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Conner left his house and rode his bike into town at 6mph. Along the way he got a flat tire so he had to turn around and walk his bike to his house traveling 3 mph. If the trip down and back took 15 hours, how far did he get before his tire went flat?Conner went ___ miles before his tire went flat.

User Muzuiget
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1 Answer

3 votes

The main point in this question, that the distance of the first part = the distance of the second part


d_1=d_2

Since the speed of the first part is 6 mph

Let the time of it be t1

Since distance = speed x time, then


\begin{gathered} d_1=v_{1_{}}* t_1_{} \\ d_1=6* t_1 \\ d_1=6t_1 \end{gathered}

Since the speed of the second part is 3 mph

Let the time of it be t2, then


\begin{gathered} d_2=3* t_2 \\ d_2=3t_2 \end{gathered}

Equate d1 and d2 to find t2 in terms of t1


3t_2=6t_1

Divide both sides by 3


\begin{gathered} (3t_2)/(3)=(6t_1)/(3) \\ t_2=2t_1\rightarrow(1) \end{gathered}

Since the total time of the two parts is 15 hours, then


t_1+t_2=15\rightarrow(2)

Substitute (1) in (2)


\begin{gathered} t_1+2t_1=15 \\ 3t_1=15 \end{gathered}

Divide both sides by 3


\begin{gathered} (3t_1)/(3)=(15)/(3) \\ t_1=5 \end{gathered}

Now, let us find d1


\begin{gathered} d_1=6*5 \\ d_1=30 \end{gathered}

Conner went 30 miles before his tire went flat

User Maelle
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