The driver of bus approaching a big wall notices that the frequency of his bus's horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms$$^{-1}$$:

The horn fixed on the bus emits a true (source) frequency of $$f = 420\text{ Hz}$$. The bus is moving straight toward a large, perfectly reflecting wall with an unknown speed $$v$$, while the velocity of sound in air is given as $$v_{s} = 330\ \text{ms}^{-1}$$.

Because of the motion, the sound undergoes two separate Doppler shifts:

1. Shift on its way to the wall (moving source, stationary observer):
We first recall the standard Doppler formula for a stationary observer and a source moving toward the observer: $$f_{1} = f\,\frac{v_{s}}{\,v_{s}-v\,}$$ Here the wall acts like the observer, so the frequency that actually strikes the wall is $$f_{1} = 420\;\text{Hz}\;\times\;\frac{330}{330 - v}\;.$$

2. Shift on the way back (stationary source, moving observer):
The wall reflects the wave without changing the frequency it has just received, therefore it re-emits (acts as a new source) with frequency $$f_{1}$$. For a stationary source and an observer moving toward the source the observed frequency is obtained from the second standard Doppler formula: $$f' = f_{1}\,\frac{v_{s} + v}{v_{s}}\;.$$ Substituting $$f_{1}$$ from the first step gives $$f' = \Bigl(420\;\text{Hz}\,\frac{330}{330 - v}\Bigr)\,\frac{330 + v}{330}\;.$$

The driver actually hears this reflected sound at the higher frequency $$f' = 490\text{ Hz}$$, so we now set up the equality

$$490 = 420\,\frac{330}{330 - v}\,\frac{330 + v}{330}\;.$$

Simplifying the right-hand side, the factor $$330$$ in numerator and denominator cancels out, leaving

$$490 = 420\,\frac{330 + v}{330 - v}\;.$$

We next isolate the fraction by dividing both sides by 420:

$$\frac{490}{420} = \frac{330 + v}{330 - v}\;.$$

The left ratio simplifies to $$\frac{490}{420} = \frac{49}{42} = \frac{7}{6}\;,$$ so we have

$$\frac{7}{6} = \frac{330 + v}{330 - v}\;.$$

Cross-multiplying gives

$$7(330 - v) = 6(330 + v)\;.$$

Expanding both brackets: $$2310 - 7v = 1980 + 6v\;.$$

Bringing like terms together, we add $$7v$$ to and subtract $$1980$$ from both sides:

$$2310 - 1980 = 6v + 7v\;$$ $$330 = 13v\;.$$

Solving for $$v$$,

$$v = \frac{330}{13}\ \text{ms}^{-1} = 25.3846\ \text{ms}^{-1}\ (\text{approximately}).$$

To express this speed in kilometres per hour we multiply by $$\frac{18}{5}$$ (since $$1\ \text{ms}^{-1} = 3.6\ \text{kmh}^{-1}$$):

$$v = 25.3846 \times 3.6 \approx 91.4\ \text{kmh}^{-1}\;.$$

Among the given options the closest (and intended) value is $$91\ \text{kmh}^{-1}$$.

Hence, the correct answer is Option A.