3.3 Future value annuities

Write down the given information and the future value formula:

\[F = \frac{x\left[(1 + i)^{n}-1\right]}{i}\] \begin{align*} x &= \text{500} \\ i &= \frac{\text{0,0596}}{4} \\ n &= 35 \times 4 = \text{140} \end{align*}

Substitute the known values and use a calculator to determine \(F\):

\begin{align*} F &= \dfrac{\text{500}\left[(1 + \frac{\text{0,0596}}{4})^{\text{140}}-1\right]}{\frac{\text{0,0596}}{4}} \\ &= \text{R}\,\text{232 539,41} \end{align*}

Calculate the total value of deposits into the account:

Shelly deposited \(\text{R}\,\text{500}\) each quarter for \(\text{35}\) years:

\begin{align*} \text{Total deposits: } &= \text{R}\,\text{500} \times 4 \times 35 \\ &= \text{R}\,\text{70 000} \end{align*}

Calculate the total interest earned:

\begin{align*} \text{Total interest } &= \text{final account balance } - \text{total value of all deposits} \\ &= \text{R}\,\text{232 539,41} - \text{R}\,\text{70 000} \\ &= \text{R}\,\text{162539,41} \end{align*}

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text{Where: } \quad x & = \text{R}\,\text{1 100} \\ i &= \text{0,089} \\ n & = \text{29} \end{align*} \begin{align*} F & = \frac{(\text{1 100}) \left[ \left(1 + \frac{\text{0,089}}{\text{12}}\right) ^{(\text{29} \times \text{12})} - 1 \right]} {\left(\frac{\text{0,089}}{\text{12}} \right)} \\ & = \text{R}\,\text{1 792 400,11} \end{align*}

After \(\text{29}\) years, Grace will have \(\text{R}\,\text{1 792 400,11}\) in her account.

The total amount of money Grace saves each year is \(\text{1 100} \times \text{12} = \text{R}\,\text{13 200}\). From that we can determine the total amount she saves by multiplying by the number of years: \(\text{13 200} \times \text{29} = \text{R}\,\text{382 800}\).

After \(\text{29}\) years, Grace deposited a total of \(\text{R}\,\text{382 800}\) into her account.

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ F & = \text{R}\,\text{750 000} \\ i & = \text{0,0613} \\ n & = \text{35} \end{align*} \begin{align*} \text{750 000} & = \frac{x \left[ \left(1 + \frac{\text{0,0613}}{\text{12}}\right) ^{(\text{35} \times \text{12})} \right]} {\left(\frac{\text{0,0613}}{\text{12}} \right)} \\ \therefore x &= \frac{ \text{750 000} \times \left(\frac{\text{0,0613}}{\text{12}} \right) }{\left[ \left(1 + \frac{\text{0,0613}}{\text{12}}\right) ^{(\text{35} \times \text{12})} \right]} \\ &= \text{510,84927} \ldots \end{align*}

In order to save \(\text{R}\,\text{750 000}\) in \(\text{35}\) years, Monique will need to save \(\text{R}\,\text{510,85}\) in her account every month.

The final amount calculated in the question above includes the money Monique deposited into the account plus the interest paid by the bank. The total amount of money Monique put into her account during the \(\text{35}\) year is the product of \(\text{12}\) payments per year, \(\text{35}\) years, and the payment amount itself:

\[\text{R}\,\text{510,85} \times 12 \times 35 = \text{R}\,\text{214 557,00}\]

After \(\text{35}\) years, Monique deposited a total of \(\text{R}\,\text{214 557}\) into her account.

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ F & = \text{R}\,\text{4 397,53} \\ x & = \text{R}\,\text{30} \\ i &= \text{0,0692} \end{align*} \begin{align*} \text{4 397,53} & = \frac{(\text{30}) \left[ \left(1 + \frac{\text{0,0692}}{\text{52}}\right) ^{(n \times \text{52})} - 1 \right]} {\left(\frac{\text{0,0692}}{\text{52}} \right)} \\ \text{4 397,53} & = \frac{(\text{30}) \left[ \left( \text{1,00133} \right) ^{\text{52}n} - 1 \right]} {\text{0,00133} \ldots} \end{align*} \begin{align*} (\text{0,00133} \ldots)(\text{4 397,53}) & = (\text{30}) \left[ \left( \text{1,00133} \ldots \right) ^{\text{52}n} - 1 \right] \\ \frac{\text{5,85209} \ldots}{\text{30}} & = \left[ \left( \text{1,00133} \ldots \right) ^{\text{52}n} - 1 \right] \end{align*} \begin{align*} \text{0,19506} \ldots + 1 & = \left( \text{1,00133} \ldots \right) ^{\text{52}n} \\ \text{1,19506} \ldots & = \left( \text{1,00133} \ldots \right) ^{\text{52}n} \\ \text{Change to logarithmic form: } \quad \text{52}n & = \log_{\text{1,00133} \ldots} (\text{1,19506} \ldots) \\ \text{52}n & = \text{134} \\ n & = \frac{\text{134}}{\text{52}} \\ n & = \text{2,57692} \ldots \end{align*}

To get to the final answer for this question, convert \(\text{2,57692} \ldots\) years into years and days.

