Topic guide 8 · Unit 4 Topic 1 · Loans, investments and annuities 1
Compound interest loans and investments
Compound interest is interest earned on interest. At the end of each period the bank adds a percentage of the current balance, so the balance is multiplied by the same number, $r = 1 + i$, every period. The recurrence relation, the compound interest formula and the effective annual rate all come from that one multiplication.
Every Paper 1 from 2020 to 2025 had a question on it, and three of its multiple choice items were answered correctly by fewer than half the state. This guide covers loans and investments that grow with no regular payments. Once a repayment comes off the balance each period, you are in the reducing balance loans guide. Tick off each skill once you can do it without help.
6 of 6
Paper 1s from 2020 to 2025 with a compound interest or effective rate question
20
Paper 1 marks on this sub-topic, 2020 to 2025: 17 short response and 3 multiple choice
34%
of students chose the right answer to 2021 Question 15, the weakest of three effective rate multiple choice results
Skills in this topic
Only tick a skill when you could do it in the exam. Your ticks are saved in this browser only.
Part 1
What the exam asks about compound interest
I count seven Paper 1 questions whose main job is this sub-topic, worth 20 marks: 2020 Q25, 2021 Q15 (multiple choice), 2022 Q17, 2023 Q19, 2024 Q14 (multiple choice) and Q18, and 2025 Q13 (multiple choice). Every paper had at least one, and five of the seven are about the effective annual rate.
Compound interest also turns up as the first stage of bigger finance questions. 2023 Paper 2 Question 7 opens with $100 000 growing at 3.8% p.a. compounding monthly for five years (2 of its 5 marks) before a perpetuity. 2021 Paper 2 Question 3 grows a one-off $10 000 contribution for two years beside an annuity, and 2025 Paper 1 Question 22 uses the effective rate on a perpetuity. All three are in the perpetuities guide. The repayment questions in Unit 4 Topic 1 are in the reducing balance loans guide.
The subject reports keep returning to the same habits. 2020 asked for the difference between the ‘compound interest formula’ and the ‘effective interest formula’, and for calculating interest rates for different compounding periods. 2024 chose an excerpt that identified the monthly rate $i$ and the number of months $n$ before substituting, and labelled the interest calculation separately. 2025 asked for formulas written and values identified before substituting, and for the effective annual rate formula to be used the same way for a perpetuity as for any other compounding investment or loan.
What the syllabus says
Unit 4 Topic 1, Loans, investments and annuities 1, gives the sub-topic Compound interest loans and investments 6 hours and four dot points:
•Use a recurrence relation to model a compound interest loan or investment.
•Use the compound interest formula to model a compound interest loan or investment.
•Calculate the effective annual rate of interest, $i_{\text{effective}}$, and use the results to compare interest on loans or investments when interest is paid or charged for different compounding periods, including daily, monthly, quarterly and six-monthly.
•Solve practical problems involving compound interest loans or investments, including determining the total amount of the loan or investment, total interest, principal, interest rate per year and per compounding period, and the effect of the interest rate and number of compounding periods on the total amount.
The last dot point is longer than it looks. “Interest rate per year and per compounding period” is 2022 Q17, where you work back from $r$ to the annual rate. “The effect of the interest rate and number of compounding periods” is the idea behind 2024 Q14.
What the formula book gives you
Formula book line
Supplied
compound interest
$A = P(1+i)^n$
effective annual rate of interest
$i_{\text{effective}} = (1+i)^k - 1$
recurrence relation for compound interest
$A_{n+1} = rA_n$
Not in the book, so supply these yourself
•$r = 1 + i$. The recurrence relation uses $r$ and the book never says what it is.
•$i$ is the rate per compounding period, as a decimal. Divide the annual rate by the number of periods in a year, and show the division.
•$k$ is the number of compounding periods per year. It has nothing to do with how long the money is invested.
•Total interest is $A - P$.
Older notation in past marking guides
The marking guides from 2020 to 2024 were written for an older formula sheet. There the effective rate is $i_e = \left(1 + \frac{i}{n}\right)^n - 1$, with $i$ the annual rate and $n$ the compounds per year, and 2022 Q17 writes $r = 1 + \frac{i}{n}$. The old $\frac{i}{n}$ is today’s $i$ and the old $n$ is today’s $k$, so the mathematics is identical.
