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Question
An excursion train has the following schedule:

9:30 AM- Departs starting point for Town A.

11:15 AM- Arrives at Town A with a 1 hr. 45 min. layover.

1:00 PM- Departs for Town B.

1:10 PM- Arrives at Town B with a 50 min. layover.

2:00 PM- Departs Town B.

3:45 PM- Arrives at starting point.

Distance:   

From starting point to Town A- 31 miles.

From Town A to Town B- 3 miles.

From Town B to starting point- 34 miles.

Total distance- 68 miles.

Calculate average speed of train.

Thanks for your assistance.

Answer
Use 24-hour notation to calculate elapsed time.

Train departs starting point at 9:30 and arrives at A at 11:15.
travel time = 11:15 - 9:30 = 1:45 = 1 hr 45 min

Train departs A at 13:00 and arrives at B at 13:10.
travel time = 13:10 - 13:00 = 0:10 = 10 min

Train departs B at 14:00 and returns to starting point at 15:45.
travel time = 15:45 - 14:00 = 1:45 = 1 hr 45 min

Total travel time = 1:45 + 0:10 + 1:45 = 3:40 = 3⅔ hr
Total distance = 31 + 3 + 34 = 68 miles
average speed = 68 miles/3⅔ hr = 18.5454... ≅ 18.55 miles per hour

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I can answer questions in Algebra, Basic Math, Calculus, Differential Equations, Geometry, Number Theory, and Word Problems. I would not feel comfortable answering questions in Probability and Statistics or Topology because I have not studied these in depth.

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