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# Future Value Annuities

## 3.3 Future value annuities (EMCFZ)

For future value annuities, we regularly save the same amount of money into an account, which earns a certain rate of compound interest, so that we have money for the future.

## Worked example 3: Future value annuities

At the end of each year for $$\text{4}$$ years, Kobus deposits $$\text{R}\,\text{500}$$ into an investment account. If the interest rate on the account is $$\text{10}\%$$ per annum compounded yearly, determine the value of his investment at the end of the $$\text{4}$$ years.

### Write down the given information and the compound interest formula

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

### Draw a timeline The first deposit in the account earns the highest amount of interest (three interest payments) and the last deposit earns the least interest (no interest payments).

We can summarize this information in the table below:

 Deposit No. of interest payments Calculation Accumulated amount Year 1 $$\text{R}\,\text{500}$$ $$\text{3}$$ $$\text{500}(1 + \text{0,1})^{3}$$ $$\text{R}\,\text{665,50}$$ Year 2 $$\text{R}\,\text{500}$$ $$\text{2}$$ $$\text{500}(1 + \text{0,1})^{2}$$ $$\text{R}\,\text{605,00}$$ Year 3 $$\text{R}\,\text{500}$$ $$\text{1}$$ $$\text{500}(1 + \text{0,1})^{1}$$ $$\text{R}\,\text{550,00}$$ Year 4 $$\text{R}\,\text{500}$$ $$\text{0}$$ $$\text{500}(1 + \text{0,1})^{0}$$ $$\text{R}\,\text{500,00}$$ Total $$\text{R}\,\text{2 320,50}$$

### Deriving the formula (EMCG2)

Note: for this section is it important to be familiar with the formulae for the sum of a geometric series (Chapter 1):

\begin{align*} S_{n} = \frac{a\left({r}^{n}-1\right)}{r-1} & \qquad \text{for } r >1 \\ S_{n} = \frac{a\left(1-{r}^{n}\right)}{1-r} & \qquad \text{for } r < 1 \end{align*}

In the worked example above, the total value of Kobus' investment at the end of the four year period is calculated by summing the accumulated amount for each deposit:

$$\begin{array}{c@{\;}l@{\;}l@{\;}l@{\;}l@{\;}} \text{R}\,\text{2 320,50} &= \text{R}\,\text{500,00} \quad + & \text{R}\,\text{550,00} \quad + & \text{R}\,\text{605,00} \quad + & \text{R}\,\text{665,50} \\ &= \text{500}(1 + \text{0,1})^{0} \enspace + & \text{500}(1 + \text{0,1})^{1} \enspace + & \text{500}(1 + \text{0,1})^{2} \enspace + & \text{500}(1 + \text{0,1})^{3} \end{array}$$

We notice that this is a geometric series with a constant ratio $$r = 1 + \text{0,1}$$.

Using the formula for the sum of a geometric series:

\begin{align*} a &= \text{500} \\ r &= \text{1,1} \\ n &= 4 \\ & \\ S_{n} &= \frac{a\left({r}^{n}-1\right)}{r-1} \\ &= \frac{\text{500}\left({\text{1,1}}^{4}-1\right)}{\text{1,1} - 1} \\ &= \text{2 320,50} \end{align*}

We can therefore use the formula for the sum of a geometric series to derive a formula for the future value ($$F$$) of a series of ($$n$$) regular payments of an amount ($$x$$) which are subject to an interest rate ($$i$$):

\begin{align*} a &= x \\ r &= 1 + i \\ & \\ S_{n} &= \frac{a\left({r}^{n}-1\right)}{r-1} \\ \therefore F &= \frac{x\left[(1 + i)^{n}-1\right]}{(1 + i)-1} \\ &= \frac{x\left[(1 + i)^{n}-1\right]}{i} \end{align*}

Future value of payments:

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$

If we are given the future value of a series of payments, then we can calculate the value of the payments by making $$x$$ the subject of the above formula.

Payment amount:

$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$

## Worked example 4: Future value annuities

Ciza decides to start saving money for the future. At the end of each month she deposits $$\text{R}\,\text{900}$$ into an account at Harringstone Mutual Bank, which earns $$\text{8,25}\%$$ interest p.a. compounded monthly.

1. Determine the balance of Ciza's account after $$\text{29}$$ years.
2. How much money did Ciza deposit into her account over the $$\text{29}$$ year period?
3. Calculate how much interest she earned over the $$\text{29}$$ year period.