Write your answers in the 2025 notation. The solutions on this page do.
Part 2
Find $i$ and $n$ first, then multiply by $r$
Compound interest has one move. At the end of each compounding period the interest is added to the balance, so the new balance is $r$ times the old one, where $r = 1 + i$. As a recurrence relation:
$$A_{n+1} = rA_n, \quad A_0 = P$$
$A_0$ is the amount invested or borrowed. One step gives $A_1$, the balance after one period, the next gives $A_2$, and so on.
Before $r$ you need $i$, and before any formula you need $i$ and $n$ to count the same periods. The rate in the question is almost always per annum. Divide it by the number of compounding periods in a year, and multiply the years by the same number.
Compounding
$i$ from an annual rate $R$
$n$ for $t$ years
Annually
$R$
$t$
Six-monthly
$R \div 2$
$2t$
Quarterly
$R \div 4$
$4t$
Monthly
$R \div 12$
$12t$
Fortnightly
$R \div 26$
$26t$
Weekly
$R \div 52$
$52t$
Daily
$R \div 365$
$365t$
A 25-year investment at 6.5% p.a. compounding monthly has $i = 0.065 \div 12$ and $n = 12 \times 25 = 300$. Change one without the other and you lose the substitution mark and everything built on it.
Write the setup on its own lines
1the principal, $P$ or $A_0$
2$i$, with the division shown
3$n$, with the multiplication shown
4the formula, then the formula with the numbers in it, then the value
In most finance questions the first mark is for correctly determining $i$ and $n$. The marking guides say it “may be implied by subsequent working”, so a correct substitution earns it anyway. Writing the values out is how you make sure the substitution is correct, and it is what the 2024 and 2025 subject reports singled out.
To write a recurrence relation, evaluate $r$ as a number and state $A_0$. Markers have accepted $(1 + i)$ when $i$ is defined, but a numeric $r$ makes the next part, iterating, mechanical. When you iterate a balance, round to the cent each period as a bank would. In a formula calculation keep the full calculator value and round only the final answer.
Questions also run this backwards. Given $r$, subtract 1 to get $i$, then multiply by the number of periods in a year to get the annual rate.
2022 Paper 1 Question 17
Paper 1 · Simple familiar · [4 marks]
An investment of $50 000 that compounds interest monthly is modelled by the recurrence relation $A_{n+1} = 1.00375A_n$ where $A_0 = 50\,000$.
a) What would be the advertised interest rate per annum, compounding monthly? [2 marks]
b) How many months would it take for the value of the investment to exceed $51 000? [2 marks]
$$\begin{aligned} r &= 1 + i \\ 1.00375 &= 1 + i \\ i &= 0.00375 \end{aligned}$$
That is the rate per month, so the annual rate is $12 \times 0.00375 = 0.045$. The advertised rate is 4.5% p.a. compounding monthly.
1 mark · correctly substitutes into an appropriate rule
1 mark · calculates annual interest rate
The guide writes $r = 1 + \frac{i}{n}$ with $i$ annual and $n = 12$, the older notation. It accepts alternative approaches and equivalent decimal forms such as 0.045. The monthly rate, 0.375%, is not the annual rate, and neither is the effective rate, 4.59%, which the question did not ask for.
b) Iterate, multiplying by 1.00375 each month, until the balance passes $51 000:
Month $n$
$A_n$
0
$50 000.00
1
$50 187.50
2
$50 375.70
3
$50 564.61
4
$50 754.23
5
$50 944.56
6
$51 135.60
Or test values of $n$ in the formula: $50\,000 \times 1.00375^5 = 50\,944.56$, still below $51 000, and $50\,000 \times 1.00375^6 = 51\,135.60$, above it. It takes 6 months.
1 mark · correctly uses an appropriate method
1 mark · determines when the investment would exceed $51 000
The guide accepts rounding errors and other methods, including logarithms. It also lists “after 5 months” as an equivalent response, read as the balance passing $51 000 during the sixth month. The 2022 subject report showed the recursion route, with the sequence of amounts written out up to and past $51 000. Write the sequence, then say 6 months.
