QUESTION 6 6.1 Six years ago, Thabo bought a phone for R13000. The value of the phone depreciated annually according 10 the reducing-balance method. The value of the phone is now R8 337,75. Calculate the annual rate of depreciation. (3) 6.2 Eric and Thandi need to save R80000 each to go on a holiday at the end of December 2027. - Thandi decides that she will start saving at the end of January 2025. She will make 36 monihly deposits into a savings account that pays interest at \( 8,6 \% \) p.a., compounded monthly. The deposit will be made at the end of each month. - Eric calculates that if he makes 48 deposits of R1 402,31, starting at the end of January 2024, he will have enough money to go on holiday. He will make his deposits into a savings account at the end of each month. The savings account pays interest at \( 8,6 \% \) p.a., compounded monthly. Calculate the difference between the total amount that Eric and Thandi will deposit into their respective savings accounts over the given period. 6.3 Lesibana was granted a loan of R225 000. The rate of interest for the loan is \( 9 \% \) p.a., compounded monthly. Lesibana will make monthly payments of RS 500 , starting exactly four months afier the loan was granted. How many payments will Lesibana make to settle the loan? (6)

Let's solve each part of the problem step by step. ### 6.1 Calculate the annual rate of depreciation We know that the value of the phone depreciated according to the reducing-balance method. The formula for the value of an asset after depreciation is given by: \[ V = P(1 - r)^t \] Where: - \( V \) = current value of the asset (R8,337.75) - \( P \) = original value of the asset (R13,000) - \( r \) = annual rate of depreciation (as a decimal) - \( t \) = number of years (6 years) We can rearrange the formula to solve for \( r \): \[ r = 1 - \left( \frac{V}{P} \right)^{\frac{1}{t}} \] Now, substituting the known values: \[ r = 1 - \left( \frac{8337.75}{13000} \right)^{\frac{1}{6}} \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(1-\left(\frac{8337.75}{13000}\right)^{\frac{1}{6}}\) - step1: Divide the terms: \(1-\left(\frac{33351}{52000}\right)^{\frac{1}{6}}\) - step2: Rewrite the expression: \(1-\frac{33351^{\frac{1}{6}}}{52000^{\frac{1}{6}}}\) - step3: Calculate: \(1-\frac{33351^{\frac{1}{6}}\times 52000^{\frac{5}{6}}}{52000}\) - step4: Reduce fractions to a common denominator: \(\frac{52000}{52000}-\frac{33351^{\frac{1}{6}}\times 52000^{\frac{5}{6}}}{52000}\) - step5: Transform the expression: \(\frac{52000-33351^{\frac{1}{6}}\times 52000^{\frac{5}{6}}}{52000}\) - step6: Simplify: \(\frac{52000-\sqrt[6]{33351}\times \sqrt[6]{52000^{5}}}{52000}\) - step7: Expand the expression: \(\frac{52000-\sqrt[6]{33351\times 52000^{5}}}{52000}\) Calculating the annual rate of depreciation: \[ r \approx 0.071352 \] To express this as a percentage, we multiply by 100: \[ r \approx 7.14\% \] Thus, the annual rate of depreciation is approximately **7.14%**. ### 6.2 Calculate the difference between the total amount deposited by Eric and Thandi **Thandi's Savings:** - Monthly deposit: \( D_T \) - Number of deposits: 36 - Interest rate: \( 8.6\% \) p.a. compounded monthly The monthly interest rate \( i_T \) is: \[ i_T = \frac{8.6\%}{12} = \frac{0.086}{12} \approx 0.00716667 \] The future value \( FV_T \) of Thandi's savings can be calculated using the future value of an annuity formula: \[ FV_T = D_T \times \frac{(1 + i_T)^n - 1}{i_T} \] Where \( n = 36 \). **Eric's Savings:** - Monthly deposit: \( D_E = 1402.31 \) - Number of deposits: 48 - Interest rate: \( 8.6\% \) p.a. compounded monthly The monthly interest rate \( i_E \) is the same as Thandi's: \[ i_E = 0.00716667 \] The future value \( FV_E \) of Eric's savings is: \[ FV_E = D_E \times \frac{(1 + i_E)^n - 1}{i_E} \] Where \( n = 48 \). Now, we need to find \( D_T \) such that \( FV_T = 80000 \) and then calculate the total deposits made by both. Let's first calculate \( D_T \) using the future value formula for Thandi's savings. Rearranging the formula gives: \[ D_T = \frac{FV_T \cdot i_T}{(1 + i_T)^n - 1} \] Substituting the values: \[ D_T = \frac{80000 \cdot 0.00716667}{(1 + 0.00716667)^{36} - 1} \] Now, let's calculate \( D_T \) and then the total deposits for both Thandi and Eric. Calculate the value by following steps: - step0: Calculate: \(1402.31\times 48\) - step1: Multiply the numbers: \(67310.88\) Calculate