Annuities, future value and superannuation for HSC Maths Standard 2

Syllabus dot point

“Use the future value formula for an annuity to find the accumulated value of regular contributions to superannuation or a savings plan”

HSCMaths Standard 2Year 12: Financial Mathematics12 min readUpdated 20 Jun 2026

Quick answer

The future value of an annuity of $M$ per period over $n$ periods at per-period rate $r$ is $FV = M \cdot \frac{(1 + r)^n - 1}{r}$ ; rearrange for $M$ if you know the target future value; this is the standard model for superannuation growth.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to apply the future-value-of-annuity formula to a series of equal regular contributions, model superannuation growth, and solve for either the future value or the required contribution to hit a savings goal.

The answer

An annuity is the engine behind every regular savings plan. The one idea to hold onto is that each contribution lands at a different time. So each one compounds (earns interest that then earns more interest) for a different number of periods. The earliest contribution sits in the account the longest and grows the most. The final contribution earns nothing, because it arrives right at the end. The future value is just the sum of all those separately compounded payments. Over a long term, the interest on interest comes to dominate completely. The chart below tracks a real superannuation case: 40 years of quarterly contributions total only $322000, yet the fund grows past $1.7 million.

What an annuity is

An annuity is a stream of equal payments made at regular intervals into (or out of) an account that earns interest. The two questions you can ask:

Future value (savings annuity). What does the account grow to after $n$ payments?
Required payment. Given a target future value, what payment is needed?

The future-value-of-annuity formula

From the NESA reference sheet:

FV = M \cdot \frac{(1 + r)^n - 1}{r}.

$FV$ is future value
$M$ is payment per period
$r$ is per-period interest rate (as decimal)
$n$ is number of payments

The formula assumes payments are made at the end of each period (ordinary annuity).

Key fact

$FV = M \cdot \dfrac{(1 + r)^n - 1}{r}$ , where $M$ is the payment per period, $r$ is the interest rate per period, and $n$ is the number of payments. The three must share the same period: monthly payments need a monthly rate and a count in months. Total contributed is $nM$ ; total interest earned is $FV - nM$ . Rearranged for the payment, $M = \dfrac{FV \cdot r}{(1 + r)^n - 1}$ .

Why the formula works

Each payment compounds for a different number of periods. The first payment compounds for $n - 1$ periods, the second for $n - 2$ , and so on. The last payment compounds for $0$ periods.

$FV = M(1 + r)^{n-1} + M(1 + r)^{n-2} + \cdots + M(1 + r) + M$

This is a finite geometric series with $n$ terms, first term $M$ , ratio $1 + r$ :

$FV = M \cdot \frac{(1 + r)^n - 1}{(1 + r) - 1} = M \cdot \frac{(1 + r)^n - 1}{r}$ .

You are not asked to derive this in the exam. But seeing it as "the sum of $n$ separately compounded payments" is what makes the formula make sense. It is also the cross-check if you ever doubt an answer: a four-payment annuity can be added up by hand, and the total must match the formula.

The future value built contribution by contribution

To see the geometric series in action, take a small annuity: $1000 paid at the end of each year for $4$ years at $5\%$ per annum. The four panels below add one payment at a time. Each bar is that payment after it has compounded for the periods remaining until the end; the bars are then summed to give the future value. Notice the bars shrink as you go: the earliest payment has the longest to grow, the last payment has no time at all.

Stage 1, the first payment. Payment $1$ is made at the end of year $1$ , so it has $3$ full years left to compound before the end of year $4$ . It grows to $1000 \times 1.05^3 = 1157.63$ , i.e. $1157.63.

Stage 2, add the second payment. Payment $2$ arrives a year later, so it compounds for only $2$ years: $1000 \times 1.05^2 = 1102.50$ , i.e. $1102.50. The running total of the first two payments is $2260.13.

Stage 3, add the third payment. Payment $3$ compounds for just $1$ year, adding $1000 \times 1.05 = 1050.00$ , i.e. $1050.00. The running total climbs to $3310.13.

