Express each of the following recurring decimals into the rational number :
`(i)6.bar(315)" "(ii)7.bar(1641)`

Explanation

To convert the recurring decimals into rational numbers, we will follow a systematic approach for each case.

### (i) Converting \( 6.\overline{315} \) into a rational number:

1. **Let \( x = 6.\overline{315} \)**.
- This means \( x = 6.315315315...\)

2. **Multiply \( x \) by \( 1000 \)** (since the repeating part has 3 digits):
\[ 1000x = 6315.315315315... \]

3. **Set up the equations**:
- Equation 1: \( x = 6.315315315... \)
- Equation 2: \( 1000x = 6315.315315315... \)

4. **Subtract Equation 1 from Equation 2**:
\[ 1000x - x = 6315.315315315... - 6.315315315... \]
This simplifies to:
\[ 999x = 6315.315315315... - 6.315315315... = 6309 \]

5. **Solve for \( x \)**:
\[ x = \frac{6309}{999} \]

6. **Simplify the fraction** (if possible):
- Both 6309 and 999 can be divided by 9:
\[ x = \frac{6309 \div 9}{999 \div 9} = \frac{701}{111} \]

Thus, the rational number representation of \( 6.\overline{315} \) is \( \frac{701}{111} \).

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### (ii) Converting \( 7.\overline{1641} \) into a rational number:

1. **Let \( x = 7.\overline{1641} \)**.
- This means \( x = 7.16411641...\)

2. **Multiply \( x \) by \( 10000 \)** (since the repeating part has 4 digits):
\[ 10000x = 71641.16411641... \]

3. **Set up the equations**:
- Equation 1: \( x = 7.16411641... \)
- Equation 2: \( 10000x = 71641.16411641... \)

4. **Subtract Equation 1 from Equation 2**:
\[ 10000x - x = 71641.16411641... - 7.16411641... \]
This simplifies to:
\[ 9999x = 71641.16411641... - 7.16411641... = 71634 \]

5. **Solve for \( x \)**:
\[ x = \frac{71634}{9999} \]

6. **Simplify the fraction** (if possible):
- Both 71634 and 9999 can be divided by 9:
\[ x = \frac{71634 \div 9}{9999 \div 9} = \frac{7960}{1111} \]

Thus, the rational number representation of \( 7.\overline{1641} \) is \( \frac{7960}{1111} \).

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