Given, frequency of sound, $f_s=5 \mathrm{kHz}$ frequency of siren records by moving train $A$,
$$
f_A=5.5 \mathrm{kHz}
$$
frequency of siren records by moving train $B$,
$$
f_B=6 \mathrm{kHz}
$$

Now, let $v_s=$ speed of sound
$\therefore$ For observer $A$, frequency of siren records by train $A$ is given as,
$$
\begin{aligned}
& f_A & =f_S\left(\frac{v_S+v_A}{v_S}\right) \\
\text { or } & 5.5 & =5\left(\frac{v_S+v_A}{v_S}\right) \\
\text { or } & 1.1 & =1+\frac{v_A}{v_S}
\end{aligned}
$$


$\therefore$ For observer $B$, frequency of siren records by train $B$ is given as,
$$
\begin{array}{cc}
& f_B=f_S\left(\frac{v_S+v_B}{v_S}\right) \\
\Rightarrow \quad 6=5\left(\frac{v_S+v_B}{v_S}\right) \text { or } 6=5\left(1+\frac{v_B}{v_S}\right) \\
\Rightarrow \quad \frac{6}{5}=1+\frac{v_B}{v_S} \\
\Rightarrow \quad \frac{6}{5}-1=\frac{v_B}{v_S} \Rightarrow \frac{6-5}{5}=\frac{v_B}{v_S} \\
\frac{1}{5}=\frac{v_B}{v_S} \Rightarrow \frac{v_B}{v_S}=0.2
\end{array}
$$

Now, from Eqs. (i) and (ii), we get or
$$
\frac{v_B}{v_A}=\frac{0.2 v_S}{0.1 v_S} \text { or } \frac{v_B}{v_A}=2
$$

So, the ratio of the speed of $\operatorname{train} B$ to that of train $A$ is $v_B: v_A=2$.