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There are three boats B1, B2, and B3 working together they carry 60 people on each trip. One day an early morning B1 carried 50 people in a few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on average 300 people cross the river using only one of the 3 boats B1, B2, and B3. How many trips it would take to B1 to carry passengers alone?

Question

There are three boats B1, B2, and B3 working together they carry 60 people on each trip. One day an early morning B1 carried 50 people in a few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on average 300 people cross the river using only one of the 3 boats B1, B2, and B3. How many trips it would take to B1 to carry passengers alone?

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Solution

To solve this problem, we need to first understand the total capacity of the three boats B1, B2, and B3.

Given that the three boats together can carry 60 people per trip, and B1 alone carried 50 people in a few trips, we can infer that B1 has a larger capacity than B2 and B3.

Since B1, B2, and B3 together carried 300 people in 10 trips, this means that they carried 30 people per trip on average.

Given that B1 carried 50 people in a few trips alone, and considering that B1, B2, and B3 together carried 30 people per trip, we can infer that B2 and B3 together have a capacity of 60 - 50 = 10 people per trip.

Now, if we want to know how many trips B1 would take to carry 300 people alone, we simply divide the total number of people by the capacity of B1.

So, 300 people / 50 people per trip = 6 trips.

Therefore, it would take B1 six trips to carry 300 people alone.

This problem has been solved

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