Mastering financial maths is an extremely important skill, not only in High School mathematics, but also in later life.
By being able to adeptly solve financial mathematics questions, students can efficiently put their Series and Sequences, AP and GP knowledge to the test.
Further, grasping the fundamentals of financial mathematics allows students also gain deeper insight into how interest, super, loans, etc., are calculated, preparing them for later life.
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NESA requires students to be proficient in the following syllabus outcomes:
Students should already be familiar with basic series and sequences concepts. This includes being able to prove an AP or GP as well as determine the common differences, ratios, sums, etc.
Students can refresh their knowledge on basic trigonometric functions in the following Year 12 Advanced Subject Guide:
Simple interest is calculated on the investment only and depends intrinsically on the amount of time the money is invested. The interest is usually calculated annually and is a percentage of the original investment.
The formula for simple interest is as follows:
\( I = Prn \)
Where \( I \) simple interest, \( P \) principal amount, \( r \) interest rate, \( n \) number of time periods.
It is important to note that \( r \) would be expressed as a decimal. I.e. if the interest rate is \( 3 \% \ p.a \), \( r = 0.03 \).
On the other hand, compound interest is the interest earned from both the principal amount as well as any past interest accumulated.
As compound interest also takes into account any past interest accumulated, it would be significantly more than simple interest calculated at the same interest rate.
The formula for compound interest is as follows:
\begin{align}
A_n = P(1 + r)^{n}
\end{align}
Where \( A_n \) amount at the end of the \( n^{th} \) period, \( P = \) principal amount, \( r = \) interest rate, \( n = \) number of time periods.
As above, it is important to note that would be expressed as a decimal. I.e. if the interest rate is \( 3 \% \ p.a., r = 0.03 \).
Further, the interest rate must correspond to the time period used.
For example, if the interest rate is given per annum, but our question requires us to compound monthly, we must then divide the interest rate by 12 (as there are 12 months) to obtain \( r \).
An annuity is a regular series of equal investments which are subject to compound interest. These investments are generally made at the end of a time period meaning that no interest is earned until the next compounding period.
With annuities, we are often required to find the final value of an annuity or the present value of an annuity.
The general formula for these two annuities is as follows:
\begin{align}
F = a \Big{(} \frac{(1 + r)^{n} – 1}{r} \Big{)}
\end{align}
Where \( F = \) the final value of the annuity at the end of the \( n^{th} \) period, \( a = \) equal investments made, \( r = \) interest rate, \( n = \) number of time periods.
\begin{align}
P = \frac{F}{(1 + r)^{k}} = \frac{a}{r} \Big{(} 1 – \frac{1}{(1 + r)^{k}} \Big{)}
\end{align}
Where \( P = \) the present/current value of the annuity at the end of a \( k^{th} \) period, \( a = \) equal investments made, \( r = \) interest rate, \( n = \) total number of time periods.
However, generally annuities are generally represented through standard tables, one for the present value of an annuity (PVA) and one for a final value of an annuity (FVA). These standard tables will use $1 as the base value.
The FVA table will represent how the final value of an annuity differs depending on the interest rate and the number of time periods.
Conversely, the PVA table will represent the current value of the annuity and how it differs depending on the interest rate and the number of time periods. Hence, the formulae described above will seldom need to be used.
Example partial tables are shown below:
| FVA | ||
| \( $1 \) | Interest Rate Per Period | |
| \( N \) | \( 1 \% \) | \( 2 \% \) |
| \( 1 \) | \( 1.00 \) | \( 1.00 \) |
| \( 2\) | \( 2.01 \) | \( 2.01 \) |
| PVA | ||
| \( $1 \) | Interest Rate Per Period | |
| \( N \) | \( 1 \% \) | \( 2 \% \) |
| \( 1 \) | \( 0.99 \) | \( 0.98 \) |
| \( 2 \) | \( 1.97 \) | \( 1.94 \) |
A recurrence relation is when the equation is expressed as a function of the preceding terms.
In most cases, the amount left at the end of a time period is expressed as a function of the previous period’s balance.
When constructing a recurrence relation, you must make sure to define the initial terms of the correctly. In particular, you have to be careful if you are analysing the amount left at the beginning or the end of the time period. In most cases, it is best to choose the end of the time period.
An example is as follows:
If I repay \( M \) on my loan each year compounded annually at \( 100r \ \% \ p.a. \):
Let \( A_{n} = \) the amount left in my loan at the end of each year and \( L = \) the original amount of the loan.
| \begin{align} A_{1} &= L – M \\ A_{2} &= A_{1}(1 + r)^{1} – M \\ &= L(1 + r) – M((1+r) + 1) \\ A_{3} &= A_{2}(1+r)^{1} – M \\ &= L(1 + r)^{2} – M \big{(} (1+r)^{2} + (1+r) + 1 \big{)} \end{align} |
Etc.
