QUESTION 4 Themba is planning a bicycle trip from Cape Town to Pretoria. The total distance cover during the trip will be 1500 km . he plans to travel 100 km on the first day. For every following day he plans to cover \( 94 \% \) of the distance he covered the previous day. 4.1 What distance will he cover on the third day of the trip? 4.2 On what day of the trip will Themba pass half way of the trip? 4.3 Themba must cover certain percentage of the previous day's distance to ensure that he eventually reach Pretoria. Calculate ALL possible value(s) of this percentage.\n(3)

**4.1 Calculation of the Distance on the Third Day** Let the distance traveled on the first day be \[ a_1 = 100 \text{ km}. \] On each subsequent day, Themba covers \( 94\% \) of the previous day’s distance. Thus, the distance on the \( n \)th day is given by \[ a_n = 100 \times 0.94^{\,n-1}. \] For the third day (\( n=3 \)): \[ a_3 = 100 \times 0.94^{\,2} = 100 \times 0.8836 = 88.36 \text{ km}. \] --- **4.2 Determining the Day When the Cumulative Distance Exceeds Half the Trip** The total distance of the trip is 1500 km, so half of the trip is \[ \frac{1500}{2} = 750 \text{ km}. \] The sum of the distances over \( n \) days is the sum of a geometric series: \[ S_n = 100 + 100(0.94) + 100(0.94)^2 + \cdots + 100(0.94)^{\,n-1}. \] The formula for the sum of the first \( n \) terms is: \[ S_n = 100 \cdot \frac{1 - (0.94)^n}{1 - 0.94} = \frac{100\bigl(1 - (0.94)^n\bigr)}{0.06}. \] We need to find the smallest integer \( n \) such that \[ S_n \ge 750. \] That is, \[ \frac{100\bigl(1 - (0.94)^n\bigr)}{0.06} \ge 750. \] Multiply both sides by \( 0.06 \): \[ 100\bigl(1 - (0.94)^n\bigr) \ge 750 \times 0.06. \] Since \( 750 \times 0.06 = 45 \): \[ 100\bigl(1 - (0.94)^n\bigr) \ge 45, \] or \[ 1 - (0.94)^n \ge 0.45. \] This leads to: \[ (0.94)^n \le 0.55. \] Taking the natural logarithm on both sides: \[ \ln (0.94)^n \le \ln 0.55. \] Using the property of logarithms, \[ n \ln 0.94 \le \ln 0.55. \] Since \( \ln 0.94 \) is negative, dividing by it reverses the inequality: \[ n \ge \frac{\ln 0.55}{\ln 0.94}. \] Evaluating the logarithms approximately: \[ \ln 0.55 \approx -0.5978,\quad \ln 0.94 \approx -0.06188, \] so \[ n \ge \frac{-0.5978}{-0.06188} \approx 9.65. \] Since \( n \) must be an integer, Themba will pass the halfway point on day \[ n = 10. \] --- **4.3 Determining All Possible Percentage Values for Eventual Arrival** Let the fraction (as a decimal) of the previous day’s distance that Themba covers be \( r \). Then the distance on the first day is still 100 km, and the distances form a geometric progression with sum \[ S_{\infty} = \frac{100}{1-r}, \] provided that \( |r| < 1 \) (i.e. when each day’s distance is reduced). For Themba to eventually reach Pretoria, he must cover at least 1500 km, so we require that \[ S_{\infty} \ge 1500. \] Thus, \[ \frac{100}{1-r} \ge 1500. \] Solving for \( r \): 1. Multiply both sides by \( 1-r \) (assuming \( 1-r>0 \)): \[ 100 \ge 1500(1-r). \] 2. Divide both sides by 1500: \[ \frac{100}{1500} \ge 1-r \quad \Longrightarrow \quad \frac{1}{15} \ge 1-r. \] 3. Rearranging gives: \[ 1 - \frac{1}{15} \le r \quad \Longrightarrow \quad r \ge \frac{14}{15} \approx 0.9333. \] Thus, when \( 0 \le r < 1 \) (i.e. a reducing daily distance), Themba will eventually reach or exceed 1500 km if and only if \[ r \ge \frac{14}{15}. \] Note that if \( r \ge 1 \), the sequence no longer represents decreasing daily distances; in fact, if \( r=1 \), he covers 100 km every day and will eventually reach Pretoria in a finite number of days, and if \( r>1 \) the traveled distance increases each day. Therefore, all percentages that ensure eventual arrival are: - For the case \( r < 1 \): Any percentage at least \( \frac{14}{15} \) (or \( 93.\overline{3}\% \)). - For \( r \ge 1 \): There is no risk of never arriving because the daily distances do not diminish, so these values also guarantee arrival. In summary, Themba will eventually reach Pretoria if the percentage \( p \) (expressed as a decimal) satisfies \[ p \ge \frac{14}{15} \quad \text{or if } p \ge 1. \] In percentage terms, any value of the percentage that is at least approximately \( 93.33\% \) (if the intention is to