Compound Interest — Practice Quiz

A Algebra cheat sheet for Compound Interest — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.

Formulas & key concepts

Compound Interest Formula

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

Where: \(A\)=Amount, \(P\)=Principal, \(r\)=Rate, \(n\)=Compounds/year, \(t\)=Time

Continuously Compounded Interest

$$A = Pe^{rt}$$

Where: \(e \approx 2.718\)

Practice quiz

  1. An initial principal of $P = 1500$ is invested at an annual interest rate of $r = 4.5\%$ compounded quarterly. What will be the total amount $A$ after $t = 6$ years? Round to two decimal places.

    • $1960.12$
    • $1965.87$
    • $1971.05$
    • $1978.33$

    Answer: $1965.87$

  2. If $P = 2500$ is invested at an annual interest rate of $r = 3.8\%$ compounded continuously, what will be the total amount $A$ after $t = 8$ years? Round to two decimal places. Use $e \approx 2.71828$.

    • $3392.15$
    • $3405.88$
    • $3418.02$
    • $3425.67$

    Answer: $3405.88$

  3. You want to have $A = 10000$ in an account after $t = 5$ years. If the account pays an annual interest rate of $r = 6\%$ compounded semi-annually, what principal $P$ must you invest today? Round to two decimal places.

    • $7440.94$
    • $7472.58$
    • $7500.00$
    • $7523.12$

    Answer: $7440.94$

  4. What principal $P$ must be invested at an annual interest rate of $r = 5.2\%$ compounded continuously to reach $A = 12000$ in $t = 7$ years? Round to two decimal places.

    • $8200.50$
    • $8250.75$
    • $8300.20$
    • $8350.10$

    Answer: $8200.50$

  5. Which investment option yields a greater amount after $t = 10$ years for an initial principal of $P = 5000$?

    • Option A: $4.0\%$ annual interest compounded monthly.
    • Option B: $3.95\%$ annual interest compounded continuously.
    • Option C: Both yield approximately the same amount.
    • Option D: The difference is negligible, less than $1.

    Answer: Option A: $4.0\%$ annual interest compounded monthly.

  6. To find the time $t$ it takes for an investment of $P = 3000$ to grow to $A = 4500$ at an annual interest rate of $r = 7\%$ compounded annually, which equation should be solved?

    • $\ln(1.5) = t \ln(1.07)$
    • $4500 = 3000(1 + 0.07)^t$
    • $t = \frac{\ln(1.5)}{0.07}$
    • Both A and B are correct.

    Answer: Both A and B are correct.

  7. Approximately how many years $t$ will it take for an investment to double if it is compounded continuously at an annual interest rate of $r = 6.5\%$? Round to one decimal place.

    • $10.7$ years
    • $11.0$ years
    • $11.3$ years
    • $11.6$ years

    Answer: $10.7$ years

  8. An investment of $P = 8000$ grows to $A = 10000$ in $t = 4$ years when compounded quarterly. What is the annual interest rate $r$? Round to two decimal places.

    • $5.50\%$
    • $5.60\%$
    • $5.70\%$
    • $5.80\%$

    Answer: $5.60\%$

  9. An investment of $P = 6000$ grows to $A = 9000$ in $t = 6$ years when compounded continuously. What is the annual interest rate $r$? Round to two decimal places.

    • $6.58\%$
    • $6.67\%$
    • $6.76\%$
    • $6.85\%$

    Answer: $6.76\%$

  10. What happens to the final amount $A$ in the compound interest formula $A = P(1 + \frac{r}{n})^{nt}$ as the number of compounding periods per year $n$ increases significantly (approaches infinity)?

    • The final amount decreases.
    • The final amount remains constant.
    • The final amount approaches the value from the continuously compounded interest formula.
    • The final amount increases indefinitely.

    Answer: The final amount approaches the value from the continuously compounded interest formula.

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