or simplify the expression \( 80000 * 0.00716667 / ((1 + 0.00716667)^36 - 1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{80000\times 0.00716667}{\left(\left(1+0.00716667\right)^{36}-1\right)}\) - step1: Remove the parentheses: \(\frac{80000\times 0.00716667}{\left(1+0.00716667\right)^{36}-1}\) - step2: Add the numbers: \(\frac{80000\times 0.00716667}{1.00716667^{36}-1}\) - step3: Convert the expressions: \(\frac{80000\times 0.00716667}{\left(\frac{100716667}{100000000}\right)^{36}-1}\) - step4: Multiply the numbers: \(\frac{573.3336}{\left(\frac{100716667}{100000000}\right)^{36}-1}\) - step5: Subtract the numbers: \(\frac{573.3336}{\frac{100716667^{36}-100000000^{36}}{100000000^{36}}}\) - step6: Convert the expressions: \(\frac{\frac{716667}{1250}}{\frac{100716667^{36}-100000000^{36}}{100000000^{36}}}\) - step7: Multiply by the reciprocal: \(\frac{716667}{1250}\times \frac{100000000^{36}}{100716667^{36}-100000000^{36}}\) - step8: Rewrite the expression: \(\frac{716667}{1250}\times \frac{1562500^{36}\times 64^{36}}{100716667^{36}-100000000^{36}}\) - step9: Rewrite the expression: \(\frac{716667}{1250}\times \frac{1250^{72}\times 64^{36}}{100716667^{36}-100000000^{36}}\) - step10: Reduce the numbers: \(716667\times \frac{1250^{71}\times 64^{36}}{100716667^{36}-100000000^{36}}\) - step11: Multiply: \(\frac{716667\times 1250^{71}\times 64^{36}}{100716667^{36}-100000000^{36}}\) Calculating Thandi's monthly deposit: \[ D_T \approx 1955.78 \] Calculating Eric's total deposits: \[ \text{Total deposits by Eric} = 1402.31 \times 48 \approx 67310.88 \] Now, we can calculate the total amount deposited by Thandi: \[ \text{Total deposits by Thandi} = D_T \times 36 \approx 1955.78 \times 36 \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(1955.78\times 36\) - step1: Multiply the numbers: \(70408.08\) Calculating Thandi's total deposits: \[ \text{Total deposits by Thandi} \approx 70408.08 \] Now, we can find the difference between the total amounts deposited by Eric and Thandi: \[ \text{Difference} = \text{Total deposits by Eric} - \text{Total deposits by Thandi} \] Substituting the values: \[ \text{Difference} = 67310.88 - 70408.08 \] Let's calculate this difference. Calculate the value by following steps: - step0: Calculate: \(67310.88-70408.08\) - step1: Subtract the numbers: \(-3097.2\) The difference between the total amounts deposited by Eric and Thandi is: \[ \text{Difference} \approx -3097.20 \] This means that Thandi will deposit approximately **R3,097.20** more than Eric. ### 6.3 Calculate how many payments Lesibana will make to settle the loan Lesibana was granted a loan of R225,000 with an interest rate of \( 9\% \) p.a., compounded monthly. The monthly interest rate \( i \) is: \[ i = \frac{9\%}{12} = \frac{0.09}{12} \approx 0.0075 \] Lesibana will make monthly payments of R500 starting four months after the loan was granted. We can use the formula for the present value of an annuity to find the number of payments \( n \): \[ PV = PMT \times \frac{1 - (1 + i)^{-n}}{i} \] Where: - \( PV = 225000 \) - \( PMT = 500 \) - \( i = 0.0075 \) Rearranging the formula to solve for \( n \): \[ n = -\frac{\log\left(1 - \frac{PV \cdot i}{PMT}\right)}{\log(1 + i)} \] Now, substituting the known values: \[ n = -\frac{\log\left(1 - \frac{225000 \cdot 0.0075}{500}\right)}{\log(1 + 0.0075)} \] Let's calculate this value. by following steps: - step0: : \(-\frac{\log_{10}{\left(1-\left(\frac{225000\times 0.0075}{500}\right)\right)}}{\log_{10}{\left(1+0.0075\right)}}\) - step1: Calculate: \(\textrm{Undefined}\) It seems that the calculation resulted in an undefined value, which typically indicates that the argument of the logarithm is less than or equal to zero. Let's check the expression inside the logarithm: \[ 1 - \frac{225000 \cdot 0.0075}{500} \] Calculating the fraction: \[ \frac{225000 \cdot 0.0075}{500} = \frac{1687.5}{500} = 3.375 \] Since \( 1 - 3.375 \) is negative, this indicates that the monthly payment of R500 is insufficient to cover the interest on the loan, leading to an infinite number of payments required to settle the loan. Thus, Lesibana will not be able to settle the loan with the current payment plan. ### Summary of Results: - **6.1**: Annual rate of depreciation: **7.14%** - **6.2**: Difference in total deposits: **R3,097.20** (Thandi deposits more) - **6.3**: Lesibana cannot settle the loan with R500 monthly payments.