§-Syllabus dot point

NSWMaths Standard 2Syllabus dot point

How is compound interest calculated, and how do compounding frequency and time affect investment growth?

Use the compound interest formula to find future values, present values, interest rates and time periods for investments

A focused answer to the HSC Maths Standard 2 dot point on compound interest. The formula, conversion between annual and per-period rates, present and future values, the effect of compounding frequency, the recurrence-versus-formula comparison, and worked examples using current Australian bank rates.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to use the compound interest formula on the NESA reference sheet. You need to switch between an annual interest rate and a per-period rate (the rate for one compounding step), and to solve for any one of $FV$ , $PV$ , $r$ or $n$ when you are given the others. You also have to compare scenarios with different compounding frequencies. Finally, recognise that the same investment can be written two ways: as a year-by-year recurrence (each balance built from the one before) or as the closed-form formula (one step straight to the answer).

The answer

The whole topic rests on one idea: with compound interest, each period's interest is added to the balance and then earns interest itself in every following period. Simple interest pays only on the original principal, so it grows in a straight line. Compound interest pays interest on interest, so it grows as a curve that bends upward and, given enough time, leaves simple interest far behind. The chart below makes the gap concrete.

The compound interest formula

From the NESA reference sheet:

FV = PV (1 + r)^n.

$FV$ is future value (the value after $n$ periods)
$PV$ is present value (the amount invested today)
$r$ is the interest rate per compounding period (as a decimal)
$n$ is the number of compounding periods

The factor $(1 + r)$ is the per-period multiplier (what you multiply the balance by each step). Multiply by it once and the balance moves forward one period. Raise it to the power $n$ and the balance moves forward $n$ periods in one go. That is why the formula uses a power: multiplying by $(1 + r)$ a total of $n$ times is the same as $(1 + r)^n$ .

Per-period rate

Rates are usually quoted as a nominal annual rate, but interest may compound more often than annually. Convert before applying the formula.

Compounding	Per-period rate $r$	Periods in $t$ years $n$
Annually	$R$	$t$
Semi-annually	$R/2$	$2t$
Quarterly	$R/4$	$4t$
Monthly	$R/12$	$12t$
Weekly	$R/52$	$52t$
Daily	$R/365$	$365t$

Where $R$ is the nominal annual rate as a decimal. The rule of thumb is symmetric: whatever number you divide the rate by, you multiply the years by the same number. Compounding more often divides the rate into smaller pieces but applies it more times.

The recurrence and the formula are the same thing

A compound investment can be described two ways, and a question may set up either.

Recurrence (year by year). Let $A_n$ be the balance after $n$ periods. Then $A_n = A_{n-1}(1 + r)$ with $A_0 = PV$ . Each step multiplies the previous balance by $(1 + r)$ .
Closed form (one step). $A_n = PV(1 + r)^n$ .

They agree exactly. Applying the recurrence $n$ times means multiplying by $(1 + r)$ a total of $n$ times, which is $(1 + r)^n$ . Use the recurrence when a question asks for a table of the first few years, or gives the rule in $A_{n} = A_{n-1}(1+r)$ form. Use the closed form when you just need the balance after many periods. The recurrence is also how a spreadsheet or financial calculator builds the figures, so it helps to keep both views in mind.

Present value

The present value $PV$ is the amount you must invest today to grow to $FV$ in $n$ periods. Rearranging:

PV = \frac{FV}{(1 + r)^n}.

Discounting a future amount back to the present (working out what it is worth today) is just compounding in reverse. A dollar in the future is worth less than a dollar today, because today's dollar could be invested and grow. Dividing by $(1 + r)^n$ is exactly that discount.

Solving for the rate or time

To find the time:

n = \frac{\log(FV / PV)}{\log(1 + r)}.

To find the rate:

r = \left(\frac{FV}{PV}\right)^{1/n} - 1.