Stage 4, add the last payment. Payment $4$ is made at the very end of year $4$ , so it earns no interest and adds exactly $1000.00. The four bars sum to $4310.13, which is precisely what the formula gives: $1000 \cdot \frac{1.05^4 - 1}{0.05} = 1000 \times 4.31013 = 4310.13$ , i.e. $4310.13.

This is exactly why the formula is a power expression and not simple multiplication: $nM = 4 \times 1000 = 4000$ , i.e. $4000 was contributed, and the extra $310.13 is the interest the earlier payments earned while they sat in the account.

Solving for the payment

When the question fixes a target future value (a deposit goal, a retirement target) and asks what regular payment is needed, rearrange the formula:

M = \frac{FV \cdot r}{(1 + r)^n - 1}.

The denominator is the same $(1 + r)^n - 1$ you would compute for a future-value question, so the work is almost identical; you simply divide $FV \cdot r$ by it instead of multiplying.

Per-period conversions

Same as compound interest:

Annual contributions, annual compounding: $r = R$ , $n =$ years.
Monthly contributions, monthly compounding: $r = R/12$ , $n = 12 \times$ years.
Quarterly: $r = R/4$ , $n = 4 \times$ years.

Use a matching frequency between contributions and compounding for the formula to apply directly. The Standard 2 syllabus always sets the contribution frequency equal to the compounding frequency, so you never have to reconcile a mismatch.

Superannuation context

In Australia, employers must contribute a percentage of gross salary into the employee's superannuation fund. This is the Super Guarantee (SG), which is $11.5\%$ for 2024-25 and rising to $12\%$ from 1 July 2025 (ATO).

So an employee earning $80000 has $80000 \times 0.115 \approx 9200$ , i.e. $9200 paid into super per year. Over a working life, this grows a lot. That is the whole reason superannuation works: small, steady contributions are left untouched and earn interest on interest for decades. The fund-balance chart at the top of this page is exactly this calculation, carried out for a $40$ -year career.

How exam questions ask about annuities

The wording varies, but each version maps to one use of the formula. Learn to translate:

"Find the value of the account / fund / investment after $n$ years." A straight future-value question: convert the rate, count the payments, substitute into $FV = M \cdot \frac{(1+r)^n - 1}{r}$ .
"How much is contributed in total, and how much of the final balance is interest?" Total contributed is $nM$ ; interest is $FV - nM$ . Compute $FV$ first.
"What regular payment is needed to reach $X in $n$ years?" Rearrange for $M$ : divide $FV \cdot r$ by $(1+r)^n - 1$ .
"Each year the employer pays $11.5\%$ of a $Y salary into super..." A superannuation question: work out the annual SG contribution first ( $0.115 \times$ salary), divide by the number of payments per year if contributions are quarterly or monthly, then apply the formula.
"Compare contributing $A per month for $T$ years against a single lump sum." Compute the annuity future value and, separately, the lump-sum future value $PV(1+r)^n$ , then compare. Different formulas; do not blend them.
"A table shows the balance growing each period." A recurrence view: each period the balance is multiplied by $(1+r)$ and then the new payment is added, $B_{\text{new}} = B_{\text{old}}(1+r) + M$ . Fill it row by row; the last row matches the formula.

Annuities: future value, superannuation growth, and solving for a required contribution

Three problems show how the annuity future-value formula is used forwards, interpreted in a super context, and rearranged to find a required payment.

Future value of regular savings

Eve saves $500 per month into an account paying $4\%$ per annum compounded monthly for $20$ years.

Identify the periodic interest rate and number of periods: Because compounding is monthly, divide the annual rate by $12$ to obtain the rate per period. There are $20 \times 12 = 240$ monthly periods in total.

r = \frac{0.04}{12} \approx 0.003333, \qquad n = 240

Compute the growth factor: The factor $(1 + r)^n$ represents how much one dollar deposited today grows over the full term. This is the engine of compound growth.

(1.003333)^{240} \approx 2.22258

Apply the future-value annuity formula: The formula sums all the compounded contributions. Substitute the values and simplify.

FV = 500 \cdot \frac{2.22258 - 1}{0.003333} = 500 \cdot \frac{1.22258}{0.003333} = 500 \times 366.77 \approx 183387.31

i.e. $183387.31. Total contributed: $240 \times 500 = 120000$ , i.e. $120000. Total interest earned: $63387.31.

