To find the time taken for a non-reactive pollutant to travel from observation well Y to observation well X under advection, we use the principles of steady radial flow in a confined aquifer.

1. Identify Parameters

• Aquifer thickness ($b$) = $30 \text{ m}$
• Hydraulic conductivity ($K$) = $5 \text{ m/day}$
• Porosity ($n$) = $0.3$
• Distance to Well X ($r_1$) = $100 \text{ m}$, Drawdown ($s_1$) = $5 \text{ m}$
• Distance to Well Y ($r_2$) = $200 \text{ m}$, Drawdown ($s_2$) = $3 \text{ m}$

2. Relate Discharge to Drawdown

For a steady-state confined aquifer, Thiem's equation relates the pumping rate ($Q$) to the drawdown at two observation wells: 
$s_1 - s_2 = \frac{Q}{2\pi Kb} \ln\left(\frac{r_2}{r_1}\right)$ 
From this, we can express the constant term $\frac{Q}{2\pi b}$ as: 
$\frac{Q}{2\pi b} = \frac{K(s_1 - s_2)}{\ln(r_2/r_1)}$

3. Determine Advection (Seepage) Velocity

The actual velocity of the pollutant (seepage velocity, $v_a$) is given by Darcy's law divided by porosity: 
$v_a = \frac{dr}{dt} = \frac{Q}{A \cdot n} = \frac{Q}{(2\pi r b) \cdot n}$ 
Rearranging to solve for the time $dt$: 
$dt = \frac{2\pi n b}{Q} r \, dr$

4. Calculate Travel Time

Integrate from the starting point ($r_2 = 200 \text{ m}$) to the destination ($r_1 = 100 \text{ m}$): 
$T = \int_{r_1}^{r_2} \frac{2\pi n b}{Q} r \, dr = \frac{2\pi n b}{Q} \left[ \frac{r^2}{2} \right]_{r_1}^{r_2} = \frac{\pi n b (r_2^2 - r_1^2)}{Q}$ 

Substitute $Q$ from the Thiem's equation derived in Step 2: 
$T = \frac{\pi n b (r_2^2 - r_1^2)}{\frac{2\pi K b (s_1 - s_2)}{\ln(r_2/r_1)}} = \frac{n (r_2^2 - r_1^2) \ln(r_2/r_1)}{2K(s_1 - s_2)}$ 

Plug in the values: 
$T = \frac{0.3 \cdot (200^2 - 100^2) \cdot \ln(200/100)}{2 \cdot 5 \cdot (5 - 3)}$ 
$T = \frac{0.3 \cdot 30000 \cdot \ln(2)}{2 \cdot 5 \cdot 2} = \frac{9000 \cdot 0.693147}{20}$ 
$T = 450 \cdot 0.693147 \approx 311.916$ days

Final Answer:

The time taken for the pollutant to reach well X is 311.92 days.