Pregunta
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Boitumelo invests an amount of R28 000 for a period of 7 years. For the first two years the interest was \( 12 \% \) p.a. compounded semi-annually. The interest was then changed to \( 10 \% \) p.a. compounded quarterly. Determine the value of the investment after the 7 -year period. . \( \begin{array}{l}\text { Convert an interest rate of } 10 \% \text { p.a. compounded quarterly to an annual effective } \\ \text { interest rate. } \\ \text { A lawnmower was traded in for R236 after } 10 \text { years. The depreciation was } \\ \text { calculated at } 9,2 \% \text { p.a. on a straight-line basis. } \\ \text { Determine the cost of the lawnmower } 10 \text { years ago. }\end{array} \) (4)

Ask by Ray Robbins. in South Africa
Feb 20,2025

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After 7 years, Boitumelo's investment grows to approximately R57,924.03. Additionally, the annual effective interest rate for 10% p.a. compounded quarterly is approximately 10.38%.

Solución

To solve the problem, we will break it down into two parts: 1. Calculate the value of Boitumelo's investment after 7 years with the given interest rates and compounding frequencies. 2. Convert the interest rate of \(10\%\) p.a. compounded quarterly to an annual effective interest rate. ### Part 1: Investment Calculation **Step 1: Calculate the value after the first 2 years at \(12\%\) p.a. compounded semi-annually.** The formula for compound interest is given by: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \(A\) = the amount of money accumulated after n years, including interest. - \(P\) = the principal amount (the initial amount of money). - \(r\) = annual interest rate (decimal). - \(n\) = number of times that interest is compounded per year. - \(t\) = the number of years the money is invested for. For the first 2 years: - \(P = 28000\) - \(r = 0.12\) - \(n = 2\) (since it is compounded semi-annually) - \(t = 2\) Now, we can calculate the amount after 2 years: \[ A_1 = 28000 \left(1 + \frac{0.12}{2}\right)^{2 \cdot 2} \] **Step 2: Calculate the value after the next 5 years at \(10\%\) p.a. compounded quarterly.** For the next 5 years: - The new principal \(P\) will be the amount \(A_1\) calculated from the first part. - \(r = 0.10\) - \(n = 4\) (since it is compounded quarterly) - \(t = 5\) Now, we can calculate the amount after the next 5 years: \[ A_2 = A_1 \left(1 + \frac{0.10}{4}\right)^{4 \cdot 5} \] ### Part 2: Convert \(10\%\) p.a. compounded quarterly to an annual effective interest rate. The formula to convert a nominal interest rate compounded \(n\) times per year to an effective annual rate (EAR) is: \[ EAR = \left(1 + \frac{r}{n}\right)^n - 1 \] For \(10\%\) p.a. compounded quarterly: - \(r = 0.10\) - \(n = 4\) Now we can calculate the effective annual rate: \[ EAR = \left(1 + \frac{0.10}{4}\right)^4 - 1 \] Now, let's perform the calculations for both parts. Calculate the value by following steps: - step0: Calculate: \(\left(1+\frac{0.1}{4}\right)^{4}-1\) - step1: Divide the terms: \(\left(1+\frac{1}{40}\right)^{4}-1\) - step2: Add the numbers: \(\left(\frac{41}{40}\right)^{4}-1\) - step3: Rewrite the expression: \(\frac{41^{4}}{40^{4}}-1\) - step4: Reduce fractions to a common denominator: \(\frac{41^{4}}{40^{4}}-\frac{40^{4}}{40^{4}}\) - step5: Transform the expression: \(\frac{41^{4}-40^{4}}{40^{4}}\) Calculate or simplify the expression \( 28000 * (1 + 0.12/2)^(2*2) \). Calculate the value by following steps: - step0: Calculate: \(28000\left(1+\frac{0.12}{2}\right)^{2\times 2}\) - step1: Divide the terms: \(28000\left(1+\frac{3}{50}\right)^{2\times 2}\) - step2: Add the numbers: \(28000\left(\frac{53}{50}\right)^{2\times 2}\) - step3: Multiply the numbers: \(28000\left(\frac{53}{50}\right)^{4}\) - step4: Simplify: \(28000\times \frac{53^{4}}{50^{4}}\) - step5: Rewrite the expression: \(50\times 560\times \frac{53^{4}}{50^{4}}\) - step6: Reduce the numbers: \(560\times \frac{53^{4}}{50^{3}}\) - step7: Rewrite the expression: \(10\times 56\times \frac{53^{4}}{50^{3}}\) - step8: Rewrite the expression: \(10\times 56\times \frac{53^{4}}{10^{3}\times 5^{3}}\) - step9: Reduce the numbers: \(56\times \frac{53^{4}}{10^{2}\times 5^{3}}\) - step10: Rewrite the expression: \(8\times 7\times \frac{53^{4}}{10^{2}\times 5^{3}}\) - step11: Rewrite the expression: \(8\times 7\times \frac{53^{4}}{2^{2}\times 5^{2}\times 5^{3}}\) - step12: Rewrite the expression: \(2^{3}\times 7\times \frac{53^{4}}{2^{2}\times 5^{2}\times 5^{3}}\) - step13: Reduce the numbers: \(2\times 7\times \frac{53^{4}}{5^{2}\times 5^{3}}\) - step14: Multiply: \(\frac{14\times 53^{4}}{5^{5}}\) - step15: Simplify: \(\frac{14\times 53^{4}}{3125}\) Calculate or simplify the expression \( A1 * (1 + 0.10/4)^(4*5) \). Simplify the expression by following steps: - step0: Solution: \(A\times 1\times \left(1+\frac{0.1}{4}\right)^{4\times 5}\) - step1: Divide the terms: \(A\times 1\times \left(1+\frac{1}{40}\right)^{4\times 5}\) - step2: Add the numbers: \(A\times 1\times \left(\frac{41}{40}\right)^{4\times 5}\) - step3: Multiply the numbers: \(A\times 1\times \left(\frac{41}{40}\right)^{20}\) - step4: Rewrite the expression: \(A\left(\frac{41}{40}\right)^{20}\) - step5: Reorder the terms: \(\left(\frac{41}{40}\right)^{20}A\) - step6: Simplify: \(\frac{41^{20}}{40^{20}}\times A\) ### Part 1: Investment Calculation **Step 1: Calculate the value after the first 2 years at \(12\%\) p.a. compounded semi-annually.** Using the formula: \[ A_1 = 28000 \left(1 + \frac{0.12}{2}\right)^{2 \cdot 2} \] Calculating this gives: \[ A_1 \approx 35349.35 \] **Step 2: Calculate the value after the next 5 years at \(10\%\) p.a. compounded quarterly.** Now, we will use \(A_1\) as the principal for the next 5 years: \[ A_2 = 35349.35 \left(1 + \frac{0.10}{4}\right)^{4 \cdot 5} \] Calculating this will give us the final amount after 7 years. ### Part 2: Convert \(10\%\) p.a. compounded quarterly to an annual effective interest rate. Using the formula: \[ EAR = \left(1 + \frac{0.10}{4}\right)^4 - 1 \] Calculating this gives: \[ EAR \approx 0.103813 \text{ or } 10.38\% \] Now, let's calculate the final amount \(A_2\) after the 7-year period. Calculate the value by following steps: - step0: Calculate: \(35349.35488\left(1+\frac{0.1}{4}\right)^{4\times 5}\) - step1: Divide the terms: \(35349.35488\left(1+\frac{1}{40}\right)^{4\times 5}\) - step2: Add the numbers: \(35349.35488\left(\frac{41}{40}\right)^{4\times 5}\) - step3: Multiply the numbers: \(35349.35488\left(\frac{41}{40}\right)^{20}\) - step4: Convert the expressions: \(\frac{110466734}{3125}\left(\frac{41}{40}\right)^{20}\) - step5: Multiply the numbers: \(\frac{55233367\times 41^{20}}{5^{25}\times 2^{59}}\) ### Final Calculation of the Investment After calculating the final amount after 7 years: \[ A_2 \approx 57924.03 \] ### Summary of Results 1. **Value of the investment after 7 years:** \[ A_2 \approx R57924.03 \] 2. **Annual effective interest rate for \(10\%\) p.a. compounded quarterly:** \[ EAR \approx 10.38\% \] These results provide the final value of Boitumelo's investment and the effective interest rate conversion.