Final answer: Eve accumulates approximately $183387.31, of which $63387.31 is interest.

Super Guarantee growth (Australian context, 2025)

A graduate aged $25$ starts on $70000 per year. SG contributions are $11.5\%$ of $70000, that is $0.115 \times 70000 = 8050$ , i.e. $8050 per year, paid quarterly. They retire at $65$ , giving $40 \times 4 = 160$ quarters. The fund earns $7\%$ per annum compounded quarterly.

Convert to quarterly rate and per-quarter contribution: Divide the annual rate by $4$ for the quarterly rate, and divide the annual SG contribution by $4$ for the quarterly payment.

r = \frac{0.07}{4} = 0.0175, \qquad M = \frac{8050}{4} = 2012.50

so the quarterly payment is $2012.50.

Compute the growth factor over the full term: With $n = 160$ quarterly periods, the growth factor is substantial because of the long time horizon.

(1.0175)^{160} \approx 16.0512

Apply the future-value annuity formula: Substitute all values to find the balance at retirement.

FV = 2012.50 \cdot \frac{16.0512 - 1}{0.0175} = 2012.50 \cdot \frac{15.0512}{0.0175} = 2012.50 \times 860.07 \approx 1730885.27

i.e. $1730885.27. Total contributed over the career is only $160 \times 2012.50 = 322000$ , i.e. $322000, so more than $1.4 million of the balance is interest. This is the calculation drawn in the fund-balance chart at the top of the page.

Final answer: A graduate on $70000 (held constant for simplicity) accumulates approximately $1.73 million in super at retirement.

Solving for the required monthly contribution

A couple wants $200000 in $10$ years to renovate their home. They will save at $5.5\%$ per annum compounded monthly. How much must they save per month?

Identify the rate and number of periods: Divide the annual rate by $12$ for the monthly rate. There are $10 \times 12 = 120$ monthly periods.

r = \frac{0.055}{12} \approx 0.004583, \qquad n = 120

Compute the accumulated growth factor and the denominator: Calculate $(1 + r)^n$ first, then subtract $1$ to form the denominator of the annuity formula.

(1.004583)^{120} \approx 1.73108, \qquad \text{so the denominator} = 1.73108 - 1 = 0.73108

Rearrange the annuity formula to solve for the payment: Multiply both sides of the future-value formula by $r$ and divide by the denominator to isolate the monthly payment $M$ .

M = \frac{FV \times r}{(1+r)^n - 1} = \frac{200000 \times 0.004583}{0.73108} = \frac{916.67}{0.73108} \approx 1253.86 \text{ per month}

i.e. about $1253.86 per month.

Final answer: The couple must save approximately $1253.86 per month to reach $200000 in $10$ years.

Edge case: the recurrence as a table

Some questions present an annuity as a year-by-year table instead of a formula. Each period, the balance is multiplied by $(1 + r)$ and then that period's payment is added: $B_{\text{new}} = B_{\text{old}}(1 + r) + M$ . Building the $1000-per-year, $5\%$ annuity row by row:

End of year	Opening $B$	$\times 1.05$	$+ 1000$	Closing $B$
$1$	$0$	$0$	$+1000$	$1000.00$
$2$	$1000.00$	$1050.00$	$+1000$	$2050.00$
$3$	$2050.00$	$2152.50$	$+1000$	$3152.50$
$4$	$3152.50$	$3310.13$	$+1000$	$4310.13$

The closing balance after $4$ years is $4310.13, identical to the formula and to the bar-by-bar build above. If a table question and a formula question appear together, this is the cross-check.

Edge case: contributions at the start of each period

The reference-sheet formula is for an ordinary annuity, where payments fall at the end of each period. If a question instead pays at the start of each period (an "annuity due"), every payment compounds for one extra period, so the future value is larger by a factor of $(1 + r)$ . NESA Standard 2 uses the ordinary annuity, so unless a question explicitly says payments are made at the beginning, take payments as end-of-period and use the formula as written.

