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End Of Chapter Exercises

End of chapter exercises

Exercise 3.6

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:

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

Wingfield school will need to replace a number of old classroom desks in \(\text{6}\) years' time. The principal has calculated that the new desks will cost \(\text{R}\,\text{44 500}\). The school establishes a sinking fund to pay for the new desks and immediately deposits an amount of \(\text{R}\,\text{6 300}\) into the fund, which accrues interest at a rate of \(\text{6,85}\%\) p.a. compounded monthly.

How much money should the school save every month so that the sinking fund will have enough money to cover the cost of the desks?

\(\text{R}\,\text{6 300}\) is deposited into the sinking fund immediately and will earn interest until the end of the \(\text{6}\) years:

\begin{align*} A & = P(1 + i)^{n} \\ & = \text{6 300} \left( 1 + \frac{\text{0,0685}}{12} \right)^{(\text{6} \times 12)} \\ & = \text{6 300} (\text{1,0057} \ldots )^{\text{72}} \\ & = \text{6 300} (\text{1,50656} \ldots) \\ & = \text{9 491,35} \end{align*}

After \(\text{6}\) years the deposit will be worth \(\text{R}\,\text{9 491,35}\).

Balance required in the sinking fund: \[\text{R}\,\text{44 500} - \text{R}\,\text{9 491,35} = \text{R}\,\text{35 008,65}\]

We calculate the monthly payments that will give a future value of \(\text{R}\,\text{35 008,65}\):

\begin{align*} F & = \frac{x \left[ \left(1 + i \right)^{n} - 1 \right]}{i} \\ \text{35 008,65} & = \frac{x \left[ \left(1 + \frac{\text{0,0685}}{12}\right) ^{(\text{6} \times 12)} - 1 \right]} {\left(\frac{\text{0,0685}}{12} \right)} \\ \therefore x & = \frac{ \text{35 008,65} \times \frac{\text{0,0685}}{12} } { \left[ \left(1 + \frac{\text{0,0685}}{12}\right) ^{(\text{6} \times 12)} - 1 \right]} \\ &= \text{394,50326} \ldots \end{align*}

Therefore, Wingfield school must deposit \(\text{R}\,\text{394,50}\) into the sinking fund each month so that there will enough money in the account to buy the new desks.

How much interest does the fund earn over the period of \(\text{6}\) years?

\begin{align*} \text{Total amount saved: } & = \text{6 300} + (\text{394,5} \times 12 \times 6 ) \\ & = \text{6 300} + \text{28 404} \\ & = \text{34 704} \end{align*}

The total amount of money the school deposits into the account is \(\text{R}\,\text{34 704,00}\).

Therefore, the amount of interest paid by the bank: \[\text{R}\,\text{44 500} - \text{R}\,\text{34 704,00} = \text{R}\,\text{9 796,00}\]

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.

Angela has just started a new job, and wants to save money for her retirement. She decides to deposit \(\text{R}\,\text{1 300}\) into a savings account once each month. Her money goes into an account at Pinelands Mutual Bank, and the account receives \(\text{6,01}\%\) interest p.a. compounded once each month.

How much money will Angela have in her account after \(\text{30}\) years?

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ x & =\text{R}\,\text{1 300} \\ i &= \text{0,0601} \\ n & = \text{30} \end{align*} \begin{align*} F & = \frac{(\text{1 300}) \left[ \left(1 + \frac{\text{0,0601}}{\text{12}}\right) ^{(\text{30} \times \text{12})} - 1 \right]} {\left(\frac{\text{0,0601}}{\text{12}} \right)} \\ & = \text{R}\,\text{1 308 370,14} \end{align*}

After \(\text{30}\) years, Angela will have \(\text{R}\,\text{1 308 370,14}\) in her account.

How much money did Angela deposit into her account after \(\text{30}\) years?

The total amount of money Angela saves each year is \(\text{1 300} \times \text{12} = \text{R}\,\text{15 600}\). From that we can determine the total amount she saves by multiplying by the number of years: \(\text{15 600} \times \text{30} = \text{R}\,\text{468 000}\).

