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# Chapter 3: Finance

• Discuss terminology.
• Very important to emphasize not rounding off calculations until final answer as this affects accuracy.
• Learners should do calculation in one step using the memory function on their calculators.
• Draw timelines showing the different time periods, interest rates and any deposits/withdrawals.
• Explain and discuss the difference between future and present value annuities.
• Learners must always check how often a given interest rate is compounded.
• Learners must be careful to calculate the correct number of payments for the term of the investment.

In earlier grades we studied simple interest and compound interest, together with the concept of depreciation. Nominal and effective interest rates were also described.

• Simple interest: $$A = P(1 + in)$$
• Compound interest: $$A = P(1 + i)^n$$
• Simple depreciation: $$A = P(1 - in)$$
• Compound depreciation: $$A = P(1 - i)^n$$
• Nominal and effective annual interest rates: $$1 + i = \left( 1 + \frac{i^{(m)}}{m} \right)^{m}$$

In this chapter we look at different types of annuities, sinking funds and pyramid schemes. We also look at how to critically analyse investment and loan options and how to make informed financial decisions.

Financial planning is very important and it allows people to achieve certain goals, such as supporting a family, attending university, buying a house, and saving enough money for retirement. Prudent financial planning includes making a budget, opening a savings account, wisely investing savings and planning for retirement.

## 3.1 Calculating the period of an investment (EMCFX)

For calculations using the simple interest formula, we solve for $$n$$, the time period of an investment or loan, by simply rearranging the formula to make $$n$$ the subject. For compound interest calculations, where $$n$$ is an exponent in the formula, we need to use our knowledge of logarithms to determine the value of $$n$$.

$A = P{\left(1+i\right)}^{n}$ \begin{align*} A &= \text{accumulated amount} \\ P &= \text{principal amount} \\ i &= \text{interest rate written as a decimal} \\ n &= \text{time period} \end{align*}

Solving for $$n$$:

\begin{align*} A & = P{\left(1+i\right)}^{n} \\ \frac{A}{P} & = {\left(1+i\right)}^{n} \\ \text{Use definition: } n & = \log_ {\left(1+i\right)}{\left( \frac{A}{P} \right)} \\ \text{Change of base: } n &= \frac{\log\left(\frac{A}{P}\right)}{\log(1 + i)} \end{align*}

## Worked example 1: Determining the value of $$n$$

Thembile invests $$\text{R}\,\text{3 500}$$ into a savings account which pays $$\text{7,5}\%$$ per annum compounded yearly. After an unknown period of time his account is worth $$\text{R}\,\text{4 044,69}$$. For how long did Thembile invest his money?

### Write down the compound interest formula and the known values

$A=P{\left(1+i\right)}^{n}$ \begin{align*} A &= \text{4 044,69} \\ P &= \text{3 500} \\ i &= \text{0,075} \end{align*}

### Substitute the values and solve for $$n$$

\begin{align*} A &=P{\left(1+i\right)}^{n} \\ \text{4 044,69} &=\text{3 500}{\left(1+\text{0,075}\right)}^{n} \\ \frac{\text{4 044,69}}{\text{3 500}} & = \text{1,075}^{n} \\ \therefore n &= \log_{(\text{1,075})} \left({\frac{\text{4 044,69}}{\text{3 500}}} \right) \\ &= \frac{\log{\frac{\text{4 044,69}}{\text{3 500}}}}{\log{\text{1,075}}} \\ &= \text{2,00} \ldots \end{align*}

The $$\text{R}\,\text{3 500}$$ was invested for $$\text{2}$$ years.

## Worked example 2: Duration of investments

Margo has $$\text{R}\,\text{12 000}$$ to invest and needs the money to grow to at least $$\text{R}\,\text{30 000}$$ to pay for her daughter's studies. If it is invested at a compound interest rate of $$\text{9}\%$$ per annum, determine how long (in full years) her money must be invested?

### Write down the compound interest formula and the known values

$A=P{\left(1+i\right)}^{n}$ \begin{align*} A &= \text{30 000} \\ P &= \text{12 000} \\ i &= \text{0,09} \end{align*}

### Substitute the values and solve for $$n$$

\begin{align*} A &= P\left( 1+i \right)^{n} \\ \text{30 000} &= \text{12 000} \left( 1+\text{0,09}\right)^{n} \\ \frac{5}{2} &= (\text{1,09})^{n} \\ \therefore n &= \log_{\text{1,09}}{\left( \frac{5}{2} \right)} \qquad (\text{use definition}) \\ &=\frac{\log{\frac{5}{2}}}{\log{\text{1,09}}} \qquad (\text{change of base}) \\ &= \text{10,632} \ldots \end{align*}

In this case we round up, because $$\text{10}$$ years will not yet deliver the required $$\text{R}\,\text{30 000}$$. Therefore the money must be invested for at least $$\text{11}$$ years.

