SOLUTION: A private plane traveled from Seattle to a rugged​ wilderness, at an average speed of 88 mph. On the return​ trip, the average speed was 154 mph. If the total traveling

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Question 1054716: A private plane traveled from Seattle to a rugged​ wilderness, at an average speed of 88 mph. On the return​ trip, the average speed was 154 mph. If the total traveling time was 8 ​hours, how far is Seattle from the​ wilderness?
Found 3 solutions by Fombitz, Boreal, MathTherapy:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Distance = Rate x Time
Let x be the time spent at 88 mph and y the time at 154 mph.
1.
2.
.
.
From 1,


Substituting,



So then,
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
v*t=distance
going there 88*t=distance
coming back, the time is (8-t) and the velocity 154, so the distance is 154(8-t)=
1232-154t. Set that equal to 88t, since the distance is the same
88t=1232-154t
242t=1232
t=5.091 hours
going over, the distance would be 448 miles
coming back, the time is 2.909 hours (add to 8 hours)*154mi/hr=447.99 miles
448 miles is the distance.
Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

A private plane traveled from Seattle to a rugged​ wilderness, at an average speed of 88 mph. On the return​ trip, the average speed was 154 mph. If the total traveling time was 8 ​hours, how far is Seattle from the​ wilderness?
Let distance be D

Then we get the following TIME equation: 

7D + 4D = 4,928 -------- Multiplying by LCD, 616

11D = 4,928

D, or distance =  

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