### Write down the given information and the future value formula

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{900} \\ i &= \frac{\text{0,0825}}{12} \\ n &= 29 \times 12 = \text{348} \end{align*}

### Substitute the known values and use a calculator to determine $$F$$

\begin{align*} F &= \dfrac{\text{900}\left[(1 + \frac{\text{0,0825}}{12})^{\text{348}}-1\right]}{\frac{\text{0,0825}}{12}} \\ &= \text{R}\,\text{1 289 665,06} \end{align*}

Remember: do not round off at any of the interim steps of a calculation as this will affect the accuracy of the final answer.

### Calculate the total value of deposits into the account

Ciza deposited $$\text{R}\,\text{900}$$ each month for $$\text{29}$$ years:

\begin{align*} \text{Total deposits: } &= \text{R}\,\text{900} \times 12 \times 29 \\ &= \text{R}\,\text{313 200} \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{1 289 665,06} - \text{R}\,\text{313 200} \\ &= \text{R}\,\text{976 465,06} \end{align*}

Useful tips for solving problems:

1. Timelines are very useful for summarising the given information in a visual way.
2. When payments are made more than once per annum, we determine the total number of payments ($$n$$) by multiplying the number of years by $$p$$:

 Term $$p$$ yearly / annually $$\text{1}$$ half-yearly / bi-annually $$\text{2}$$ quarterly $$\text{4}$$ monthly $$\text{12}$$ weekly $$\text{52}$$ daily $$\text{365}$$

3. If a nominal interest rate $$\left( i^{(m)} \right)$$ is given, then use the following formula to convert it to an effective interest rate:

$1 + i = \left( 1 + \frac{i^{(m)}}{m} \right)^{m}$

## Worked example 5: Calculating the monthly payments

Kosma is planning a trip to Canada to visit her friend in two years' time. She makes an itinerary for her holiday and she expects that the trip will cost $$\text{R}\,\text{25 000}$$. How much must she save at the end of every month if her savings account earns an interest rate of $$\text{10,7}\%$$ per annum compounded monthly?

### Write down the given information and the future value formula

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$

To determine the monthly payment amount, we make $$x$$ the subject of the formula:

$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{25 000} \\ i &= \frac{\text{0,107}}{12} \\ n &= 2 \times 12 = \text{24} \end{align*}

### Substitute the known values and calculate $$x$$

\begin{align*} x &= \dfrac{\text{25 000} \times \frac{\text{0,107}}{12}}{\left[(1 + \frac{\text{0,107}}{12})^{24}-1\right]} \\ &= \text{R}\,\text{938,80} \end{align*}

Kosma must save $$\text{R}\,\text{938,80}$$ each month so that she can afford her holiday.

## Worked example 6: Determining the value of an investment

Simon starts to save for his retirement. He opens an investment account and immediately deposits $$\text{R}\,\text{800}$$ into the account, which earns $$\text{12,5}\%$$ per annum compounded monthly. Thereafter, he deposits $$\text{R}\,\text{800}$$ at the end of each month for $$\text{20}$$ years. What is the value of his retirement savings at the end of the $$\text{20}$$ year period?

### Write down the given information and the future value formula

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{800} \\ i &= \frac{\text{0,125}}{12} \\ n &= 1 + (20 \times 12) = \text{241} \end{align*}

Note that we added one extra month to the $$\text{20}$$ years because Simon deposited $$\text{R}\,\text{800}$$ immediately.

### Substitute the known values and calculate $$F$$

\begin{align*} F &= \frac{\text{800}\left[(1 + \frac{\text{0,125}}{12})^{\text{241}}-1\right]}{ \frac{\text{0,125}}{12}} \\ &= \text{R}\,\text{856 415,66} \end{align*}

Simon will have saved $$\text{R}\,\text{856 415,66}$$ for his retirement.

# Success in Maths and Science unlocks opportunities

## Future value annuities

Exercise 3.2

Shelly decides to start saving money for her son's future. At the end of each month she deposits $$\text{R}\,\text{500}$$ into an account at Durban Trust Bank, which earns an interest rate of $$\text{5,96}\%$$ per annum compounded quarterly.

Determine the balance of Shelly's account after $$\text{35}$$ years.

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*}

How much money did Shelly deposit into her account over the $$\text{35}$$ year period?

Calculate the total value of deposits into the account:

Shelly deposited $$\text{R}\,\text{500}$$ each month for $$\text{35}$$ years:

\begin{align*} \text{Total deposits: } &= \text{R}\,\text{500} \times 12 \times 35 \\ &= \text{R}\,\text{210 000} \end{align*}

Calculate how much interest she earned over the $$\text{35}$$ year period.

