Two trains, Train A and Train B, are traveling on parallel tracks. Train A's speed is 135 km/hr, and Train B's speed is 32.5 m/s. If they cross each other in 20 seconds when traveling in opposite dire

The speed of the first train is $S_1 = 135 \text{ km/hr}$.
To convert km/hr to m/s, we multiply by $\frac{5}{18}$.

$$S_1 = 135 \times \frac{5}{18} \text{ m/s} = \frac{675}{18} \text{ m/s} = 37.5 \text{ m/s}$$

The speed of the second train is $S_2 = 32.5 \text{ m/s}$.
Comparing the speeds, $S_1 = 37.5 \text{ m/s}$ is the faster train.

When two trains travel in opposite directions, their relative speed is the sum of their individual speeds.

$$S_{rel, opp} = S_1 + S_2 = 37.5 \text{ m/s} + 32.5 \text{ m/s} = 70 \text{ m/s}$$

When two trains cross each other, the total distance covered is the sum of their lengths ($L_1 + L_2$).
Given that they take 20 seconds to cross each other in opposite directions:

$$L_1 + L_2 = S_{rel, opp} \times T_{opp}$$ $$L_1 + L_2 = 70 \text{ m/s} \times 20 \text{ s} = 1400 \text{ m}$$

The ratio of lengths $2:3$ is consistent with this sum, but we only need the total length for the next step.

When the faster train crosses the slower train from behind (traveling in the same direction), their relative speed is the difference between their speeds.

$$S_{rel, same} = S_{faster} - S_{slower} = S_1 - S_2 = 37.5 \text{ m/s} - 32.5 \text{ m/s} = 5 \text{ m/s}$$

The distance to be covered for one train to completely cross the other is still the sum of their lengths, which is $L_1 + L_2 = 1400 \text{ m}$.

$$T_{same} = \frac{L_1 + L_2}{S_{rel, same}}$$ $$T_{same} = \frac{1400 \text{ m}}{5 \text{ m/s}} = 280 \text{ s}$$

The time taken by the faster train to completely cross the other train from behind while traveling in the same direction is $\boxed{\text{280 seconds}}$.