Loans and Mortgages (Amortization) — Hard Practice Quiz

A Financial Math cheat sheet for Loans and Mortgages (Amortization) — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Formulas & key concepts

The level periodic payment that fully repays a loan or mortgage of principal \(P\). \(i = r/n\) is the periodic rate and \(N = nt\) the total number of payments.

$$PMT = \dfrac{P \times i}{1 - (1 + i)^{-N}} \quad\text{or}\quad PMT = \dfrac{P \times \frac{r}{n}}{1 - \left(1 + \frac{r}{n}\right)^{-nt}}$$

The loan amount that corresponds to a known periodic payment \(PMT\), where \(i = r/n\) is the periodic rate and \(N = nt\) the total number of payments.

$$P = PMT \times \dfrac{1 - (1 + i)^{-N}}{i} \quad\text{or}\quad P = PMT \times \dfrac{1 - \left(1 + \frac{r}{n}\right)^{-nt}}{\frac{r}{n}}$$

The outstanding balance after \(k\) payments have been made, with \(N - k\) payments still remaining. \(i = r/n\) is the periodic rate and \(N = nt\) the total number of payments.

$$B_{k} = PMT \times \dfrac{1 - (1 + i)^{-(N - k)}}{i} \quad\text{or}\quad B_{k} = PMT \times \dfrac{1 - \left(1 + \frac{r}{n}\right)^{-(nt - k)}}{\frac{r}{n}}$$

Practice quiz

  1. A homeowner takes out a \$300,000 mortgage at an annual interest rate of $4.5\%$ compounded monthly over $30$ years. What is the total amount of interest paid over the life of the loan? Round to the nearest cent.

    • \$247,220.13
    • \$405,000.00
    • \$547,220.13
    • \$300,000.00

    Answer: \$247,220.13

  2. Consider a loan with principal $P$, periodic interest rate $i$, and total number of payments $N$. If the total number of payments $N$ is doubled, while $P$ and $i$ remain constant, which of the following statements is true regarding the periodic payment $PMT$ and the total interest paid over the life of the loan?

    • The $PMT$ will be less than half, and the total interest paid will more than double.
    • The $PMT$ will be exactly half, and the total interest paid will exactly double.
    • The $PMT$ will be more than half, and the total interest paid will more than double.
    • The $PMT$ will be more than half, and the total interest paid will be less than double.

    Answer: The $PMT$ will be more than half, and the total interest paid will more than double.

  3. A loan of \$250,000 is taken out at an annual interest rate of $6\%$ compounded monthly for $15$ years. After $5$ years, the borrower decides to refinance the remaining balance at a new annual interest rate of $4\%$ compounded monthly for $10$ years. What will be the new monthly payment? Round to the nearest cent.

    • \$1,923.89
    • \$2,531.13
    • \$2,109.64
    • \$1,405.58

    Answer: \$1,923.89

  4. A borrower takes out a loan of \$150,000 at an annual interest rate of $5\%$ compounded monthly. They can afford to pay \$1,200 per month. How many full months will it take to pay off the loan? Round to the nearest whole month.

    • 177 months
    • 195 months
    • -177 months
    • 125 months

    Answer: 177 months

  5. A loan has a principal $P$, periodic interest rate $i$, and total number of payments $N$. If the principal $P$ is doubled, and the periodic interest rate $i$ and the total number of payments $N$ remain constant, how does the periodic payment $PMT$ change?

    • The $PMT$ will double.
    • The $PMT$ will more than double due to compounding effects.
    • The $PMT$ will less than double.
    • The $PMT$ will remain the same, but the loan will be paid off faster.

    Answer: The $PMT$ will double.

  6. A loan of \$200,000 is taken out at an annual interest rate of $4.8\%$ compounded monthly for $20$ years. The borrower decides to pay an extra \$100 each month in addition to the calculated monthly payment. How many months earlier will the loan be paid off compared to the original schedule? Round to the nearest whole month.

    • 27 months
    • 213 months
    • 97 months
    • 143 months

    Answer: 27 months

  7. A loan of \$180,000 is taken out at an annual interest rate of $3.6\%$ compounded monthly for $25$ years. What is the total interest paid during the $10^{th}$ year of the loan? Round to the nearest cent.

    • \$4,677.85
    • \$6,405.84
    • \$93,241.46
    • \$6,251.81

    Answer: \$4,677.85

  8. For a fully amortizing loan, which of the following statements is true regarding the impact of a higher periodic interest rate $i$ (assuming $P$ and $N$ are fixed and $PMT$ is calculated)?

    • A higher $i$ causes a larger portion of early payments to go towards principal, accelerating the principal reduction.
    • A higher $i$ causes a larger portion of early payments to go towards interest, slowing down the principal reduction.
    • A higher $i$ has no effect on the proportion of $PMT$ allocated to principal versus interest, only on the total $PMT$.
    • A higher $i$ causes the loan to be paid off faster if $PMT$ remains constant.

    Answer: A higher $i$ causes a larger portion of early payments to go towards interest, slowing down the principal reduction.

  9. A person is considering two loan options for \$220,000:\nOption A: $30$-year loan at an annual interest rate of $4.2\%$ compounded monthly.\nOption B: $15$-year loan at an annual interest rate of $3.8\%$ compounded monthly.\nHow much more total interest will be paid with Option A compared to Option B? Round to the nearest cent.

    • \$98,338.22
    • \$-529.51
    • \$108,172.04
    • \$-98,338.22

    Answer: \$98,338.22

  10. Given the formula for the outstanding balance after $k$ payments, $B_k = PMT \times \frac{1 - (1 + i)^{-(N - k)}}{i}$, and the formula for the periodic payment, $PMT = \frac{P \times i}{1 - (1 + i)^{-N}}$, derive an expression for the original principal $P$ in terms of $B_k$, $i$, $N$, and $k$.

    • $P = B_k \times \frac{1 - (1 + i)^{-N}}{1 - (1 + i)^{-(N - k)}}$
    • $P = B_k \times \frac{1 - (1 + i)^{-(N - k)}}{1 - (1 + i)^{-N}}$
    • $P = B_k \times \frac{i}{1 - (1 + i)^{-N}}$
    • $P = B_k \times \frac{1 - (1 + i)^{-N}}{i}$

    Answer: $P = B_k \times \frac{1 - (1 + i)^{-N}}{1 - (1 + i)^{-(N - k)}}$

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