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Jaron made a trip of 450 miles in 8hours. Before noon he averaged 60 miles per hour , and afternoon he averaged 50 miles per hour. At what time did he begin his trip and when did he end it?

Sagot :

Data:

Total distance: 450 miles

Total time: 8 h

Average 60 mi/h before noon

Average 50 mi/h afternoon

The relationship between the time, speed (average) and distance is drescribed in the next equations:

[tex]\begin{gathered} s=\frac{d}{t} \\ \\ d=s\times t \\ \\ \end{gathered}[/tex]

Then, if you multiply the speed and the time you get the distance:

time before noon: b

time afternoon: a

[tex](60\times b)+(50\times a)=450[/tex]

The sum of a and b is the total time:

[tex]a+b=8[/tex]

Use the next system of equations to find a and b:

[tex]\begin{gathered} 60b+50a=450 \\ a+b=8 \end{gathered}[/tex]

1. Solve a in the second equation:

[tex]\begin{gathered} \text{Subtract b in both sides of the equation:} \\ a+b-b=8-b \\ \\ a=8-b \end{gathered}[/tex]

2. Substitute the a in first equation by the value you get in first step:

[tex]60b+50(8-b)=450[/tex]

3. Solve b:

[tex]\begin{gathered} 60b+400-50b=450 \\ 10b+400=450 \\ \\ \text{Subtract 400 in both sides of the equation:} \\ 10b+400-400=450-400 \\ 10b=50 \\ \\ \text{Divide both sides of the equation into 10:} \\ \frac{10}{10}b=\frac{50}{10} \\ \\ b=5 \end{gathered}[/tex]

4. Use the value of b=5 to solve a:

[tex]\begin{gathered} a=8-b \\ a=8-5 \\ a=3 \end{gathered}[/tex]Then, Jaron begin his trip 5 hours before noon ( at 7:00) and end it 3 hours afternoon (at 15:00)