SOLUTION: Randy rode out of town on the bus at an average speed of 24 miles per hour and walked back at an average speed of 3 mile sper hour. How far did he go if he was gone for six hours?

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Question 749902: Randy rode out of town on the bus at an average speed of 24 miles per hour and walked back at an average speed of 3 mile sper hour. How far did he go if he was gone for six hours? (Having trouble setting it up)
Found 2 solutions by timvanswearingen, MathTherapy:
Answer by timvanswearingen(106) About Me  (Show Source):
You can put this solution on YOUR website!
Use distance=rate*time. You can set up a couple equations for d and set them equal to find t. Then you'll plug back in to find d.
d=24mi%2Fh%2At
The time t spent on the bus will be subtracted from 6 (since the total time was 6 hours). This will give the time spent walking:
d=3mi%2Fh%2A%286-t%29
Now set the two equal and solve for t, then plug the t back into one of the previous equations to find the distance:
24mi%2Fh%2At=3mi%2Fh%2A%286-t%29

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

Randy rode out of town on the bus at an average speed of 24 miles per hour and walked back at an average speed of 3 mile sper hour. How far did he go if he was gone for six hours? (Having trouble setting it up)

Let distance to out-of-town destination, be D
Then time it took him to get to out-of-town destination = D%2F24
Also, time it took him to return home = D%2F3
Since it took him a total of 6 hours to complete the journey, then: D%2F24+%2B+D%2F3+=+6
D + 8D = 144 ------- Multiplying by LCD, 24
9D = 144
D, or distance traveled to out-of-town destination = 144%2F9, or highlight_green%2816%29 miles
You can do the check!!

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