In this example,

We’re looking for P₀, 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 $11,120 loan.

You will pay a total of $12,000 ($200 per month for 60 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 $12,000-$11,120 = $880 interest total.

In this example, we’re looking for d.

In this case, we’re going to have to set up the equation, and solve for d.

[latex]\begin{align}&140,000=\frac{d\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&140,000=\frac{d\left(1-{{\left(1.005\right)}^{-360}}\right)}{\left(0.005\right)}\\&140,000=d(166.792)\\&d=\frac{140,000}{166.792}=\$839.37 \\\end{align}[/latex]

You will make payments of $839.37 per month for 30 years.

You’re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 – $140,000 = $162,173.20 in interest over the life of the loan.

To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we’re looking for P₀ when

[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 $90,073.45.

First we will calculate their monthly payments.

We’re looking for d.

We set up the equation and solve for d.

[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 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 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 5 years, with 25 years remaining on the loan, will be $162,758.21.

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