An observer moves towards a stationary source of sound with a velocity of one fifth of the velocity of sound. The percentage increase in the apparent frequency is

Doppler effect for observer moving toward stationary source: apparent frequency $f' = f \left( \frac{v + v_o}{v} \right)$, where $f$ is source frequency, $v$ is sound velocity, and $v_o$ is observer velocity.

Given observer velocity $v_o = \frac{v}{5}$, substitute to obtain:

$f' = f \left( \frac{v + \frac{v}{5}}{v} \right) = f \left( \frac{\frac{6v}{5}}{v} \right) = \frac{6}{5}f = 1.2f$

The percentage increase is:

$\frac{f' - f}{f} \times 100\% = \frac{1.2f - f}{f} \times 100\% = 0.2 \times 100\% = 20\%$

The apparent frequency increases by $\boxed{20\%}$.