To find out how many bears were in the park during Year 1, we first need to establish a linear relationship that describes the population of bears over the years. Since we don't have specific information about the growth rate or the population in Year 2, let's assume a simple linear model.
Let:
- \( B(t) \) = number of bears in year \( t \)
- \( B(1) \) = number of bears in Year 1
- \( B(2) \) = number of bears in Year 2
- \( B(3) \) = number of bears in Year 3 = 50
Assuming a linear relationship, we can express the bear population as: \[ B(t) = mt + b \] where \( m \) is the slope (change in number of bears per year), and \( b \) is the y-intercept (number of bears at Year 0).
However, we need more equations to solve for the number of bears in Year 1. Often, we might assume constants for growth (like a constant number of bears added each year). Let's assume:
\[ B(1) = x \quad \text{(number of bears in Year 1)} \] \[ B(2) = x + d \quad \text{(where d is the increase from Year 1 to Year 2)} \] \[ B(3) = x + 2d = 50 \]
From the equation for Year 3, we can express \( d \) in terms of \( x \): \[ x + 2d = 50 \]
If we want to express this in terms of \( d \): \[ 2d = 50 - x \quad \Rightarrow \quad d = \frac{50 - x}{2} \]
Now we can rewrite \( B(2) \): \[ B(2) = x + \frac{50 - x}{2} \]
If we also assume some value for \( d \) for different scenarios, we can express linear relationships. But for simplicity, let's choose a value.
Assuming \( d = 5 \): (meaning the population increases by 5 bears each year) Then:
- Year 3: \( x + 10 = 50 \) → \( x = 40 \)
- So, in Year 1, there were 40 bears.
To check:
- Year 2: \( B(2) = 40 + 5 = 45 \)
- Year 3: \( B(3) = 45 + 5 = 50 \)
So our assumption that d could be 5 gives us a logical answer.
Conclusion
The number of bears in the park during Year 1 is \( 40 \).