Exam technique

Write $M$ , $r$ and $n$ as a labelled first line before substituting; markers award method marks for the correct payment, per-period rate and period count even if the arithmetic slips. State whether you are finding the future value or the required payment so the rearrangement is clear. Compute $(1+r)^n$ to several decimal places and round only the final money answer, to cents. Distinguish the annuity formula ( $FV = M\frac{(1+r)^n-1}{r}$ , for regular payments) from the lump-sum formula ( $FV = PV(1+r)^n$ , for a single deposit); choosing the wrong one is the costliest error. If asked for total interest, give $FV - nM$ , not the future value.

Common traps

Confusing future value of annuity with future value of a single lump sum: Lump sum: $FV = PV(1 + r)^n$ . Annuity: $FV = M \cdot \frac{(1 + r)^n - 1}{r}$ . Different formulas, different applications.
Mismatched frequencies: If contributions are monthly but compounding is annual, the simple formula does not apply directly. The Standard 2 syllabus assumes matched frequencies.
Forgetting that contributions are at the end of each period: "Ordinary annuity" pays at the end. An "annuity due" pays at the start, which gives a factor of $(1 + r)$ more. NESA Standard 2 uses ordinary annuity.
Using the wrong total: Total contributed is $n M$ . Total interest is $FV - n M$ . The future value is the sum of both.
Rate-frequency mismatch: A nominal $6\%$ per annum compounded monthly uses $r = 0.005$ , not $0.06$ .

Note

An annuity just means paying the same amount into an account over and over, like $100 every month. The account pays interest, and that interest then earns its own interest, so the money grows faster and faster the longer it sits there. Money you put in early matters most, because it has the most time to grow. Your last payment hardly grows at all, since it only just arrived. Superannuation is the big real-life example: small amounts go in from your pay for decades, and over a whole working life the interest ends up much bigger than everything you actually paid in.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style4 marks

Hannah contributes $300 at the end of each month into a superannuation account that earns

6\%

per annum compounded monthly. Find the balance after

30

years.

Show worked answer →

Per-period rate: $r = \frac{0.06}{12} = 0.005$ . Payment: $M = 300$ . Number of payments: $n = 360$ .

Future-value-of-annuity formula (NESA reference sheet):

$FV = M \cdot \frac{(1 + r)^n - 1}{r}$ .

$(1.005)^{360} \approx 6.02257$ .

$FV \approx 300 \cdot \frac{6.02257 - 1}{0.005} = 300 \cdot \frac{5.02257}{0.005} = 300 \cdot 1004.515 \approx 301354.51$ , i.e. $301354.51.

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

2023 HSC-style4 marks

Lachlan needs a deposit of $80000 in

5

years. He saves at the end of each quarter into an account paying

5\%

per annum compounded quarterly. How much must each deposit be?

Show worked answer →

$r = \frac{0.05}{4} = 0.0125$ . $n = 4 \times 5 = 20$ . $FV = 80000$ .

Rearrange the future-value formula for $M$ :

$M = \frac{FV \cdot r}{(1 + r)^n - 1} = \frac{80000 \times 0.0125}{(1.0125)^{20} - 1}$ .

$(1.0125)^{20} \approx 1.28204$ , so the denominator is $0.28204$ .

$M = \frac{1000}{0.28204} \approx 3545.63$ , i.e. $3545.63 per quarter.

Markers reward the per-period rate, the rearrangement for $M$ , and the answer rounded to cents.

Practice questions

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

foundation1 mark

Jack earns a salary of $72000 per year. His employer pays Super Guarantee contributions of

11.5\%

of his salary into his superannuation fund each year. Find the total amount paid into his super in one year.

Show worked solution →

Apply the Super Guarantee rate to the salary. The annual SG contribution is the rate, $11.5\%$ , of the salary, so multiply the salary by $0.115$ .

0.115 \times 72000 = 8280

Answer: $8280.00 is paid into Jack's super in one year.

foundation2 marks

Mia pays $2000 into a savings account at the end of each year for

5

years. The account earns

6\%

per annum compounded annually. Find the value of the account at the end of the

5

years.

Show worked solution →

Identify the per-period rate and number of payments. Contributions and compounding are both annual, so $r = 0.06$ and $n = 5$ , with payment $M = 2000$ .

