Question:
In a class of 150 students, the average weight is 42 kg. If the number of girls increases by 50% and the number of boys becomes two-thirds of its original value, while keeping the total number of students the same, the new average weight of the class becomes 43 kg. It is also observed that the average weight of boys and girls remains the same as before. Find the sum of the average weights of boys and girls.
In a class of 150 students, the average weight is 42 kg. If the number of girls increases by 50% and the number of boys becomes two-thirds of its original value, while keeping the total number of students the same, the new average weight of the class becomes 43 kg. It is also observed that the average weight of boys and girls remains the same as before. Find the sum of the average weights of boys and girls.
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In weighted average problems, always start by setting up equations for the total count and the total sum (or weight). When dealing with changes, create a new set of equations and solve the system. It's often helpful to simplify equations by dividing by a common factor.
Updated On: Oct 2, 2025
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The Correct Option is C
Solution and Explanation
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.
\[ W_B + W_G = 40 + 45 = 85 \text{ kg} \]
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.
\[ W_B + W_G = 40 + 45 = 85 \text{ kg} \]
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