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Three cars A, B and C are moving around ...

Three cars A, B and C are moving around a circular path. In one hour, car A completes 3 rounds, car B completes 2 rounds and car C completes 4 rounds. If they start from the same point at the same time, after how much time will they meet again at the same point from where they started out?

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To find out when the three cars A, B, and C will meet again at the same point after starting together, we need to determine the time it takes for each car to complete one round and then find the least common multiple (LCM) of those times. ### Step-by-Step Solution: 1. **Determine the time taken for each car to complete one round:** - Car A completes 3 rounds in 1 hour. Therefore, the time taken for Car A to complete 1 round is: \[ \text{Time for A} = \frac{1 \text{ hour}}{3} = \frac{1}{3} \text{ hour} \] - Car B completes 2 rounds in 1 hour. Therefore, the time taken for Car B to complete 1 round is: \[ \text{Time for B} = \frac{1 \text{ hour}}{2} = \frac{1}{2} \text{ hour} \] - Car C completes 4 rounds in 1 hour. Therefore, the time taken for Car C to complete 1 round is: \[ \text{Time for C} = \frac{1 \text{ hour}}{4} = \frac{1}{4} \text{ hour} \] 2. **Convert the times into a common format:** - The times are: - Car A: \( \frac{1}{3} \) hour - Car B: \( \frac{1}{2} \) hour - Car C: \( \frac{1}{4} \) hour 3. **Find the least common multiple (LCM) of the denominators:** - The denominators are 3, 2, and 4. - The LCM of 3, 2, and 4 can be calculated: - The multiples of 3: 3, 6, 9, 12, 15, ... - The multiples of 2: 2, 4, 6, 8, 10, 12, ... - The multiples of 4: 4, 8, 12, 16, ... - The smallest common multiple is 12. 4. **Convert the LCM back to hours:** - Since the LCM is 12, we convert it back to hours: - \( 12 \) is the LCM of the denominators, so we need to convert it to the time taken in hours: - Car A: \( 12 \div 3 = 4 \) hours - Car B: \( 12 \div 2 = 6 \) hours - Car C: \( 12 \div 4 = 3 \) hours - All cars will meet again after 12 hours. ### Final Answer: The three cars A, B, and C will meet again at the same point after **12 hours**.
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