Mpumelelo deposits \(\text{R}\,\text{500}\) into a savings account, which earns interest at
\(\text{6,81}\%\) p.a. compounded quarterly. How long will it take for the savings account to have a balance
of \(\text{R}\,\text{749,77}\)?

\begin{align*}
A & = P(1+i)^n \\
\text{Where: } & \\
A & = \text{749,77} \\
P & = \text{500} \\
i & = \text{0,0681}
\end{align*}

In this question the interest is payable quarterly, so \(i \rightarrow \frac{\text{0,0681}}{\text{4}}\) and
\(n \rightarrow (n \times \text{4})\). In this case, \(n\) represents the number of years; the product \((n
\times \text{4})\) represents the number of times the bank pays interest into the account.

\begin{align*}
\text{749,77} & = \text{500} \left( 1 + \frac{\text{0,0681}}{\text{4}} \right) ^ {(n \times \text{4})}
\end{align*}
\begin{align*}
\text{749,77} & = \text{500}(\text{1,01702} \ldots)^{\text{4}n} \\
\frac{\text{749,77}}{\text{500}} & = (\text{1,01702} \ldots)^{\text{4}n} \\
\text{1,49954} \ldots & = (\text{1,01702} \ldots)^{\text{4}n}
\end{align*}

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

\begin{align*}
\text{4}n & = \log_{\text{1,017025}} (\text{1,49954} \ldots) \\
\text{4}n & = \frac{\log{\text{1,49954} \ldots}}{\log{\text{1,017025}} } \qquad (\text{change of base}) \\
\text{4}n & = \text{24}
\end{align*}

To find the number of years, we solve for \(n\):

\begin{align*}
\text{4}n & = \text{24} \\
n & = \frac{\text{24}}{\text{4}} \\
n & = \text{6}
\end{align*}

Therefore, it will take \(\text{6}\) years.

How much interest will Gavin pay on a loan of \(\text{R}\,\text{360 000}\) for \(\text{5}\) years at
\(\text{10,3}\%\) per annum compounded monthly?

\[P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}\]
\begin{align*}
\text{360 000} & = \frac{x \left[ 1 - \left(1 + \frac{\text{0,103}}{12}\right) ^{-(\text{5} \times
12)} \right]} {\left(\frac{\text{0,103}}{12} \right)} \\
\therefore x & = \frac{ \text{360 000} \times \left(\frac{\text{0,103}}{12} \right)}{ \left[ 1 -
\left(1 + \frac{\text{0,103}}{12}\right) ^{-\text{60}} \right]} \\
&= \text{7 702,184} \ldots
\end{align*}

Monthly payments are \(\text{R}\,\text{7 702,18}\).

Total loan:

\[\text{R}\,\text{7 702,18} \times\text{12} \times \text{5} = \text{R}\,\text{462 130,80}\]

Interest:

\[\text{R}\,\text{462 130,80} - \text{R}\,\text{360 000} = \text{R}\,\text{102 130,80}\]

Determine how many years (to the nearest integer) it will take for the value of a motor vehicle to decrease
to \(\text{25}\%\) of its original value if the rate of depreciation, based on the reducing-balance
method, is \(\text{21}\%\) per annum.

Let the value of the vehicle be \(x\).

\begin{align*}
A & = P(1 - i)^{n} \\
x \times \frac{25}{100} & = x\left(1 - \frac{21}{100}\right)^{n} \\
\text{0,25} & = \left(1 - \frac{21}{100}\right)^{n} \\
\text{0,25} & = ( \text{0,79})^{n} \\
\therefore n &= \log_{\text{0,79} }{\text{0,25} } \qquad (\text{use definition}) \\
&= \frac{\log{\text{0,25}}}{\log{\text{0,79} }} \qquad (\text{change of base}) \\
&= \text{5,881} \ldots
\end{align*}

Therefore, it will take about \(\text{6}\) years.

