Three boats B1, B2 and B3, when working together, can carry 60 people in a single trip. One day, in the early morning, boat B1 alone carries 50 people in a few trips and then stops. After that, B2 and B3 together carry the remaining people. In total, 10 trips are made by B1, B2 and B3 to carry 300 people across the river. It is known that on an average day, 300 people cross the river using only one of the three boats. How many trips would B1 need to make to carry 150 passengers alone?

Introduction / Context:
This puzzle-style time and work question is framed in terms of boats carrying people across a river. Instead of work done per day, we talk about passengers carried per trip. You are given how many people the boats carry together, how many people B1 carries before stopping, and the total number of trips made to move 300 people. You must deduce the capacity of boat B1 and then compute the number of trips B1 alone would need to carry 150 people.

Given Data / Assumptions:
Boats B1, B2 and B3, working together in a single trip, carry 60 people. One day, B1 alone first carries 50 people and then stops. After that, boats B2 and B3 together carry the remaining passengers. The total number of trips made by all three boats (counting each time any boat or combination crosses with passengers) is 10. In total, 300 people are transported that day. We assume that each boat has a fixed capacity per trip and that every trip is fully utilized up to that boat’s capacity as implied by the numbers.

Concept / Approach:
Let the per-trip capacity of B1 be b1, and the combined per-trip capacity of B2 and B3 be b23. We know that when all three work together, they carry 60 people per trip, so b1 + b23 = 60. On the specific day described, B1 alone carries 50 people in n1 trips, so n1 * b1 = 50. Then B2 and B3 together carry the remaining 250 people in n2 trips, so n2 * b23 = 250. We are also told that the total number of trips is n1 + n2 = 10. We must find integer solutions for b1, n1, b23, and n2. Once we know b1, we can determine how many trips B1 alone needs to carry 150 people: trips = 150 / b1.

Step-by-Step Solution:
Step 1: Let b1 be the number of people B1 can carry in one trip. Let b23 be the number of people B2 and B3 together can carry in one trip. Step 2: Since together B1, B2, and B3 can carry 60 people per trip, we have b1 + b23 = 60. Step 3: Suppose B1 makes n1 trips alone. Then n1 * b1 = 50, since B1 alone carries 50 people at the start. Step 4: After B1 stops, the remaining number of people to be transported is 300 - 50 = 250. Step 5: Let B2 and B3 together make n2 trips. Then n2 * b23 = 250. Step 6: Total number of trips is n1 + n2 = 10. Step 7: From n1 * b1 = 50, because capacities and trips are integer quantities, b1 must be a factor of 50. So b1 can be 1, 2, 5, 10, 25, or 50. Step 8: For each possible b1, compute n1 = 50 / b1 and then b23 = 60 - b1. Check whether 250 is divisible by b23 to get an integer n2, and whether n1 + n2 = 10. Step 9: Taking b1 = 10 gives n1 = 50 / 10 = 5. Then b23 = 60 - 10 = 50. This implies n2 = 250 / 50 = 5, and n1 + n2 = 5 + 5 = 10, which matches the condition. Step 10: Thus, the only consistent solution is b1 = 10 passengers per trip and b23 = 50 passengers per trip. Step 11: To find the number of trips B1 alone would need to carry 150 passengers, we compute 150 / b1 = 150 / 10 = 15 trips.

Verification / Alternative check:
We can verify the scenario using the found capacities. B1’s capacity is 10 per trip, so in 5 trips B1 carries 5 * 10 = 50 people and stops. Then B2 and B3 together, with a combined capacity of 50, carry 50 people per trip for 5 trips, transporting 5 * 50 = 250 people. In total, 50 + 250 = 300 people are carried, and exactly 5 + 5 = 10 trips are made. This confirms that b1 = 10 is correct. Therefore, for 150 people, 150 / 10 = 15 trips is consistent.

Why Other Options Are Wrong:
30 and 25 trips would imply much smaller capacities for B1, which would not satisfy the given constraints and integer conditions for carrying 50 people in whole trips. 10 trips would mean B1 carries 15 passengers per trip to reach 150, which conflicts with the unique solution b1 = 10 that is consistent with all constraints. 12 trips corresponds to a capacity of 12.5 passengers per trip, which is not allowed as we are dealing with whole passengers per trip.

Common Pitfalls:
A common mistake is to assume equal capacities or to divide the total passenger count incorrectly among the boats without considering integer constraints. Some learners overlook that the number of trips and the number of passengers must both be integers, which severely restricts possible capacities. Another pitfall is not using the equation b1 + b23 = 60, which connects the individual and combined capacities and is crucial to solving the system.

Final Answer:
Boat B1 would need to make 15 trips to carry 150 passengers alone.