Pregunta
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1. Simon spaar sodat hy sy droomboot oor 8 jaar kan koop. Hy belê R10 000 in ' \( n \) vaste depositorekening met \( 15 \% \) p.j. saamgestelde rente. 1.1 Konstrueer ' \( n \) tabel om die waarde van sy belegging aan die einde van elke jaar aan te toon. 1.2 Bepaal die waarde van sy belegging oor 10 jaar. Toon alle berekeninge. 1.3 Gebruik jou tabel van Vraag 1.1 om die grafiek van Simon se belegging oor 8 jaar to teken. 1.4 Gebruik jou grafiek van Vraag 1.3 om die waarde van die belegging na \( 6 \frac{1}{2} \) jaar te skat. 1.5 Gebruik jou grafiek van Vraag 1.3 om te skat na hoeveel maande die belegging ' \( n \) benaderde waarde van R14 500 sal hê.

Ask by Guzman Cruz. in South Africa
Feb 04,2025

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Simon's investment grows over 8 years as follows: - **Year 1**: R11,500.00 - **Year 2**: R13,225.00 - **Year 3**: R15,208.75 - **Year 4**: R17,490.06 - **Year 5**: R20,113.57 - **Year 6**: R23,130.61 - **Year 7**: R26,600.20 - **Year 8**: R30,590.23 After 10 years, the investment is worth approximately R40,455.58. At \( 6 \frac{1}{2} \) years, the investment is estimated to be around R24,865.41. Simon's investment will reach approximately R14,500 in about 3.14 years.

