Finite Mathematics Quiz: Compound Interest

14 questions · exam conditions

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Compound InterestQuestion 1 of 14

A savings account compounds interest continuously at 3.6% annual rate. Another account compounds quarterly at what annual rate to produce the same effective annual yield?

3.655%

3.661%

3.674%

3.682%

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Finite Mathematics Quiz

Finite Mathematics Quiz: Compound Interest

Practice Compound Interest in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Compound Interest, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A savings account compounds interest continuously at 3.6% annual rate. Another account compounds quarterly at what annual rate to produce the same effective annual yield?

3.655%
3.661% (correct answer)
3.674%
3.682%

Explanation: The effective annual rate for continuous compounding at 3.6% is e^0.036 - 1 ≈ 0.036656 or 3.6656%. For quarterly compounding to have the same effective rate: (1 + r/4)^4 = 1.036656, so 1 + r/4 = (1.036656)^(1/4) ≈ 1.009154, giving r/4 ≈ 0.009154, so r ≈ 0.03661 or 3.661%. Choice A uses an approximation formula. Choice C uses wrong conversion. Choice D assumes monthly instead of quarterly.

Question 2

An account pays 6% annual interest. If $1,000 is invested with quarterly compounding for the first 2 years, then the entire balance is moved to monthly compounding for the next 3 years, what is the final account balance?

$1,343.92 (correct answer)
$1,348.18
$1,352.55
$1,357.91

Explanation: After 2 years with quarterly compounding: A₁ = $1,000(1 + 0.06/4)^8 = $1,000(1.015)^8 ≈ $1,126.49. This amount then compounds monthly for 3 years: A₂ = $1,126.49(1 + 0.06/12)^36 = $1,126.49(1.005)^36 ≈ $1,343.92. Choice B uses 5.9% rate incorrectly. Choice C compounds quarterly for all 5 years. Choice D uses continuous compounding for the second period.

Question 3

Two investment accounts offer the same annual percentage rate of 4.8%. Account A compounds monthly, while Account B compounds daily (assume 365 days per year). If you invest $15,000 in each account, what is the difference in their values after exactly 2 years, with Account B's value minus Account A's value?

$2.47 (correct answer)
$3.12
$4.89
$5.21

Explanation: Account A: A = 15000(1 + 0.048/12)^(12×2) = 15000(1.004)^24 ≈ $16,509.84. Account B: A = 15000(1 + 0.048/365)^(365×2) = 15000(1 + 0.048/365)^730 ≈ $16,512.31. The difference is $16,512.31 - $16,509.84 = $2.47. Choice B incorrectly uses 360 days. Choice C uses wrong compounding periods. Choice D compounds one account incorrectly.

Question 4

An investor is considering two five-year investment options, both with an initial principal of $10,000. Option X offers a 4.0% nominal annual interest rate compounded semi-annually. Option Y offers a 3.9% nominal annual interest rate compounded quarterly. What is the absolute difference between the final values of these two investments after five years?

$11.96
$18.79
$41.28 (correct answer)
$101.34

Explanation: To solve this, we must calculate the future value (A) for each option using the formula

A = P(1 + r/n)^{nt}

and then find the difference.\nFor Option X:

P = 10000

r = 0.04

n = 2

t = 5

. So,

A_X = 10000(1 + 0.04/2)^{2*5} = 10000(1.02)^{10} \approx \

12,189.94

.\nFor Option Y:

P = 10000

,

r = 0.039

,

n = 4

,

t = 5

. So,

A_Y = 10000(1 + 0.039/4)^{4*5} = 10000(1.00975)^{20} \approx $12,148.66

.\nThe absolute difference is

|A_X - A_Y| = |$12,189.94 - $12,148.66| = $41.28$.

Question 5

An investment firm advertises two savings plans. Plan A offers a 6.20% nominal annual rate compounded monthly. Plan B offers a 6.25% nominal annual rate compounded semi-annually. Which statement correctly compares the effective annual rates (APY) of the two plans?

Plan B has a higher APY, exceeding Plan A's APY by 0.050%.
Plan A has a higher APY, exceeding Plan B's APY by 0.032%. (correct answer)
Plan B has a higher APY, exceeding Plan A's APY by 0.018%.
The APYs are effectively equal, with a difference of less than 0.001%.

Explanation: The effective annual rate (APY) is calculated by the formula

APY = (1 + r/n)^n - 1

.\nFor Plan A:

r = 0.062

n = 12

APY_A = (1 + 0.062/12)^{12} - 1 \approx (1.0051667)^{12} - 1 \approx 1.06380 - 1 = 6.380\%

.\nFor Plan B:

r = 0.0625

n = 2

APY_B = (1 + 0.0625/2)^2 - 1 = (1.03125)^2 - 1 = 1.06348 - 1 = 6.348\%

.\nComparing the two, Plan A has a higher APY. The difference is

6.380\% - 6.348\% = 0.032\%

Question 6

A principal of $5,000 is invested in an account with a nominal annual interest rate of 4.8%. For the first 3 years, the interest is compounded quarterly. For the next 4 years, the interest is compounded monthly. What is the total value of the investment after the full 7 years?

