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.
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.
