Respuesta :
Let the distance from his home to the park be d .
The time it took him to bike home in hours is given by
[tex]\frac{10}{60}=\frac{1}{6}\text{hours}[/tex]The time it took him to walk home in hours is given by
[tex]\frac{25}{60}=\frac{5}{12}\text{ hours}[/tex][tex]\text{speed }=\frac{dis\tan ce}{time}[/tex]Hence, the speed with which he biked home is given by
[tex]d\div\frac{1}{6}=d\times6=6d\text{ mph}[/tex]The speed with which he walk home is given by
[tex]d\div\frac{5}{12}=d\times\frac{12}{5}=\frac{12d\text{ }}{5}\text{mph}[/tex]Since, he can bike 6 mph faster than he can walk, then
[tex]6d=\frac{12d}{5}+6[/tex]Subtracting 12d/5 from both sides we have
[tex]\begin{gathered} 6d-\frac{12d}{5}=6 \\ \frac{18d}{5}=6 \end{gathered}[/tex]Multiplying both sides by 5/18 we have
[tex]\begin{gathered} \frac{18d}{5}\times\frac{5}{18}=\frac{6}{1}\times\frac{5}{18} \\ d=\frac{5}{3}=1\frac{2}{3} \end{gathered}[/tex]Therefore,
[tex]6d=\frac{6}{1}\times\frac{5}{3}=2\times5=10[/tex]Hence, the speed of the bike is 10 mph
