Question

Two buses leave a station at the same time and travel in opposite directions. One bus travels 19Km/hapart after 4 hours, what is the rate of each bus?kmRate of the faster bus:음hХ$?kmRate of the slower bus:0V

Answer 1

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:

`\begin{gathered} \text{distance = sp}eed\text{ }*\text{ time} \\ =\text{ x}*4 \\ =\text{ 4x km} \end{gathered}`

After 4 hrs, the distance covered by B:

`\begin{gathered} \text{distance = sp}eed\text{ }*\text{ time} \\ =\text{ (x + 19)}*4 \\ =\text{ (4x + 76) km} \end{gathered}`

The distance between them is 804km. Hence, the sum of the distance covered by A and B is 804km:

`\begin{gathered} 4x\text{ + (4x + 76) = 804} \\ \text{Collect like terms} \\ 8x\text{ = 804- 76} \\ 8x\text{ = 728} \\ \text{Divide both sides by 8:} \\ (8x)/(8)=\text{ }(728)/(8) \\ x\text{ =91} \end{gathered}`

This implies that the speed of A is 91 km/hr. The speed of B would be:

`\begin{gathered} =\text{ 91 + 19} \\ =\text{ 110 km/hr} \end{gathered}`

Answer:

Rate of faster bus = 110 km/hr

Rate of slower bus = 91 km/hr