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The figure shows traffic entoring and exiting a road system (the units are vehides perhour)Equations:At node A: \quad x_{1}+x_{2}=200At node B : x_{1}=x_{3}+x_{4}At nodo E: \quad x_{2}+x_{3}=x_{5}At node D : x_{4}+x_{5}=200\begin{aligned}\text { Total time } & =k x_{1}+2 k x_{2}+k x_{3}+2 k x_{4}+k x_{5} \\& =k\left[x_{1}+2 x_{2}+x_{3}+2 x_{4}+x_{5}\right] \\& =k\left[\left(x_{1}+x_{2}\right)+\left(x_{2}+x_{3}\right)+2 x_{4}+x_{5}\right] \\& =k\left[\left(x_{1}+x_{2}\right)+x_{5}+2 x_{4}+x_{5}\right]\left\{\left(x_{2}+x_{3}=x_{5}\right)\right\} \\& =k\left[\left(x_{1}+x_{2}\right)+2 x_{4}+2 k_{5}\right] \\& =k\left[\left(x_{1}+x_{2}\right)+2\left(x_{4}+x_{5}\right)\right] \\& =k[200+2(200)]=600 k\end{aligned}If k=4, Then total time =600(4)=2400 minutesHonce Total time =2400 minutes or 40 hours.\begin{aligned}\text { Average timo for each car } & =\frac{\text { Tutal time }}{\text { numberof cars }} \\& =\frac{2400}{200}=12 \text { minutes }\end{aligned}Arerage time for each car =12 minutes ...