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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Here are the solutions to the questions: Question 6.1.1 Patrick deposited an amount of R18 500 into an account earning r\% interest p.a., compounded monthly. After 6 months, his balance was R19 319.48. Calculate the value of r. The formula for compound interest is A = P(1 + (r)/(m))^n, where A is the future value, P is the principal, r is the annual interest rate, m is the number of compounding periods per year, and n is the total number of compounding periods. Step 1: Identify the given values and substitute them into the formula. Given: P = R18\,500, A = R19\,319.48, n = 6 months, m = 12 (compounded monthly). 19319.48 = 18500(1 + (r)/(12))^6 Step 2: Isolate the term containing r. (19319.48)/(18500) = (1 + (r)/(12))^6 1.044296216 ≈ (1 + (r)/(12))^6 Step 3: Take the 6th root of both sides. [6]1.044296216 = 1 + (r)/(12) 1.007233333 ≈ 1 + (r)/(12) Step 4: Solve for r. 0.007233333 ≈ (r)/(12) r ≈ 0.007233333 × 12 r ≈ 0.0868 To express r as a percentage: r ≈ 0.0868 × 100\% 8.68\% Question 6.1.2 Calculate the effective interest rate. The formula for the effective interest rate is i_eff = (1 + i_nomm)^m - 1, where i_nom is the nominal annual interest rate and m is the number of compounding periods per year. Step 1: Substitute the nominal interest rate (r ≈ 0.0868) and compounding frequency (m=12) into the formula. i_eff = (1 + (0.0868)/(12))^12 - 1 Step 2: Calculate the effective interest rate. i_eff = (1 + 0.007233333)^12 - 1 i_eff = (1.007233333)^12 - 1 i_eff ≈ 1.0905 - 1 i_eff ≈ 0.0905 To express as a percentage: i_eff ≈ 0.0905 × 100\% 9.05\% Question 6.2.1 The value of the old laptop depreciates annually at a rate of 20% p.a. according to the straight-line method. After how many years will the laptop have a value of R0? The formula for straight-line depreciation is A = P(1 - in), where A is the future value, P is the present value, i is the annual depreciation rate, and n is the number of years. Step 1: Identify the given values and substitute them into the formula. Given: P = R10\,000, i = 20\% = 0.20, A = R0. 0 = 10000(1 - 0.20n) Step 2: Solve for n. 0 = 1 - 0.20n 0.20n = 1 n = (1)/(0.20) n = 5 The laptop will have a value of R0 after: 5 years Question 6.2.2 Kuda will buy a laptop that costs R20 000. In order to cover the cost price, he made his first monthly deposit into a savings account on 28 February 2019. He will make his 60th monthly deposit on 31 January 2024. The savings account pays interest at 8.7% p.a., compounded monthly. Calculate Kuda's monthly deposit into this account. This is a future value of an ordinary annuity problem. The formula is FV = x ([(1+i)^n - 1])/(i), where FV is the future value, x is the monthly deposit, i is the monthly interest rate, and n is the number of deposits. Step 1: Identify the given values and calculate the monthly interest rate. Given: FV = R20\,000, annual interest rate = 8.7\% = 0.087, n = 60 deposits. i = (0.087)/(12) = 0.00725 Step 2: Substitute the known values into the future value of an annuity formula. 20000 = x [(1+0.00725)^60 - 1]0.00725 Step 3: Calculate the annuity factor. (1.00725)^60 ≈ 1.5434805 ([1.5434805 - 1])/(0.00725) = (0.5434805)/(0.00725) ≈ 74.962827586 Step 4: Solve for x. 20000 = x × 74.962827586 x = (20000)/(74.962827586) x ≈ 266.79 Kuda's monthly deposit is: R266.79 Question 6.3 Tino wins a jackpot of R1 600 000. He invests all of his winnings in a fund that earns interest of 11.2% p.a., compounded monthly. He withdraws R20 000 from the fund at the end of each month. His first withdrawal is exactly 1 month after his initial investment. How many withdrawals of R20 000 will Tino be able to make from this fund? This is a present value of an ordinary annuity problem. The formula is PV = x [1 - (1+i)^-n]i, where PV is the present value, x is the withdrawal amount, i is the monthly interest rate, and n is the number of withdrawals. Step 1: Identify the given values and calculate the monthly interest rate. Given: PV = R1\,600\,000, x = R20\,000, annual interest rate = 11.2\% = 0.112. i = (0.112)/(12) = 0.009333333 Step 2: Substitute the known values into the present value of an annuity formula. 1600000 = 20000 [1 - (1+0.009333333)^-n]0.009333333 Step 3: Isolate the term containing n. (1600000 × 0.009333333)/(20000) = 1 - (1.009333333)^-n (14933.33328)/(20000) = 1 - (1.009333333)^-n 0.746666664 = 1 - (1.009333333)^-n (1.009333333)^-n = 1 - 0.746666664 (1.009333333)^-n = 0.253333336 Step 4: Use logarithms to solve for n. -n (1.009333333) = (0.253333336) -n = ((0.253333336))/((1.009333333)) -n ≈ (-0.596321)/(-0.004037) -n ≈ -147.71 n ≈ 147.71 Since Tino can only make full withdrawals, we round down to the nearest whole number. Tino will be able to make: 147 withdrawals That's 2 down. 3 left today — send the next one.