# Success in Maths and Science unlocks opportunities

## Determining the period of an investment

Exercise 3.1

Nzuzo invests $$\text{R}\,\text{80 000}$$ at an interest rate of $$\text{7,5}\%$$ per annum compounded yearly. How long will it take for his investment to grow to $$\text{R}\,\text{100 000}$$?

\begin{align*} A & = P(1 + i)^{n} \\ \text{100 000} & = \text{80 000} \left(1 + \frac{\text{7,5}}{100} \right)^{n} \\ \frac{\text{100 000}}{\text{80 000}} &= (\text{1,075})^{n} \\ \frac{\text{5}}{\text{4}} &= (\text{1,075})^{n} \\ \therefore n &= \log_{\text{1,075}}{\text{1,25}} \\ &= \frac{\log{\text{1,25}}}{\log{\text{1,075}}} \\ &= \text{3,09} \ldots \end{align*}

It will take just over $$\text{3}$$ years.

Sally invests $$\text{R}\,\text{120 000}$$ at an interest rate of $$\text{12}\%$$ per annum compounded quarterly. How long will it take for her investment to double?

\begin{align*} A & = P(1 + i)^{n} \\ \text{240 000} & = \text{120 000} \left(1 + \frac{\text{12}}{100 \times 4} \right)^{4n} \\ \frac{\text{240 000}}{\text{120 000}} &= (\text{1,03})^{4n} \\ 2 &= (\text{1,03})^{4n} \\ \therefore 4n &= \log_{\text{1,03}}{\text{2}} \\ &= \frac{\log{\text{2}}}{\log{\text{1,03}}} \\ &= \text{23,449} \ldots \\ \therefore n &= \text{5,86} \ldots \end{align*}

It will take just over $$\text{5}$$ years and $$\text{10}$$ months.

When Banele was still in high school he deposited $$\text{R}\,\text{2 250}$$ into a savings account with an interest rate of $$\text{6,99}\%$$ per annum compounded yearly. How long ago did Banele open the account if the balance is now $$\text{R}\,\text{2 882,53}$$? Write the answer as a combination of years and months.

This is a compound interest problem:

\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{2 882,53} \\ P & = \text{2 250} \\ i & = \text{0,0699} \end{align*}

Now substitute the known values and solve for $$n$$:

\begin{align*} \text{2 882,53} & = \text{2 250} \left( 1 + \text{0,0699} \right) ^ {n} \\ \text{2 882,53} & = \text{2 250}(\text{1,0699})^{n} \\ \frac{\text{2 882,53}}{\text{2 250}} & = (\text{1,0699})^{n} \end{align*}

Change to logarithmic form:

\begin{align*} & n = \log_{\text{1,0699}} \left(\frac{\text{2 882,53}}{\text{2 250}} \right) \\ & n = \text{3,6666} \ldots \end{align*}

Banele left the money in the account for about $$\text{3,67}$$ years.

However, we must give our answer in terms of years and months, not as a decimal number of years. $$\text{3,6666} \ldots$$ years means $$\text{3}$$ years and some number of months; to figure out how many months, we need to convert $$\text{0,6666} \ldots$$ years into months.

We know that there are $$\text{12}$$ months in a year.

To convert $$\text{0,6666} \ldots$$ years into months, do the following:

$(\text{0,6666} \ldots )\text{ year} \times \left( \frac{12\text{ months}}{\text{year}} \right) = \text{8} \text{ months}$

Banele deposited the money into the account $$\text{3}$$ years and $$\text{8}$$ months ago.

The annual rate of depreciation of a vehicle is $$\text{15}\%$$. A new vehicle costs $$\text{R}\,\text{122 000}$$. After how many years will the vehicle be worth less than $$\text{R}\,\text{40 000}$$?

\begin{align*} A & = P(1 - i)^{n} \\ \text{40 000} & = \text{122 000} \left(1 - \frac{15}{100} \right)^{n} \\ \frac{\text{40 000}}{\text{122 000}} &= (\text{0,85})^{n} \\ \frac{\text{20}}{\text{61}} &= (\text{0,85})^{n} \\ \therefore n &= \log_{\text{0,85}}{\frac{\text{20}}{\text{61}} } \\ &= \frac{\log{\frac{\text{20}}{\text{61}} }}{\log{\text{0,85}}} \\ &= \text{6,86} \ldots \end{align*}

The vehicle will be worth less than $$\text{R}\,\text{40 000}$$ after about $$\text{7}$$ years.

Some time ago, a man opened a savings account at KMT South Bank and deposited an amount of $$\text{R}\,\text{2 100}$$. The balance of his account is now $$\text{R}\,\text{3 160,59}$$. If the account gets $$\text{8,52}\%$$ compound interest p.a., determine how many years ago the man made the deposit.

