2 Omphile wishes to have R10 000 in five years. He can invest R2 000 now. Determine the interest rate he will need to receive compounded annually. At what compound interest rate do I need to invest R12000 for my investment to grow to R 16500 in seven years? Palesa invests R2 000 in an account that offers her a \( 9.5 \% \) compound interest rate p.a. Alter two years she moves her money to an investment bank that can offer her an \( 11.2 \% \mathrm{growth} \). compounded annually, on her investment. How much is Palesa's investment worth after 8 years? Ethan invests R 10000 in a saving account, where interest is calculated at 8,75\% p.a. compounded annually. After two years he invests a further R 2000 in the account and three years after that he withdraws R 4500 . Determine how much money Ethan will have at the end of 8 years.

Calculate the value by following steps: - step0: Calculate: \(2000\left(1+0.095\right)^{2}\left(1+0.112\right)^{6}\) - step1: Add the numbers: \(2000\times 1.095^{2}\left(1+0.112\right)^{6}\) - step2: Add the numbers: \(2000\times 1.095^{2}\times 1.112^{6}\) - step3: Convert the expressions: \(2000\left(\frac{219}{200}\right)^{2}\times 1.112^{6}\) - step4: Convert the expressions: \(2000\left(\frac{219}{200}\right)^{2}\left(\frac{139}{125}\right)^{6}\) - step5: Multiply the terms: \(\frac{47961}{20}\left(\frac{139}{125}\right)^{6}\) - step6: Evaluate the power: \(\frac{47961}{20}\times \frac{139^{6}}{125^{6}}\) - step7: Multiply the fractions: \(\frac{47961\times 139^{6}}{20\times 125^{6}}\) Solve the equation \( 12000(1+r)^{7}=16500 \). Solve the equation by following steps: - step0: Solve for \(r\): \(12000\left(1+r\right)^{7}=16500\) - step1: Divide both sides: \(\frac{12000\left(1+r\right)^{7}}{12000}=\frac{16500}{12000}\) - step2: Divide the numbers: \(\left(1+r\right)^{7}=\frac{11}{8}\) - step3: Simplify the equation: \(\sqrt[7]{\left(1+r\right)^{7}}=\sqrt[7]{\frac{11}{8}}\) - step4: Calculate: \(1+r=\sqrt[7]{\frac{11}{8}}\) - step5: Simplify the root: \(1+r=\frac{\sqrt[7]{176}}{2}\) - step6: Move the constant to the right side: \(r=\frac{\sqrt[7]{176}}{2}-1\) - step7: Subtract the numbers: \(r=\frac{\sqrt[7]{176}-2}{2}\) Calculate or simplify the expression \( 10000(1+0.0875)^{2}+2000(1+0.0875)^{6}-4500 \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.0875\right)^{2}+2000\left(1+0.0875\right)^{6}-4500\) - step1: Add the numbers: \(10000\times 1.0875^{2}+2000\left(1+0.0875\right)^{6}-4500\) - step2: Add the numbers: \(10000\times 1.0875^{2}+2000\times 1.0875^{6}-4500\) - step3: Convert the expressions: \(10000\left(\frac{87}{80}\right)^{2}+2000\times 1.0875^{6}-4500\) - step4: Convert the expressions: \(10000\left(\frac{87}{80}\right)^{2}+2000\left(\frac{87}{80}\right)^{6}-4500\) - step5: Multiply the terms: \(\frac{189225}{16}+2000\left(\frac{87}{80}\right)^{6}-4500\) - step6: Simplify: \(\frac{189225}{16}+2000\times \frac{87^{6}}{80^{6}}-4500\) - step7: Rewrite the expression: \(\frac{189225}{16}+\frac{2000\times 87^{6}}{80^{6}}-4500\) - step8: Reduce fractions to a common denominator: \(\frac{189225\times 128000^{2}}{16\times 128000^{2}}+\frac{2000\times 87^{6}}{80^{6}}-\frac{4500\times 16\times 128000^{2}}{16\times 128000^{2}}\) - step9: Rewrite the expression: \(\frac{189225\times 