Question 299160: a plane flew 500 km from atlanta to louisville. when returning to atlanta the flight took 1/2 hour less time.if the rate to Louisville was 50km/hr faster than the rate returning, find the two rates in km/hr.
Found 4 solutions by mananth, MathTherapy, ikleyn, josgarithmetic:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! a plane flew 500 km from atlanta to louisville. when returning to atlanta the flight took 1/2 hour less time.if the rate to Louisville was 50km/hr faster than the rate returning, find the two rates in km/hr.
let the returning rate be x kmh
time taken while returning = 500/x
the rate to louisville will be x+50 mph
the time taken while going is 500/ x+50
the distance traveled each way is 500 km
500/x -500/ x+50 = 1/2
500(x+50) - 500x= 1/2 * x(x+50)
500x+25000 -500x = x^2 +50x /2
2*500x +2 * 25000 = x^2 +50x
1000x +50000 = x^2 +50x
x^2-950x -50000=0
x^2-1000x+50x-50000=0
x(x-1000)+50(x-1000)=0
(x-1000)(x+50)=0
x=1000 km / hour returning rate
1050 km / hr is the going rate to louisville
Answer by MathTherapy(10839)  (Show Source):
You can put this solution on YOUR website! a plane flew 500 km from atlanta to louisville. when returning to atlanta the flight took 1/2 hour less time.if the rate to Louisville
was 50km/hr faster than the rate returning, find the two rates in km/hr.
This doesn't make sense!!
How can the return trip take less time than the outgoing one when the plane is traveling at a slower speed on the return trip? This is
not practical!!
It has to be that for the plane to take less time on the return trip, it has to be traveling at a faster speed than it did on the outgoing
trip to Louisville, not a slower one.
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
a plane flew 500 km from Atlanta to Louisville. when returning to Atlanta the flight took 1/2 hour less time.
if the rate to Louisville was 50km/hr faster than the rate returning, find the two rates in km/hr.
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After reading this problem and its solution in the post by @mananth, I have two notices.
Firstly, the problem's formulation in the post is absurdist and self-contradictory.
Indeed, it says that the flight to Atlanta took 1/2 hour less time
and the rate to Louisville was 50 km/h faster than the rate returning.
So, the problem's creator was so hurry that missed circumstances.
I will edit the problem as it is shown below, to make sense from nonsense:
A plane flew 500 km from Atlanta to Louisville. When returning to Atlanta the flight took 1/2 hour  time.
If the rate to Louisville was 50km/h faster than the rate returning, find the two rates in km/hr.
Secondly, in the post by @mananth, both the solution and the answer are incorrect due to arithmetic errors.
I came to bring a correct solution to this my edited formulation.
Let the rate to Louisville be x km/h.
Time of this flight is  hours.
The rate to Atlanta (returning) will be (x-50) km/h.
The time of this returning flight is  hours.
Time equation is
- =  of an hour. <<<---=== It says literally what the problem says
Simplify and find x
2*500*x - 2*500*(x-50) = x*(x-50),
1000x - 1000x + 50000 = x^2 - 50x,
x^2 - 50x + 50000 = 0,
(x-250)*(x+200) = 0,
x = 250 or x = -200.
We choose the positive root and reject the negative one.
ANSWER. The rate from Atlanta to Louisville was 250 km/h. The rate of the returning trip from Louisville to Atlanta was 250-50 = 200 km/h.
CHECK. The tome flying from Atlanta to Louisville was  = 2 hours.
The time flying from Louisville to Atlanta (returning) was  = 2.5 hours, in 0.5 hours longer.
Precisely correct.
The problem edited and converted from self-contradictory to self-consistent
and then solved in a right way.
Answer by josgarithmetic(39823)  (Show Source):
You can put this solution on YOUR website!
SPEED TIME DIST.
A to L r+50 500/(r+50) 500
L to A r 500/r 500
DIFF. 1/2
...
Suspecting that part of the description wording is mixed-up.
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