Question

John drove to Daytona Beach, Florida, in hours. When he returned, there was less traffic, and the trip took only hours. If John averaged mph faster on the return trip, how fast did he drive each way?

Answer 1

Question: John drove to a distant city in 5 hours.

When he returned, there was less traffic and the trip took only 3 hours.

If John averaged 26 mph faster on the return trip, how fast did he drive each way

Answer:

For the first trip he drove at a speed of 39 mph

For the second trip he drove at a speed of 65 mph

Step-by-step explanation:

Let the distance for both journey be Z because they are equal.

Let the speed for the first journey be X

Let the speed of the second journey be Y

Formula for speed = distance ÷ time

For the first journey the speed X = Z ÷ 5

For the second journey the speed Y = Z ÷ 3

Since John averaged 26 mph faster in the second trip than the first trip due to traffic, it means that the difference in speed between the first & second trip is 26 mph

Difference in speed = (Z÷3) - (Z÷5) = 26

subtracting both results to  (5Z-3Z) ÷ 15 = 26

Upon cross multiplication

2Z = 390

Z = 390÷2 = 195 miles

Therefore speed for first journey = 195 ÷ 5 = 39 mph

Speed for second journey = 195 ÷ 3 = 65 mph

To verify, 65 - 39 = 26 mph