After \(\text{30}\) years, Angela deposited a total of \(\text{R}\,\text{468 000}\) into her account.

Nicky has been working at Meyer and Associates for \(\text{5}\) years and gets an increase in her salary. She opens a savings account at Langebaan Bank and begins making deposits of \(\text{R}\,\text{350}\) every month. The account earns \(\text{5,53}\%\) interest p.a., compounded monthly. Her plan is to continue saving on a monthly schedule until she retires. However, after \(\text{8}\) years she stops making the monthly payments and leaves the account to continue growing.

How much money will Nicky have in her account \(\text{29}\) years after she first opened it?

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ x & = \text{R}\,\text{350} \\ i & = \text{0,0553} \\ n & = \text{8} \end{align*} \begin{align*} F & = \frac{(\text{350}) \left[ \left(1 + \frac{\text{0,0553}}{12}\right) ^{(\text{8} \times 12)} - 1 \right]} {\left(\frac{\text{0,0553}}{12} \right)} \\ & = \text{R}\,\text{42 141,06} \end{align*}

This calculation shows that the account will have a balance of \(\text{R}\,\text{42 141,06}\) after \(\text{8}\) years, which is when Nicky stops making monthly deposits.

From this point on, the account grows from \(\text{R}\,\text{42 141,06}\) with compound interest only (no more monthly payments). This will continue for \(\text{29} - \text{8} = \text{21}\) years.

\begin{align*} A & = P(1 + i)^{n} \\ & = \text{42 141,06} \left( 1 + \frac{\text{0,0553}}{12} \right)^{\text{21} \times 12} \\ & = \text{134 243,45} \end{align*}

The total amount of money in the account \(\text{29}\) years after Nicky opens the account is \(\text{R}\,\text{134 243,45}\).

Calculate the difference between the total deposits made into the account and the amount of interest paid by the bank.

The total deposits: \[\text{R}\,\text{350} \times 12 \times 8 = \text{R}\,\text{33 600}\]

At the end of the period, the interest earned is: \[\quad \text{R}\,\text{134 243,45} - \text{33 600} = \text{R}\,\text{100 643,45}\]

Difference: \[\quad \text{R}\,\text{100 643,45} - \text{R}\,\text{33 600} = \text{R}\,\text{67 043,45}\]

Every three months Louis puts \(\text{R}\,\text{500}\) into an annuity. His account earns an interest rate of \(\text{7,51}\%\) p.a. compounded quarterly. How long will it take Louis's account to reach a balance of \(\text{R}\,\text{13 465,87}\)?

\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ F & = \text{R}\,\text{13 465,87} \\ x & = \text{R}\,\text{500} \\ i & = \text{0,0751} \end{align*} \begin{align*} \text{13 465,87} & = \frac{\text{500} \left[ \left(1 + \frac{\text{0,0751}}{\text{4}}\right) ^{(n \times \text{4})} - 1 \right]} {\left(\frac{\text{0,0751}}{\text{4}} \right)} \\ \text{13 465,87} & = \frac{\text{500} \left[ \left( \text{1,01877} \ldots \right) ^{\text{4}n} - 1 \right]} {\text{0,01877} \ldots} \end{align*} \begin{align*} (\text{0,01877} \ldots)(\text{13 465,87}) & = (\text{500}) \left[ \left( \text{1,01877} \ldots \right) ^{\text{4}n} - 1 \right] \\ \frac{\text{252,82172} \ldots}{\text{500}} & = \left[ \left( \text{1,01877} \ldots \right) ^{\text{4}n} - 1 \right] \end{align*}

Now that the square bracket is alone, work inside of it: add one to both sides, and then change the equation to log-form to get the exponent.

\begin{align*} \text{0,50564} \ldots + 1 & = \left( \text{1,01877} \ldots \right) ^{\text{4}n} \\ \text{1,50564} \ldots & = \left( \text{1,01877} \ldots \right) ^{\text{4}n} \\ \text{Change into logarithmic form: } \quad \text{4}n & = \log_{\text{1,01877} \ldots} (\text{1,50564} \ldots) \\ \text{4}n & = \text{22} \\ n & = \frac{\text{22}}{\text{4}} \\ n & = \text{5,5} \end{align*}

It will take about \(\text{5,5}\) years.