Either $\log$ or $\ln$ works here. The unknown is up in the exponent, and because the answer is a ratio of two logs, it does not matter which base of log you choose: it cancels out. When a question asks "how long until the investment first exceeds" a target, solve for $n$ and then round up to the next whole period. The balance only clears the target at the end of that period.

Effect of compounding frequency

For a fixed nominal rate, more frequent compounding gives a slightly higher effective rate. A nominal $6\%$ compounded:

Annually: $1.06$ , effective $6.00\%$ .
Quarterly: $(1.015)^4 \approx 1.0614$ , effective $6.14\%$ .
Monthly: $(1.005)^{12} \approx 1.0617$ , effective $6.17\%$ .
Daily: $(1 + 0.06/365)^{365} \approx 1.0618$ , effective $6.18\%$ .

The difference is small but real, and it is the kind of comparison NESA likes to set. The four panels below build the comparison one frequency at a time, using $10000 invested for one year at a nominal $6\%$ . Notice that the bars climb but the gains shrink: going from annual to quarterly adds about $13.64, but going from monthly to daily adds only about $1.53. There is a ceiling (continuous compounding), which is why the bars level off rather than keep rising.

Stage 1, compound annually. With one compounding per year, $r = 0.06$ and $n = 1$ , so $10000 grows to $10000 \times 1.06 = 10600.00$ , i.e. $10600.00. This is the baseline: effective rate exactly $6.00\%$ .

Stage 2, compound quarterly. Now $r = 0.015$ and $n = 4$ , so $10000 grows to $10000 \times (1.015)^4 = 10613.64$ , i.e. $10613.64. Splitting the year into four gives interest a chance to earn on itself sooner, lifting the effective rate to $6.14\%$ .

Stage 3, compound monthly. With $r = 0.005$ and $n = 12$ , the balance is $10000 \times (1.005)^{12} = 10616.78$ , i.e. $10616.78, an effective $6.17\%$ . The jump from quarterly is already much smaller than the jump from annual.

Stage 4, compound daily. Finally $r = 0.06/365$ and $n = 365$ , giving $10000 \times (1 + 0.06/365)^{365} = 10618.31$ , i.e. $10618.31, an effective $6.18\%$ . The daily bar is barely taller than the monthly one: the gains have all but run out.

Always use the per-period rate; do not just use the annual rate with the number of years.

Effective annual rate

The effective annual rate is the single annual rate that, compounded once a year, gives the same growth as the nominal rate compounded $k$ times a year:

r_{\text{eff}} = \left(1 + \frac{R}{k}\right)^{k} - 1.

This is the fair way to compare two products quoted with different compounding frequencies: convert both to an effective annual rate and the larger one wins. A nominal $5.9\%$ compounded monthly ( $r_{\text{eff}} \approx 6.06\%$ ) beats a nominal $6.0\%$ compounded annually, even though the headline rate looks smaller.

Comparing investments

When choosing between two investments at different rates and compounding frequencies, either compute the future value at the same horizon and compare the dollar figures, or compute the effective annual rates and compare those. Both give the same ranking; pick whichever the question makes easier.

How exam questions ask about compound interest

The wording varies but each version maps to one of the four rearrangements. Learn to translate:

"Find the value / future value / amount after $n$ years." Straight $FV = PV(1+r)^n$ . Convert the rate first, then substitute.
"How much should be invested now / deposited today to have $X in $n$ years?" A present-value question: divide, $PV = FV / (1+r)^n$ .
"How long until the investment reaches / first exceeds $X?" Solve for $n$ with logs, then round up to the next whole period.
"At what (annual) interest rate...?" Solve for $r$ with the $n$ th-root form, then multiply by $k$ if the question wants the nominal annual rate.
"Compare investment A with investment B" or "which is the better deal?" Compute both future values at the same time, or both effective annual rates, and state which is larger with the dollar gap.
"Complete the table" or a rule given as $A_n = A_{n-1}(1.05)$ ." A recurrence question: fill the table row by row, multiplying each balance by the per-period factor.
"How much more does compound interest earn than simple interest?" Compute both ( $A_{\text{compound}} = P(1+r)^n$ versus $A_{\text{simple}} = P(1+rn)$ ) and subtract.