Answer
a) 4.5% p.a. compounding monthly · b) 6 months
Part 3
The compound interest formula jumps straight to the end
Iterating 216 times is not an option in an exam. Multiplying by $r$ again and again is a power, so after $n$ periods:
$$A = P(1+i)^n$$
$A$ is the total amount, $P$ the principal, $i$ the rate per compounding period and $n$ the number of periods. The interest is what you end with minus what you started with, $I = A - P$. Questions ask for the total, the interest or the principal, so read which one before you stop.
Finding the principal
If you know the target and need the starting amount, divide instead of multiplying. For $15 000 in four years at 4.4% p.a. compounding monthly, $i = 0.044 \div 12$ and $n = 12 \times 4 = 48$:
So $12 583.32 invested now grows to $15 000, and $2416.68 of that is interest. The principal must be smaller than the target. If yours is bigger, you multiplied.
Finding the number of periods
For the number of periods, test values of $n$ in the formula (or iterate) until the amount passes the target, as in 2022 Q17 b). The answer is the first whole period that gets there, and it is worth writing the value just below the target and the value just above it.
What changes the total
The syllabus asks for the effect of the interest rate and the number of compounding periods on the total amount. Here is $10 000 invested for five years at 6% p.a.:
Compounding
Total after 5 years
Annually
13 382.26
Six-monthly
13 439.16
Quarterly
13 468.55
Monthly
13 488.50
Daily
13 498.26
Amounts in dollars. The interest is the total minus 10 000, from 3382.26 annually to 3498.26 daily. $n$ is 5, 10, 20, 60 and 1825 periods.
More frequent compounding gives more, but the gains shrink. Going from monthly to daily adds less than $10 over five years. The rate matters far more: at 7% p.a. compounding monthly the same $10 000 reaches $14 176.25.
2024 Paper 1 Question 18
Paper 1 · Simple familiar · [4 marks]
When a child is born, their parent deposits $3000 to open an investment account earning interest at 4.2% p.a. compounding monthly. If there are no further transactions and the interest rate does not change, calculate the amount of interest earned by the child’s 18th birthday.
$$\begin{aligned} I &= A - P \\ &= 6380.79 - 3000 \\ &= 3380.79 \end{aligned}$$
1 mark · determines amount of interest earned
The amount of interest earned is $3380.79.
The first mark may be implied by later working, and the guide allows follow-through and alternative approaches. For the value of the investment it accepts rounding to the nearest cent, nearest 5 cents or nearest dollar. The 2024 subject report chose an excerpt that identified $i$ and $n$ before substituting and labelled the interest calculation separately.
Two traps. The question asks for the interest, so $6380.79 on its own is not the answer. And using 0.042 with $n = 216$ gives over $21 million, which should stop you.
Answer
$3380.79
Try it yourself
Grow an investment with $A = P(1+i)^n$
Set the principal, the rate, the term and how often the interest compounds. The working shows $i$ and $n$ first, then the substitution, then the interest, which is the order the marking guides reward. The bars show the balance at the end of each year, and the orange part is the interest earned so far.
Change the compounding and watch the table: more frequent compounding always gives a little more, but changing the rate moves the total far more.
principalinterest earned so farBalance at the end of each year
Compounding
Load an example
Find $i$ and $n$ first
Substitute
Interest
Answer
Total , of which is interest
The same principal, rate and term with every compounding period. Your choice is in orange.
Compounding
$k$
$n$
Total
Part 4
The effective rate counts compounds per year, never the term
Two rates with different compounding periods cannot be compared by their nominal rates. The effective annual rate turns each one into the single rate, compounded once a year, that gives the same growth over one year:
$$i_{\text{effective}} = (1+i)^k - 1$$
Here $i$ is the rate per compounding period and $k$ is the number of compounding periods per year. At 6% p.a. compounding monthly, $i = 0.005$ and $k = 12$, so $i_{\text{effective}} = 1.005^{12} - 1 = 0.0617$, or 6.17%. Money at 6% compounding monthly grows exactly as fast as money at 6.17% compounding once a year.
