Smith borrows 27,000 to purchase a new car. The car dealer finances the purchase with a loan that will require level monthly payments at the end of each month for 4 years, starting at the end of the month in which the car is purchased (assume the car is purchased on the 1st of the month). The loan has 0% interest rate for the first year followed by 5% annual nominal interest rate, compounded monthly, for the following three years. Find the outstanding balance on the loan at the end of the first year.
I thought I knew how to do this, but I got it wrong, so can I get corrected on what's going wrong please? Here is what I've got: 27,000 is the present value of all of the payments together. The payment is always the same amount, so I'll call that $x$. The first year has no interest to account for, so the outstanding balance is just $27,000 – 12*x$. I think I can solve for $x$ by creating an equation where the loan amount is equal to the first 12 payments plus the present value of the following three years of payments, sent back a year with the discount factor.
$27,000 = 12x + x*\frac{1-\frac{1}{1+.05/12}^{12*3}}{.05/12}*\frac{1}{.05/12}^{12}$
$x\approx 617.258$
$27,000-617.258*12 \approx 19592.89$
Best Answer
The interest is only ever charged on the remaining balance, which complicates things somewhat.
I'm going to assume that every month from the second year onward, the balance shifts by $b_i \to b_i - x$, and then an interest charge of $(0.05/12)(b_i-x)$ is added. Hence $b_{i+1} = (1+0.05/12)(b_i-x)$.
You can solve $b_{i+1} = c(b_i - x)$ to find $$b_n = b_0 c^n - \frac{c(c^n-1)x}{c-1}$$ (For example, you could write $$b_n = c(b_{n-1} - x) = c^2 b_{n-2} - cx - c^2x = c^n b_0 - cx(1 + c + c^2 + \cdots + c^{n-1})$$ and sum the geometric series.)
Substituting in $b_{36} = 0$ and $b_0 = 27000-12x$, we can solve for $x$. I'll let you finish it off.
I get the numerical answer for monthly repayments $x\approx593.35$ with the above assumptions. Two possible variations: you could use the $(1+0.05)^{1/12}$ formula instead for the interest rate; and you could apply interest before making a repayment.
The former only changes the value of $c$. ($x\approx592.658$.) The latter changes $cx\to x$ in the solution. ($x\approx595.16$ with the $c$ value above or $x\approx594.44$ with the alternate one. These are the values of Zhuoran He.)