Question 925529: Hello, I have solved this problem, but I was wondering if I had the correct answer to the problem. The question is," Lisa made a trip to the lake and back. The trip there took four hours and the trip back took six hours. She averaged 25 km/h faster on the trip there than on the return trip. What was Lisa's average speed on the outbound trip?
This is what I have right now. I wanted to make sure that this was correct.
6x=4(x+25)
6x=4x+100
2x=100
x=50
Thanks in advance!
Found 3 solutions by TimothyLamb, josgarithmetic, MathTherapy:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! x = outbound speed
y = inbound speed
x = y + 25 kph
---
s = d/t
d = st
---
outbound:
x = d/4
d = 4x
---
inbound:
y = d/6
d = 6y
---
4x = 6y
4(y + 25) = 6y
4y + 100 = 6y
2y = 100
---
answer:
y = inbound speed = 50 kph
x = outbound speed = 75 kph
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Answer by josgarithmetic(39620)  (Show Source):
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
Hello, I have solved this problem, but I was wondering if I had the correct answer to the problem. The question is," Lisa made a trip to the lake and back. The trip there took four hours and the trip back took six hours. She averaged 25 km/h faster on the trip there than on the return trip. What was Lisa's average speed on the outbound trip?
This is what I have right now. I wanted to make sure that this was correct.
6x=4(x+25)
6x=4x+100
2x=100
x=50
Thanks in advance!
Yes, that's CORRECT!!
However, you used the variable x for the INBOUND speed, and with the INBOUND time being 6 hours, the distance
covered on the INBOUND trip was 6x, which you also have. From your equation you determined x, which was the INBOUND
speed. You'll then need to ADD 25 km/h to the INBOUND speed to get the OUTBOUND speed, since the OUTBOUND speed
was 25 km/h greater than the INBOUND speed, and as you might've noticed, the OUTBOUND speed is what's being sought.
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