Finance loan interest rate homework

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Homework Statement:
A loan of $30,000 is paid back after 6 years with a final value of $36,295. At what interest rate, compounding quarterly, has this money been invested?
Relevant Equations:
Finance solver
I got 6.33% but apparently it's 3.2%
 

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  • #2
phinds
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And are we supposed to guess how you got that? I suppose we could just speculate about where you went wrong (or didn't) but it seems like a waste of time. How about you show your work?
 
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  • #5
pasmith
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Problem Statement: A loan of $30,000 is paid back after 6 years with a final value of $36,295. At what interest rate, compounding quarterly, has this money been invested?
Relevant Equations: Finance solver

I got 6.33% but apparently it's 3.2%

If an amount [itex]A[/itex] is invested for [itex]N[/itex] years at a rate [itex]r\,\%[/itex] and compounded quarterly, then the final value is [tex]
F = A\left(1 + \frac{r}{400}\right)^{4N}.[/tex] You are given [itex]F[/itex], [itex]N[/itex] and [itex]A[/itex] and asked to solve for [itex]r[/itex].
 
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  • #6
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@paddo, what does the solver do if you enter 6 for N instead of 24? I'm thinking that N represents the number of years, not the number of payments. There is already a field for the number of payments per year (PpY). The interest rate that I get by direct calculation is a little under 3.2%.
 
  • #7
pasmith
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@paddo, what does the solver do if you enter 6 for N instead of 24? I'm thinking that N represents the number of years, not the number of payments. There is already a field for the number of payments per year (PpY). The interest rate that I get by direct calculation is a little under 3.2%.

I don't think that solver is designed to deal with the situation in the OP in any event.

I think it's designed to deal with the situation where regular repayments are made throughout the term, thereby reducing the balance on which interest is charged. The OP suggests instead a single payment at the end of the term.

If an amount [itex]P[/itex] is lent at a rate of [itex]r\,\%[/itex] to repaid by [itex]n[/itex] equal installments of [itex]A[/itex] per year over [itex]Y[/itex] years, then the balance outstanding after [itex]k+1[/itex] periods is [tex]
B_{k+1} = B_k\left(1 + \frac{r}{100n}\right) - A[/tex] on the assumption that the interest is calculated before the payment is deducted (reasonable, since it allows the lender to charge more interest). This recurrence relation can be solved subject to [itex]B_0 = P[/itex] to yield [tex]
B_k =\left(P - \frac{100nA}{r}\right)\left(1 + \frac{r}{100n}\right)^k + \frac{100nA}{r}.[/tex] Now after [itex]Yn[/itex] periods the loan should be fully repaid, so [tex]B_{Yn} =
\left(P - \frac{100nA}{r}\right)\left(1 + \frac{r}{100n}\right)^{Yn} + \frac{100nA}{r} = 0[/tex] which in terms of the total amount repaid [itex]T = nYA[/itex] is [tex]\left(P - \frac{100T}{Yr}\right)\left(1 + \frac{r}{100n}\right)^{Yn} + \frac{100T}{Yr} = 0.[/tex] This is much more difficult to solve for [itex]r[/itex] than the equation which holds where the entire amount outstanding is repaid at the end of the term:
[tex]
T - P\left(1 + \frac{r}{100n}\right)^{Yn} = 0[/tex]
 

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