SOLUTION: 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 h
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-> SOLUTION: 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 h
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Question 334331: 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?
It seems that something is missing from the problem. Help please. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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?
:
We don't know the distance of the trips, but we know they are equal.
:
Let s = speed to the city
then
(s+26 = return speed
:
Write a distance equation: dist = time * speed
:
To dist = return dist
5s = 3(s+26)
5s = 3s + 78
5s - 3s = 78
2s = 78
s =
s = 39 mph to the city
and
39 + 26 = 65 mph back
;
;
We can confirm our solution by finding the distances now
5*39 = 195 mi
3*65 = 195 mi also
:
:
Pretty simple, right?
