Asked by Hank
The distance between 2 towns is 330km. A train travels between these two towns and returns on the same day. On the return journey, the train left late and so the driver decided to drive the train 5km/h fasterthan its inward journey, and it took 30 minutes less time to return. What is the speed on the return journey?
I used the folowing equation:
Let inward speed be x,
330/x - 330/x+5 = 1/2
However, I keep getting a squared x?
I used the folowing equation:
Let inward speed be x,
330/x - 330/x+5 = 1/2
However, I keep getting a squared x?
Answers
Answered by
Reiny
330/x - 330/(x+5) = 1/2
That's ok, we end up with a quadratic.
Multiply each term by 2x(x+5)
660(x+5) - 660x = x(x+5)
660x + 3300 - 660x = x^2 + 5x
x^2 + 5x - 3300 = 0
(x-55)(x+60) = 0
x = 55 or x = -60, we would reject the -60
So your inward speed is 55 km/h, the return is 60 km/h
check:
330/55 = 6 hrs
330/60 = 5.5 hrs, which is 1/2 hour less.
That's ok, we end up with a quadratic.
Multiply each term by 2x(x+5)
660(x+5) - 660x = x(x+5)
660x + 3300 - 660x = x^2 + 5x
x^2 + 5x - 3300 = 0
(x-55)(x+60) = 0
x = 55 or x = -60, we would reject the -60
So your inward speed is 55 km/h, the return is 60 km/h
check:
330/55 = 6 hrs
330/60 = 5.5 hrs, which is 1/2 hour less.
Answered by
Hank
Thanks Reiny, appreciate it
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.