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

When observer moves towards stationary source, then the apparent frequency is given by 
$$\begin{gathered} n^{\prime }=n\left(\frac{v+v_O}{v}\right)\text{ or }\frac{n^{\prime }}{n}=\left(\frac{v+v_O}{v}\right) \\ \text{ Here }{v}_o=\frac{V}{5}=\frac{320}{5}=64m/s \\ \frac{n^{\prime }}{n}=\left(\frac{320+64}{320}\right)=\frac{384}{320} \\ erefore, percentage incrcas \\ e\left(\frac{n^{\prime }-n}{n}\right)=\left(\frac{n^{\prime }}{n}-1\right)=\left(\frac{384}{320}-1\right)\times 100 \\ \begin{array}{r}=\left(\frac{384-320}{320}\right)\times 100=\left(\frac{64}{320}\times 100\right)\\ =20\%\end{array} \end{gathered}$$