The two sheets are placed side-by-side at the same angle to form two inclined planes, one with a dip and one with a hump. Two identical marbles are released from the top of the planes and roll to the bottom without slipping or jumping. Which marble reaches the bottom first, or is it simultaneous? Can you show why? Please feel free to assume any dimensions needed to provide a simple proof.

Now consider a sheet with a dip followed by an identical hump. Again the cut is made and one half turned over. Now one piece has a dip followed by a hump, and the other has a hump followed by a dip. When the two marbles run their race this time, which will reach the bottom first?

Source: Several places. I originally saw it in a UCD Physics 8A course.
(I've reworded this slightly, due to the answers received.)

1)
To find the solution we try to reduce the complexity... ;)
We assume that the marble has a size of 0. (this is needed because the
marble may 'skip' a part because it's bigger than the 'edge' in the far
end of the dip!) Assuming this will make sure the marble 'touches' the
whole plane-line it travels down.. Next we assume that during the travel
of the flat part (of the hump and the dip) the marble gets no extra
acceleration and during the travel of the steep parts it's accelerated.

Assuming that the time to travel a distance is: 
          the distance
----------------------------------
(Starting speed + ending Speed)/2)
Now on with the solution:
Assume that x = the distance of one end of the dip or hump.

When the marbles reach the edge of the dip and hump they have the same
speed. (assume the speed is S)

Now the marble which travels down the dip will get an extra
acceleration, an thus will have a speed of (S + SAD) (the extra speed is
called SAD).
Assume other marble will get no extra acceleration so it's speed at the 
top of the hump is S.

The time which is used to reach the bottom of the dip is shorter than
the time it takes to reach the top of the hump. So the dip-marble will
start his travel down the 'flat-part' of the dip earlier than the
hump-marble starts his travel down the steep part of the hump. (****
this is of no conseqents for the result !!!!***)
Now first continue with the hump-marble. 
When it starts its travel down the steep part of the hump it will get an
equal acceleration as the dip-marble got first. Thus making it's speed
S+SAD. The time it takes to travel the steep-hump-part is equal as the
time the dip-marble took to travel his steep-dip-part. 
At the end op his hump he has a speed of S+SAD.
Making the formula for his time to travel the hump is:
hump-marble: >>   x/S + x/((S + S+SAD)/2)   <<


During his travel down the flat-part of the dip the dip-marble keeps the
same speed (as did the hump-marble when it travel up the hump because
their angel is the same), this makes that the dip-marble will less time
to travel the flat-part because it has a bigger speed ***** this is the
important part...******
Time used to travel the dip is:
dip-marble : >>   x/((S + S+SAD)/2) + x/(S+SAD)   <<

The speed of both marbles at the end of their hump and dip is equal!!!!
This makes that the dip-marble will reach the bottom of the plane first.


2)
Now consider that they will immediatly travel down the opposite form...

This will result in the following formula's:
(where SAD is the extra-speed after one steep side,
 and SAD2 is the extra-speed after a steep side when the speed
 of the  marble is S+SAD at the beginning of that steep side,
 this SAD2 is smaller than the SAD because the time the acceleration
 takes place is shorter)

The dip-hump-marble will need time:
x/((S + S+SAD)/2) + (2 * x)/(S+SAD) + x/((S+SAD + S+SAD+SAD2)/2)

The hump-dip-marble:
x/S + x/((S + S+SAD)/2) + x/((S+SAD + S+SAD+SAD2)/2) + 
   x/(S+SAD+SAD2)

At the end of both dip-hump and hump-dip the speed of the both marbles
is equal. (thus the time to travel the rest of the plate is equal..)

--->

dip-hump     : hump-dip
2/(S+SAD)    <  1/S + 1/(S+SAD+SAD2)

This is always true for positive numbers!!!

---->

the dip-hump time is ALWAYS shorter than the hump-dip.. thus the
dip-hump marble is always down earlier..

1. The length of the marble's path on each plane is the same, and the velocity at any point is determined by the height (potential energy changing to kinetic energy), so the velocity before and after the dip/hump is the same for each marble. So the only difference that could affect the marbles is their velocity through the variation.

   The marble going through the dip has a higher velocity throughout its travel than the marble going over the hump. With D=RT, the time for the marble through the dip is less than that for the hump, so the marble through the dip arrives first.

2. Through the first dip/hump, the above analysis holds, and the time through the dip on the first ramp is less than the time over the hump on the second ramp: Th2>Td1. A similar argument shows: Th1>Td2. But the speed at the onset of the second dip/hump is higher than for the first, so the difference in time is less than for the first: (Th1-Td2)<(Th2-Td1). So the marble that goes through the dip first still reaches the bottom first.