A cyclist is riding over a symmetric mountain. On the way up she moves 10 km/h (kilometers per hour), and on the way down she moves 40 km/h.
What is her average speed?
The answer is supposed to be...
16 km/h
... however I can't find a solution which gives me this answer.
On
Consider the cyclist goes $100$km up and $100$km down. So, total time taken is $$\left(\frac{100}{10}+\frac{100}{40}\right)\text{hr}=12.5\text{hr}$$ So, she travelled $200$km in $12.5$hr. So, average speed is $$\frac{200\text{km}}{12.5\text{hr}}=16\text{km/hr}$$ This completes the solution.
Following DJohn's advice, we can note that the choice of 100 km determined the size of the 12.5 hours and the size of the 200 km total distance- dividing the two numbers eliminates the contribution of the arbitrary choice.
Approach 1: Let $d$ be the distance going up the mountain in km. The total distance is $2d$. The total time is $\frac{d}{10} + \frac{d}{40}$ hours. So the average velocity is the ratio of these two quantities: $$\frac{2d}{\frac{d}{10} + \frac{d}{40}} = \frac{2}{\frac{1}{10} + \frac{1}{40}} = 16.$$ As Ned noted in the comments, this is the harmonic mean of the two speeds.
Approach 2: Let $t_1$ be the time going up the mountain and $t_2$ being the time going down. The distances going up and going down are $10t_1$ and $40t_2$ respectively; since the mountain is symmetric, these two quantities are equal and thus $t_1=4t_2$ and in particular $\frac{t_1}{t_1 + t_2} = \frac{4}{5}$. The average velocity is total distance over total time: $$\frac{10t_1 + 40t_2}{t_1 + t_2} = \frac{t_1}{t_1 + t_2} \cdot 10 + \frac{t_2}{t_1 + t_2} \cdot 40 = \frac{4}{5} \cdot 10 + \frac{1}{5} \cdot 40 = 16.$$ This formalizes the intuition behind my first comment regarding a weighted average of the two speeds $10$ and $40$.
For general speeds $v_1$ and $v_2$, the first approach is $\frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$ while the second approach can be shown to be $\frac{v_2}{v_1 + v_2} v_1 + \frac{v_1}{v_1 + v_2} v_2$. You can check that these two quantities are equivalent.