Bus A and Bus B leave the bus depot at 7 am. Bus A takes 25 minutes to do its route and bus B takes 40 minutes to complete its route. At what time are they both back at the bus depot together? Give your answer as a 12-hour clock time.

Explanation:

List out the multiples of 25 to get {25, 50, 75, 100, 125, 150, 175, 200}

Do the same for the multiples of 40 to get {40, 80, 120, 160, 200}

We see that the lowest common multiple (LCM) is 200

Another way to get the LCM is to multiply 25 and 40 to get 25*40 = 1000. Then divide this over 5 as this is the GCF between the two original numbers. We end up with the same LCM since 1000/5 = 200.

There's a third way to get the LCM

List out the prime factorization of each value

25 = 5*5

40 = 2*3*5

Note how we have the unique factors of 2, 3 and 5. Circle the unique factors such that we highlight the ones that occur the most. We have 2 occur once, 3 occur once, and 5 occurs twice. We have circled one 2, one 3, and two 5's. They multiply out to 2*3*5*5 = 8*25 = 200

Whichever method works for you, the LCM is 200. This means that every 200 minutes, we have bus A and bus B arrive at the depot at the same time. Their cycles line up at the 200 minute mark.

Convert this to hours,minutes format.

200 minutes = 180 minutes + 20 minutes

200 minutes = 3 hours + 20 minutes

200 minutes = 3 hrs, 20 min

We can represent "3 hrs, 20 min" in 12-hour clock notation by writing 3:20

Add this onto 7:00, which is 7 AM and we get

(7:00) + (3:20) = (7+3):(00+20) = 10:20 AM

We have started at 7 AM and fast forwarded 3 hours and 20 minutes to arrive at 10:20 AM.

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