The average age of a class is 35 yr. 6 new students with an average age of 33 yr joined in that class, there by decreasing the average by half year. The original strength of the class was

Options

A. 14
B. 18
C. 16
D. 20

Answer

To solve the problem step by step, we need to find the original strength of the class before the 6 new students joined.

### Step 1: Define Variables
Let the original number of students in the class be \( n \).

### Step 2: Calculate the Total Age of the Original Class
The average age of the class is 35 years. Therefore, the total age of the original class can be calculated as:
\[ \text{Total age of original class} = \text{Average age} \times \text{Number of students} = 35 \times n \]

### Step 3: Calculate the Total Age of the New Students
There are 6 new students with an average age of 33 years. Thus, the total age of the new students is:
\[ \text{Total age of new students} = \text{Average age} \times \text{Number of students} = 33 \times 6 = 198 \]

### Step 4: Calculate the New Total Age of the Class
After the new students join, the total age of the class becomes:
\[ \text{New total age} = \text{Total age of original class} + \text{Total age of new students} = 35n + 198 \]

### Step 5: Calculate the New Average Age of the Class
The new number of students in the class is \( n + 6 \). The new average age is given as:
\[ \text{New average age} = \frac{\text{New total age}}{\text{New number of students}} = \frac{35n + 198}{n + 6} \]
According to the problem, this new average age is 34.5 years (which is 35 - 0.5).

### Step 6: Set Up the Equation
We can set up the equation:
\[ \frac{35n + 198}{n + 6} = 34.5 \]

### Step 7: Cross-Multiply to Solve for \( n \)
Cross-multiplying gives:
\[ 35n + 198 = 34.5(n + 6) \]
Expanding the right-hand side:
\[ 35n + 198 = 34.5n + 207 \]

### Step 8: Rearrange the Equation
Rearranging the equation to isolate \( n \):
\[ 35n - 34.5n = 207 - 198 \]
\[ 0.5n = 9 \]

### Step 9: Solve for \( n \)
Dividing both sides by 0.5 gives:
\[ n = \frac{9}{0.5} = 18 \]

### Conclusion
The original strength of the class was \( n = 18 \).

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