Solution 1: Let $d$ denote the distance travelled down the hill. Let $t_1$ denote the time it takes Charlie to ride down the hill, and let $t_2$ denote the time it takes Charlie to ride up the hill. Then using the formula distance = rate $\times$ time, we get that $d = 30t_1 = 6t_2$ . The total time for the whole trip is $t = t_1 + t_2 = \frac{d}{30} + \frac{d}{6} = \frac{d}{5}.$ Thus, Charlie's average speed for the trip is $\frac{2d}{t_1+t_2} = \frac{2d}{\frac{d}{5}} = 10$ kilometers per hour.

Solution 2: Since Charlie spends $5$ times as long going up the hill than he spends riding down the hill, his average speed for the whole trip is the average of $30$ , $6$ , $6$ , $6$ , $6$ and $6$ , which is $\frac{60}{6} = 10$ kilometers per hour.

Harmonic Mean

We can use the harmonic mean of $a$ and $b$ . The harmonic mean is defined as $2ab \div (a + b)$ . Note that this only works when the distance traveled for both rates are equal. We let $2ab$ be the distance and $a + b$ as the time traveled.

Solving the Problem

Using this formula, $360 \div 36$ gives us $10$ .