3.64% p.a. compounding quarterly, so $i = 0.0091$ and each step multiplies by 1.0091. The effective annual rate is the growth over four steps, one year: 3.69%. Eight steps give the growth over two years, 7.52%, which is not an annual rate. The animation adds one quarter at a time.
Because the effective rate is about one year, the term of the investment never enters it. All three effective rate multiple choice items were answered correctly by fewer than half the state, and each one went wrong in a different way.
Question
Correct
The idea that cost marks
2021 Q15
34.0%
More frequent compounding always wins
2024 Q14
44.3%
Changing when interest is charged leaves the rate alone
2025 Q13
44.9%
The term of the investment belongs in the calculation
Before you substitute, write $k =$ and the number of compounds per year: 2 for six-monthly, 4 for quarterly, 12 for monthly, 26 for fortnightly. 2021 Q15 is in Part 5 with the other comparisons.
2025 Paper 1 Question 13
Paper 1 · Multiple choice · [1 mark]
Determine the effective annual rate of interest for a two-year investment, for which the interest compounds quarterly at 3.64% p.a.
(A) uses $k = 2$, as if the interest compounded six-monthly. (C) doubles 3.64% for two years with no compounding, and it was the most common wrong answer (20.8%). (D) uses $k = 8$, which is the growth over the whole two years, not an annual rate. 44.9% of students chose (B).
Answer
(B) 3.69%
2024 Paper 1 Question 14
Paper 1 · Multiple choice · [1 mark]
Which option will not change the effective annual rate of interest for a loan?
$(1+i)^k - 1$ has two inputs, the rate per period $i$ and the number of compounding periods per year $k$. (A) changes $i$. (B) and (D) are the same change in different words: charging interest monthly instead of quarterly changes $k$, and $i$ with it. At 6% p.a., monthly gives 6.17% and quarterly gives 6.14%.
The repayment appears nowhere in the formula. Changing it changes how quickly the loan is paid off, but not the effective rate. 44.3% of students chose (C), and 31.5% chose (B).
Every comparison question has the same three moves: calculate, compare, conclude. Calculate the effective rate of each option, compare the two numbers with a $\gt$ or $\lt$, then write a sentence that names the option and gives the reason. For an investment the higher effective rate is better. For a loan the lower one is.
Keep at least four significant figures in each effective rate. The options are usually close. In 2023 the gap was 0.0064 of a percentage point, and rounded to one decimal place both options were 5.7%.
Never decide from the nominal rates. In 2020 the higher nominal rate won, in 2023 the lower one did, and in 2021 neither the highest rate nor the most frequent compounding came out on top.
2020 Paper 1 Question 25
Paper 1 · Simple familiar · [5 marks]
A financial institution offers two investment options:
Option 1: 7% p.a. compounding quarterly
Option 2: 6.8% p.a. compounding monthly
Use the effective interest rate formula to determine the option that will provide the better return.
1 mark · correctly substitutes into appropriate rule (Option 2)
1 mark · calculates the effective interest rate for Option 2
$0.07186 \gt 0.07016$, so Option 1 provides the better return because it has the higher effective interest rate.
1 mark · states better option
Every calculation mark allows follow-through from an error in substitution, and the last mark allows follow-through for an option linked to your earlier working. The question names the effective interest rate formula, so choosing Option 1 because 7% is bigger than 6.8% does not do what was asked.
Answer
Option 1 (7.19% against 7.02%)
2023 Paper 1 Question 19
Paper 1 · Simple familiar · [4 marks]
Ngarra compares two investment options and decides option A will provide the better return.
•Option A: 5.60% p.a. compounding monthly
•Option B: 5.62% p.a. compounding quarterly
Use the effective annual rate of interest formula to evaluate the reasonableness of Ngarra’s decision.
1 mark · correctly substitutes into appropriate rule for either option
1 mark · calculates effective interest rate for option A
1 mark · calculates effective interest rate for option B
$0.05746 \gt 0.05740$. Ngarra’s decision is reasonable because option A has a higher effective interest rate, even though its nominal rate is lower.
1 mark · provides a statement of reasonableness linked to effective interest rate
The guide writes the rates as 0.05745... and 0.05739..., cut off rather than rounded, and allows follow-through. The 2023 subject report chose an excerpt that compared the two calculated rates to identify the higher one before stating reasonableness. “Evaluate the reasonableness” wants a verdict, reasonable or not, and the reason.
