Question

In a class with a certain number of students, if one student weighing 50 kg is added, then the average weight of the class increases by 1 kg. If one more student weighing 50 kg is added, then the average weight of the class increases by 1.5 kg over the original average. What is the original average weight (in kg) of the class?

• A. 46
• B. 4
• C. 2
• D. 47

Hint:

When dealing with average-related problems, always use the formula for the average and create equations based on the changes in the total and number of items.

Answer 1

The Correct Option is D

Let \( x \) be the original average weight and \( n \) be the number of students in the class. Let \( x_1, x_2, x_3, \dots, x_n \) represent the weights of the \( n \) students respectively. The total weight of the class is given by: \[ x_1 + x_2 + x_3 + \dots + x_n = n \cdot x \] When one student weighing 50 kg is added, the new average increases by 1 kg, so: \[ \frac{n \cdot x + 50}{n + 1} = x + 1 \quad \text{(1)} \] Now, adding another student weighing 50 kg, the average increases by 1.5 kg, so: \[ \frac{n \cdot x + 100}{n + 2} = x + 1.5 \quad \text{(2)} \] Solving equation (1) for \( n \), we get: \[ n \cdot x + 50 = (n + 1)(x + 1) \] Expanding: \[ n \cdot x + 50 = n \cdot x + x + n + 1 \] Simplifying: \[ 50 = x + n + 1 \] \[ x + n = 49 \quad \text{(3)} \] Solving equation (2) for \( n \), we get: \[ n \cdot x + 100 = (n + 2)(x + 1.5) \] Expanding: \[ n \cdot x + 100 = (n + 2)(x + 1.5) \] Simplifying: \[ n \cdot x + 100 = n \cdot x + 1.5n + 3 + 2x + 3 \] Simplifying further: \[ 100 = 1.5n + 2x + 6 \] \[ 94 = 1.5n + 2x \quad \text{(4)} \] Solving equations (3) and (4), we get: \[ x = 47 \] Thus, the original average weight of the class is \( 47 \) kg.