SOLUTION: 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 t

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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