How much interest will Louis receive from his investment?

The total amount of deposits:

\[\quad \text{500} \times \text{4} \times \text{5,5} = \text{R}\,\text{11 000,00}\]

The total amount of interest: \[\text{R}\,\text{13 465,87} - \text{R}\,\text{11 000,00} = \text{R}\,\text{2 465,87}\]

A dairy farmer named Kayla needs to buy new equipment for her dairy farm which costs \(\text{R}\,\text{200 450}\). She bought her old equipment \(\text{12}\) years ago for \(\text{R}\,\text{167 000}\). The value of the old equipment depreciates at a rate of \(\text{12,2}\%\) per year on a reducing balance. Kayla will need to arrange a bond for the remaining cost of the new equipment.

An agency which supports farmers offers bonds at a special interest rate of \(\text{10,01}\%\) p.a. compounded monthly for any loan up to \(\text{R}\,\text{175 000}\) and \(\text{9,61}\%\) p.a. compounded monthly for a loan above that amount. Kayla arranges a bond such that she will not need to make any payments on the loan in the first six months (called a 'grace period') and she must pay the loan back over \(\text{20}\) years.

Determine the monthly payment.

Old equipment:

\begin{align*} A & = P(1 - i)^{n} \\ & = \text{167 000}(1 - \text{0,122} )^{\text{12}} \\ &= \text{35 046,98494} \ldots \end{align*}

The value of the old equipment is \(\text{R}\,\text{35 046,98}\).

Bond amount:

\[\text{R}\,\text{200 450} - \text{R}\,\text{35 046,98} = \text{R}\,\text{165 403,02}\] \begin{align*} A & = P(1 + i)^{n} \\ & = \text{165 403,02} \left( 1 + \frac{\text{0,1001}}{12} \right)^{\text{6}} \\ &= \text{173 856,01291} \ldots \end{align*}

After the first six months of the loan period, during which she makes no payments, the amount she owes increases to \(\text{R}\,\text{173 856,01}\).

Now we can use the present value formula to solve for the value of \(x\). Remember that the repayment period is \(\text{19,5}\) years.

\begin{align*} \text{173 856,01} & = \frac{x \left[ 1 - \left(1 + \frac{\text{0,1001}}{12}\right) ^{-(\text{19,5} \times 12)} \right]} {\left(\frac{\text{0,1001}}{12} \right)} \\ \therefore x & = \frac{ \text{173 856,01} \times \left(\frac{\text{0,1001}}{12} \right)}{x \left[ 1 - \left(1 + \frac{\text{0,1001}}{12}\right) ^{-(\text{19,5} \times 12)} \right]} \\ &= \text{1 692,53772} \ldots \end{align*}

Kayla must pay \(\text{R}\,\text{1 692,54}\) each month.

What is the total amount of interest Kayla will pay for the bond?

Total cost of the loan:

\(\quad \text{R}\,\text{1 692,54} \times 12 \times \text{19,5} = \text{R}\,\text{396 054,36}\)

Interest:

\(\quad \text{R}\,\text{396 054,36} - \text{R}\,\text{165 403,02} = \text{R}\,\text{230 651,34}\)

The total amount of interest Kayla paid is \(\text{R}\,\text{230 651,34}\).

By what factor is the interest she pays greater than the value of the loan? Give the answer correct to one decimal place.

\[k = \frac{\text{interest paid}}{\text{amount of the loan}} = \frac{\text{230 651,34}}{\text{165 403,02}} = \text{1,39448}\]

The interest is greater than the loan amount by a factor of \(\text{1,4}\).

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*}