$6,877.05
$6,978.86 (correct answer)
$7,000.95
$7,003.55

Explanation: This is a two-step calculation. First, find the value after the first 3 years with quarterly compounding.\nStep 1:

P = 5000

r = 0.048

n = 4

t = 3

. The value is

A_1 = 5000(1 + 0.048/4)^{4*3} = 5000(1.012)^{12} \approx \

5,769.34

.\nStep 2: This amount,

A_1

, becomes the principal for the next 4 years with monthly compounding.

P = 5769.34

,

r = 0.048

,

n = 12

,

t = 4

. The final value is

A_2 = 5769.34(1 + 0.048/12)^{12*4} = 5769.34(1.004)^{48} \approx $6,978.86$.\nDistractor C calculates the value assuming quarterly compounding for all 7 years. Distractor D assumes monthly compounding for all 7 years. Distractor A incorrectly uses simple interest for the second period.

Question 7

Bank A offers a savings account with a 5.10% nominal annual rate compounded daily (using 365 days/year). Bank B offers an account with a 5.08% nominal annual rate compounded continuously. Which bank offers a better return, and by approximately what percentage difference in their effective annual yields (APY)?

Bank A, by approximately 0.021%. (correct answer)
Bank B, by approximately 0.020%.
Bank A, by approximately 0.020%, which is the difference in their nominal rates.
Bank B, as continuous compounding always yields more than daily compounding.

Explanation: We must compare the effective annual yields (APY). For Bank A (daily compounding),

APY_A = (1 + r/n)^n - 1 = (1 + 0.051/365)^{365} - 1 \approx 1.05232 - 1 = 0.05232

5.232\%

. For Bank B (continuous compounding),

APY_B = e^r - 1 = e^{0.0508} - 1 \approx 1.05211 - 1 = 0.05211

5.211\%

. Bank A's APY is higher. The difference is

5.232\% - 5.211\% = 0.021\%

. Distractor D states a common misconception; while continuous compounding is the limit of discrete compounding for the same nominal rate, a different (lower) rate may result in a lower yield.

Question 8

An initial principal of $20,000 is invested in an account with a 6% nominal annual interest rate, compounded quarterly. How much interest is earned specifically during the fourth year of the investment?

$1,200.00
$1,434.74
$1,482.34 (correct answer)
$5,394.70

Explanation: To find the interest earned during the fourth year, we must calculate the account balance at the end of year 4 and subtract the balance at the end of year 3.\nBalance at end of year 3 (

t=3

A_3 = 20000(1 + 0.06/4)^{4*3} = 20000(1.015)^{12} \approx \

23,912.35

.\nBalance at end of year 4 (

t=4

):

A_4 = 20000(1 + 0.06/4)^{4*4} = 20000(1.015)^{16} \approx $25,394.70

.\nInterest earned in the fourth year =

A_4 - A_3 = $25,394.70 - $23,912.35 = $1,482.35

. The closest answer is

1,482.34

.\nDistractor A is simple interest on the original principal for one year. Distractor B is simple interest on the balance at the start of year 4 (

A_3 \times 0.06$). Distractor D is the total interest earned over all four years.

Question 9

An investment of $15,000 is made for a period of 42 months in an account that pays a 3.6% nominal annual interest rate, compounded semi-annually. What is the total amount of interest earned over the entire period?

$1,701.81
$1,962.37
$1,988.66 (correct answer)
$2,293.42

Explanation: First, convert the time period from months to years:

t = 42

months /

12

months/year

= 3.5

years. The interest is compounded semi-annually, so

n=2

. The rate is

r=0.036

. The total number of compounding periods is

nt = 2 \times 3.5 = 7

. The future value is

A = 15000(1 + 0.036/2)^{7} = 15000(1.018)^7 \approx \

16,988.66

. The interest earned is the future value minus the principal:

I = A - P = $16,988.66 - $15,000 = $1,988.66

. Distractor B miscalculates by using annual compounding (

n=1$) with a fractional exponent. Distractors A and D result from incorrectly rounding the time to 3 and 4 years, respectively.

Question 10

A person takes out a $20,000 loan for 5 years at a nominal annual interest rate of 8.4%. How much more total interest is paid over the life of the loan if the interest is compounded monthly versus compounded quarterly?

$158.17 (correct answer)
$280.00
$10,252.37
$10,410.54

Explanation: We must calculate the total amount owed (

A

) for each compounding frequency and then find the difference in the interest paid (

A-P

).\nFor monthly compounding (

n=12

A_{monthly} = 20000(1 + 0.084/12)^{12*5} = 20000(1.007)^{60} \approx \

30,410.54

. Interest paid is

10,410.54.\nFor quarterly compounding (

n=4

A_{quarterly} = 20000(1 + 0.084/4)^{4*5} = 20000(1.021)^{20} \approx \

30,252.37

. Interest paid is $10,252.37.\nThe difference in interest paid is $\

10,410.54 - $10,252.37 = $158.17$. Distractors C and D represent the total interest paid for quarterly and monthly compounding, respectively, not the difference between them.

