Question \( 6 \mathbf{c} \) (WASSCE 2008 May/June) A cyclist starts a journey at \( 10: 30 \mathrm{am} \) and lans to get home 45 km away by 3.00 pm . At irst, he traveled for 2 hours at \( 7 \frac{1}{2} \mathrm{kmh}^{-1} \) and hen \( x \mathrm{kmh}^{-1} \) for the rest the time in order to rrive promptly at home. Find the value of \( x \).

To solve this problem, we first calculate the total time available for the journey. The cyclist starts at 10:30 am and wants to arrive by 3:00 pm, giving him a total of 4.5 hours (from 10:30 am to 3:00 pm) to complete the 45 km trip. Next, we calculate the distance traveled during the first part of the journey. The cyclist rides for 2 hours at a speed of \( 7.5 \, \text{km/h} \). **Distance traveled in 2 hours:** \[ \text{Distance} = \text{Speed} \times \text{Time} = 7.5 \, \text{km/h} \times 2 \, \text{h} = 15 \, \text{km} \] After 2 hours, the cyclist has 30 km left to travel (since \( 45 \, \text{km} - 15 \, \text{km} = 30 \, \text{km} \)). **Time left to travel the remaining distance:** Since the total journey is 4.5 hours and 2 hours have already been used, the cyclist has: \[ \text{Time left} = 4.5 \, \text{h} - 2 \, \text{h} = 2.5 \, \text{h} \] Now, we need to find the speed \( x \, \text{km/h} \) for the remaining distance of 30 km: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{30 \, \text{km}}{2.5 \, \text{h}} = 12 \, \text{km/h} \] Thus, the value of \( x \) is \( 12 \, \text{km/h} \). The cyclist must ride at a speed of \( 12 \, \text{km/h} \) for the remaining 2.5 hours to reach home on time!