Bus A and Bus B leave the bus depot at 7 am. Bus A takes 30 minutes to complete its route once and bus B takes 50 minutes to complete its route once.
If both buses continue to repeat their route, at what time will they be back at the bus depot together?
Assume the buses have no breaks in between routes.
Give your answer as a 12-hour clock time.

9:30 am

Step-by-step explanation:

We can make a chart and keep filling it in until we see the same time in both columns.

Bus A          Bus B

7:00             7:00

7:30             7:50

8:00            8:40

8:30           9:30

9:00            10:20

9:30

9:30 am

We can also use the least common multiple of 30 and 50.

We need to find the smallest number that is a multiple of both 30 and 50.

30 = 2 * 3 * 5

50 = 2 * 5^2

LCD = 2 * 3 * 5^2 = 150

The first time the buses will be at the depot together after 7:00 am is 150 minutes later. 150 minutes = 2 hours 30 minutes.

7:00 am + 2 hours and 30 minutes = 9:30 am

9:30 am

they never get back together bus b is always behind

Step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

step-by-step explanation:

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