Relative Velocity

Considering the length of train A = La, length of train B = Lb.

The speed of train A be 5*x, speed of train B be 3*x.

From the information provided :

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.

Hence (8*x)*(46) = Lb.

In the information provided :

It took another 69 seconds for the rear ends of the trains to cross each other.

In the next 69 seconds 

The train B traveled a distance equivalent to the length of train A in this 69 seconds.

Hence (8*x)*(69) = La.

La/Lb = 69/46 = 3/2 = 3 : 2

Considering the distance travelled by Mira in one minute = M,

The distance traveled by Amal in one minute = A.

Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)

C is the circumference of the circle.

Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :

Hence 45*(A-M) = 3C. (2)

Multiplying (1)*15 we have :

45A + 45M = 15C.

45A - 45M = 3C.

Adding the two we have A = $$\frac{18C}{90}$$

Subtracting the two M = $$\frac{12C}{90}$$

Since Mira travels $$\frac{12C}{90}$$ in one minute, in one hour she travels :$$\frac{12C}{90}\cdot60\ =\ 8C$$

Hence a total of 8 rounds.

Alternatively, 

Let the length of track be L 
and velocity of Mira be a and Amal be b 
Now when they meet after 45 minutes Amal completes 3 more rounds than Mira
so we can say they met for the 3rd time moving in the same direction
so we can say they met for the first time after 15 minutes
So we know Time to meet = Relative distance /Relative velocity
so we get $$\frac{15}{60}=\frac{L}{a-b}$$     (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $$\frac{3}{60}=\frac{L}{a+b}$$    (2)
Dividing (1) and (2)
we get $$\frac{\left(a+b\right)}{\left(a-b\right)}=5$$
or 4a =6b
or a = 3b/2
Now substituting in (1)
we get :
$$\frac{L}{b}\times\ 2=\ \frac{15}{60}$$
so $$\frac{L}{b}\ =\frac{1}{8}$$
So we can say 1 round is covered in $$\frac{1}{8}$$ hours
so in 1-hour total rounds covered = 8.

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 - 8 = 16

In triangle OFS, $$\cos60^0\ =\ \frac{OF}{OS}$$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 - 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.