We write the compound interest formula and the given information:

\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{3 160,59} \\ P & = \text{2 100} \\ i & = \text{0,0852} \end{align*}

We know everything except for the value of $$n$$. Substitute and then solve for $$n$$.

\begin{align*} \text{3 160,59} & = \text{2 100} \left( 1 + \text{0,0852} \right) ^ {n} \\ \text{3 160,59} & = \text{2 100}(\text{1,0852})^{n} \\ \frac{\text{3 160,59}}{\text{2 100}} & = (\text{1,0852})^{n} \end{align*}

Use the definition of a logarithm to solve for $$n$$:

$n = \log_{\text{1,0852}} \left(\frac{\text{3 160,59}}{\text{2 100}} \right)$

Use a calculator to evaluate the $$\log$$:

$n = \text{4,999} \ldots$

The man made the deposit $$\text{5}$$ years ago.

Mr. and Mrs. Dlamini want to save money for their son's university fees. They deposit $$\text{R}\,\text{7 000}$$ in a savings account with a fixed interest rate of $$\text{6,5}\%$$ per year compounded annually. How long will take for this deposit to double in value?
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{14 000} \\ P & = \text{7 000} \\ i & = \text{0,065} \end{align*} \begin{align*} \text{14 000} &= \text{7 000} \left( 1 + \frac{\text{6,5}}{100} \right)^{n} \\ \text{2} &= \left( \text{1,065} \right)^{n} \\ \therefore n &= \log_{\text{1,065}}{\text{2}} \\ &= \frac{\log{\text{2}}}{\log{\text{1,065}}} \\ &= \text{11,00} \ldots \\ &= \text{11} \end{align*}

It will take $$\text{11}$$ years for their deposit to double in value.

A university lecturer retires at the age of $$\text{60}$$. She has saved $$\text{R}\,\text{300 000}$$ over the years.

She decides not to let her savings decrease at a rate faster than $$\text{15}\%$$ per year. How old will she be when the value of her savings is less than $$\text{R}\,\text{50 000}$$?
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{50 000} \\ P & = \text{300 000} \\ i & = \text{0,15} \end{align*} \begin{align*} \text{50 000} &= \text{300 000}\left( 1 - \frac{15}{100}\right)^{n} \\ \frac{1}{6} &= \left( \text{0,85}\right)^{n} \\ \therefore n &= \frac{\log{\frac{1}{6}}}{\log{\text{0,85}}} \\ &= \text{11,024} \ldots \end{align*}

If she manages her money carefully, she will be $$\text{71}$$ years or older.

If she doesn't use her savings and invests all her money in an investment account that earns a fixed interest rate of $$\text{5,95}\%$$ per annum, how long will it take for her investment to grow to $$\text{R}\,\text{390 000}$$?
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{390 000} \\ P & = \text{300 000} \\ i & = \text{0,0595} \end{align*} \begin{align*} \text{390 000} &= \text{300 000}\left( 1 + \frac{\text{5,95}}{100}\right)^{n} \\ \text{1,3} &= \left( \text{1,0595}\right)^{n} \\ \therefore n &= \frac{\log{ \text{1,3}}}{\log{\text{1,0595}}} \\ &= \text{4,539} \ldots \end{align*}

It will take less than $$\text{5}$$ years.

Simosethu puts $$\text{R}\,\text{450}$$ into a bank account at the Bank of Upington. Simosethu's account pays interest at a rate of $$\text{7,11}\%$$ p.a. compounded monthly. After how many years will the bank account have a balance of $$\text{R}\,\text{619,09}$$?

Use the compound interest formula and determine the value of $$n$$.

\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{619,09} \\ P & = \text{450} \\ i & = \text{0,0711} \end{align*}

In this question the interest is payable every month. Therefore $$i \rightarrow \frac{\text{0,0711}}{\text{12}}$$ and $$n \rightarrow (n \times \text{12})$$. In this case, $$n$$ represents the number of years; the product $$(n \times \text{12})$$ represents the number of times the bank pays interest into the account.

\begin{align*} \text{619,09} & = \text{450} \left( 1 + \frac{\text{0,0711}}{\text{12}} \right) ^ {(n \times \text{12})} \end{align*} \begin{align*} \text{619,09} & = \text{450}(\text{1,00592} \ldots)^{\text{12}n} \\ \frac{\text{619,09}}{\text{450}} & = (\text{1,00592} \ldots)^{\text{12}n} \\ \text{1,37575} \ldots & = (\text{1,00592} \ldots )^{\text{12}n} \end{align*}

At this point we must change the equation to logarithmic form:

\begin{align*} \text{12}n & = \log_{\text{1,005925}} (\text{1,37575} \ldots ) \\ \text{12}n & = \text{54} \end{align*}

To find the number of years, we solve for $$n$$:

\begin{align*} \text{12}n & = \text{54} \\ n & = \frac{\text{54}}{\text{12}} \\ n & = \text{4,5} \end{align*}

The money has been in the Simosethu's account for $$\text{4,5}$$ years.