Everything covered in this lesson, in one place - useful for revision or printing.
\[ \begin{gathered} A = P(1 \pm in) \\ A = P(1 \pm i)^{n} \\ 1 + i_{\text{eff}} = \left(1 + \tfrac{i^{(m)}}{m}\right)^{m} \end{gathered} \]
Lesson 1 gave you the formulas. This lesson works through real exam questions, step by step, the way the memo awards the marks.
\[ \begin{gathered} 277\,216 = P(1 - 0{,}0825)^{3} \\ P = \dfrac{277\,216}{(0{,}9175)^{3}} \\ P = R\,358\,922{,}00 \end{gathered} \]
Miss Coetzee's new car depreciates at 8,25% p.a. on the reducing-balance method. After 3 years its book value is R277 216. Calculate the purchase price.
\[ \begin{gathered} A = P(1 - i)^{n} \\ A = 250\,000(1 - 0{,}13)^{6} \\ A = R\,108\,406{,}55 \end{gathered} \]
A motorbike costs R250 000 and depreciates by 13% p.a. on the reducing-balance method. What is it worth after 6 years?
\[ \begin{gathered} 79\,120 = 175\,000(1 - i)^{4} \\ 1 - i = \sqrt[4]{\tfrac{79\,120}{175\,000}} \\ i = 0{,}18 = 18\% \end{gathered} \]
A car bought for R175 000 is sold after 4 years for R79 120. Find the annual reducing-balance rate of depreciation.
\[ \begin{gathered} 1 + i_{\text{eff}} = \left(1 + \tfrac{0{,}09}{4}\right)^{4} \\ i_{\text{eff}} = 0{,}0931 = 9{,}31\% \end{gathered} \]
An investment earns 9% p.a. compounded quarterly. Calculate the effective interest rate.
\[ \begin{gathered} \left(1 + \tfrac{0{,}09}{2}\right)^{2} - 1 = 9{,}20\% \\ \left(1 + \tfrac{0{,}23}{12}\right)^{12} - 1 = 25{,}59\% \end{gathered} \]
Top: 9% p.a. compounded half-yearly. Bottom: a credit card charging 23% p.a. compounded monthly.
\[ \begin{gathered} 1 + 0{,}12 = \left(1 + \tfrac{i}{4}\right)^{4} \\ \tfrac{i}{4} = \sqrt[4]{1{,}12} - 1 \\ i = 4\left(\sqrt[4]{1{,}12} - 1\right) = 11{,}49\% \end{gathered} \]
XY Bank offers loans at an effective rate of 12% p.a. What is the nominal rate if interest is compounded quarterly?
\[ \begin{gathered} i = \tfrac{0{,}0925}{2} \qquad n = 19 \\ A = 25\,000\left(1 + \tfrac{0{,}0925}{2}\right)^{19} \\ A = R\,59\,022{,}00 \end{gathered} \]
Mr Dlezi invested R25 000 at 9,25% p.a. compounded half-yearly. Calculate the amount at the end of nineteen half-years.
\[ \begin{gathered} A = 12\,000\left(1 + \tfrac{0{,}1135}{m}\right)^{20} \\ m = 1: \ R\,103\,035{,}21 \\ m = 2: \ R\,36\,193{,}15 \\ m = 4: \ R\,20\,999{,}62 \end{gathered} \]
Mr Mavundla invested R12 000 at 11,35% p.a. but forgot how often interest was added. After 20 interest payments he had R36 193,15. How many years did it take?
Seven years ago Krish deposited R25 000. The bank paid 8% p.a. compounded monthly for the first 4 years and then 12% p.a. compounded quarterly for 3 years. He withdrew R10 000 three years after the deposit. How much does he have today?
\[ \begin{gathered} A_{3} = 25\,000\left(1 + \tfrac{0{,}08}{12}\right)^{36} \\ A_{3} = R\,31\,755{,}93 \\ A_{4} = 21\,755{,}93\left(1 + \tfrac{0{,}08}{12}\right)^{12} \\ A_{4} = R\,23\,561{,}66 \\ A_{7} = A_{4}\left(1 + \tfrac{0{,}12}{4}\right)^{12} \\ A_{7} = R\,33\,593{,}29 \end{gathered} \]
Grow to the withdrawal, take it off, grow to the rate change, then switch rates.
\[ \begin{gathered} A_{2} = 25\,000\left(1 + \tfrac{0{,}04}{12}\right)^{24} \\ A_{2} = R\,27\,078{,}57 \\ A_{4} = (A_{2} + 10\,000)(1{,}035)^{4} \\ A_{4} = R\,42\,548{,}52 \\ A_{5} = (A_{4} - 5\,000)(1{,}035)^{2} \\ A_{5} = R\,40\,222{,}91 \end{gathered} \]
Rodney deposits R25 000. Two years later he deposits R10 000, and four years after the start he withdraws R5 000. Interest: 4% p.a. compounded monthly for 2 years, then 7% p.a. compounded semi-annually. Find the amount 5 years after the first deposit.
\[ \begin{gathered} A_{2} = 300\,000(1{,}0225)^{8} - 175\,000 \\ A_{2} = R\,183\,449{,}34 \\ A_{4} = A_{2}(1{,}0225)^{8} \\ A_{4} = R\,219\,190{,}99 \\ A_{6} = A_{4}\left(1 + \tfrac{0{,}085}{12}\right)^{24} \\ A_{6} = R\,259\,652{,}50 \end{gathered} \]
Sam invests R300 000 at 9% p.a. compounded quarterly for 4 years, then 8,5% p.a. compounded monthly. After 2 years he withdraws R175 000 to buy his first son a car. How much is there after 6 years for his second son?
\[ \begin{gathered} 175\,000(1 + 0{,}08)^{4} \\ = R\,238\,085{,}57 \\ 259\,652{,}50 - 238\,085{,}57 \\ = R\,21\,566{,}93 \end{gathered} \]
Inflation is 8% p.a. Sam's first son got a R175 000 car. Would giving the second son R259 652,50, four years later, be fair?
\[ \begin{gathered} 3\,500\,000 - 2\,250\,000 \\ = R\,1\,250\,000 \\ \text{New: } 2\,250\,000(1{,}04)^{6} \\ = R\,2\,846\,967{,}79 \\ \text{Old: } 2\,250\,000(0{,}86)^{6} \\ = R\,910\,276{,}28 \end{gathered} \]
Peter makes a profit of R3,5 million, buys a tractor for R2 250 000 and invests the rest. In 6 years he must replace it. Inflation is 4% p.a.; the old tractor depreciates at 14% p.a. reducing balance.
\[ \begin{gathered} 1\,250\,000\left(1 + \tfrac{0{,}09}{12}\right)^{48} \\ \times \left(1 + \tfrac{0{,}085}{4}\right)^{8} \\ -\ 100\,000\left(1 + \tfrac{0{,}085}{4}\right)^{4} \\ = R\,2\,008\,265{,}87 \end{gathered} \]
He invests R1 250 000 at 9% p.a. compounded monthly for 4 years, then 8,5% p.a. compounded quarterly. One year before the end he withdraws R100 000. Will he have enough?
\[ \begin{gathered} A = P(1 \pm i)^{n} \\ P = \dfrac{A}{(1 \pm i)^{n}} \\ 1 + i_{\text{eff}} = \left(1 + \tfrac{i^{(m)}}{m}\right)^{m} \end{gathered} \]
The exam moves, on one board.