Question 558130
. 
A passenger plane flew to Moscow and back. The trip there took five hours and the trip back took nine hours. 
It averaged 144 km/h faster on the trip there than on the return trip. 
Find the passenger plane's average speed on the outbound trip.
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As I read the post by @Theo, I clearly see that his solution is overcomplicated
and his setup is over-sophisticated. Actually, this problem admits much more simple treatment.
It is not about a choice of the solution method - it is about how to teach 
young students in a right way. It is impossible to teach them presenting Theo' approach. 
Therefore, I developed and present here a simple solution, which is a standard
method (or one of standards methods) for this problem.



<pre>
Outbound trip = the trip back (from Moscow).

Trip to Moscow is 5 hours.

The trip back is 9 hours.


Let x be the average speed on the outbound trip (from Moscow), in km/h.

It is the value, which is under the question of the problem.


Then the average speed on the trip to Moscow was (x-144) km/h, according to the problem.


The distance for the flight to Moscow is 5x kilometers  (the product of the time and the speed).

The distance for the returning flight is 9(x-144) kilometers, similarly..


This distances are equal - so, we can write this equation

    5x = 9(x-144).    (1)


Simplify and find x

    5x = 9x - 9*144

    9*144 = 9x - 5x

    9*144 = 4x

    x = (9*144)/4 = 9*36 = 324 kilometers per hour.


<U>ANSWER</U>.  The average speed on the outbound trip was 324 km/h.
</pre>

Solved.


The solution, which I presented in this my post, follows a standard approach to the problem 
without any sophistications. It provides the most simple and the most straightforward treatment of the problem.