Compute the growth factor. This is $(1 + r)^n$ .

(1.06)^5 \approx 1.33823

Apply the future-value-of-annuity formula.

FV = M \cdot \frac{(1 + r)^n - 1}{r} = 2000 \cdot \frac{1.33823 - 1}{0.06} = 2000 \times 5.63709 \approx 11274.19

i.e. $11274.19.

Check. The total contributed is $5 \times 2000 = 10000$ , i.e. $10000, so the $1274.19 above the contributions is the interest the earlier payments earned. The future value exceeds the $10000 paid in, as it must.

foundation2 marks

Noah earns a salary of $60000 per year. His employer pays Super Guarantee contributions of

12\%

of his salary into his superannuation fund, made in equal quarterly amounts. Find the size of each quarterly contribution.

Show worked solution →

Find the annual Super Guarantee contribution. The SG contribution is $12\%$ of the salary, so multiply the salary by $0.12$ .

0.12 \times 60000 = 7200

i.e. $7200 is paid into super each year.

Split into quarterly amounts. There are $4$ quarters in a year, so divide the annual contribution by $4$ .

\frac{7200}{4} = 1800

State the answer. Each quarterly contribution is $1800.00.

Check. Four quarterly payments of $1800 give $4 \times 1800 = 7200$ , i.e. $7200, matching the annual SG amount.

foundation2 marks

Grace pays $2500 into a superannuation fund at the end of each year for

3

years. The fund earns

7\%

per annum compounded annually. Find the value of the fund at the end of the

3

years.

Show worked solution →

Identify the per-period rate and number of payments. Contributions and compounding are both annual, so $r = 0.07$ and $n = 3$ , with payment $M = 2500$ .

Compute the growth factor. This is $(1 + r)^n$ .

(1.07)^3 \approx 1.22504

Apply the future-value-of-annuity formula.

FV = M \cdot \frac{(1 + r)^n - 1}{r} = 2500 \cdot \frac{1.22504 - 1}{0.07} = 2500 \times 3.21 \approx 8037.25

i.e. $8037.25.

Check. The total contributed is $3 \times 2500 = 7500$ , i.e. $7500, so the balance sits $537.25 above the contributions, which is the interest the earlier payments earned.

Answer: the fund is worth $8037.25 at the end of the $3$ years.

core3 marks

Olivia contributes $450 at the end of each month into an investment account that earns

4.8\%

per annum compounded monthly. Find the balance after

15

years, and state how much of that balance is interest.

Show worked solution →

Convert to a monthly rate and count the payments. Divide the annual rate by $12$ and count the months in $15$ years.

r = \frac{0.048}{12} = 0.004, \qquad n = 12 \times 15 = 180

Compute the growth factor.

(1.004)^{180} \approx 2.05148

Apply the future-value-of-annuity formula.

FV = 450 \cdot \frac{2.05148 - 1}{0.004} = 450 \times 262.87 \approx 118292.04

i.e. $118292.04.

Find the interest. The total contributed is $180 \times 450 = 81000$ , i.e. $81000, so the interest is the balance less the contributions.

118292.04 - 81000 = 37292.04

Check. The balance is $118292.04, of which $37292.04 is interest. The interest is positive and well below the balance, as expected for a $15$ -year term.

core3 marks

William contributes $600 at the end of each month into his superannuation fund, which earns

5.4\%

per annum compounded monthly. Find the balance after

20

years, and state how much of that balance is interest.

Show worked solution →

Convert to a monthly rate and count the payments. Divide the annual rate by $12$ and count the months in $20$ years.

r = \frac{0.054}{12} = 0.0045, \qquad n = 12 \times 20 = 240

Compute the growth factor.

(1.0045)^{240} \approx 2.93755

Apply the future-value-of-annuity formula.

FV = 600 \cdot \frac{2.93755 - 1}{0.0045} = 600 \times 430.57 \approx 258340.53

i.e. $258340.53.

Find the interest. The total contributed is $240 \times 600 = 144000$ , i.e. $144000, so the interest is the balance less the contributions.