128000^{2}}{80^{6}}+\frac{2000\times 87^{6}}{80^{6}}-\frac{4500\times 16\times 128000^{2}}{80^{6}}\) - step10: Transform the expression: \(\frac{189225\times 128000^{2}+2000\times 87^{6}-4500\times 16\times 128000^{2}}{80^{6}}\) - step11: Multiply the terms: \(\frac{55680000^{2}+2000\times 87^{6}-4500\times 16\times 128000^{2}}{80^{6}}\) - step12: Multiply the terms: \(\frac{55680000^{2}+2000\times 87^{6}-72000\times 128000^{2}}{80^{6}}\) - step13: Subtract the numbers: \(\frac{1920614400000000+2000\times 87^{6}}{80^{6}}\) - step14: Reduce the fraction: \(\frac{960307200000+9\times 219501^{2}}{131072000}\) Solve the equation \( 2000(1+r)^{5}=10000 \). Solve the equation by following steps: - step0: Solve for \(r\): \(2000\left(1+r\right)^{5}=10000\) - step1: Divide both sides: \(\frac{2000\left(1+r\right)^{5}}{2000}=\frac{10000}{2000}\) - step2: Divide the numbers: \(\left(1+r\right)^{5}=5\) - step3: Simplify the equation: \(\sqrt[5]{\left(1+r\right)^{5}}=\sqrt[5]{5}\) - step4: Calculate: \(1+r=\sqrt[5]{5}\) - step5: Move the constant to the right side: \(r=\sqrt[5]{5}-1\) Let's break down each part of the problem step by step. ### 1. Interest Rate for R10,000 in 5 Years Omphile wants to have R10,000 in 5 years by investing R2,000 now. The formula for compound interest is: \[ A = P(1 + r)^n \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of years the money is invested or borrowed. Substituting the known values: \[ 10000 = 2000(1 + r)^5 \] To find \( r \), we can rearrange the equation: \[ (1 + r)^5 = \frac{10000}{2000} = 5 \] Taking the fifth root: \[ 1 + r = \sqrt[5]{5} \] Thus, the interest rate \( r \) is: \[ r = \sqrt[5]{5} - 1 \] ### 2. Interest Rate for R12,000 to Grow to R16,500 in 7 Years We need to find the interest rate \( r \) for R12,000 to grow to R16,500 in 7 years. Using the same formula: \[ 16500 = 12000(1 + r)^7 \] Rearranging gives: \[ (1 + r)^7 = \frac{16500}{12000} = 1.375 \] Taking the seventh root: \[ 1 + r = \sqrt[7]{1.375} \] Thus, the interest rate \( r \) is: \[ r = \sqrt[7]{1.375} - 1 \] ### 3. Palesa's Investment Worth After 8 Years Palesa invests R2,000 at a \( 9.5\% \) interest rate for 2 years, then moves it to an \( 11.2\% \) interest rate for the next 6 years. Calculating the value after 2 years: \[ A_1 = 2000(1 + 0.095)^2 \] Then, this amount is invested for another 6 years at \( 11.2\% \): \[ A_2 = A_1(1 + 0.112)^6 \] Calculating this gives us the total value of Palesa's investment after 8 years. ### 4. Ethan's Investment After 8 Years Ethan invests R10,000 at \( 8.75\% \) for 2 years, then adds R2,000. After 3 more years, he withdraws R4,500. Calculating the value after 2 years: \[ A_1 = 10000(1 + 0.0875)^2 \] Adding R2,000 after 2 years: \[ A_2 = A_1 + 2000 \] Calculating the value after 5 years (3 years after the additional investment): \[ A_3 = A_2(1 + 0.0875)^3 \] Finally, subtracting the withdrawal: \[ A_4 = A_3 - 4500 \] Now, let's calculate the final values for Palesa and Ethan's investments. Calculate the value by following steps: - step0: Calculate: \(2000\left(1+0.095\right)^{2}\left(1+0.112\right)^{6}\) - step1: Add the numbers: \(2000\times 1.095^{2}\left(1+0.112\right)^{6}\) - step2: Add the numbers: \(2000\times 1.095^{2}\times 1.112^{6}\) - step3: Convert the