Answer :
Let A and B represent the two buses.
Given:
Let the speed of A be x km/h
The speed of B is (x + 19) km/h
Distance between A and B = 804 km
After 4 hrs, the distance covered by A:
[tex]\begin{gathered} \text{distance = sp}eed\text{ }\times\text{ time} \\ =\text{ x}\times4 \\ =\text{ 4x km} \end{gathered}[/tex]After 4 hrs, the distance covered by B:
[tex]\begin{gathered} \text{distance = sp}eed\text{ }\times\text{ time} \\ =\text{ (x + 19)}\times4 \\ =\text{ (4x + 76) km} \end{gathered}[/tex]The distance between them is 804km. Hence, the sum of the distance covered by A and B is 804km:
[tex]\begin{gathered} 4x\text{ + (4x + 76) = 804} \\ \text{Collect like terms} \\ 8x\text{ = 804- 76} \\ 8x\text{ = 728} \\ \text{Divide both sides by 8:} \\ \frac{8x}{8}=\text{ }\frac{728}{8} \\ x\text{ =91} \end{gathered}[/tex]This implies that the speed of A is 91 km/hr. The speed of B would be:
[tex]\begin{gathered} =\text{ 91 + 19} \\ =\text{ 110 km/hr} \end{gathered}[/tex]Answer:
Rate of faster bus = 110 km/hr
Rate of slower bus = 91 km/hr