Linear Tracks

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$$

It is given,

$$\dfrac{D}{a}=10$$ and $$\dfrac{x}{b}=9$$

$$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$$

$$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$$

$$10x^2=\ \ 9D^2-9Dx$$

$$10x^2+9Dx-9D^2=\ 0$$

Solving, we get $$x=\dfrac{3D}{5}$$

$$\dfrac{x}{b}=9$$

$$\dfrac{3D}{b\times5}=9$$

$$\dfrac{D}{b}=15$$ 

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $$t^2=6\times\ 24$$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

We can use the formula for when two people moving towards each other and meeting in a straight line. $$t^2=t_1\times\ t_2$$
Where $$t$$ is time taken for them to meet each other. $$t_1$$ is the time taken by person 1 to reach the destination after meeting and 
$$t_2$$ is the time taken by person 2 to reach the destination after meeting

We are told they meet each other in 1 and a half hours, that is 90 minutes. 
And if Sunil takes X minutes to reach A, Anil will take X+75 minutes
Since it is given that, Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A

$$90^2=x\left(x+75\right)$$
$$8100=x^2+75x$$
$$x^2+75x-8100=0$$
$$\dfrac{\left(-75\pm\sqrt{5625+32400}\right)}{2}$$
$$\dfrac{\left(-75+195\right)}{2}$$
$$x=60$$

Total time of travel of Anil is, $$90\ \min\ +60\ \min\ +75\ \min$$
Total of 225 minutes. 
Time in hours will be 3.75 hours. 
Speed is $$\dfrac{45}{3.75}=12\ \dfrac{km}{hr}$$