Step 1: Understanding the Concept:
This problem involves the concept of weighted averages. We need to set up a system of linear equations based on the information given for the initial and final states of the class composition and average weights.
Step 2: Detailed Explanation:
Let the initial number of boys be \(B\) and girls be \(G\).
Total students = 150, so \(B + G = 150\) --- (1)
After the change:
New number of boys, \(B' = \frac{2}{3}B\).
New number of girls, \(G' = G + 0.5G = 1.5G\).
The total number of students remains the same.
\[ B' + G' = \frac{2}{3}B + 1.5G = 150 \]
\[ \frac{2}{3}B + \frac{3}{2}G = 150 \]
Multiplying by 6 to clear the fractions:
\[ 4B + 9G = 900 \text{ --- (2)} \]
Now, we solve the system of equations (1) and (2).
From (1), \(B = 150 - G\). Substitute this into (2):
\[ 4(150 - G) + 9G = 900 \]
\[ 600 - 4G + 9G = 900 \]
\[ 5G = 300 \implies G = 60 \]
Then, \(B = 150 - 60 = 90\).
So, initially, there were 90 boys and 60 girls.
Let \(W_B\) be the average weight of boys and \(W_G\) be the average weight of girls. These remain constant.
Initial total weight = \(150 \times 42 = 6300\) kg.
\[ 90W_B + 60W_G = 6300 \]
Dividing by 30, we get:
\[ 3W_B + 2W_G = 210 \text{ --- (3)} \]
New number of students: \(B' = \frac{2}{3}(90) = 60\), \(G' = 1.5(60) = 90\).
New average weight of the class is 43 kg.
New total weight = \(150 \times 43 = 6450\) kg.
\[ 60W_B + 90W_G = 6450 \]
Dividing by 30, we get:
\[ 2W_B + 3W_G = 215 \text{ --- (4)} \]
Now we solve the system of equations (3) and (4) for \(W_B\) and \(W_G\).
Multiply (3) by 3 and (4) by 2:
\[ 9W_B + 6W_G = 630 \]
\[ 4W_B + 6W_G = 430 \]
Subtracting the second new equation from the first:
\[ 5W_B = 200 \implies W_B = 40 \text{ kg} \]
Substitute \(W_B = 40\) into (3):
\[ 3(40) + 2W_G = 210 \]
\[ 120 + 2W_G = 210 \]
\[ 2W_G = 90 \implies W_G = 45 \text{ kg} \]
Step 3: Final Answer:
The question asks for the sum of the average weights of boys and girls.
\[ W_B + W_G = 40 + 45 = 85 \text{ kg} \]