Step 1: Understanding the Concept:
This is a weighted average or mixture problem. We can solve it by setting up a system of linear equations representing the total number of tickets and the total revenue.
Step 2: Detailed Explanation:
Let \(A\) be the number of adult tickets sold and \(C\) be the number of children's tickets sold.
Equation for the total number of tickets:
\[ A + C = 1000 \]
Equation for the total revenue:
The total revenue is the average price per ticket multiplied by the total number of tickets.
\[ \text{Total Revenue} = \$42 \times 1000 = \$42,000 \]
The total revenue can also be expressed as the sum of the revenue from adult and children tickets:
\[ 50A + 30C = 42,000 \]
Now we have a system of two equations:
1) \( A + C = 1000 \)
2) \( 50A + 30C = 42,000 \)
We need to solve for \(C\). From equation (1), express \(A\) in terms of \(C\):
\[ A = 1000 - C \]
Substitute this expression for \(A\) into equation (2):
\[ 50(1000 - C) + 30C = 42,000 \]
\[ 50,000 - 50C + 30C = 42,000 \]
\[ 50,000 - 20C = 42,000 \]
Subtract 42,000 from both sides:
\[ 8,000 = 20C \]
Divide by 20:
\[ C = \frac{8000}{20} = 400 \]
Step 3: Final Answer:
There were 400 tickets sold for children. This corresponds to option (C).