SOLUTION: John drove from his house to Kingston in 2 hours. One hour later, he returned hom at a speed 20km/h less than his speed going to Kingston. If John took a total of 6 hours for his t

Algebra ->  Finance -> SOLUTION: John drove from his house to Kingston in 2 hours. One hour later, he returned hom at a speed 20km/h less than his speed going to Kingston. If John took a total of 6 hours for his t      Log On


   

Question 1101055: John drove from his house to Kingston in 2 hours. One hour later, he returned hom at a speed 20km/h less than his speed going to Kingston. If John took a total of 6 hours for his trip (including a one hour stop in Kingston), how fast did he travel on each leg of the trip?
Let speed in km/h going to Kingston be x.
Found 4 solutions by jorel1380, greenestamps, ikleyn, richwmiller:
Answer by jorel1380(3719) About Me  (Show Source):
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If the speed going to Kingston is x, then his speed coming back is x-20. So:
2(x)= 3(x-20)
x=60
His speed going to Kingston was 60 kph; his speed going back was 40 kph
☺☺☺☺

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The solution by tutor jorell380 is incorrect. His initial equation says that the return trip took 3 hours; it should say 4 hours.

You can look at his solution and modify the calculations to get the right answer.

Or perhaps he will see this post and correct his solution....

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
The solution by the turor @jorell380 is correct.


Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
I think Ikleyn and Jorel are interpreting the problem such that:
The trip to Kingston took 2 hours
He spent an hour at his destination doing something.
The problem clearly states, at least now, that the one hour stop is included in the total trip time of 6 hours.
He took 3 hours traveling back home.
2+1+3=6
So Jorel's solution seems right to me.
If the speed going to Kingston is x, then his speed coming back is x-20. So:
2(x)= 3(x-20)
x=60
His speed going to Kingston was 60 kph; his speed going back was 40 kph
60 kph is only 37.2823 mph
40 kph is only 24.8548 mph
He was out for a Sunday drive in the country.