In this example,

[latex]d = $200[/latex]	the monthly loan payment
[latex]r= 0.03[/latex]	[latex]3\%[/latex] annual rate
[latex]n = 12[/latex]	since we’re doing monthly payments, we’ll compound monthly
[latex]t = 5[/latex]	since we’re making monthly payments for [latex]5[/latex] years

 

We’re looking for [latex]P_0[/latex], the starting amount of the loan.

[latex]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\\end{align}[/latex]

You can afford a [latex]$11,120[/latex] loan.

You will pay a total of [latex]$12,000[/latex] ([latex]$200[/latex] per month for [latex]60[/latex] months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying [latex]$12,000-$11,120 = $880[/latex] interest total.

Details of this example are examined in this video.

You can view the transcript for “Car loan” here (opens in new window).

Note: This video uses [latex]k[/latex] for [latex]n[/latex] and [latex]N[/latex] for [latex]t[/latex].

To determine this, we are looking for the amount of the loan that can be paid off by [latex]$1,000[/latex] a month payments in [latex]10[/latex] years. In other words, we’re looking for [latex]P_0[/latex] when

[latex]d = $1,000[/latex]	the monthly loan payment
[latex]r= 0.06[/latex]	[latex]6\%[/latex] annual rate
[latex]n = 12[/latex]	since we’re doing monthly payments, we’ll compound monthly
[latex]t = 10[/latex]	since we’re making monthly payments for [latex]10[/latex] more years

[latex]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-10(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\left(1-{{\left(1.005\right)}^{-120}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\left(1-0.5496\right)}{\left(0.005\right)}=\$90,073.45 \\\end{align}[/latex]

The loan balance with 10 years remaining on the loan will be [latex]$90,073.45[/latex].

This example is explained in the following video:

You can view the transcript for “Calculate remaining balance on loan from payment” here (opens in new window).

Note: This video uses [latex]k[/latex] for [latex]n[/latex] and [latex]N[/latex] for [latex]t[/latex].

First we will calculate their monthly payments. We’re looking for [latex]d[/latex].

[latex]r = 0.04[/latex]	[latex]4\%[/latex] annual rate
[latex]n = 12[/latex]	since they’re paying monthly
[latex]t = 30[/latex]	[latex]30[/latex] years
[latex]P_0 = $180,000[/latex]	the starting loan amount

 

We set up the equation and solve for [latex]d[/latex].

[latex]\begin{align}&180,000=\frac{d\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&180,000=\frac{d\left(1-{{\left(1.00333\right)}^{-360}}\right)}{\left(0.00333\right)}\\&180,000=d(209.562)\\&d=\frac{180,000}{209.562}=\$858.93 \\\end{align}[/latex]

Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after [latex]5[/latex] years, when [latex]25[/latex] years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those [latex]25[/latex] years.

[latex]d = $858.93[/latex]	the monthly loan payment we calculated above
[latex]r= 0.04[/latex]	[latex]4\%[/latex] annual rate
[latex]n = 12[/latex]	since they’re doing monthly payments
[latex]t = 25[/latex]	since they’d be making monthly payments for [latex]25[/latex] more years

[latex]\begin{align}&{{P}_{0}}=\frac{858.93\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-25(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&{{P}_{0}}=\frac{858.93\left(1-{{\left(1.00333\right)}^{-300}}\right)}{\left(0.00333\right)}\\&{{P}_{0}}=\frac{858.93\left(1-0.369\right)}{\left(0.00333\right)}=\$162,758.21 \\\end{align}[/latex]

The loan balance after [latex]5[/latex] years, with [latex]25[/latex] years remaining on the loan, will be [latex]$162,758.21[/latex].

Over that [latex]5[/latex] years, the couple has paid off [latex]$180,000 - $162,758.21 = $17,241.79[/latex] of the loan balance. They have paid a total of [latex]$858.93[/latex] a month for [latex]5[/latex] years ([latex]60[/latex] months), for a total of [latex]$51,535.80[/latex], so [latex]$51,535.80 - $17,241.79 = $34,294.01[/latex] of what they have paid so far has been interest.

[latex]\begin{array}{r@{\hfill}l}
d=\$30 && \text{The monthly payments} \\
r=0.12 && \text{12% annual rate} \\
n=12 && \text{since we're making monthly payments} \\
P_0=\$1,000 && \text{we're starting with a } \$1,000 \text{ loan} \\
\end{array}[/latex]

We are solving for [latex]t[/latex], the time to pay off the loan.

[latex]1000=\frac{30\left(1-\left(1+\frac{0.12}{12}\right)^{-N*12}\right)}{\frac{0.12}{12}}[/latex]

Solving for [latex]t[/latex] gives [latex]3.396[/latex]. It will take about [latex]3.4[/latex] years to pay off the purchase.