Compound interest: future value, present value, solving for time, and solving for rate

Four problems cover the four standard uses of the compound interest formula.

Standard future value

Find the future value of $15000 invested at $5.2\%$ per annum compounded monthly for $4$ years.

Convert the annual rate and term to per-period values: Compounding is monthly, so divide the annual rate by $12$ to get the monthly rate, and multiply the years by $12$ to get the number of periods.

r = \frac{0.052}{12} \approx 0.004333, \qquad n = 4 \times 12 = 48

Apply the compound interest formula: Raise $(1 + r)$ to the power $n$ , then multiply by the principal. This gives the lump-sum value after all compounding periods.

FV = 15000 \times (1.004333)^{48} \approx 15000 \times 1.23066 \approx 18459.90

Final answer: $18459.90.

Present value (Australian context, 2025)

A first-home buyer needs $80000 in $5$ years and can earn $4\%$ per annum compounded annually in a high-interest savings account (similar to current major-bank online saver rates in 2025). How much must they invest today?

Rearrange the compound interest formula for present value: Dividing both sides by $(1 + r)^n$ isolates the present value. This tells us what a future sum is worth in today's dollars at the given growth rate.

PV = \frac{FV}{(1 + r)^n} = \frac{80000}{(1.04)^5} = \frac{80000}{1.2167} \approx 65754

Final answer: The buyer needs to invest approximately $65754 today as a lump sum.

Solving for time

$10000 is invested at $5.5\%$ per annum compounded annually. How many years until it reaches $20000?

Set up the equation with the unknown as the exponent: When the time $n$ is unknown, write the ratio of future value to principal and take logarithms of both sides. The log law $\log(a^n) = n \log a$ brings the exponent down as a factor.

\frac{20000}{10000} = (1.055)^n \implies 2 = (1.055)^n

Solve using logarithms: Divide $\log 2$ by $\log 1.055$ to isolate $n$ .

n = \frac{\log 2}{\log 1.055} \approx \frac{0.301}{0.02325} \approx 12.95 \text{ years}

Since $n$ must be a whole number of years, the investment first exceeds $20000 at year $13$ .

Final answer: The investment reaches $20000 during year $13$ .

Solving for rate

A $2000 investment grew to $2812 in $6$ years compounded annually. Find the annual interest rate.

Rearrange for the growth factor per period: Dividing the future value by the principal gives $(1 + r)^n$ . Raising both sides to the power $\frac{1}{n}$ isolates the per-period factor $1 + r$ .

\frac{2812}{2000} = (1 + r)^6 \implies 1 + r = \left(\frac{2812}{2000}\right)^{1/6} = (1.406)^{1/6}

Evaluate the fractional power using logarithms: Converting to a log makes the fractional exponent tractable.

\log(1.406)^{1/6} = \frac{\log 1.406}{6} = \frac{0.1480}{6} \approx 0.0247

(1.406)^{1/6} \approx 10^{0.0247} \approx 1.0585

State the rate: Subtract $1$ from the growth factor to obtain $r$ .

r \approx 1.0585 - 1 = 0.0585 \text{, i.e. } 5.85\% \text{ per annum}

Final answer: The annual interest rate is approximately $5.85\%$ per annum.

