Question 2 2.1 Rewrite 92 and 146 as products of their prime factors. (2) 2.2. Write down the LCM and HCF of 92 and 146. (2) 2.3. Divide 240 g in the ratio \( 5: 3: 4 \). (3) 24. Thoriso can either walk to school a \( 5 \mathrm{~km} / \mathrm{h} \) or ride her bicycle at \( 15 \mathrm{~km} / \mathrm{h} \). 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) 2.4.2. Is this direct or indirect proportion? Give a reason for your answer. (2) 2.5. Brian wants to exchange South-African rand for British pound. If R1 is worth 0,075199 pound, how many pounds will he get for R2 100 if he must pay an agent commission of \( 1,5 \% \) ? (3) 2.7. The price of petrol is increased from R12,58 per litre to R13,28 per litre. Determine the percentage increase in the price. (2) 2.8. 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 distance in 2 hours 35 minutes? (4) \( [20] \)

Let's solve the problems step by step. ### 2.1 Rewrite 92 and 146 as products of their prime factors. **Step 1: Factor 92** - 92 is even, so divide by 2: \[ 92 \div 2 = 46 \] - 46 is also even, divide by 2 again: \[ 46 \div 2 = 23 \] - 23 is a prime number. Thus, the prime factorization of 92 is: \[ 92 = 2^2 \times 23 \] **Step 2: Factor 146** - 146 is even, so divide by 2: \[ 146 \div 2 = 73 \] - 73 is a prime number. Thus, the prime factorization of 146 is: \[ 146 = 2^1 \times 73 \] ### 2.2 Write down the LCM and HCF of 92 and 146. **Step 1: Find the HCF (Highest Common Factor)** - The common prime factor is \( 2 \). - The lowest power of 2 in both factorizations is \( 2^1 \). Thus, the HCF is: \[ \text{HCF} = 2^1 = 2 \] **Step 2: Find the LCM (Lowest Common Multiple)** - The LCM is found by taking the highest power of all prime factors: \[ \text{LCM} = 2^2 \times 23^1 \times 73^1 \] Calculating this: \[ \text{LCM} = 4 \times 23 \times 73 \] Calculating \( 4 \times 23 = 92 \) and then \( 92 \times 73 = 6716 \). Thus, the LCM is: \[ \text{LCM} = 6716 \] ### 2.3 Divide 240 g in the ratio \( 5: 3: 4 \). **Step 1: Calculate the total parts in the ratio** \[ 5 + 3 + 4 = 12 \text{ parts} \] **Step 2: Calculate the value of one part** \[ \text{Value of one part} = \frac{240}{12} = 20 \text{ g} \] **Step 3: Calculate each part** - First part: \( 5 \times 20 = 100 \text{ g} \) - Second part: \( 3 \times 20 = 60 \text{ g} \) - Third part: \( 4 \times 20 = 80 \text{ g} \) Thus, the division is: \[ 100 \text{ g}, 60 \text{ g}, 80 \text{ g} \] ### 2.4.1 How long will it take her if she walks to school? **Step 1: Calculate the distance to school** - If she rides her bicycle at \( 15 \text{ km/h} \) for \( 10 \text{ minutes} \): \[ \text{Distance} = \text{Speed} \times \text{Time} = 15 \times \frac{10}{60} = 2.5 \text{ km} \] **Step 2: Calculate the time to walk** - Walking speed is \( 5 \text{ km/h} \): \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{2.5}{5} = 0.5 \text{ hours} = 30 \text{ minutes} \] ### 2.4.2 Is this direct or indirect proportion? Give a reason for your answer. This is an **indirect proportion** because as the speed increases (bicycling), the time taken decreases. The relationship between speed and time is inversely proportional. ### 2.5 Calculate how many pounds Brian will get for R2,100 after paying a commission of \( 1.5\% \). **Step 1: Calculate the commission** \[ \text{Commission} = 2100 \times \frac{1.5}{100} = 31.5 \] **Step 2: Calculate the amount after commission** \[ \text{Amount after commission} = 2100 - 31.5 = 2068.5 \] **Step 3: Convert to pounds** \[ \text{Pounds} = 2068.5 \times 0.075199 \] Calculating this: \[ \text{Pounds} \approx 155.0 \] ### 2.7 Determine the percentage increase in the price of petrol. **Step 1: Calculate the increase in price** \[ \text{Increase} = 13.28 - 12.58 = 0.70 \] **Step 2: Calculate the percentage increase** \[ \text{Percentage Increase} = \left( \frac{0.70}{12.58} \right) \times 100 \] Calculating this: \[ \text{Percentage Increase} \approx 5.57\% \] ### 2.8 Calculate the constant speed for the car to travel the same distance in 2 hours 35 minutes. **Step 1: Calculate the distance traveled in 2 hours 45 minutes** \[ \text{Distance} = 106 \times 2.75 = 291.5 \text{ km} \] **Step 2: Calculate the speed for 2 hours 35 minutes** \[ \text{Time} = 2 + \frac{35}{60} = \frac{155}{60} \text{ hours} \] **Step 3: Calculate the required speed** \[ \text{Speed} = \frac{291.5}{\frac{155}{60}} = 291.5 \times \frac{60}{155} \] Calculating this: \[ \text{Speed} \approx 112.0 \text{ km/h} \] ### Summary of Answers: - 2.1: \( 92 = 2^2 \times 23 \), \( 146 = 2^1 \times 73 \) - 2.2: HCF = 2, LCM = 6716 - 2.3: 100 g, 60 g, 80 g - 2.4.1: 30 minutes - 2.4.2: Indirect proportion - 2.5: Approximately 155.0 pounds - 2.7: Approximately 5.57% - 2.8: Approximately 112.0 km/h