\(\qquad (\text{0,57692} \ldots) \times \frac{\text{365}}{\text{year}} = \text{210,577}\) days

Harold's investment takes \(\text{2}\) years and \(\text{211}\) days to reach the final value of \(\text{R}\,\text{4 397,53}\).

The total amount Harold invests is as follows:

\(\quad \text{30} \times \text{52} \times \text{2,57692} \ldots = \text{R}\,\text{4 020,00}\)

Therefore, the total amount of interest paid by the bank: \(\text{R}\,\text{4 397,53} - \text{R}\,\text{4 020,00} = \text{R}\,\text{377,53}\).

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text {Where: } \quad & \\ x & = \text{3 100} \\ i & = \text{0,053} \\ n & = \text{6} \end{align*}

Interest is compounded monthly: \(\; i = \text{0,053} \rightarrow \frac{\text{0,053}}{12} \;\) and \(\; n = \text{6} \rightarrow \text{6} \times (12) \;\).

\begin{align*} F & = \frac{(\text{3 100}) \left[ \left(1 + \frac{\text{0,053}}{12}\right) ^{(\text{6} \times 12)} - 1 \right]} {\left(\frac{\text{0,053}}{12} \right)} \\ & = \text{R}\,\text{262 094,55} \end{align*}

After \(\text{6}\) years, Mfethu will have \(\text{R}\,\text{262 094,55}\) in his sinking fund.

No, Mfethu does not have enough money in his account:

\[\text{R}\,\text{285 000} - \text{R}\,\text{262 094,55} = \text{R}\,\text{22 905,45}\]

\begin{align*} P &= \text{265 000} \\ i &= \text{0,17} \\ n &= \text{5} \\ & \\ A &= P(1 - i)^{n} \\ &= \text{265 000}(1 - \text{0,17})^{\text{5}} \\ &= \text{R}\,\text{104 384,58} \end{align*}

\begin{align*} P &= \text{265 000} \\ i &= \text{0,12} \\ n &= \text{5} \\ & \\ A &= P(1 + i)^{n} \\ &= \text{265 000}(1 + \text{0,12})^{\text{5}} \\ &= \text{R}\,\text{467 020,55} \end{align*}

Therefore, the balance of the sinking fund (\(F\)) must be greater than the cost of a new van in five years' time less the money from the sale of the old van:

\begin{align*} F &= \text{R}\,\text{467 020,55} - \text{R}\,\text{104 384,58} \\ &= \text{R}\,\text{362 635,97} \end{align*}

Calculate the required monthly payment into the sinking fund:

\[x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}\] \begin{align*} F &= \text{362 635,97} \\ i &= \frac{\text{0,11}}{12} \\ n &= \text{5} \times \text{12} = \text{60} \end{align*}

Substitute the values and calculate \(x\):

\begin{align*} x &= \dfrac{\text{362 635,97} \times \frac{\text{0,11}}{12}}{\left[(1 + \frac{\text{0,11}}{12})^{60}-1\right]} \\ &= \text{R}\,\text{4 560,42} \end{align*}

Therefore, the company must deposit \(\text{R}\,\text{4 560,42}\) each month.

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text{Where: } \quad & \\ F & = \text{8 450} \\ i & = \text{0,0767} \\ n & = \text{7} \end{align*}

Interest is compounded per quarter, therefore \(\; i = \text{0,0767} \rightarrow \frac{\text{0,0767}}{\text{4}}\) and \(n = \text{7} \rightarrow \text{7} \times \text{4}\)

\begin{align*} \text{8 450} & = \frac{x \left[ \left(1 + \frac{\text{0,0767}}{\text{4}}\right) ^{(\text{7} \times \text{4})} - 1 \right]} {\left(\frac{\text{0,0767}}{\text{4}} \right)} \\ \therefore x & = \frac{\left( \text{8 450} \times \frac{\text{0,0767}}{\text{4}} \right)}{ \left[ \left(1 + \frac{\text{0,0767}}{\text{4}}\right) ^{(\text{7} \times \text{4})} - 1 \right]} \\ &= \text{230,80273} \ldots \end{align*}

Tonya must deposit \(\text{R}\,\text{230,80}\) into the sinking fund quarterly.

Total savings:

\[\text{R}\,\text{230,80} \times \text{4} \times \text{7} = \text{R}\,\text{6 462,40}\]

Interest earned:

\[\quad \text{R}\,\text{8 450} - \text{R}\,\text{6 462,40} = \text{R}\,\text{1 987,60}\]

To the nearest rand, the bank paid \(\text{R}\,\text{1 988}\) into the account.