A and B start moving towards each other from places X and Y, respectively, at the same time. The speed of A is 20% more than that of B. After meeting on the way, A and B take  $$2\frac{1}{2}$$ hours and x hours, now to reach Y and X, respectively. What is the value of x ?

Let the speed of B be v.

Speed of A = 1.2v

Let the distance be d.

Relative speed = v + 1.2v = 2.2v

Time taken to meet = $$\frac{d}{2.2v}$$

Time taken by A to cover distance = $$\frac{d}{2.2v} + 2\frac{1}{2}$$ = $$\frac{d}{2.2v} + \frac{5}{2}$$

$$\frac{d}{1.2v} = \frac{d}{2.2v} + \frac{5}{2}$$

$$\frac{dv}{2.64v^2} = \frac{5}{2}$$

$$\frac{d}{v} = \frac{13.2}{2}$$ ---(1)

Time taken by B to cover distance = $$\frac{d}{2.2v} + x$$ 

$$\frac{d}{v} = \frac{d}{2.2v} + x$$

$$\frac{d}{v} - \frac{d}{2.2v} = x$$

$$\frac{1.2dv}{2.2v^2} = x$$

$$\frac{d}{v} = \frac{11x}{6}$$

From eq(1),

$$\frac{13.2}{2} =  \frac{11x}{6}$$

x = 3.6 =$$ \frac{36}{10} = 3\frac{3}{5}$$