A train passes through a station with a constant speed. A stationary observer at the station platform measures the tone of the train whistle as 484 Hz when it approaches the station and  442 Hz when it leaves the station. If the sound velocity in air is  330m/s, then the tone of the whistle and the speed of the train are

When train approaches the station, the frequency heard by the observer
${\upsilon }_1=\upsilon \frac{v}{v-v_s}=\upsilon \left(\frac{330}{330-v_s}\right)$
Here,  $v=330\text{\hspace{0.17em}m}/\text{s}$
$\upsilon$ is the actual frequency of the whistle
$484=\upsilon \left(\frac{330}{330-v_s}\right)\mathrm{.....}\left(1\right)$
When the train leaves the station
${\upsilon }_2=\upsilon \frac{v}{v+v_s}=\upsilon \left(\frac{330}{330+v_s}\right)\mathrm{....}\left(2\right)$
Divide Equations (1) & (2), we get
$\Rightarrow \frac{484}{442}=\frac{330+v_s}{330-v_s}$
$\Rightarrow 1.09=\frac{\left(330+v_s\right)}{\left(330-v_s\right)}$
$\Rightarrow v_s=\frac{31.35}{2.09}$
$=15\text{\hspace{0.17em}m}/\text{s=15 ×}\frac{18}{5}\text{=54kmph}$
Substituting  $v_s$ in equation gives
$\Rightarrow 484=\upsilon \left(\frac{330}{330-15}\right)$
$=\upsilon \left(\frac{330}{315}\right)$
$\Rightarrow \upsilon =\frac{484\times 21}{22}$
$\therefore \upsilon =462\text{\hspace{0.17em}Hz}$
Therefore, the correct answer is (A).