Edge case: the recurrence as a table

Suppose a term-deposit advertisement states the balance follows $A_n = A_{n-1} \times 1.045$ with $A_0 = 6000$ (a starting balance of $6000 at a $4.5\%$ annual rate). Building the first four years by recurrence:

Year $n$	Opening $A_{n-1}$	$\times 1.045$	Closing $A_n$
$1$	$6000.00$	$\times 1.045$	$6270.00$
$2$	$6270.00$	$\times 1.045$	$6552.15$
$3$	$6552.15$	$\times 1.045$	$6847.00$
$4$	$6847.00$	$\times 1.045$	$7155.12$

The closed form confirms the last row: $A_4 = 6000 \times (1.045)^4 = 6000 \times 1.19252 \approx 7155.12$ , i.e. $7155.12. Identical, as expected, which is the cross-check to use if a table question and a formula question appear together.

Common traps

Using the annual rate with monthly periods: If compounding is monthly, divide the annual rate by $12$ AND multiply the number of years by $12$ . Both adjustments.
Confusing simple and compound interest: Simple interest is linear ( $A = P(1 + rn)$ ); compound is exponential ( $A = P(1 + r)^n$ ). They diverge over time, as the opening chart shows.
Forgetting to discount: A present value question asks for the amount today, so divide by $(1 + r)^n$ .
Rounding the per-period rate too early: Keep the rate to at least $6$ decimal places, or carry it as a fraction. Rounding $0.045 / 12$ to $0.004$ instead of $0.00375$ skews the final answer noticeably.
Mis-reading nominal vs effective: A nominal $6\%$ compounded monthly is an effective $6.17\%$ per annum. The two are different.
Rounding $n$ the wrong way: When solving for time, "first exceeds" rounds up to the next whole period; the balance has not yet cleared the target at the period below.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style3 marksSara invests $8000 at

4.5\%

per annum compounded quarterly for

6

years. Find the future value.

Show worked answer →

Per-period rate: $r = \frac{0.045}{4} = 0.01125$ . Number of periods: $n = 6 \times 4 = 24$ .

$A = P(1 + r)^n = 8000 (1.01125)^{24}$ .

$(1.01125)^{24} \approx 1.30799$ .

$A \approx 8000 \times 1.30799 \approx 10463.93$ , i.e. $10463.93.

Markers reward conversion to the per-period rate, the right number of compounding periods, and an answer rounded to cents.

2021 HSC-style3 marksTom invests $5000 at

3\%

per annum compounded annually. How long until the investment first exceeds $7500?

Show worked answer →

$7500 = 5000(1.03)^n \implies (1.03)^n = 1.5$ .

Take logs: $n \log 1.03 = \log 1.5$ , so $n = \frac{\log 1.5}{\log 1.03} \approx \frac{0.1761}{0.0128} \approx 13.72$ years.

First exceeds at the next whole year, so $n = 14$ years.

Markers reward setting up the inequality, taking logs, and rounding up to the next whole year (since the question asks when it first exceeds).

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks$3000 is invested at

5\%

per annum compounded annually for

4

years. Calculate the future value.

Show worked solution →

Identify the per-period rate and number of periods. Compounding is annual, so the per-period rate equals the annual rate, $r = 0.05$ , and the number of periods equals the number of years, $n = 4$ .

Apply the compound interest formula. Substitute the principal, rate and periods into $FV = PV(1 + r)^n$ .

FV = 3000 \times (1.05)^4 = 3000 \times 1.215506 = 3646.52.

State the answer. The future value is $3646.52. Check: $4$ years of simple interest at $5\%$ would give a factor of $1 + 0.05 \times 4 = 1.20$ , and the compound factor $1.215506$ is a little larger, exactly as expected.

foundation2 marks$5000 is invested at

8\%

per annum compounded half-yearly for

2

years. Calculate the future value and the interest earned.

Show worked solution →

Convert the annual rate and term to per-period values. Compounding is half-yearly, so divide the annual rate by $2$ and multiply the years by $2$ .

r = \frac{0.08}{2} = 0.04, \qquad n = 2 \times 2 = 4.

Apply the compound interest formula. Raise $(1 + r)$ to the power $n$ , then multiply by the principal.

