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Helen borrows a sum of money from a bank at 12% convertible monthly and wishes to repay it by 24 monthly payments. In total, she will pay 584 of interest. Determine the size of the loan.

I have started by doing this:

The total amount paid back is given by $Pi(1+i)^n/(1+i)^n-1$ so the total interest would be this minus principal $P$. Given $i=0.12$ and given $n=2$. I substitute into our equation and get: $$584=((P(0.12)(1+0.12)^2)/(1+0.12)^2-1)-P$$ but I am not sure how to go farther with this, or if I am doing it right. I got $-1430.31$.

Caleb Stanford
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Tiny-E
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  • $i$ is the interest per period, here $0.01$ because the payments are monthly. $n$ is the number of periods, here $24$. I think you are missing a factor $n$ in the total amount paid back. – Ross Millikan Dec 07 '17 at 02:59
  • so if I insert i-0.01 instead and I use n=24. will that get me the correct answer? Anything else I need to do? Am I using the right equation even? @RossMillikan – Tiny-E Dec 07 '17 at 03:04
  • I believe that equation is the individual payment, not the total. That is where the $n$ factor comes in. Aside from the $n$ you did it for two annual payments, not $24$ monthly ones. That shouldn't make much difference, but the loan should not be negative. – Ross Millikan Dec 07 '17 at 03:05

1 Answers1

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The monthly interest rate is $i=\frac{i^{(12)}}{12}=\frac{12\%}{12}=1\%$.

The total interest is $I=nP-L$, where $P$ is the monthly payment, $n$ is the number of months and $L$ is the loan.

Then $$\left\{ \begin{align} I&=P(n-a_{\overline{n}|i})\\ L&=Pa_{\overline{n}|i} \end{align}\right.\qquad \Longrightarrow\quad \boxed{L=I\cdot\frac{a_{\overline{n}|i}}{n-a_{\overline{n}|i}}=584\cdot\frac{a_{\overline{24}|1\%}}{24-a_{\overline{24}|1\%}}\approx 4,500.5} $$ where $a_{\overline{n}|i}=\frac{1-(1+i)^{-n}}{i}$.

We can also find $P=\frac{L}{a_{\overline{n}|i}}=\frac{4,500.5}{21.24}\approx 211.85$.

alexjo
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