expressions: \(2000\left(\frac{219}{200}\right)^{2}\times 1.112^{6}\) - step4: Convert the expressions: \(2000\left(\frac{219}{200}\right)^{2}\left(\frac{139}{125}\right)^{6}\) - step5: Multiply the terms: \(\frac{47961}{20}\left(\frac{139}{125}\right)^{6}\) - step6: Evaluate the power: \(\frac{47961}{20}\times \frac{139^{6}}{125^{6}}\) - step7: Multiply the fractions: \(\frac{47961\times 139^{6}}{20\times 125^{6}}\) Calculate or simplify the expression \( 10000*(1+0.0875)^{2}+2000-4500*(1+0.0875)^{3} \). Calculate the value by following steps: - step0: Calculate: \(10000\left(1+0.0875\right)^{2}+2000-4500\left(1+0.0875\right)^{3}\) - step1: Add the numbers: \(10000\times 1.0875^{2}+2000-4500\left(1+0.0875\right)^{3}\) - step2: Add the numbers: \(10000\times 1.0875^{2}+2000-4500\times 1.0875^{3}\) - step3: Convert the expressions: \(10000\left(\frac{87}{80}\right)^{2}+2000-4500\times 1.0875^{3}\) - step4: Convert the expressions: \(10000\left(\frac{87}{80}\right)^{2}+2000-4500\left(\frac{87}{80}\right)^{3}\) - step5: Multiply the terms: \(\frac{189225}{16}+2000-4500\left(\frac{87}{80}\right)^{3}\) - step6: Simplify: \(\frac{189225}{16}+2000-4500\times \frac{87^{3}}{80^{3}}\) - step7: Rewrite the expression: \(\frac{189225}{16}+2000-\frac{4500\times 87^{3}}{80^{3}}\) - step8: Reduce fractions to a common denominator: \(\frac{189225\times 32000}{16\times 32000}+\frac{2000\times 16\times 32000}{16\times 32000}-\frac{4500\times 87^{3}}{80^{3}}\) - step9: Multiply the numbers: \(\frac{189225\times 32000}{512000}+\frac{2000\times 16\times 32000}{16\times 32000}-\frac{4500\times 87^{3}}{80^{3}}\) - step10: Multiply the numbers: \(\frac{189225\times 32000}{512000}+\frac{2000\times 16\times 32000}{512000}-\frac{4500\times 87^{3}}{80^{3}}\) - step11: Rewrite the expression: \(\frac{189225\times 32000}{80^{3}}+\frac{2000\times 16\times 32000}{80^{3}}-\frac{4500\times 87^{3}}{80^{3}}\) - step12: Transform the expression: \(\frac{189225\times 32000+2000\times 16\times 32000-4500\times 87^{3}}{80^{3}}\) - step13: Multiply the numbers: \(\frac{6055200000+2000\times 16\times 32000-4500\times 87^{3}}{80^{3}}\) - step14: Multiply the terms: \(\frac{6055200000+1024000000-4500\times 87^{3}}{80^{3}}\) - step15: Add the numbers: \(\frac{7079200000-4500\times 87^{3}}{80^{3}}\) - step16: Rewrite the expression: \(\frac{500\left(14158400-9\times 87^{3}\right)}{500\times 1024}\) - step17: Reduce the fraction: \(\frac{14158400-9\times 87^{3}}{1024}\) ### Summary of Results 1. **Interest Rate for R10,000 in 5 Years:** \[ r = \sqrt[5]{5} - 1 \approx 1.37973 - 1 \approx 0.37973 \text{ or } 37.973\% \] 2. **Interest Rate for R12,000 to Grow to R16,500 in 7 Years:** \[ r = \sqrt[7]{1.375} - 1 \approx 0.0487 \text{ or } 4.87\% \] 3. **Palesa's Investment Worth After 8 Years:** \[ A \approx 4534.06 \] 4. **Ethan's Investment After 8 Years:** \[ A \approx 8038.94 \] ### Final Values - Omphile needs an interest rate of approximately **37.973%** compounded annually to reach R10,000 in 5 years. - To grow R12,000 to R16,500 in 7 years, the required interest rate is approximately **4.87%** compounded annually. - Palesa's investment will be worth approximately **R4,534.06** after 8 years. - Ethan will have approximately **R8,038.94** at the end of 8 years.