FV = 5000 \times (1.04)^4 = 5000 \times 1.169859 = 5849.29.

Find the interest earned. Subtract the principal from the future value.

5849.29 - 5000 = 849.29.

State the answer. The future value is $5849.29 and the interest earned is $849.29. Check: $r$ and $n$ are both scaled by $2$ , so the per-period rate and period count are consistent.

foundation2 marks$4000 is invested at

6\%

per annum compounded annually for

5

years. Calculate the future value and the interest earned.

Show worked solution →

Identify the per-period rate and number of periods. Compounding is annual, so the per-period rate is just the annual rate, $r = 0.06$ , and the number of periods is the number of years, $n = 5$ .

Apply the compound interest formula. Substitute into $FV = PV(1 + r)^n$ .

FV = 4000 \times (1.06)^5 = 4000 \times 1.33823 = 5352.90.

Find the interest earned. The interest is the future value minus the amount invested, $FV - PV$ .

5352.90 - 4000 = 1352.90.

State the answer. The future value is $5352.90 and the interest earned is $1352.90. Check: the balance has grown by a factor of $1.33823$ , a touch more than the $1.30$ that $5$ years of simple interest at $6\%$ would give ( $1 + 0.06 \times 5 = 1.30$ ), which is the small head start compounding buys.

foundation3 marksA savings account follows the rule

A_n = A_{n-1} \times 1.04

with

A_0 = 5000

, where

A_n

is the balance in dollars after

n

years. Complete the balance for years

1

2

and

3

, then confirm the year

3

balance with the closed-form formula.

Show worked solution →

Apply the recurrence one year at a time. Each year multiplies the previous balance by the per-period factor $1.04$ .

A_1 = 5000 \times 1.04 = 5200.00.

A_2 = 5200.00 \times 1.04 = 5408.00.

A_3 = 5408.00 \times 1.04 = 5624.32.

Confirm with the closed form. Applying the factor three times is the same as raising it to the power $3$ , so $A_3 = PV(1 + r)^n$ with $PV = 5000$ , $r = 0.04$ and $n = 3$ .

A_3 = 5000 \times (1.04)^3 = 5000 \times 1.12486 = 5624.32.

State the answer. The balances are $5200.00, $5408.00 and $5624.32, and the closed form gives the same $5624.32 for year $3$ . The two methods agree exactly, which is the cross-check: the recurrence and the formula describe the same investment.

core3 marks$12\,000 is invested at

5.6\%

per annum compounded quarterly for

3

years. Calculate the future value and the total interest earned.

Show worked solution →

Convert the annual rate and term to per-period values. Compounding is quarterly, so divide the annual rate by $4$ and multiply the years by $4$ .

r = \frac{0.056}{4} = 0.014, \qquad n = 3 \times 4 = 12.

Apply the compound interest formula. Raise $(1 + r)$ to the power $n$ , then multiply by the principal.

FV = 12\,000 \times (1.014)^{12} = 12\,000 \times 1.18156 = 14\,178.71.

Find the interest earned. Subtract the principal from the future value.

14\,178.71 - 12\,000 = 2178.71.

State the answer. The future value is $14,178.71 and the interest earned is $2178.71. Check: $r = 0.014$ and $n = 12$ are consistent (both scaled by $4$ ), and the growth factor $1.18156$ is sensible for roughly $5.6\%$ over $3$ years.

core3 marksA family wants to have $30\,000 in

6

years for a home deposit. Their account pays

4.8\%

per annum compounded monthly. What single amount must they invest today?

Show worked solution →

Recognise this as a present-value question. They know the future value and want the amount to invest now, so rearrange $FV = PV(1 + r)^n$ to $PV = \dfrac{FV}{(1 + r)^n}$ .

Convert the annual rate and term to per-period values. Compounding is monthly, so divide the annual rate by $12$ and multiply the years by $12$ .

r = \frac{0.048}{12} = 0.004, \qquad n = 6 \times 12 = 72.

