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
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