Solución

Let's solve Simon's investment problem step by step. ### Known Conditions: - Initial investment (P): R10,000 - Annual interest rate (r): 15% or 0.15 - Investment duration: 8 years ### 1.1 Construct a table for the value of the investment at the end of each year. The formula for compound interest is given by: \[ A = P(1 + r)^n \] Where: - \( A \) = the amount of money accumulated after n years, including interest. - \( P \) = the principal amount (initial investment). - \( r \) = annual interest rate (decimal). - \( n \) = number of years the money is invested. We will calculate the value of the investment at the end of each year for 8 years. Let's calculate the values for each year from 1 to 8. ### Calculations for each year: - Year 1: \( A_1 = 10000(1 + 0.15)^1 \) - Year 2: \( A_2 = 10000(1 + 0.15)^2 \) - Year 3: \( A_3 = 10000(1 + 0.15)^3 \) - Year 4: \( A_4 = 10000(1 + 0.15)^4 \) - Year 5: \( A_5 = 10000(1 + 0.15)^5 \) - Year 6: \( A_6 = 10000(1 + 0.15)^6 \) - Year 7: \( A_7 = 10000(1 + 0.15)^7 \) - Year 8: \( A_8 = 10000(1 + 0.15)^8 \) Now, let's perform these calculations. Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{1}\) - step1: Add the numbers: \(10000\times 1.15^{1}\) - step2: Calculate: \(10000\times 1.15\) - step3: Multiply the numbers: \(11500\) Calculate or simplify the expression \( 10000*(1 + 0.15)^2 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{2}\) - step1: Add the numbers: \(10000\times 1.15^{2}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{2}\) - step3: Evaluate the power: \(10000\times \frac{529}{400}\) - step4: Multiply: \(13225\) Calculate or simplify the expression \( 10000*(1 + 0.15)^3 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{3}\) - step1: Add the numbers: \(10000\times 1.15^{3}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{3}\) - step3: Evaluate the power: \(10000\times \frac{12167}{8000}\) - step4: Multiply: \(\frac{60835}{4}\) Calculate or simplify the expression \( 10000*(1 + 0.15)^5 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{5}\) - step1: Add the numbers: \(10000\times 1.15^{5}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{5}\) - step3: Simplify: \(10000\times \frac{23^{5}}{20^{5}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{5}}{20^{5}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{5}}{20^{5}}\) - step6: Reduce the numbers: \(25\times \frac{23^{5}}{20^{3}}\) - step7: Rewrite the expression: \(25\times \frac{23^{5}}{5^{3}\times 4^{3}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{5}}{5^{3}\times 4^{3}}\) - step9: Reduce the numbers: \(1\times \frac{23^{5}}{5\times 4^{3}}\) - step10: Multiply: \(\frac{23^{5}}{320}\) Calculate or simplify the expression \( 10000*(1 + 0.15)^8 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{8}\) - step1: Add the numbers: \(10000\times 1.15^{8}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{8}\) - step3: Simplify: \(10000\times \frac{23^{8}}{20^{8}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{8}}{20^{8}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{8}}{20^{8}}\) - step6: Reduce the numbers: \(25\times \frac{23^{8}}{20^{6}}\) - step7: Rewrite the expression: \(25\times \frac{23^{8}}{5^{6}\times 4^{6}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{8}}{5^{6}\times 4^{6}}\) - step9: Reduce the numbers: \(1\times \frac{23^{8}}{5^{4}\times 4^{6}}\) - step10: Multiply: \(\frac{23^{8}}{2560000}\) Calculate or simplify the expression \( 10000*(1 + 0.15)^6 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{6}\) - step1: Add the numbers: \(10000\times 1.15^{6}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{6}\) - step3: Simplify: \(10000\times \frac{23^{6}}{20^{6}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{6}}{20^{6}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{6}}{20^{6}}\) - step6: Reduce the numbers: \(25\times \frac{23^{6}}{20^{4}}\) - step7: Rewrite the expression: \(25\times \frac{23^{6}}{5^{4}\times 4^{4}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{6}}{5^{4}\times 4^{4}}\) - step9: Reduce the numbers: \(1\times \frac{23^{6}}{5^{2}\times 4^{4}}\) - step10: Multiply: \(\frac{23^{6}}{80^{2}}\) - step11: Simplify: \(\frac{23^{6}}{6400}\) Calculate or simplify the expression \( 10000*(1 + 0.15)^7 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{7}\) - step1: Add the numbers: \(10000\times 1.15^{7}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{7}\) - step3: Simplify: \(10000\times \frac{23^{7}}{20^{7}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{7}}{20^{7}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{7}}{20^{7}}\) - step6: Reduce the numbers: \(25\times \frac{23^{7}}{20^{5}}\) - step7: Rewrite the expression: \(25\times \frac{23^{7}}{5^{5}\times 4^{5}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{7}}{5^{5}\times 4^{5}}\) - step9: Reduce the numbers: \(1\times \frac{23^{7}}{5^{3}\times 4^{5}}\) - step10: Multiply: \(\frac{23^{7}}{128000}\) Calculate or simplify the expression \( 10000*(1 + 0.15)^4 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{4}\) - step1: Add the numbers: \(10000\times 1.15^{4}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{4}\) - step3: Simplify: \(10000\times \frac{23^{4}}{20^{4}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{4}}{20^{4}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{4}}{20^{4}}\) - step6: