The arithmetic mean of the marks scored by student of a class is 58. 20 % of them secured a mean score of 60. The mean score of another 30% of the student was 40. The mean score of the remaining student is

Options

A. A. 65
B. B. 66
C. C. 68
D. D. 70

Answer

To solve the problem step by step, we will use the concept of weighted averages and the arithmetic mean.

### Step 1: Understand the given information
We know:
- The overall mean score of the class is 58.
- 20% of the students have a mean score of 60.
- 30% of the students have a mean score of 40.
- We need to find the mean score of the remaining students.

### Step 2: Define the percentages
Let:
- Total students = 100% (or 1)
- Students with mean score of 60 = 20% = 0.20
- Students with mean score of 40 = 30% = 0.30
- Remaining students = 100% - 20% - 30% = 50% = 0.50

### Step 3: Set up the equation using the mean
The overall mean can be expressed as:
\[ \text{Overall Mean} = (w_1 \cdot m_1 + w_2 \cdot m_2 + w_3 \cdot m_3) \]
Where:
- \(w_1 = 0.20\), \(m_1 = 60\)
- \(w_2 = 0.30\), \(m_2 = 40\)
- \(w_3 = 0.50\), \(m_3 = m\) (mean score of the remaining students)

Thus, we can write:
\[ 58 = (0.20 \cdot 60) + (0.30 \cdot 40) + (0.50 \cdot m) \]

### Step 4: Calculate the contributions of known groups
Calculate the contributions:
- Contribution from 20% of students:
\[ 0.20 \cdot 60 = 12 \]
- Contribution from 30% of students:
\[ 0.30 \cdot 40 = 12 \]

### Step 5: Substitute into the equation
Now substituting these values into the equation:
\[ 58 = 12 + 12 + (0.50 \cdot m) \]
\[ 58 = 24 + 0.50m \]

### Step 6: Solve for m
Rearranging the equation gives:
\[ 0.50m = 58 - 24 \]
\[ 0.50m = 34 \]
Now, divide both sides by 0.50:
\[ m = \frac{34}{0.50} = 68 \]

### Conclusion
The mean score of the remaining students is **68**.

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