Answer
Reasonable: A 5.746% against B 5.740%
2021 Paper 1 Question 15
Paper 1 · Multiple choice · [1 mark]
Which of the following investment options gives the best return?
The nominal rates are only 0.06 of a percentage point apart, so every option has to be calculated:
Option
$k$
$i_{\text{effective}}$
(A)
365
6.1088%
(B)
12
6.1150%
(C)
4
6.1050%
(D)
2
6.0797%
In each row $i$ is the nominal rate divided by $k$, for example $0.0593 \div 365$ for (A).
(B) is highest, by 0.006 of a percentage point over (A). Daily compounding on the lowest rate does not beat monthly compounding on a slightly higher one, and the highest nominal rate, (D), comes last. 35.4% of students chose (A), more than the 34.0% who chose (B).
Answer
(B) 5.95% p.a. compounding monthly
The same comparison runs on perpetuities in 2025 Paper 1 Question 22, and a compound interest balance funds a perpetuity in 2023 Paper 2 Question 7. Both are in the perpetuities guide.
Try it yourself
Calculate, compare, conclude
Set up two options. The page calculates each effective annual rate with $k$ as the compounding periods per year, compares them with $\gt$ or $\lt$, and writes the conclusion. Switch between an investment and a loan to see the decision flip: for a loan the lower effective rate wins.
The graph shows one year of growth for each option. Its height at one year is $1 + i_{\text{effective}}$, and nothing after one year matters.
start6 months1 year
Option AOption BValue of 1 dollar over one year
Option ACompounding
Option BCompounding
Load a past exam question
Option A
Option B
Compare
Conclude
Part 6
Earning the marks in finance questions
The marking guides reward the same things every year. If I had to guess where you’ll lose marks, it is the conclusion sentence and the question that asked for interest rather than the total.
•$i$ and $n$ are a mark. 2024 Q18 gives its first mark for determining them, and so do the repayment questions 2020 Q20, 2021 Q17, 2022 Q25 and 2023 Q17. The guides say it can be implied by working, but only by working that is right.
•The substitution is its own mark. In 2020 Q25 and 2023 Q19 the substituted formula earned a mark before any value was calculated. Write the formula, then the line with the numbers in it, then the value.
•Follow-through is generous. Almost every mark after the first allows for errors in prior working, so a wrong $i$ used correctly still earns later marks. That only works if the marker can see what you did.
•The decision needs its reason. The last mark in 2020 Q25 is “states better option”, and in 2023 Q19 it is “a statement of reasonableness linked to effective interest rate”. Justify, don’t assert: write the two rates with $\gt$ or $\lt$ between them, then the sentence.
•Answer the question asked. 2024 Q18 wanted the interest, not the balance. 2022 Q17 a) wanted the annual rate, not the monthly one.
•Rounding. Money to the cent unless the question says otherwise. 2024 Q18 accepted the nearest cent, 5 cents or dollar for the balance, and the 2020 subject report gave $2363.225 ≈ $2363.23 as its example of rounding to the nearest cent. Effective rates need enough figures to separate the options.
•Units and context. The 2020 subject report asked for units on quantities such as $8000. State a rate with its compounding, as the 2022 Q17 sample does: “4.5% p.a. compounding monthly”.
Game plan
Your game plan for compound interest
1Decide the model. Interest added with no regular payments is compound interest. A payment coming off every period is a reducing balance loan, in the reducing balance loans guide.
2Write $P$, then $i$ with the division shown, then $n$ with the multiplication shown. Check that $i$ and $n$ count the same periods.
3For a recurrence relation, write $r = 1 + i$ as a number and state $A_0$. Iterate one line per period, rounding to the cent.
4For a total after many periods, use $A = P(1+i)^n$. For the interest, subtract $P$. For the principal, divide the target by $(1+i)^n$. For the number of periods, test values of $n$ either side of the target.
5For an effective rate, write $k =$ compounds per year before you substitute. The term of the investment never enters it.