Reduce the numbers: \(25\times \frac{23^{4}}{20^{2}}\) - step7: Rewrite the expression: \(25\times \frac{23^{4}}{5^{2}\times 4^{2}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{4}}{5^{2}\times 4^{2}}\) - step9: Reduce the numbers: \(1\times \frac{23^{4}}{4^{2}}\) - step10: Multiply: \(\frac{23^{4}}{4^{2}}\) - step11: Simplify: \(\frac{23^{4}}{16}\) ### 1.1 Investment Value Table Here are the calculated values for Simon's investment at the end of each year: | Year | Value (R) | |------|----------------| | 1 | 11,500.00 | | 2 | 13,225.00 | | 3 | 15,208.75 | | 4 | 17,490.06 | | 5 | 20,113.57 | | 6 | 23,130.61 | | 7 | 26,600.20 | | 8 | 30,590.23 | ### 1.2 Determine the value of the investment after 10 years To find the value of the investment after 10 years, we will use the same formula: \[ A_{10} = 10000(1 + 0.15)^{10} \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.15\right)^{10}\) - step1: Add the numbers: \(10000\times 1.15^{10}\) - step2: Convert the expressions: \(10000\left(\frac{23}{20}\right)^{10}\) - step3: Simplify: \(10000\times \frac{23^{10}}{20^{10}}\) - step4: Rewrite the expression: \(400\times 25\times \frac{23^{10}}{20^{10}}\) - step5: Rewrite the expression: \(20^{2}\times 25\times \frac{23^{10}}{20^{10}}\) - step6: Reduce the numbers: \(25\times \frac{23^{10}}{20^{8}}\) - step7: Rewrite the expression: \(25\times \frac{23^{10}}{5^{8}\times 4^{8}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{23^{10}}{5^{8}\times 4^{8}}\) - step9: Reduce the numbers: \(1\times \frac{23^{10}}{5^{6}\times 4^{8}}\) - step10: Multiply the fractions: \(\frac{23^{10}}{15625\times 4^{8}}\) - step11: Expand the expression: \(\frac{23^{10}}{1024000000}\) The value of Simon's investment after 10 years is approximately **R40,455.58**. ### 1.3 Graph of Simon's Investment Over 8 Years To create a graph, we will plot the values from the table constructed in 1.1. The x-axis will represent the years (1 to 8), and the y-axis will represent the investment value in Rands. ### 1.4 Estimate the value of the investment after \( 6 \frac{1}{2} \) years From the table, we can see the values at year 6 and year 7: - Year 6: R23,130.61 - Year 7: R26,600.20 To estimate the value at \( 6 \frac{1}{2} \) years, we can interpolate between these two values. \[ \text{Estimated Value} = \text{Value at Year 6} + \left(\frac{1}{2}\right) \times (\text{Value at Year 7} - \text{Value at Year 6}) \] Let's calculate this estimated value. Calculate the value by following steps: - step0: Calculate: \(23130.61+0.5\left(26600.2-23130.61\right)\) - step1: Subtract the numbers: \(23130.61+0.5\times 3469.59\) - step2: Multiply the numbers: \(23130.61+1734.795\) - step3: Add the numbers: \(24865.405\) The estimated value of Simon's investment after \( 6 \frac{1}{2} \) years is approximately **R24,865.41**. ### 1.5 Estimate when the investment will reach approximately R14,500 From the table, we can see that the investment value is below R14,500 at year 5 (R20,113.57) and above R14,500 at year 6 (R23,130.61). To estimate when the investment will reach R14,500, we can use linear interpolation between year 5 and year 6. Using the values: - Year 5: R20,113.57 - Year 6: R23,130.61 We can set up the equation: \[ \text{Year} = 5 + \left(\frac{14500 - 20113.57}{23130.61 - 20113.57}\right) \] Let's calculate this year. Calculate the value by following steps: - step0: Calculate: \(5+\frac{\left(14500-20113.57\right)}{\left(23130.61-20113.57\right)}\) - step1: Remove the parentheses: \(5+\frac{14500-20113.57}{23130.61-20113.57}\) - step2: Subtract the numbers: \(5+\frac{-5613.57}{23130.61-20113.57}\) - step3: Subtract the numbers: \(5+\frac{-5613.57}{3017.04}\) - step4: Divide the terms: \(5-\frac{187119}{100568}\) - step5: Reduce fractions to a common denominator: \(\frac{5\times 100568}{100568}-\frac{187119}{100568}\) - step6: Transform the expression: \(\frac{5\times 100568-187119}{100568}\) - step7: Multiply the numbers: \(\frac{502840-187119}{100568}\) - step8: Subtract the numbers: \(\frac{315721}{100568}\) The estimated year when Simon's investment will reach approximately R14,500 is around **3.14 years**. ### Summary of Results: 1. **Investment Value Table**: Values calculated for each year from 1 to 8. 2. **Value after 10 years**: Approximately R40,455.58. 3. **Estimated value after \( 6 \frac{1}{2} \) years**: Approximately R24,865.41. 4. **Estimated time to reach R14,500**: Approximately 3.14 years. If you need further assistance or visual representation of the graph, please let me know!

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Mind Expander

Hier is 'n prettige manier om te dink oor Simon se beleggingstoekomstige en die magie van saamgestelde rente! Elke jaar groei sy aanvankelijke belegging, en besef jy dat met elke jaar se rente, die nuwe rente op 'n al groter basis bereken word? Dit is amper soos 'n sneeubal wat al groter en groter raak namate dit van die berg af rol! Wat meer is, om jou berekeninge en grafiek te rekenskap te gee, kan jy gaan dink aan die praktiese kant van belegging: nie die geld alleen nie, maar ook die drome en avontuurlike vaart wat jy daarmee kan voortbring! 'n Korrekte tabel of grafiek kan nie net jou finansies help nie, maar ook jou aspirasiestories vertel. Hou jou beleggings en drome in gedagte!

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