Solution:

Concept: Average speed is total distance over total time. Let total distance be \( D \). The time taken for each one-third distance is \( t_1 = \frac{D/3}{10} = \frac{D}{30} \), \( t_2 = \frac{D/3}{20} = \frac{D}{60} \), \( t_3 = \frac{D/3}{60} = \frac{D}{180} \). Total time \( T = t_1+t_2+t_3 = D \left( \frac{6+3+1}{180} \right) = \frac{D}{18} \). Average speed \( = \frac{D}{T} = \frac{D}{D/18} = 18 \text{ km/h} \).