Recurrence relations are an integral part of a many financial mathematics questions but are most commonly used to determine repayments needed to be made on a loan, etc. Other types of questions which will use recurrence relations include:
Superannuation questions involve regular investments made into a fund for time periods. The fund pays interest which is compounded every period.
These sorts of questions often want us to determine the amount left in the account at the end of \( n \) time periods, i.e. we need to find what \( A_{n} \) is.
To do this we use the recurrence relation described above.
Example 1
Sally opens a new super account and deposits of \( $885 \). At the beginning of each month, her employer matches this deposit. The interest for her account is compounded monthly at a rate of \( 2.65 \ \% \ p.a. \) After 35 years, how much will Sally have in her account?
Solution 1
\begin{align}
A_{1} &= 885 \\
A_{2} &= 885 \Bigg{(} 1 + \Big{(} 1 + \frac{2.65}{100 \times 12} \Big{)} \Bigg{)}
\end{align}
Following this recurrence relation:
| \begin{align} A_{420} &= 885 \Bigg{(} 1 + \Big{(} 1 + \frac{2.65}{100 \times 12} \Big{)} + … + \Big{(} 1 + \frac{2.65}{100 \times 12} \Big{)}^{419} \Bigg{)} \\ &= 885 \Bigg{[} \frac{\big{(} 1 + \frac{2.65}{100 \times 12} \big{)}^{419} – 1}{ \big{(} 1 + \frac{2.65}{100 \times 12} \big{)} – 1} \Bigg{]} \\ &= \$ 611 \ 390.11 \end{align} |
Questions which involve repayments on loans also use the recurrence relation but are a little bit more difficult when compared to superannuation problems. These repayments are made on a timely, regular basis (most often monthly) and are equal.
When solving these questions, we must take into account the interest left on the previous balance of our loan. I.e. the loan keeps growing in value the longer we take to repay — it does not stagnant as the original amount.
Thus, in general, after repayments of \( \$ M \) on a loan worth \( \$ L \) :
| \begin{align} A_{n} &= A_{n-1} (1 + r)^{1} – M \\ &= L(1+r)^{n} – M \big{(} (1 + r)^{n – 1} + … + (1 + r) + 1 \big{)} \end{align} |
Where \( A_{n} \) the balance left in my loan at the end of each time period.
Example 2
Roger borrows $300,000 at an interest rate of \( 5.35 \% \ p.a \). He borrowed the money at the beginning of January, 2012. At the end of each month he pays back an instalment of \( \$ M \). What is the value of \( M \) if the loan is repaid after 20 years?
Solution 2
| \begin{align} A_{1} &= \$ 300 \ 000 \Big{(} 1 + \frac{5.35}{100 \times 12} \Big{)} – M \\ A_{2} &= \$ 300 \ 000 \Big{(} 1 + \frac{5.35}{100 \times 12} \Big{)}^{2} – M \Bigg{(} 1 + \Big{(} 1 + \frac{5.35}{100 \times 12} \Big{)} \Bigg{)} \end{align} |
Following this recurrence relation:
| \begin{align} A_{420} = 300 \ 000 \Big{(} 1 + \frac{5.35}{100 \times 12}^{239} \Big{)} – M \Bigg{(} 1 + \Big{(} 1 + \frac{5.35}{100 \times 12} \Big{)} + … + \Big{(} 1 + \frac{5.35}{100 \times 12} \Big{)}^{239} \Bigg{)} \end{align} |
When \( n = 240, \ A_{n} = 0 \). Therefore, we rearrange and solve for \( M \):
\begin{align}
M &= \frac{ \big{(} 1 + \frac{5.35}{100 \times 12} \big{)}^{240} (300 \ 000)}{ \Bigg{[} \frac{\big{(} 1 + \frac{5.35}{100 \times 12} \big{)}^{240} – 1}{ \big{(} 1 + \frac{5.35}{100 \times 12} \big{)} – 1} \Bigg{]}} \\
&= \$ 2038.33
\end{align}
1. The total value of an investment after earning simple interest for 7 years is $30,598. If original investment was $25,000, what was the interest rate?
2. Veronica opens a savings account and deposits $798 fortnightly. The bank gives Veronica an interest rate of \( 1.15 \% \ p.a. \) Calculate the amount, to the nearest dollar, left in her account after 3 years if the account collects compound interest calculated monthly.
3. Tim’s employer deposits $700 into a super account at the beginning of every month. The interest earned on the account is \( 4.65 \% \ p.a. \) compounding quarterly. If Tim works for this employer for 24 years, how much will he collect when he retires (to the nearest dollar)?
1. \( 3.20 \% \)
2. \( \$ 63 \ 299 \)
3. \( \$122 \ 419 \)
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