Question 2 2.1. Rewrite 92 and 146 as products of their prime factors. 2.2. Write down the LCM and HCF of 92 and 146 . 2.3. Divide 240 g in the ratio \( 5: 3: 4 \). 2.4. Thoriso can either walk to school at \( 5 \mathrm{~km} / \mathrm{h} \) or ride her bicycle at 15 km . If she rides her bicycle, it takes her 10 minutes to get to school. 2.4.1. How long will it take her if she walks to school? 2.4.2. Is this direct or indirect proportion? Give a reason for your answer. 2.5. The price of petrol is increased from R12, 58 per litre to R13, 28 per litre. Determine the percentage increase in the price. 2.6. A motor car drives at an average speed of \( 106 \mathrm{~km} / \mathrm{h} \) for 2 hours 45 minutes. At which constant speed must the car drive to travel the same dista in 2 hours 35 minutes? (4)

**2.1. Prime Factorisation** - For \(92\): \[ 92 = 2 \times 46 = 2 \times (2 \times 23) = 2^2 \times 23. \] - For \(146\): \[ 146 = 2 \times 73. \] **2.2. LCM and HCF** - The prime factors are: - \(92 = 2^2 \times 23\) - \(146 = 2 \times 73\) - The common prime factor is \(2\). - **HCF** (Highest Common Factor) is the product of the smallest powers of common factors: \[ \text{HCF} = 2. \] - **LCM** (Least Common Multiple) is the product of the highest powers of all prime factors present: \[ \text{LCM} = 2^2 \times 23 \times 73. \] Calculate: \[ 2^2 = 4 \quad \text{and} \quad 23 \times 73 = 1679 \quad (\text{since } 23\times70=1610 \text{ and } 23\times3=69), \] so, \[ \text{LCM}= 4 \times 1679 = 6716. \] **2.3. Dividing 240 g in the Ratio \(5:3:4\)** - Total parts: \[ 5+3+4 = 12. \] - The shares are: - First share: \[ \frac{5}{12} \times 240 = 100 \text{ g}. \] - Second share: \[ \frac{3}{12} \times 240 = 60 \text{ g}. \] - Third share: \[ \frac{4}{12} \times 240 = 80 \text{ g}. \] **2.4. Thoriso's Journey** Given: - Walking speed: \(5\) km/h - Bicycle speed: \(15\) km/h - Time by bicycle: \(10\) minutes 1. **Finding the Distance:** Convert 10 minutes to hours: \[ 10 \text{ minutes} = \frac{10}{60} = \frac{1}{6} \text{ hour}. \] Distance traveled by bicycle: \[ \text{Distance} = 15 \times \frac{1}{6} = 2.5 \text{ km}. \] 2. **2.4.1. Walking Time:** Time when walking: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{2.5}{5} = 0.5 \text{ hour} = 30 \text{ minutes}. \] 3. **2.4.2. Proportionality:** When speed increases, the time required decreases for a fixed distance. This is an example of **indirect (inverse) proportionality**. **2.5. Percentage Increase in Petrol Price** - Original price: \(R12.58\) per litre - New price: \(R13.28\) per litre Increase in price: \[ \text{Increase} = 13.28 - 12.58 = 0.70 \text{ R}. \] Percentage increase: \[ \text{Percentage increase} = \left(\frac{0.70}{12.58}\right) \times 100 \approx 5.56\%. \] **2.6. Calculating the Required Speed** - Original journey: - Speed: \(106\) km/h - Time: \(2\) hours \(45\) minutes, where: \[ 45 \text{ minutes} = 0.75 \text{ hour} \quad \Rightarrow \quad \text{Total time} = 2.75 \text{ hours}. \] - Distance: \[ \text{Distance} = 106 \times 2.75 = 291.5 \text{ km}. \] - New journey time: - \(2\) hours \(35\) minutes, where: \[ 35 \text{ minutes} = \frac{35}{60} \approx 0.5833 \text{ hour} \quad \Rightarrow \quad \text{Total time} \approx 2.5833 \text{ hours}. \] - Required speed: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{291.5}{2.5833} \approx 112.8 \text{ km/h}. \]