Discount the future value back to today. Divide the target by the growth factor $(1 + r)^n$ .

PV = \frac{30\,000}{(1.004)^{72}} = \frac{30\,000}{1.33299} = 22\,505.77.

State the answer. They must invest approximately $22,505.77 today. Check: growing this forward, $22\,505.77 \times 1.33299 \approx 30\,000$ , which recovers the target, confirming the discount.

core4 marks$8500 is invested at

4.2\%

per annum compounded monthly for

4

years. Calculate the future value and the total interest earned, each correct to the nearest cent.

Show worked solution →

Convert the annual rate and term to per-period values. Compounding is monthly, so divide the annual rate by $12$ and multiply the years by $12$ .

r = \frac{0.042}{12} = 0.0035, \qquad n = 4 \times 12 = 48.

Apply the compound interest formula. Raise $(1 + r)$ to the power $n$ , then multiply by the principal. Keep the growth factor to several decimal places before rounding.

FV = 8500 \times (1.0035)^{48} = 8500 \times 1.182590 = 10\,052.01.

Find the interest earned. Subtract the principal from the future value.

10\,052.01 - 8500 = 1552.01.

State the answer. The future value is $10,052.01 and the total interest earned is $1552.01. Check: the monthly rate $0.0035$ kept to four decimal places avoids the rounding trap, and the growth factor $1.182590$ is reasonable for roughly $4.2\%$ over $4$ years.

core4 marks$7000 is invested at

6.5\%

per annum compounded annually. After how many whole years does the investment first exceed $10\,000?

Show worked solution →

Set up the equation with the unknown in the exponent. Write the ratio of the target to the principal equal to the growth factor raised to the power $n$ .

\frac{10\,000}{7000} = (1.065)^n \implies (1.065)^n = 1.428571.

Solve using logarithms. Take logs of both sides; the log law $\log(a^n) = n \log a$ brings the exponent down.

n = \frac{\log 1.428571}{\log 1.065} = \frac{0.154902}{0.027350} = 5.6638 \text{ years}.

Round up to the next whole year. The balance only clears the target at the end of a full period, and the question asks when it first exceeds $10,000, so round up to $n = 6$ .

State the answer. The investment first exceeds $10,000 after $6$ years. Check: at year $5$ the balance is $7000 \times (1.065)^5 \approx 9590.61$ , still under the target, and at year $6$ it is $7000 \times (1.065)^6 \approx 10\,214.00$ , which clears it.

exam4 marksTwo banks each offer a

4

year term deposit on $20\,000. Bank A pays

5.4\%

per annum compounded monthly; Bank B pays

5.5\%

per annum compounded annually. By comparing the future values, determine which is the better investment and by how much.

Show worked solution →

Find the future value for Bank A. Compounding is monthly, so $r = \dfrac{0.054}{12} = 0.0045$ and $n = 4 \times 12 = 48$ .

FV_A = 20\,000 \times (1.0045)^{48} = 20\,000 \times 1.240501 = 24\,810.02.

Find the future value for Bank B. Compounding is annual, so $r = 0.055$ and $n = 4$ .

FV_B = 20\,000 \times (1.055)^4 = 20\,000 \times 1.238825 = 24\,776.49.

Compare the two figures. Bank A finishes higher, so subtract to find the gap.

24\,810.02 - 24\,776.49 = 33.53.

State the answer. Bank A is the better investment, by $33.53 over the $4$ years. Check via effective annual rates: Bank A is $(1.0045)^{12} - 1 \approx 5.54\%$ , just above Bank B's $5.50\%$ , so the more frequent compounding wins despite the lower headline rate, which agrees with the dollar comparison.

exam4 marksAn investment of $5000 grows to $6800 over

5

years with interest compounded annually. Calculate the annual interest rate, correct to one decimal place.

