Explanation

(d) Given,
$v_{o} = \frac{v}{5}$
$\Longrightarrow \ \ v_{o} = \frac{320}{5} = 64ms^{- 1}$
When observers moves towards the stationary source, then
$n^{'} = \left( \frac{v + v_{o}}{v} \right)n$
$n^{'} = \left( \frac{320 + 64}{320} \right)n$
$n^{'} = \left( \frac{384}{320} \right)n$
$\frac{n^{'}}{n} = \frac{384}{320}$
Hence, percentage increases
$\left( \frac{n^{'} - n}{n} \right) = \left( \frac{384 - 320}{320} \times 100 \right)\%$
$= \left( \frac{64}{320} \times 100 \right)\% = 20\%$