Average speed

Translating the language directly into math yields the following ( $t$ is the total time and $d$ is the total distance)

$\large{t = \frac{1}{5}t + \frac{3}{4} \frac{d-\frac{v_1 t}{5}}{v_2} + \frac{1}{4} \frac{d-\frac{v_1 t}{5}}{v_3}}$

Grouping $t$ and $d$ terms gives:

$\large{t\Big(\frac{4}{5} + \frac{3v_1}{20v_2} + \frac{v_1}{20v_3}\Big) = d\Big(\frac{3}{4v_2} + \frac{1}{4v_3}\Big)}$

Multiplying both sides by $20v_2 v_3$ yields:

$\large{t(16v_2 v_3 + 3 v_1 v_3 + v_1 v_2) = d(15 v_3 + 5 v_2)}$

The average speed is therefore:

$\large{v_{av} = \frac{d}{t} = \frac{16v_2 v_3 + 3 v_1 v_3 + v_1 v_2}{15 v_3 + 5 v_2}}$