Thabo invests \(\text{R}\,\text{8 500}\) in a special banking product which will pay \(\text{1}\%\) per
annum for 1 month, then \(\text{2}\%\) per annum for the next \(\text{2}\) months, then \(\text{3}\%\) per
annum for the next 3 months, \(\text{4}\%\) per annum for the next \(\text{4}\) months, and \(\text{0}\%\)
for the rest of the year. If the bank charges him \(\text{R}\,\text{75}\) to open the account, how much can
he expect to get back at the end of the year?

Subtract account fee from the investment amount: \(\text{R}\,\text{8 500}\) - \(\text{R}\,\text{75}\) =
\(\text{R}\,\text{8 425}\)

\begin{align*}
A &= P(1 + i)^{n} \\
& \\
\text{At } \enspace T_{0}: \enspace A &= \text{8 425} \\
\text{At } \enspace T_{1}: \enspace A &= \text{8 425}\left( 1 + \frac{\text{0,01}}{12} \right)^{1} \\
\text{At } \enspace T_{3}: \enspace A &= \text{8 425}\left( 1 + \frac{\text{0,01}}{12} \right)^{1}
\left( 1 + \frac{\text{0,02}}{12} \right)^{2} \\
\text{At } \enspace T_{6}: \enspace A &= \text{8 425}\left( 1 + \frac{\text{0,01}}{12} \right)^{1}
\left( 1 + \frac{\text{0,02}}{12} \right)^{2} \left( 1 + \frac{\text{0,03}}{12} \right)^{3} \\
\text{At } \enspace T_{10}: \enspace A &= \text{8 425}\left( 1 + \frac{\text{0,01}}{12} \right)^{1}
\left( 1 + \frac{\text{0,02}}{12} \right)^{2} \left( 1 + \frac{\text{0,03}}{12} \right)^{3} \left( 1 +
\frac{\text{0,04}}{12} \right)^{4} \\
\therefore \text{Final amount }&= \text{R}\,\text{8 637,98}
\end{align*}

Thabani and Lungelo are both using Harper Bank for their savings. Lungelo makes a deposit of \(x\) at an
interest rate of \(i\) for six years. Three years after Lungelo made his first deposit, Thabani makes a
deposit of \(3x\) at an interest rate of \(\text{8}\%\) per annum. If after \(\text{6}\) years their
investments are equal, calculate the value of \(i\) (correct to three decimal places). If the sum of their
investment is \(\text{R}\,\text{20 000}\), determine how much Thabani earned in \(\text{6}\) years.

\begin{align*}
A & = P(1 + i)^{n} \\
& \\
x(1 + i)^{6} & = 3x\left(1 + \frac{8}{100}\right)^{3} \\
x(1 + i)^{6} - 3x(\text{1,08})^{3} & = 0 \\
x \left( (1 + i)^{6} - 3(\text{1,08})^{3} \right) & = 0 \\
\therefore x = 0 &\text{ or } (1 + i)^{6} - 3(\text{1,08})^{3} = 0 \\
\text{If } x = 0, & \enspace i \text{ can be any value,} \\
\therefore & \text{ not a valid solution} \\
\text{If } (1 + i)^{6} - 3(\text{1,08})^{3} &= 0 \\
(1 + i)^{6} &= 3(\text{1,08})^{3} \\
1 + i &= \sqrt[6]{3(\text{1,08})^{3}} \\
\therefore i &= \sqrt[6]{3(\text{1,08})^{3}} - 1 \\
&= \text{0,248} \ldots \\
\therefore i &= \text{24,8}\%
\end{align*}
\begin{align*}
x + 3x & = \text{R}\,\text{20 000} \\
4x & = \text{R}\,\text{20 000} \\
\therefore x & = \text{R}\,\text{5 000} \\
& \\
A & = \text{15 000}\left(1 + \frac{\text{8}}{100}\right)^{3} \\
& = \text{15 000}(\text{1,08})^{3} \\
& = \text{18 895,68} \\
\text{Interest earned:} &= \text{R}\,\text{18 895,68} - \text{R}\,\text{15 000} \\
&= \text{R}\,\text{3 895,68}
\end{align*}