Riddle: can you swim faster upstream than downstream (with respect to the water)? A friend of mine posed a riddle to me:
A man swims upstream in a river, which is flowing at an unknown rate. He is wearing swimming goggles. At a certain point he loses his goggles. 10 minutes later he realizes he lost them, so he immediately turns around and swims back downstream to get them. When he finds his goggles, floating in the water, he finds himself at a point 500 meters downstream from the point where he lost his goggles (with respect to the ground, of course, not the water).
Question: At what speed (km/h) is the river flowing?

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My solution is one and a half kilometres per hour, but another friend does not agree with me, he says swimming upstream is more efficient (with respect to the water) than swimming downstream.
Is he correct? If so, why?
 A: Since this is not marked as homework...
An equation for this problem can be created as follows:
Velocity of the swimmer is: $V_{sw}$
Velocity of the stream is: $V_{st}$
The ratio of velocity of swimmer and stream is: $k = \dfrac{V_{sw}}{V_{st}}$
The time the swimmer goes upstream is: $t_{u}$
The time the swimmer goes downstream is: $t_{d}$
The distance traveled by the goggles is: $d_{g}$
The distance the swimmer travels upstream is: $(k-1)V_{st}t_u$
The distance the swimmer travels downstream is: $(k+1)V_{st}t_d$
The equation then is:$$(k+1)V_{st}t_d - (k-1)V_{st}t_u = d_g$$
If one sets $k = 1$ and $t_u = t_d = \dfrac{1}{6}hr$ and $d_g = 0.5km$ then one gets the solution of $V_{st} = 1.5 \dfrac{km}{hr}$
For the sake of argument, lets set $k=2$ and $d_g = 0.5km$ and $t_u =  \dfrac{1}{6}hr$. Let's first rewrite the equation: $$(k+1)t_d - (k-1)t_u = \dfrac{d_g}{V_{st}}$$
then as: $$(k+1)t_d - (k-1)\dfrac{1}{6}hr = \dfrac{0.5km}{V_{st}}$$
and: $$6t_d - \dfrac{1}{3} = \dfrac{1}{V_{st}}$$
This can be plugged into wolfram alpha to find the set of solutions.
Update for Berhard's sake (where now all you need to do is input the values in for $t_d$ and $V_{st}$ since I have already their units):
$$6t_d\dfrac{hr}{km} - \dfrac{1}{3}\dfrac{hr}{km}  = \dfrac{1}{V_{st}}\dfrac{hr}{km} $$
Note: For future ref, the general form of the equation is:$$x + y + k(x-y) = \dfrac{d}{z}$$
A: 
 he says swimming upstream is more efficient (with respect to the water) than swimming downstream.

Go to a swimming pool and try swimming in various directions, the water in any pool on planet Earth is moving at 67,000 miles per hour around the sun. If it is easier "upstream" you should soon find out.
If you move the water in this pool to the middle of the Amazon river, the body of water in which you are swimming is still travelling at about the same 67,000 MPH around the sun.
If there is some property of the water's motion that makes swimming in one direction more efficient, your friend should explain why it doesn't apply to movement of that water around the Sun
A: You are also travelling at the same speed as the water around the sun so it becomes more to do with the motion of the water itself 'on top' of the 67,000 miles an hour 
If you had to swim against a current you are clearly going to cover less distance going against the current than with it 
A river with a heavy current could easily carry someone downstream without them even needing to swim so then you add their swimming motion to that and they will travel faster than the river's actual current 
Swim against the current and you will have a force against the body of the swimmer so if they use the same swimming motion then they will not get as far 
Simple 
A: I think it depends on what your friend means by "with respect to the water"
When I hear that I think spatially.  
sample constants: you travel 1 meter per stroke, 1 second per stroke, water traveling 1 meter per second.
When swimming downstream you'll cover 2 meters with respect to the ground in 1 second (yay for triathlons that are downstream). However, from the point you start your next cycle you're only 1 meter away from the point in the water you started your previous stroke. (why you don't pull too quickly when swimming you don't want to lose your grip on the water and you what to have the water that you pulled be behind you for your next stroke). 
"With respect to the water" you traveled 1 meter.
If you're swimming upstream you take one stroke at that same time the water is going the opposite direction (lets forget about drag.. it is a riddle after all) so you'd be 1 meter from where you were but at the same time you'll get pushed back that 1 meter by the current. You've gone nowhere with respect to the ground. However, from the point of your hand entry to your next stroke you'll have 2 meters of distance between your hand and it's previous entry. 
"With respect to the water" you traveled 2 meters. 
With respect to the water you are faster swimming upstream! But that's not how races are measured so I want no part of it.
If that's not what your friend means by "With respect to the water" I take it all back!
A: Swimming upstream is less efficient than downstream.
When swimming upstream, you are also swimming uphill. Part of the energy you use for the swimming goes into potential gravitational energy. Therefore you'll have a lower swimming speed w.r.t. the water when swimming upstream.
That potential energy is released again when swimming downstream/downhill, and so your swimming speed w.r.t. the water will be higher downstream.
The mathematical answer to the question will depend on many unknown factors, like the relation between the descent rate of the river and the velocity of the water, and the energy loss due to friction at various swimming speeds. But it will be more 1.5 km/h
