\begin{tabular}{|l|}\hline A car is travelling at \( 100 \mathrm{~km} / \mathrm{h} \). How long will it take to travel \( 300 \mathrm{~km} ? \) \\ \hline \( \begin{array}{l}\text { Carol invests R18 } 000 \text { for } 6 \text { years at rate of } 15 \% \text { p.a. compounded } \\ \text { annually. What will be the total value of Carol's investment after } 6 \\ \text { years? }\end{array} \) \\ \hline\end{tabular}

To solve the first problem, we need to calculate the time it will take for a car traveling at 100 km/h to cover a distance of 300 km. The formula to calculate time is: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] Given: - Distance = 300 km - Speed = 100 km/h Substitute the values into the formula: \[ \text{Time} = \frac{300}{100} \] Now, let's calculate the time it will take for the car to travel 300 km. Calculate the value by following steps: - step0: Calculate: \(\frac{300}{100}\) - step1: Reduce the numbers: \(\frac{3}{1}\) - step2: Calculate: \(3\) The time it will take for the car to travel 300 km is 3 hours. Now, let's move on to the second problem. We need to calculate the total value of Carol's investment after 6 years, given that she invests R18,000 for 6 years at a rate of 15% per annum compounded annually. The formula to calculate the total value of an investment with compound interest is: \[ A = P(1 + r)^n \] Where: - \( A \) is the total value of the investment - \( P \) is the principal amount (initial investment) - \( r \) is the annual interest rate - \( n \) is the number of years Given: - Principal amount (P) = R18,000 - Annual interest rate (r) = 15% = 0.15 - Number of years (n) = 6 Substitute the values into the formula: \[ A = 18000(1 + 0.15)^6 \] Now, let's calculate the total value of Carol's investment after 6 years. Calculate the value by following steps: - step0: Calculate: \(18000\left(1+0.15\right)^{6}\) - step1: Add the numbers: \(18000\times 1.15^{6}\) - step2: Convert the expressions: \(18000\left(\frac{23}{20}\right)^{6}\) - step3: Simplify: \(18000\times \frac{23^{6}}{20^{6}}\) - step4: Rewrite the expression: \(400\times 45\times \frac{23^{6}}{20^{6}}\) - step5: Rewrite the expression: \(20^{2}\times 45\times \frac{23^{6}}{20^{6}}\) - step6: Reduce the numbers: \(45\times \frac{23^{6}}{20^{4}}\) - step7: Rewrite the expression: \(5\times 9\times \frac{23^{6}}{20^{4}}\) - step8: Rewrite the expression: \(5\times 9\times \frac{23^{6}}{5^{4}\times 4^{4}}\) - step9: Reduce the numbers: \(9\times \frac{23^{6}}{5^{3}\times 4^{4}}\) - step10: Multiply: \(\frac{9\times 23^{6}}{32000}\) The total value of Carol's investment after 6 years is approximately R41,635.09.