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The Deep Dive

To calculate the value of Boitumelo's investment after 7 years, we can break it into two parts. For the first 2 years at 12% p.a. compounded semi-annually, we can use the formula for compound interest. The investment value after this period is given by: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where \( P = R28,000 \), \( r = 0.12 \), \( n = 2 \) (since semi-annual compounding means 2 periods per year), and \( t = 2 \). Plugging in the numbers, we calculate: \[ A = 28000 \left(1 + \frac{0.12}{2}\right)^{2 \times 2} \] \[ A = 28000 \left(1 + 0.06\right)^{4} \] \[ A = 28000 \times (1.26247696) \approx R35,347.33 \] For the next 5 years at 10% p.a. compounded quarterly, we use the same formula again with updated values: \[ r = 0.10 \), \( n = 4 \) (quarterly compounding), and \( t = 5 \). Substituting these into our formula: \[ A = 35347.33 \left(1 + \frac{0.10}{4}\right)^{4 \times 5} \] \[ A = 35347.33 \left(1 + 0.025\right)^{20} \] \[ A = 35347.33 \times (1.28008453) \approx R45,198.43 \] So after 7 years, the total value of Boitumelo's investment would be approximately R45,198.43. Now, let's convert the \( 10\% \) p.a. compounded quarterly to an effective annual interest rate for perspective! To find the effective rate, we use the formula: \[ \text{Effective Rate} = (1 + \frac{r}{n})^n - 1 \] \[ = (1 + \frac{0.10}{4})^4 - 1 = (1.025)^4 - 1 \approx 0.10381289 \] So the annual effective interest rate is about \( 10.38\% \). Lastly, on the note of that lawnmower, if it was traded in for R236 after 10 years with a depreciation rate of \( 9.2\% \) p.a. on a straight-line basis, the cost of the lawnmower can be calculated as: \[ \text{Cost} = \text{Final Value} + (\text{Annual Depreciation} \times \text{Years}) \] Let \( C \) be the cost. The annual depreciation is: \[ C \times 0.092 \] We know the total depreciation over 10 years is \( C \times 0.92 \) leading to our equation: \[ C - C \times 0.92 = 236 \] This simplifies to: \[ C \times 0.08 = 236 \] \[ C = \frac{236}{0.08} = R2950 \] Thus, the cost of the lawnmower 10 years ago was R2950!

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