When a car is approaching the observer, the frequency of horn is $$100 \text{ Hz}$$. After passing the observer, it is $$50 \text{ Hz}$$. If the observer moves with the car, the frequency will be $$\dfrac{x}{3} \text{ Hz}$$ where $$x =$$ ______.

When the car approaches the observer, the apparent frequency is 100 Hz. When the car moves away, the apparent frequency is 50 Hz.

We start by setting up the Doppler effect equations.

When the source approaches:

$$f_{\text{app}} = \dfrac{f_0 v}{v - v_s} = 100 \quad \text{...(i)}$$

When the source recedes:

$$f_{\text{rec}} = \dfrac{f_0 v}{v + v_s} = 50 \quad \text{...(ii)}$$

Next, dividing equation (i) by equation (ii) yields:

$$\dfrac{v + v_s}{v - v_s} = \dfrac{100}{50} = 2$$

$$v + v_s = 2v - 2v_s$$

$$3v_s = v$$

$$v_s = \dfrac{v}{3}$$

Now we substitute $$v_s = v/3$$ in equation (i):

$$f_0 = \dfrac{100(v - v/3)}{v} = \dfrac{100 \times 2v/3}{v} = \dfrac{200}{3} \text{ Hz}$$

Since the observer moves with the car (same velocity as the source), there is no relative motion between them, so the observer hears the actual frequency of the horn:

$$f = f_0 = \dfrac{200}{3} \text{ Hz}$$

Comparing with $$\dfrac{x}{3}$$ Hz, we get $$x = \boxed{200}$$.

Therefore, the value of $$x$$ is 200.