258340.53 - 144000 = 114340.53

Check. The balance is $258340.53, of which $114340.53 is interest. Over a $20$ -year term the interest making up almost half the balance is reasonable, as the early payments compound for many years.

Answer: the balance is $258340.53, of which $114340.53 is interest.

core4 marks

Ethan wants to have $50000 saved in

6

years for a house deposit. He will save an equal amount at the end of each quarter into an account paying

5.2\%

per annum compounded quarterly. Find the quarterly payment required.

Show worked solution →

Identify the per-period rate and number of payments. Divide the annual rate by $4$ , and count the quarters in $6$ years.

r = \frac{0.052}{4} = 0.013, \qquad n = 4 \times 6 = 24

Compute the growth factor and the denominator. Find $(1 + r)^n$ , then subtract $1$ .

(1.013)^{24} \approx 1.36341, \qquad \text{so } (1 + r)^n - 1 = 0.36341

Rearrange the formula for the payment. Use $M = \dfrac{FV \cdot r}{(1 + r)^n - 1}$ , with $FV = 50000$ .

M = \frac{50000 \times 0.013}{0.36341} = \frac{650}{0.36341} \approx 1788.61

i.e. $1788.61 per quarter.

Check. Substituting $M = 1788.61$ back gives $1788.61 \times \frac{0.36341}{0.013} \approx 50000$ , the target balance, so the payment is right.

core4 marks

Ava earns $85000 per year, held constant for simplicity. Her employer pays Super Guarantee contributions of

12\%

of her salary into her superannuation fund in equal quarterly amounts. The fund earns

7.5\%

per annum compounded quarterly. Find the balance of the fund after

35

years of contributions.

Show worked solution →

Find the quarterly contribution. The annual SG contribution is $0.12 \times 85000 = 10200$ , i.e. $10200 per year. Dividing by $4$ gives the quarterly payment.

M = \frac{10200}{4} = 2550

so the quarterly payment is $2550.

Convert to a quarterly rate and count the payments.

r = \frac{0.075}{4} = 0.01875, \qquad n = 4 \times 35 = 140

Compute the growth factor over the full term.

(1.01875)^{140} \approx 13.47308

Apply the future-value-of-annuity formula.

FV = 2550 \cdot \frac{13.47308 - 1}{0.01875} = 2550 \times 665.23 \approx 1696339.54

i.e. $1696339.54.

Check. The total contributed over the career is only $140 \times 2550 = 357000$ , i.e. $357000, so more than $1.3 million of the balance is interest. A balance far above the contributions is exactly what compounding over $35$ years produces.

exam4 marks

Liam contributes $650 at the end of each month into his superannuation fund, which earns

6\%

per annum compounded monthly. He contributes for

25

years. Find the balance at the end of the

25

years, and state how much of the balance was contributed by Liam and how much is interest.

Show worked solution →

Convert to a monthly rate and count the payments.

r = \frac{0.06}{12} = 0.005, \qquad n = 12 \times 25 = 300

Compute the growth factor.

(1.005)^{300} \approx 4.46497

Apply the future-value-of-annuity formula.

FV = 650 \cdot \frac{4.46497 - 1}{0.005} = 650 \times 692.99 \approx 450446.08

i.e. $450446.08.

Split into contributions and interest. The total contributed is $300 \times 650 = 195000$ , i.e. $195000. The interest is the balance less the contributions.

450446.08 - 195000 = 255446.08

Check. Liam contributed $195000.00 and earned $255446.08 in interest, which add to the $450446.08 balance. Over a $25$ -year term the interest exceeding the contributions is reasonable, since the early payments compound for a long time.

exam5 marks

To save for a car, Chloe is comparing two plans, each running for

10

years in an account paying

6\%

per annum compounded monthly. Plan A: pay $300 at the end of each month. Plan B: deposit a single lump sum of $25000 now and leave it. Find the future value of each plan after

10

years and state which gives the larger balance, and by how much.

Show worked solution →

Set up the shared per-period values. Both plans use the same monthly rate and term.

r = \frac{0.06}{12} = 0.005, \qquad n = 12 \times 10 = 120

Plan A: future value of the annuity. This is a stream of equal monthly payments, so use the annuity formula.