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{210 000} \\ &= \text{R}\,\text{22 539,41} \end{align*}

Gerald wants to buy a new guitar worth $$\text{R}\,\text{7 400}$$ in a year's time. How much must he deposit at the end of each month into his savings account, which earns a interest rate of $$\text{9,5}\%$$ p.a. compounded monthly?

Write down the given information and the future value formula:

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$

To determine the monthly payment amount, we make $$x$$ the subject of the formula:

$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{7 400} \\ i &= \frac{\text{0,095}}{12} \\ n &= 1 \times 12 = \text{12} \end{align*}

Substitute the known values and calculate $$x$$:

\begin{align*} x &= \dfrac{\text{7 400} \times \frac{\text{0,095}}{12}}{\left[(1 + \frac{\text{0,095}}{12})^{12}-1\right]} \\ &= \text{R}\,\text{590,27} \end{align*}

Gerald must deposit $$\text{R}\,\text{590,27}$$ each month so that he can afford his guitar.

A young woman named Grace has just started a new job, and wants to save money for the future. She decides to deposit $$\text{R}\,\text{1 100}$$ into a savings account every month. Her money goes into an account at First Mutual Bank, and the account earns $$\text{8,9}\%$$ interest p.a. compounded every month.

How much money will Grace have in her account after $$\text{29}$$ years?

\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.

How much money did Grace deposit into her account by the end of the $$\text{29}$$ year period?

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.

Ruth decides to save for her retirement so she opens a savings account and immediately deposits $$\text{R}\,\text{450}$$ into the account. Her savings account earns $$\text{12}\%$$ per annum compounded monthly. She then deposits $$\text{R}\,\text{450}$$ at the end of each month for $$\text{35}$$ years. What is the value of her retirement savings at the end of the $$\text{35}$$ year period?

Write down the given information and the future value formula:

$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{450} \\ i &= \frac{\text{0,12}}{12} \\ n &= 1 + (35 \times 12) = \text{421} \end{align*}

Substitute the known values and calculate $$F$$:

\begin{align*} F &= \frac{\text{450}\left[(1 + \frac{\text{0,12}}{12})^{\text{421}}-1\right]}{ \frac{\text{0,12}}{12}} \\ &= \text{R}\,\text{2 923 321,08} \end{align*}

Ruth will have saved $$\text{R}\,\text{2 923 321,08}$$ for her retirement.

Musina MoneyLenders offer a savings account with an interest rate of $$\text{6,13}\%$$ p.a. compounded monthly. Monique wants to save money so that she can buy a house when she retires. She decides to open an account and make regular monthly deposits. Her goal is to end up with $$\text{R}\,\text{750 000}$$ in her account after $$\text{35}$$ years.

How much must Monique deposit into her account each month in order to reach her goal?

\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.

How much money, to the nearest rand, did Monique deposit into her account by the end of the $$\text{35}$$ year period?

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.

Lerato plans to buy a car in five and a half years' time. She has saved $$\text{R}\,\text{30 000}$$ in a separate investment account which earns $$\text{13}\%$$ per annum compound interest. If she doesn't want to spend more than $$\text{R}\,\text{160 000}$$ on a vehicle and her savings account earns an interest rate of $$\text{11}\%$$ p.a. compounded monthly, how much must she deposit into her savings account each month?

First calculate the accumulated amount for the $$\text{R}\,\text{30 000}$$ in Lerato's investment account:

$A = P(1 + i)^{n}$ \begin{align*} P &= \text{30 000} \\ i &= \text{0,13} \\ n &= \text{5,5} \end{align*} \begin{align*} A &= \text{30 000}(1 + \text{0,13})^{\text{5,5}}\\ &= \text{R}\,\text{58 756,06} \end{align*}

In five and a half years' time, Lerato needs to have saved $$\text{R}\,\text{160 000} - \text{R}\,\text{58 756,06} = \text{R}\,\text{101 243,94}$$.

$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{101 243,94} \\ i &= \frac{\text{0,11}}{12} \\ n &= \text{5,5} \times 12 = \text{66} \end{align*}

Substitute the known values and calculate $$x$$:

\begin{align*} x &= \dfrac{\text{101 243,94} \times \frac{\text{0,11}}{12}}{\left[(1 + \frac{\text{0,11}}{12})^{66}-1\right]} \\ &= \text{R}\,\text{1 123,28} \end{align*}

Lerato must deposit $$\text{R}\,\text{1 123,28}$$ each month into her savings account.

Every Monday Harold puts $$\text{R}\,\text{30}$$ into a savings account at the King Bank, which accrues interest of $$\text{6,92}\%$$ p.a. compounded weekly. How long will it take Harold's account to reach a balance of $$\text{R}\,\text{4 397,53}$$. Give the answer as a number of years and days to the nearest integer.

\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 $$\text{R}\,\text{4 397,53}$$.

How much interest will Harold receive from the bank during the period of his investment?

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: $$\text{R}\,\text{4 397,53} - \text{R}\,\text{4 020,00} = \text{R}\,\text{377,53}$$.

### Sinking funds (EMCG3)

Vehicles, equipment, machinery and other similar assets, all depreciate in value as a result of usage and age. Businesses often set aside money for replacing outdated equipment or old vehicles in accounts called sinking funds. Regular deposits, and sometimes lump sum deposits, are made into these accounts so that enough money will have accumulated by the time a new machine or vehicle needs to be purchased.

## Worked example 7: Sinking funds

Wellington Courier Company buys a delivery truck for $$\text{R}\,\text{296 000}$$. The value of the truck depreciates on a reducing-balance basis at $$\text{18}\%$$ per annum. The company plans to replace this truck in seven years' time and they expect the price of a new truck to increase annually by $$\text{9}\%$$.

1. Calculate the book value of the delivery truck in seven years' time.
2. Determine the minimum balance of the sinking fund in order for the company to afford a new truck in seven years' time.
3. Calculate the required monthly deposits if the sinking fund earns an interest rate of $$\text{13}\%$$ per annum compounded monthly.

### Determine the book value of the truck in seven years' time

\begin{align*} P &= \text{296 000} \\ i &= \text{0,18} \\ n &= \text{7} \\ & \\ A &= P(1 - i)^{n} \\ &= \text{296 000}(1 - \text{0,18})^{\text{7}} \\ &= \text{R}\,\text{73 788,50} \end{align*}

### Determine the minimum balance of the sinking fund

Calculate the price of a new truck in seven years' time:

\begin{align*} P &= \text{296 000} \\ i &= \text{0,09} \\ n &= \text{7} \\ & \\ A &= P(1 + i)^{n} \\ &= \text{296 000}(1 + \text{0,09})^{\text{7}} \\ &= \text{R}\,\text{541 099,58} \end{align*}

Therefore, the balance of the sinking fund ($$F$$) must be greater than the cost of a new truck in seven years' time minus the money from the sale of the old truck:

\begin{align*} F &= \text{R}\,\text{541 099,58} - \text{R}\,\text{73 788,50} \\ &= \text{R}\,\text{467 311,08} \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{467 311,08} \\ i &= \frac{\text{0,13}}{12} \\ n &= \text{7} \times 12 = \text{84} \end{align*}

Substitute the values and calculate $$x$$:

\begin{align*} x &= \dfrac{\text{467 311,08} \times \frac{\text{0,13}}{12}}{\left[(1 + \frac{\text{0,13}}{12})^{84}-1\right]} \\ &= \text{R}\,\text{3 438,77} \end{align*}

Therefore, the company must deposit $$\text{R}\,\text{3 438,77}$$ each month.

# Success in Maths and Science unlocks opportunities

## Sinking funds

Exercise 3.3

Mfethu owns his own delivery business and he will need to replace his truck in $$\text{6}$$ years' time. Mfethu deposits $$\text{R}\,\text{3 100}$$ into a sinking fund each month, which earns $$\text{5,3}\%$$ interest p.a. compounded monthly.

How much money will be in the fund in $$\text{6}$$ years' time, when Mfethu wants to buy the new truck?

\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.

If a new truck costs $$\text{R}\,\text{285 000}$$ in $$\text{6}$$ years' time, will Mfethu have enough money to buy it?

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

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

Atlantic Transport Company buys a van for $$\text{R}\,\text{265 000}$$. The value of the van depreciates on a reducing-balance basis at $$\text{17}\%$$ per annum. The company plans to replace this van in five years' time and they expect the price of a new van to increase annually by $$\text{12}\%$$.

Calculate the book value of the van in five years' time.
\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*}
Determine the amount of money needed in the sinking fund for the company to be able to afford a new van in five years' time.
\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 deposits if the sinking fund earns an interest rate of $$\text{11}\%$$ per annum compounded monthly.

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.

Tonya owns Freeman Travel Company and she will need to replace her computer in $$\text{7}$$ years' time. Tonya creates a sinking fund so that she will be able to afford a new computer, which will cost $$\text{R}\,\text{8 450}$$. The sinking fund earns interest at a rate of $$\text{7,67}\%$$ p.a. compounded each quarter.

How much money must Tonya save quarterly so that there will be enough money in the account to buy the new computer?

\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.

How much interest (to the nearest rand) does the bank pay into the account by the end of the $$\text{7}$$ year period?

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.