Question 1047695: Juan Carlos traveled to the town square and back. The trip there took 3 hours and the trip back took 4 hours. If he averages 32 mph on the trip back, what speed of the average on the trip there?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! Trip going was at a faster speed, R. Trip back was at lower speed r. Variables also shown as assigned assumed.
SPEED TIME DISTANCE
GOING R x d
RETURN r y d
Basic constant travel rate rule is RT=D relating speed, time, distance.
R=unknownSpeedGoing
d=unknownDistanceOneWay
r=32
x=3
y=4
 ------because you could simply equate two expressions for d.
ONE STEP ONLY
 ----------symbolic answer.
Substitute the given values and evaluate R.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Juan Carlos traveled to the town square and back. The trip there took 3 hours and the trip back took 4 hours. If he averages 32 mph on the trip back, what speed of the average on the trip there?
The return trip took 4 hours, and at a speed of 32 mph, distance = 4(32), or 132 miles
Average speed on outbound trip: 
No need to go all crazy! It's that SIMPLE! Very, very easy to figure out!
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