There are three boats `B_1, B_2` and `B_3` working together they carry 60 people in each trip. One day an early morning carried 50 people in few trips alone. When it stopped carrying the passengers `B_2` and `B_3` started carrying the people together. It took a total of 10 trips to carry 300 people by `B_1, B_2` and `B_3`. It is known that each day on an average 300 people cross the river using only one of the 3 boats `B_1,B_2` and `B_3`. How many trips it would take to `B_1` to carry 150 passengers alone?

Options

A. 15
B. 30
C. 25
D. 10

Answer

To solve the problem step by step, we will break it down into manageable parts:

### Step 1: Understand the problem
We have three boats, \( B_1, B_2, \) and \( B_3 \), which together can carry 60 people in one trip. On a particular day, \( B_1 \) carried 50 people alone for some trips, and then \( B_2 \) and \( B_3 \) carried the remaining passengers together. In total, they carried 300 people in 10 trips.

### Step 2: Determine the contribution of each boat
Let’s denote the number of trips \( B_1 \) made as \( x \). Therefore, the number of trips made by \( B_2 \) and \( B_3 \) together will be \( 10 - x \).

- In \( x \) trips, \( B_1 \) carried 50 people. Thus, the average number of people carried by \( B_1 \) per trip is:
\[ \text{Average by } B_1 = \frac{50}{x} \]

- In \( 10 - x \) trips, \( B_2 \) and \( B_3 \) carried the remaining 250 people. Thus, the average number of people carried by \( B_2 \) and \( B_3 \) per trip is:
\[ \text{Average by } B_2 + B_3 = \frac{250}{10 - x} \]

### Step 3: Set up the equation
When all three boats work together, they carry an average of 60 people per trip. Therefore, we can set up the equation:
\[ \frac{50}{x} + \frac{250}{10 - x} = 60 \]

### Step 4: Solve the equation
To solve for \( x \), we will multiply through by \( x(10 - x) \) to eliminate the denominators:
\[ 50(10 - x) + 250x = 60x(10 - x) \]
Expanding both sides:
\[ 500 - 50x + 250x = 600x - 60x^2 \]
Combining like terms:
\[ 500 + 200x = 600x - 60x^2 \]
Rearranging gives:
\[ 60x^2 - 400x + 500 = 0 \]

### Step 5: Simplify the quadratic equation
Dividing the entire equation by 20:
\[ 3x^2 - 20x + 25 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3, b = -20, c = 25 \)
\[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot 25}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{20 \pm \sqrt{400 - 300}}{6} = \frac{20 \pm \sqrt{100}}{6} = \frac{20 \pm 10}{6} \]
This gives us two possible values for \( x \):
\[ x = \frac{30}{6} = 5 \quad \text{or} \quad x = \frac{10}{6} \approx 1.67 \text{ (not feasible)} \]

### Step 7: Calculate average passengers carried by \( B_1 \)
With \( x = 5 \):
\[ \text{Average passengers carried by } B_1 = \frac{50}{5} = 10 \text{ passengers per trip} \]

### Step 8: Determine trips needed for 150 passengers
To find out how many trips \( B_1 \) needs to carry 150 passengers:
\[ \text{Number of trips} = \frac{150}{10} = 15 \text{ trips} \]

### Final Answer
Thus, \( B_1 \) would take **15 trips** to carry 150 passengers alone.

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