# Part 7: Financial Mathematics | Free Worksheet

Has Financial Maths not accrued your interest? Well, after you read this article, it will have! In this article, we dsicuss everything you need to know to master financial maths, including types of interest, modelling investments and loans and harder questions.

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.

## NESA Syllabus Outcomes: Financial Maths

NESA requires students to be proficient in the following syllabus outcomes:

• Solve compound interest problems involving financial decisions, including a home loan, a savings account, a car loan or superannuation
• Use geometric sequences to model and analyse practical problems involving exponential growth and decay (ACMMM076) o Calculate the effective annual rate of interest and use results to compare investment returns and cost of loans when interest is paid or charged daily, monthly, quarterly or six-monthly (ACMGM095)
• Solve problems involving compound interest loans or investments, eg determining the future value of an investment or loan, the number of compounding periods for an investment to exceed a given value and/or the interest rate needed for an investment to exceed a given value (ACMGM096)
• Recognise a reducing balance loan as a compound interest loan with periodic repayments, and solve problems including the amount owing on a reducing balance loan after each payment is made
•  Solve problems involving financial decisions, including a home loan, a savings account, a car loan or superannuation

## Assumed Knowledge

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:

• Series and Sequences

## Types of Interest

### Simple Interest

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 $$3 \% \ p.a$$, $$r = 0.03$$.

### Compound Interest

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 $$r$$.

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## Modelling Investment and Loans

### Annuity

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.

### Home Loans/Time Repayments

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}

## Concept Check Questions

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)?

## Concept Check Solutions

1. $$3.20 \%$$

2. $$\ 63 \ 299$$

3. $$\122 \ 419$$

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