Question

How many cars would be asked to take the route A-N-B, that is Akala-Nanur-Bakala route, by the police department?

• A. 1
• B. 0.1
• C. 0.2
• D. 0.9
• A. 26
• B. 32
• C. 29.9
• D. 30

Answer 1

Correct Answer: 2

Step 1: Problem setup

We have 4 cars that must travel from point A to point B. The two possible routes are:

• Route 1: A → M → B
• Route 2: A → N → B

The travel times for both routes are almost identical under normal traffic conditions.

Step 2: Recommended strategy

To ensure that:

1. No car can improve its travel time by violating assigned routes.
2. Traffic load is evenly distributed between the two routes.

The optimal strategy is: \[ \boxed{\text{Assign 2 cars to A–M–B and 2 cars to A–N–B.}} \]

Step 3: Why not assign 3 cars to one route?

If 3 cars are directed along the same route, congestion will slightly increase that route’s travel time. One of the cars might then choose to break the police order and switch to the less congested route, thereby reducing its travel time and violating the fairness condition.

Step 4: Conclusion

By splitting the cars evenly — 2 on each route — we prevent any incentive to disobey orders and maintain near-equal travel times for all drivers.

Answer 2

The problem involves two possible routes from Akala (A) to Bakala (B): via Mamur (M) and via Nanur (N). Let's calculate the travel time for each route for one car.

• Route A-N-B:
  • A to N: One car takes 20 minutes, and each additional car increases the time by 1 minute.
  • N to B: One car takes 6 minutes, with each additional car increasing time by 3 minutes.
  • Assuming one car takes route A-N-B, it spends 20 + 6 = 26 minutes.
• Route A-M-B:
  • A to M: One car takes 6 minutes, and each additional car increases the time by 3 minutes.
  • M to B: One car takes 20 minutes, with each additional car increasing time by 0.9 minutes.
  • Assuming one car takes route A-M-B, it spends 6 + 20 = 26 minutes.

The time spent for both routes initially is 26 minutes. Now, consider congestion when more cars are on the road:

• From the Comprehension, two cars taking each route directly influences travel time:
• Route A-N-B with 2 cars:
  • A to N: 20 + 1 = 21 minutes per car.
  • N to B: 6 + 3 = 9 minutes per car.
  • Total: 21 + 9 = 30 minutes per car.
• Route A-M-B with 2 cars:
  • A to M: 6 + 3 = 9 minutes per car.
  • M to B: 20 + 0.9 = 20.9 minutes per car.
  • Total: 9 + 20.9 = 29.9 minutes per car.

The difference in travel time between routes A-N-B and A-M-B when congestion is considered is 30 - 29.9 = 0.1 minutes. This matches the correct answer provided.

Answer 3

Traffic Routing Logic 

Step 1: Problem conditions

We have the following constraints:

• No car should be able to reduce its travel time by disobeying the police order.
• Not all cars can take the same route.

Thus, either 2 or 3 cars should go through the A–M route.

Step 2: Case — Two cars take M–B route

If two cars go via M–B, one car could disobey and take M–N instead, saving time: \[ \text{Time via A–M–B (two cars)} = 12 + 20.9 = 32.9\ \text{minutes} \] \[ \text{Time via A–M–N–B} = 12 + 8 + 8 = 28\ \text{minutes} \] So, violation leads to faster travel — not acceptable.

Step 3: Case — Two cars take A–M route

If two cars take A–M and one is sent through M–N, a car from the A–N route could violate: \[ \text{Original (A–N–B)} = 21 + 12 = 33\ \text{minutes} \] \[ \text{New (A–M–B)} = 12 + 20.9 = 32.9\ \text{minutes} \] Again, violating reduces travel time — not acceptable.

Step 4: Case — Three cars via A–M

Two subcases:

• One car through M–N: A car from M–B could violate by going through M–N: \[ \text{Original (A–M–B)} = 12 + 20.9 = 32.9 \] \[ \text{New (A–M–N–B)} = 12 + 8 + 12 = 32.0 \] Violation saves 0.9 minutes — not acceptable.
• Two cars through M–N: No route change results in a faster time, so the rule holds.

Step 5: Conclusion

To ensure no car can reduce travel time by violating the order: \[ \boxed{\text{Two cars must be directed through M–N}} \]

Answer 4

To find the minimum travel time from A to B, we need to distribute the four cars across the three possible routes: A-M-B, A-N-B, and A-M-N-B. We start by understanding the travel times for each segment: 

• A-M (each car):
  • 1 car: 6 minutes
  • +1 car increases 3 minutes per car
• A-N (each car):
  • 1 car: 20 minutes
  • +1 car increases 1 minute per car
• M-B (each car):
  • 1 car: 20 minutes
  • +1 car increases 0.9 minutes per car
• M-N (each car):
  • 1 car: 7 minutes
  • +1 car increases 1 minute per car
• N-B (each car and same time as A-N):
  • Same as the A-N travel time
  •  

We will analyze each route configuration to find the best distribution of four cars:

Route	Cars	Total Time
A-M-B	2	(6 + 3×1) + (20 + 0.9×1) = 30.9 min
A-N-B	2	(20 + 1×1) + 0 = 22 min
A-M-N-B	0	N/A

With 2 cars on A-M-B and 2 on A-N-B, we evaluate: A-M-B takes 30.9 min and A-N-B takes 22 min. The M-N segment is not required here. We find the maximum among these as all cars must reach B, ensuring no car can reduce its time:

Max time = 30.9 minutes

We consider another configuration with 1 car on each route:

Route	Cars	Total Time
A-M-B	1	6 + 20 = 26 min
A-N-B	1	20 min
A-M-N-B	2	(6 + 3×1) + (7 + 1×1) + (20 + 1×1) = 32 min

In this configuration, each possible path takes:

Max time = 32 minutes

After evaluating configurations, the best distribution minimizes the time to 32 minutes without placing all cars on one route. Hence, the minimum travel time from A to B is 32 minutes. By considering all constraints, this optimizes travel while preventing route congestion.