The average score of S students of a class in a Physics test was 2S . If the lowest three scores were excluded, the average score of the class would have increased by three. Which of the following gives the correct expression of the average score of these three students?

Options

A. S - 3
B. S + 3
C. 2S -3
D. 2S + 3

Answer

To solve the problem step by step, we will derive the average score of the three lowest students based on the information provided.

### Step 1: Understand the initial conditions
We know that the average score of S students is given as \(2S\). Therefore, the total score of all students can be expressed as:
\[ \text{Total Score} = \text{Average} \times \text{Number of Students} = 2S \times S = 2S^2 \]

### Step 2: Exclude the lowest three scores
Let’s denote the scores of the students as \(X_1, X_2, X_3, \ldots, X_S\). The lowest three scores are \(X_1, X_2, X_3\). When we exclude these three scores, the total score of the remaining students becomes:
\[ \text{Total Score after excluding three} = 2S^2 - (X_1 + X_2 + X_3) \]
The number of remaining students is \(S - 3\).

### Step 3: Determine the new average
According to the problem, the new average after excluding the three lowest scores increases by 3. Therefore, the new average can be expressed as:
\[ \text{New Average} = 2S + 3 \]
Using the formula for average, we have:
\[ \frac{2S^2 - (X_1 + X_2 + X_3)}{S - 3} = 2S + 3 \]

### Step 4: Set up the equation
Now, we can set up the equation based on the new average:
\[ 2S^2 - (X_1 + X_2 + X_3) = (2S + 3)(S - 3) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side gives:
\[ (2S + 3)(S - 3) = 2S^2 - 6S + 3S - 9 = 2S^2 - 3S - 9 \]
Thus, our equation now looks like:
\[ 2S^2 - (X_1 + X_2 + X_3) = 2S^2 - 3S - 9 \]

### Step 6: Solve for the sum of the lowest three scores
Rearranging the equation to isolate \(X_1 + X_2 + X_3\):
\[ -(X_1 + X_2 + X_3) = -3S - 9 \]
This simplifies to:
\[ X_1 + X_2 + X_3 = 3S + 9 \]

### Step 7: Calculate the average of the three lowest scores
To find the average score of the three lowest students, we divide their total score by 3:
\[ \text{Average of } X_1, X_2, X_3 = \frac{X_1 + X_2 + X_3}{3} = \frac{3S + 9}{3} = S + 3 \]

### Final Answer
The average score of the three lowest students is:
\[ \boxed{S + 3} \]