Pregunta
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3.3 Two buses, bus \( A \) and bus \( B \), both use the same bus - Bus A runs every 10 minutes. - Bus B runs every 14 minutes. Both buses are at the bus stop at 11 am. What time will both buses next be at the bus stop?

Ask by Fowler Erickson. in South Africa
Mar 10,2025

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Both buses will next be at the bus stop together at 12:10 PM.

Solución

1. We need to find the time at which both buses are at the bus stop simultaneously again. Since bus \( A \) runs every 10 minutes and bus \( B \) every 14 minutes, the buses will coincide after a time equal to the least common multiple (LCM) of 10 and 14. 2. First, factorize the numbers: - \( 10 = 2 \times 5 \) - \( 14 = 2 \times 7 \) 3. The LCM is obtained by taking the highest powers of all prime factors: \[ \text{LCM}(10, 14) = 2 \times 5 \times 7 = 70 \, \text{minutes} \] 4. Both buses are at the bus stop at 11:00 AM. Therefore, they will next be at the bus stop together after 70 minutes: \[ 11:00 \, \text{AM} + 70 \, \text{minutes} = 12:10 \, \text{PM} \] The buses will be next at the bus stop together at \( 12:10 \, \text{PM} \).

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Beyond the Answer

To find out when both buses next arrive at the stop together after 11 am, we need to calculate the least common multiple (LCM) of their schedules. Bus A runs every 10 minutes, and Bus B runs every 14 minutes. First, let's break down each bus's schedule: - Bus A: 10, 20, 30, 40, 50, 60 minutes (which is 11:10, 11:20, 11:30, 11:40, 11:50, 12:00) - Bus B: 14, 28, 42, 56, 70, 84 minutes (which is 11:14, 11:28, 11:42, 11:56, 12:10, 12:24) Now, we are looking for the smallest number that both schedules share. The multiples of 10 are 10, 20, 30, 40, 50, 60, 70, etc., and the multiples of 14 are 14, 28, 42, 56, 70, etc. The first common multiple is 70 minutes. Adding 70 minutes to the starting time of 11 am gives us: 11:00 + 1 hour 10 minutes = 12:10 pm. So, both buses will next be at the stop at **12:10 pm**.

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