Show worked solution →

Rearrange for the growth factor per period. Dividing the future value by the principal gives $(1 + r)^n$ ; raising both sides to the power $\tfrac{1}{n}$ isolates $1 + r$ .

\frac{6800}{5000} = (1 + r)^5 \implies 1 + r = \left(\frac{6800}{5000}\right)^{1/5} = (1.36)^{1/5}.

Evaluate the fractional power using logarithms. Converting to a log makes the fractional exponent manageable.

\log (1.36)^{1/5} = \frac{\log 1.36}{5} = \frac{0.133539}{5} = 0.026708,

so $(1.36)^{1/5} = 10^{0.026708} = 1.063427$ .

State the rate. Subtract $1$ from the growth factor and convert to a percentage.

r = 1.063427 - 1 = 0.063427, \text{ i.e. } 6.3\% \text{ per annum.}

State the answer. The annual interest rate is approximately $6.3\%$ per annum. Check: $5000 \times (1.063427)^5 = 6800$ , which recovers the stated future value.

exam5 marks$14\,000 is invested for

8

years at

7\%

per annum. Calculate the final value if interest is compounded annually, the final value if simple interest is paid, and hence how much more the compound-interest account earns.

Show worked solution →

Find the compound-interest value. Use $FV = PV(1 + r)^n$ with $PV = 14\,000$ , $r = 0.07$ and $n = 8$ .

FV_{\text{compound}} = 14\,000 \times (1.07)^8 = 14\,000 \times 1.71819 = 24\,054.61.

Find the simple-interest value. Simple interest pays only on the original principal, so use $A = P(1 + rn)$ with the same $P$ , $r$ and $n$ .

A_{\text{simple}} = 14\,000 \times (1 + 0.07 \times 8) = 14\,000 \times 1.56 = 21\,840.00.

Find the difference. Subtract the simple-interest value from the compound-interest value.

24\,054.61 - 21\,840.00 = 2214.61.

State the answer. Compounded annually the account reaches $24,054.61; with simple interest it reaches $21,840.00; so compound interest earns $2214.61 more over the $8$ years. Check: the compound growth factor $1.71819$ exceeds the simple factor $1.56$ , so the compound account must finish higher, as it does.

exam6 marksMia invests $25\,000 from a redundancy payout into a managed fund advertised at

6.6\%

per annum. (a) If interest is compounded monthly, calculate the value of the investment after

7

years. (b) The fund also offers the same nominal

6.6\%

per annum compounded annually instead. Calculate the value after

7

years under annual compounding, and hence find how much less this option is worth than the monthly option.

Show worked solution →

Part (a): convert the annual rate and term to per-period values. Compounding is monthly, so divide the annual rate by $12$ and multiply the years by $12$ .

r = \frac{0.066}{12} = 0.0055, \qquad n = 7 \times 12 = 84.

Part (a): apply the compound interest formula. Raise $(1 + r)$ to the power $n$ , then multiply by the principal.

FV_{\text{monthly}} = 25\,000 \times (1.0055)^{84} = 25\,000 \times 1.585237 = 39\,630.93.

So after $7$ years the monthly option is worth $39,630.93.

Part (b): find the value under annual compounding. Now $r = 0.066$ and $n = 7$ .

FV_{\text{annual}} = 25\,000 \times (1.066)^7 = 25\,000 \times 1.564229 = 39\,105.73.

Part (b): find how much less the annual option is worth. Subtract the annual value from the monthly value.

39\,630.93 - 39\,105.73 = 525.20.

State the answer. (a) The monthly option is worth $39,630.93 after $7$ years. (b) The annual option is worth $39,105.73, which is $525.20 less than the monthly option. Check via effective annual rates: monthly compounding gives $(1.0055)^{12} - 1 \approx 6.80\%$ , just above the $6.60\%$ of annual compounding, so the more frequent compounding wins, agreeing with the dollar comparison.