Given that there are four cars, and the time through each route is nearly the same, it is recommended to divide the cars into two groups. Two cars should take the route A-M-B, and the other two should take the route A-N-B. If, for some reason, three cars are directed to go through any of the routes, one of the three cars can choose to break the police order and reduce its travel time.
According to the police order, two cars each would pass through A – M – B and A – N – B. The time taken through A – M – B is 29.9, and the time taken through A – N – B is 30.0. Therefore, the difference is 0.1.
No car should be able to decrease its travel time by disobeying the order, and all cars cannot take the same route. Therefore, either two or three cars should go through A-M. If two cars take the M-B route, one car can violate the police order and opt for M-N, reaching B in 28 minutes compared to the 29.9 minutes required if both had gone through A-M-B. If two cars choose the A-M route and one is directed to go through M-N, one of the cars directed through A-N can violate the police order and go through A-M-B, saving time as follows:
Original time (A-N-B) = 21 + 12 = (three cars) = 33
New time = 12 (three cars) + 20.9 = 32.9
The police department cannot direct both cars to go through M-N, as in that case, all four cars would go through N-B. In case three cars are directed to go through A-M, either one car can be directed through M-N or two cars can be directed through M-N. If one car is directed through M-N, one of the two cars directed through M-B can violate the police order and go through M-N, saving time, as shown.
Original time (A-M-B) = 12 (three cars) + 20.9 = 32.9
New time (A-M-N-B) = 12 + 8 + 12 = 32 minutes.
Therefore, two cars must be directed through M-N to ensure that any car violating the police order cannot reduce the travel time.
When all cars follow the police order, the time taken would be: