If the vulgar fraction of `0.bar(41)` is given by `(41)/(999....9(" n times"))` then find the value of n?

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem, we need to convert the repeating decimal \(0.\overline{41}\) into a vulgar fraction and determine how many times the digit '9' appears in the denominator.

### Step-by-Step Solution:

1. **Let \( x = 0.\overline{41} \)**:
- We define \( x \) as the repeating decimal we want to convert.

2. **Multiply by 100**:
- Since the repeating part (41) has 2 digits, we multiply \( x \) by \( 100 \) to shift the decimal point two places to the right.
- Thus, \( 100x = 41.\overline{41} \).

3. **Set up the equations**:
- We now have two equations:
- \( x = 0.\overline{41} \) (Equation 1)
- \( 100x = 41.\overline{41} \) (Equation 2)

4. **Subtract Equation 1 from Equation 2**:
- We subtract Equation 1 from Equation 2:
\[ 100x - x = 41.\overline{41} - 0.\overline{41} \]
- This simplifies to:
\[ 99x = 41 \]

5. **Solve for \( x \)**:
- Divide both sides by 99:
\[ x = \frac{41}{99} \]

6. **Identify the denominator**:
- The problem states that the vulgar fraction of \( 0.\overline{41} \) is given by \( \frac{41}{999\ldots9} \) (with 'n' nines).
- We have found that \( x = \frac{41}{99} \).

7. **Determine the value of \( n \)**:
- The denominator \( 99 \) can be expressed as \( 999\ldots9 \) with 'n' nines.
- We know that \( 99 = 100 - 1 = 10^2 - 1 \), which means it has 2 nines.
- Therefore, \( n = 2 \).

### Final Answer:
The value of \( n \) is \( 2 \).