The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are interchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is

The correct option is 5: 7.

Let the distance from A to B be denoted as d.
According to the problem, the distance from B to C is thrice that from A to B. Therefore, the distance from B to C is:

$$d_{BC}=3d$$

Now, let us define the speeds of the two trains.
While traveling from A to B, the speed of train 2 is double that of train 1.
Let the speed of train 1 from A to B be v. Therefore, the speed of train 2 from A to B is:

$$2v$$

While traveling from B to C, their speeds are interchanged.
This means that from B to C, the speed of train 1 becomes 2v, and the speed of train 2 becomes v.

We can now calculate the time taken by each train to travel from A to C. The total time is the sum of the time taken to travel from A to B and the time taken to travel from B to C.

For Train 1:
Time taken from A to B is:

$$t_{1,AB}=\frac{d}{v}$$

Time taken from B to C is:

$$t_{1,BC}=\frac{3d}{2v}$$

The total time taken by Train 1, T₁, is:

$$T_1=\frac{d}{v}+\frac{3d}{2v}=\frac{2d+3d}{2v}=\frac{5d}{2v}$$

For Train 2:
Time taken from A to B is:

$$t_{2,AB}=\frac{d}{2v}$$

Time taken from B to C is:

$$t_{2,BC}=\frac{3d}{v}$$

The total time taken by Train 2, T₂, is:

$$T_2=\frac{d}{2v}+\frac{3d}{v}=\frac{d+6d}{2v}=\frac{7d}{2v}$$

Finally, we find the ratio of the time taken by Train 1 to that taken by Train 2:

$$\frac{T_1}{T_2}=\frac{\left(\frac{5d}{2v}\right)}{\left(\frac{7d}{2v}\right)}=\frac{5}{7}$$

Thus, the ratio of the time taken by train 1 to that taken by train 2 is 5:7.