We make the fractions improper so:
[tex]9 \frac{3}{4} = \frac{39}{4} \\ 15 \frac{2}{4} =15 \frac{1}{2}=\frac{31}{2} [/tex]
Their rates is the reciprocal of the time they take so the difference is:
[tex] \frac{4}{39} - \frac{2}{31} = \frac{124-78}{1209} = \frac{46}{1209} [/tex]
Therefore the first train is [tex] \frac{46}{1209} [/tex] units per hour faster than the second
