If you cover the same distance by bike, on the one hand a flat road, on the other seriously downhill and then back up, which one can you cover the fastest?

Answer

You have to be careful when you average out such speeds. The average of the numbers 10 and 40 is indeed 25, but you have used that 40 km/h for a shorter time than the 10 km/h, and therefore that 40 actually weighs less than the 10.

Let us do the following:

1) you drive D km on a flat road at a constant speed v. So you are on the road D/v seconds

2) Now suppose that the road descends the first D/2 meter, (so D/2 is the actual distance along the descending road, not the horizontal distance) and then rises again. So the road has the shape of a letter V. Each leg of the V has length D/2.
Let us also assume that the extra speed you can gain when descending is equal to the loss of speed when ascending. OK, that’s a choice I make, and that’s the simplest and a good first order approach.
So you descend at a speed v + dv, and rise at v – dv.
How long have you been traveling now? again we do time = distance traveled / speed. So in total:

0.5 D / (v + dv) + 0.5 D / ( v – dv) = D v / (v² – dv²) = D / [ v . ( 1 – dv²/v² ) ]

and that is MORE than the flat road, because you are now dividing D by a factor smaller than v, namely
v . (1 – something),
and as a result you get a larger break, and therefore a longer travel time than with the flat road.

That is the effect of driving at a higher speed for a shorter time and at a slower speed for a longer period of time. The slower speed therefore weighs more in the average, and means that you need more time on the V-road than on the flat road.