Question 11

An individual needs to have a balance of exactly $25,000 in a savings account in 8 years. The account offers a nominal annual interest rate of 5.4%, compounded monthly. Assuming no other deposits or withdrawals are made, what is the principal amount that must be invested today to achieve this goal?

$16,248.55 (correct answer)
$16,383.34
$17,458.10
$24,119.89

Explanation: This requires solving for the principal,

P

, in the compound interest formula

A = P(1 + r/n)^{nt}

. We can rearrange this to

P = A / (1 + r/n)^{nt}

.\nGiven:

A = 25000

r = 0.054

n = 12

, and

t = 8

. The total number of compounding periods is

nt = 12 * 8 = 96

.\nThe periodic interest rate is

r/n = 0.054 / 12 = 0.0045

.\nSo,

P = 25000 / (1 + 0.0045)^{96} = 25000 / (1.0045)^{96} \approx 25000 / 1.538615 \approx \

16,248.55

.\nDistractor B uses annual compounding (

n=1

). Distractor C uses the simple interest formula. Distractor D incorrectly uses

t=8

instead of

nt=96$ in the exponent.

Question 12

An investor wants to grow an initial sum of $7,000 to a future value of $10,000 over a period of 6 years. If the investment compounds interest monthly, what nominal annual interest rate is required to achieve this goal?

0.50%
5.97% (correct answer)
6.12%
7.14%

Explanation: We use the formula

A = P(1 + r/n)^{nt}

and solve for

r

. Given

A=10000, P=7000, t=6, n=12

.\n

10000 = 7000(1 + r/12)^{12*6}

10/7 = (1 + r/12)^{72}

\nTo isolate

r

, we first take the 72nd root of both sides:

(10/7)^{1/72} = 1 + r/12

. Then,

(10/7)^{1/72} - 1 = r/12

. Finally,

r = 12 \left[ (10/7)^{1/72} - 1 \right]

.\nCalculating the value:

r \approx 12 \left[ (1.42857)^{0.013889} - 1 \right] \approx 12[1.004975 - 1] = 12(0.004975) \approx 0.0597

. This is a rate of 5.97%.\nDistractor A forgets to multiply by

n=12

at the end. Distractor C uses annual compounding (

n=1

). Distractor D is the rate required if simple interest were used.

Question 13

Approximately how many years will it take for an initial investment of $8,000 to grow to at least $12,000 if it is invested in an account that pays a 5% nominal annual interest rate compounded quarterly? Round your answer to the nearest tenth of a year.

8.0 years
8.2 years (correct answer)
10.0 years
32.6 years

Explanation: We use the formula

A = P(1 + r/n)^{nt}

and solve for

t

. Here,

A=12000

P=8000

r=0.05

, and

n=4

.\n

12000 = 8000(1 + 0.05/4)^{4t}

1.5 = (1.0125)^{4t}

\nTo solve for

t

, we use logarithms:

\ln(1.5) = \ln((1.0125)^{4t})

\ln(1.5) = 4t \cdot \ln(1.0125)

t = \ln(1.5) / (4 \cdot \ln(1.0125)) \approx 0.405465 / (4 \cdot 0.0124225) \approx 0.405465 / 0.04969 \approx 8.16

years. Rounded to the nearest tenth, this is 8.2 years.\nDistractor C is the result of using simple interest. Distractor D is the number of quarters (

nt

), not years, a common error made by forgetting to divide by

n=4

Question 14

An amount $P$ is invested for $t$ years at a nominal annual rate $r$ . Let $A(n)$ be the future value if the interest is compounded $n$ times per year. Assume $P, r,$ and $t$ are positive constants. Which statement best describes the relationship between the number of compounding periods $n$ and the future value $A(n)$ ?

$A(n)$ increases as $n$ increases, and its value approaches a specific finite limit as $n$ becomes very large. (correct answer)
$A(n)$ increases as $n$ increases, and its value will grow infinitely large as $n$ becomes very large.
$A(n)$ is directly proportional to $n$ , so doubling the compounding frequency doubles the interest earned.
$A(n)$ is independent of $n$ because the nominal annual rate $r$ is held constant for all compounding frequencies.

Explanation: The formula for the future value is

A(n) = P(1 + r/n)^{nt}

. As

n

(the number of compounding periods per year) increases, the value of

A(n)

also increases because interest is being calculated on previously earned interest more often. However, this growth is not limitless. As

n

approaches infinity, the expression

(1 + r/n)^{nt}

approaches

e^{rt}

. Therefore, the future value

A(n)

approaches the finite limit

Pe^{rt}

, which is the formula for continuously compounded interest. Distractor B is incorrect because the value is bounded. Distractor C is incorrect because the relationship is not linear. Distractor D is incorrect because the compounding frequency is a key factor in the formula and affects the effective interest rate.