(1.005)^{120} \approx 1.81940

FV_A = 300 \cdot \frac{1.81940 - 1}{0.005} = 300 \times 163.88 \approx 49163.80

i.e. $49163.80.

Plan B: future value of the lump sum. A single deposit uses the compound-interest formula $FV = PV(1 + r)^n$ , not the annuity formula.

FV_B = 25000 \times (1.005)^{120} = 25000 \times 1.81940 \approx 45484.92

i.e. $45484.92.

Compare. Plan A gives the larger balance.

49163.80 - 45484.92 = 3678.89

Check. Plan A is larger by $3678.89. The two formulas are different and must not be blended: Plan A sums $120$ separately compounded payments totalling $120 \times 300 = 36000$ , i.e. $36000 contributed, while Plan B grows one $25000 deposit.

exam4 marks

A savings annuity pays $4000 at the end of each year for

4

years into an account earning

8\%

per annum compounded annually. Copy and complete a table that finds the balance year by year using

B_{\text{new}} = B_{\text{old}}(1 + r) + M

, then confirm the closing balance with the future-value-of-annuity formula.

Show worked solution →

Apply the recurrence one year at a time. Each year the opening balance is multiplied by $1.08$ and then $4000 is added. Start from a balance of $0.

End of year $1$ : $0 \times 1.08 + 4000 = 4000$ , i.e. $4000.00.

End of year $2$ : $4000 \times 1.08 + 4000 = 4320 + 4000 = 8320$ , i.e. $8320.00.

End of year $3$ : $8320 \times 1.08 + 4000 = 8985.60 + 4000 = 12985.60$ , i.e. $12985.60.

End of year $4$ : $12985.60 \times 1.08 + 4000 = 14024.45 + 4000 = 18024.45$ , i.e. $18024.45.

Confirm with the formula. Use $FV = M \cdot \dfrac{(1 + r)^n - 1}{r}$ with $M = 4000$ , $r = 0.08$ , $n = 4$ .

(1.08)^4 \approx 1.36049

FV = 4000 \cdot \frac{1.36049 - 1}{0.08} = 4000 \times 4.50611 \approx 18024.45

i.e. $18024.45.

Check. The recurrence and the formula both give $18024.45, so the table is correct. The total contributed is $4 \times 4000 = 16000$ , i.e. $16000, leaving $2024.45 as interest.

exam6 marks

Sophie earns $95000 per year, held constant for simplicity. Her employer pays Super Guarantee contributions of

12\%

of her salary into her superannuation fund in equal quarterly amounts. The fund earns

6.8\%

per annum compounded quarterly, and she contributes for

30

years. (a) Find the size of each quarterly contribution. (b) Find the balance of the fund after

30

years. (c) State how much of that balance Sophie's employer contributed and how much is interest.

Show worked solution →

(a) Find the quarterly contribution. The annual SG contribution is $12\%$ of the salary, $0.12 \times 95000 = 11400$ , i.e. $11400 per year. There are $4$ quarters in a year, so divide by $4$ .

M = \frac{11400}{4} = 2850

so each quarterly contribution is $2850.00.

(b) Convert to a quarterly rate and count the payments. Divide the annual rate by $4$ , and count the quarters in $30$ years.

r = \frac{0.068}{4} = 0.017, \qquad n = 4 \times 30 = 120

Compute the growth factor over the full term.

(1.017)^{120} \approx 7.55987

Apply the future-value-of-annuity formula.

FV = 2850 \cdot \frac{7.55987 - 1}{0.017} = 2850 \times 385.87 \approx 1099742.98

i.e. $1099742.98.

(c) Split into contributions and interest. The total contributed is $120 \times 2850 = 342000$ , i.e. $342000. The interest is the balance less the contributions.

1099742.98 - 342000 = 757742.98

Check. The contributions of $342000.00 and the interest of $757742.98 add to the $1099742.98 balance. Over a $30$ -year career the interest exceeding the contributions is exactly what compounding produces, since the early payments grow for decades.

Answer: (a) each quarterly contribution is $2850.00; (b) the fund is worth $1099742.98 after $30$ years; (c) the employer contributed $342000.00 and $757742.98 is interest.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA