X takes z hours more that …

Let the speeds of walking of A and B be x km/hour and y km/hour respectively.
Time taken by a to walk a distance of 30 km {tex}{\text{ = }}\frac{{{\text{30}}}}{{\text{x}}}{\text{hours}}.....{\text{time = }}\frac{{{\text{distance}}}}{{{\text{speed}}}}{/tex}
Time taken by B to walk a distance of 30 km {tex}{\text{ = }}\frac{{{\text{30}}}}{{\text{y}}}{\text{hours}}.....{\text{time = }}\frac{{{\text{distance}}}}{{{\text{speed}}}}{/tex}
According to the question, {tex}\frac{{30}}{x} = \frac{{30}}{y} + 3 \Rightarrow \frac{{30}}{x} - \frac{{30}}{y} = 3{/tex}
{tex}\Rightarrow \frac{1}{x} - \frac{1}{y} = \frac{3}{{30}}{/tex} ....Dividing throughout by 30
{tex}\Rightarrow \frac{1}{x} - \frac{1}{y} = \frac{1}{{10}}{/tex} ....(1)
When A double his pace (speed), then speed of A = 2x km/hour.
Now,
Time taken by A to walk a distance of 30 km {tex} = \frac{{30}}{{2x}}.....time = \frac{{{\text{distance}}}}{{{\text{speed}}}}{/tex}
{tex}= \frac{{15}}{x}hours{/tex}
According to the question, {tex}\frac{{15}}{x} + 1\frac{1}{2} = \frac{{30}}{y}{/tex}
{tex}\Rightarrow \;\frac{{15}}{x} = \frac{{30}}{y} - \frac{3}{2} \Rightarrow \frac{{15}}{x} - \frac{{30}}{y} = - \frac{3}{2}{/tex}
{tex}\Rightarrow \frac{1}{x} - \frac{2}{y} = - \frac{1}{{10}}{/tex} ....(2) [Dividing throughout by 15]
Subtracting equation (2) from equation (1), we get {tex}\frac{1}{y} = \frac{1}{5} \Rightarrow{/tex} y = 5
Substituting y = 5 in equation (1), we get {tex}\frac{1}{x} - \frac{1}{5} = \frac{1}{{10}}{/tex}
{tex}\Rightarrow \;\frac{1}{x} = \frac{1}{5} + \frac{1}{{10}} \Rightarrow \frac{1}{x} = \frac{3}{{10}} \Rightarrow x = \frac{{10}}{3}{/tex}
So the solution of the equations (1) and (2) is {tex}x = \frac{{10}}{3}{/tex} and y = 5.
Hence, their speed of walking is {tex}\frac{{10}}{3}{/tex} km/hour and 5 km/hour respectively.
Verification. Substituting {tex}x = \frac{{10}}{3}{/tex} and y = 5,
We find that both the equations (1) and (2) are satisfied as shown below:
{tex}\frac{1}{x} - \frac{1}{y} = \frac{1}{{(10/3)}} - \frac{1}{5} = \frac{3}{{10}} - \frac{1}{5} = \frac{1}{{10}}{/tex}
{tex}\frac{1}{x} - \frac{2}{y} = \frac{1}{{10/3}} - \frac{2}{5} = \frac{3}{{10}} - \frac{2}{5} = - \frac{1}{{10}}{/tex}
This verifies the solution.