X takes 3 hours more than …

Let the speed of X and Y be x km/hr and y km/hr respectively.

We know that the time taken to cover 'd' km with speed 's' km/hr is {tex}\frac ds{/tex} hr.

Now, using above formula for time 
Time taken by X to cover 30 km {tex}= \frac { 30 } { x } \mathrm { hrs }{/tex}
and, Time taken by Y to cover 30 km {tex}= \frac { 30 } { y } \mathrm { hrs }{/tex}
By the given condition in 1st half of problem, we have 
{tex}\frac { 30 } { x } - \frac { 30 } { y } = 3 \Rightarrow \frac { 10 } { x } - \frac { 10 } { y } = 1{/tex} ................................(i)
If X doubles his pace, then speed of X becomes 2x km/hr 
{tex}{/tex}Time taken by X to cover 30 km {tex}= \frac { 30 } { 2 x } \mathrm { hrs }{/tex}
&, Time taken by Y to cover 30 km {tex}= \frac { 30 } { y } \mathrm { hrs }{/tex} (speed of Y remains constant)
According to the given condition in 2nd half of problem, we have
{tex}\frac { 30 } { y } - \frac { 30 } { 2 x } = 1 \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad \frac { 30 } { y } - \frac { 30 } { 2 x } = \frac { 3 } { 2 }{/tex}
{tex}\Rightarrow \quad \frac { 10 } { y } - \frac { 5 } { x } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad - \frac { 10 } { x } + \frac { 20 } { y } = 1{/tex}.................................(ii)
Putting {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v,{/tex}in equations (i) and (ii) we get
10{tex}u{/tex} -10{tex}v{/tex} =1 {tex}\Rightarrow{/tex}10{tex}u{/tex}- 10{tex}v{/tex}-1=0 ........(iii)
-10{tex}u{/tex}+ 20{tex}v{/tex} = 1 {tex}\Rightarrow{/tex} -10{tex}u{/tex} + 20{tex}v{/tex} -1=0 ........(iv)
Adding equations (iii) and (iv), we get
{tex}10 v - 2 = 0 \Rightarrow v = \frac { 1 } { 5 }{/tex}
Putting {tex}v = \frac { 1 } { 5 }{/tex} in equation (iii), we get
{tex}10 u - 3 = 0 \Rightarrow u = \frac { 3 } { 10 }{/tex}
Now, {tex}u = \frac { 3 } { 10 } \Rightarrow \frac { 1 } { x } = \frac { 3 } { 10 } \Rightarrow x = \frac { 10 } { 3 }{/tex}and {tex}v = \frac { 1 } { 5 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 5 } \Rightarrow y = 5{/tex}
Hence, X's speed {tex}= \frac { 10 } { 3 } \mathrm { km } / \mathrm { hr }{/tex} and, Y's speed =5km/hr.