swerve wrote:Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes a different path for onward and returns journey - the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?
A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph
The information in red is irrelevant.
Only the following facts matter:
Ross travels onward at a rate of 1 mile every 12 minutes.
He travels home for twice the onward distance at half the onward speed.
Onward speed = 1 mile per 12 minutes = 5 miles per 60 minutes = 5 mph.
Let the onward distance = 5 miles.
Time to travel 5 miles onward at a rate of 5 mph = d/r = 5/5 = 1 hour.
Time to travel 10 miles home at a rate of 2.5 mph = d/r = 10/2.5 = 100/25 = 4 hours.
Since the total time = 1+4 = 5 hours, the average speed for the entire 15-mile trip = (total distance)/(total time) = 15/5 = 3 mph.
The correct answer is
A.
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