Train station

A man walks from his house to the station. If he walks at 5 km/h, he misses a train by 7 minutes. However, if he walks at 6 km/h, he reaches the station 5 minutes before the train departs. Find the distance from his house to the station.

Final Answer:

x =  6 km

Step-by-step explanation:

$$\begin{gathered} t_1=0:7=0hr7min=0+\frac{7}{60}=0.1167hr\doteq 0.1167 \\ {v}_1=5\text{km/h} \\ {t}_2=0:5=0hr5min=0+\frac{5}{60}=0.0833hr\doteq 0.0833 \\ {v}_2=6\text{km/h} \\ x=s_1=s_2 \\ x=v_1\cdot (t+t_1) \\ x=v_2\cdot (t-t_2) \\ {v}_1\cdot (t+t_1)=v_2\cdot (t-t_2) \\ 5\cdot (t+0.11666666666667)=6\cdot (t-0.083333333333333) \\ t=1.083333 \\ t=\frac{1.08333333}{1}=1.08333333 \\ t=\frac{13}{12}\doteq 1.083333 \\ x=v_1\cdot (t+t_1)=5\cdot (1.0833+0.1167)=6\text{km} \end{gathered}$$

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