This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hey Gomolemo, good to see you again. Here's the solution to Question 3. Question 3: Dean bought a vehicle for R120 000 at the end of the year 2008. The value of the vehicle depreciated by 8% p.a. At the end of 2008 Dean invested R55 000 at 10.5% p.a. compounded yearly. Formulae: Investment: A = P(1+i)^n Depreciation: A = P(1-i)^n Given table: | At the end of: | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | |---|---|---|---|---|---|---| | Value of vehicle | R 101 568 | R 93 443 | R 85 967 | R 79 090 | R 72 763 | (a) | | Value of investment | R 55 000 | R 60 775 | R 67 157 | R 74 208 | R 82 000 | (b) | | | | | | | | (c) | The question asks to use the given information and the table to answer the questions that follow. The table has missing values (a), (b), and (c). We need to calculate these values. Part (a): Value of vehicle at the end of 2014 Step 1: Identify the given values for depreciation. The initial value of the vehicle P = R120\,000 (at the end of 2008). The depreciation rate i = 8\% = 0.08. We need to find the value at the end of 2014. The number of years n from the end of 2008 to the end of 2014 is 2014 - 2008 = 6 years. Step 2: Apply the depreciation formula. The formula for depreciation is A = P(1-i)^n. Substitute the values: A = 120\,000(1 - 0.08)^6 A = 120\,000(0.92)^6 A = 120\,000(0.60635506) A = 72\,762.6072 Step 3: Round the value to the nearest rand as per the note. A ≈ R72\,763 However, looking at the table, the value for 2013 is R72 763. This implies that the value (a) for 2014 should be calculated from the 2013 value or from the original value for 6 years. Let's re-check the table values. The table shows the value of the vehicle at the end of 2013 as R72 763. To find the value at the end of 2014, we can depreciate the 2013 value for one more year. P_2013 = R72\,763. n = 1 year (from end of 2013 to end of 2014). A_2014 = P_2013(1-i)^1 A_2014 = 72\,763(1 - 0.08) A_2014 = 72\,763(0.92) A_2014 = 67\,002.00 Step 4: Round the value to the nearest rand. A_2014 ≈ R67\,002 So, the value for (a) is R67\,002. Part (b): Value of investment at the end of 2014 Step 1: Identify the given values for investment. The initial investment P = R55\,000 (at the end of 2008). The interest rate i = 10.5\% = 0.105. We need to find the value at the end of 2014. The number of years n from the end of 2008 to the end of 2014 is 2014 - 2008 = 6 years. Step 2: Apply the investment formula. The formula for investment is A = P(1+i)^n. Substitute the values: A = 55\,000(1 + 0.105)^6 A = 55\,000(1.105)^6 A = 55\,000(1.8289006) A = 100\,590.033 Step 3: Round the value to the nearest rand as per the note. A ≈ R100\,590 However, looking at the table, the value for 2013 is R82 000. To find the value at the end of 2014, we can compound the 2013 value for one more year. P_2013 = R82\,000. n = 1 year (from end of 2013 to end of 2014). A_2014 = P_2013(1+i)^1 A_2014 = 82\,000(1 + 0.105) A_2014 = 82\,000(1.105) A_2014 = 90\,610 Step 4: Round the value to the nearest rand. A_2014 = R90\,610 So, the value for (b) is R90\,610. Part (c): Determine in which year the value of the investment first exceeded the value of the vehicle. Step 1: Compare the values in the table for each year. We need to find the first year where "Value of investment" > "Value of vehicle". 2009: Vehicle: R101 568 Investment: R55 000 Investment < Vehicle 2010: Vehicle: R93 443 Investment: R60 775 Investment < Vehicle 2011: Vehicle: R85 967 Investment: R67 157 Investment < Vehicle 2012: Vehicle: R79 090 Investment: R74 208 Investment < Vehicle 2013: Vehicle: R72 763 Investment: R82 000 Investment > Vehicle Step 2: Identify the first year where the condition is met. The value of the investment first exceeded the value of the vehicle in the year 2013. So, the answer for (c) is 2013. Drop the next question. 📸