{10em} \begin{cases} P= \begin{array}{llll} \textit{original amount deposited}\\ \end{array}\dotfill & \begin{array}{llll} 91000 \end{array}\\ pymt=\textit{periodic payments}\\ r=rate\to 3.75\%\to (3.75)/(100)\dotfill &0.0375\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &30 \end{cases}


pymt=91000\left[ \cfrac{(0.0375)/(12)}{1-\left((12)/(12+0.0375)\right)^(12\cdot 30)} \right] \implies pymt=91000\left[ \cfrac{0.003125}{1-\left((12)/(12.0375)\right)^(360)} \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill pymt\approx 421.44~\hfill

User Jaquelina
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