6To compare, calculate each effective rate to at least four significant figures, compare them with $\gt$ or $\lt$, and conclude in a sentence that names the option.
7Check the answer is the quantity asked for (total, interest, principal, annual rate or number of periods), rounded as asked, with its unit.
I wrote these four to the QCAA format and checked every answer in Python. Questions 1 and 2 cover the recurrence relation and the compound interest formula. Questions 3 and 4 are effective rate questions.
Question 1 of 4
Skill 1: Rate per period and the recurrence relation · Simple familiar · [5 marks]
Ari invests $6000 in an account earning 3.6% p.a. compounding monthly. There are no further deposits or withdrawals.
a) Use the monthly interest rate to write a recurrence relation for the value of the investment after $n$ months. [2 marks]
b) Calculate the value of the investment after three months. [1 mark]
c) How many months would it take for the value of the investment to exceed $6500? [2 marks]
a)
$$\begin{aligned} i &= \frac{0.036}{12} = 0.003 \\ r &= 1 + 0.003 = 1.003 \end{aligned}$$
1 mark · correctly calculates the monthly interest rate
After 26 months the balance is still below $6500, and after 27 months it is above, so it takes 27 months.
1 mark · determines the number of months
Answer
a) $A_{n+1} = 1.003A_n$, $A_0 = 6000$ · b) $6054.16 · c) 27 months
Question 2 of 4
Skill 2: The compound interest formula · Simple familiar · [4 marks]
Mei wants $20 000 for a house deposit in six years. Her savings account earns 5.2% p.a. compounding quarterly.
a) Determine the amount Mei must invest now, with no further deposits, to have $20 000 in six years. [3 marks]
b) Calculate the total interest Mei will earn. [1 mark]
a)
$$\begin{aligned} A &= 20\,000 \\ i &= \frac{0.052}{4} = 0.013 \\ n &= 6 \times 4 = 24 \end{aligned}$$
1 mark · correctly determines the $i$ and $n$ values
$$\begin{aligned} A &= P(1+i)^n \\ 20\,000 &= P(1.013)^{24} \\ P &= \frac{20\,000}{1.013^{24}} \end{aligned}$$
1 mark · substitutes into the compound interest formula
$$P = 14\,669.09$$
1 mark · determines the amount to invest
Mei must invest $14 669.09.
b)
$$\begin{aligned} I &= A - P \\ &= 20\,000 - 14\,669.09 \\ &= 5330.91 \end{aligned}$$
1 mark · determines the total interest
Check it forwards: $14\,669.09 \times 1.013^{24} \approx 20\,000$. Multiplying by $1.013^{24}$ instead of dividing gives $27 268.21, more than the target, which cannot be right.
(A) is the nominal rate, with no compounding at all. (C) multiplies 6% by five years. (D) uses $k = 60$, the compounds over the whole term, which is the trap in 2025 Q13.
Answer
(B) 6.17%
Question 4 of 4
Skill 4: Compare options with the effective rate · Simple familiar · [4 marks]
Marika compares two savings accounts and decides that account P will provide the better return.
•Account P: 4.80% p.a. compounding monthly
•Account Q: 4.85% p.a. compounding quarterly
Use the effective annual rate of interest formula to evaluate the reasonableness of Marika’s decision.
Account P compounds monthly, so $i = 0.048 \div 12 = 0.004$ and $k = 12$.
1 mark · calculates the effective rate for account Q
$0.04907 \lt 0.04939$. Marika’s decision is not reasonable, because account Q has the higher effective annual rate, by about 0.03 of a percentage point.
1 mark · provides a statement of reasonableness linked to the effective rates
This is the reverse of 2023 Q19, where the monthly option won, so compute both every time. Keep at least four significant figures: rounded to one decimal place, both are 4.9%.
Answer
Not reasonable: Q 4.939% against P 4.907%
Sources: QCAA General Mathematics question and response books, multiple choice books and marking guides, 2020 to 2025. General Mathematics 2025 v1.3 general senior syllabus, Unit 4 Topic 1. QCAA General Mathematics subject reports, 2020 to 2025. General Mathematics formula book 2025, and the earlier formula sheet used in the 2020 to 2024 marking guides. Every numerical answer was checked in Python against the published marking guides.
