Let S be sample space for 5 digit numbers . If p is probability of a number being randomly selected which is multiple of 7 but not divisible by 5, then `9p` is equal to:

Options

A. 1.0146
B. 1.2085
C. 1.0285
D. 1.1521

Answer

To solve the problem, we need to find the probability \( p \) of selecting a 5-digit number that is a multiple of 7 but not divisible by 5. Then, we will calculate \( 9p \).

### Step-by-Step Solution:

1. **Identify the Range of 5-Digit Numbers**:
The smallest 5-digit number is 10,000 and the largest is 99,999.

2. **Calculate the Total Number of 5-Digit Numbers**:
\[ \text{Total 5-digit numbers} = 99,999 - 10,000 + 1 = 90,000 \]

3. **Find the Number of 5-Digit Multiples of 7**:
- The smallest multiple of 7 that is a 5-digit number:
\[ \text{Smallest} = \lceil \frac{10000}{7} \rceil \times 7 = 1429 \times 7 = 10,003 \]
- The largest multiple of 7 that is a 5-digit number:
\[ \text{Largest} = \lfloor \frac{99999}{7} \rfloor \times 7 = 14285 \times 7 = 99,995 \]
- The number of multiples of 7:
\[ n_7 = \frac{99995 - 10003}{7} + 1 = \frac{89992}{7} + 1 = 12856 + 1 = 12857 \]

4. **Find the Number of 5-Digit Multiples of 35** (which are multiples of both 7 and 5):
- The smallest multiple of 35 that is a 5-digit number:
\[ \text{Smallest} = \lceil \frac{10000}{35} \rceil \times 35 = 286 \times 35 = 10,010 \]
- The largest multiple of 35 that is a 5-digit number:
\[ \text{Largest} = \lfloor \frac{99999}{35} \rfloor \times 35 = 2857 \times 35 = 99,995 \]
- The number of multiples of 35:
\[ n_{35} = \frac{99995 - 10010}{35} + 1 = \frac{89985}{35} + 1 = 2571 + 1 = 2572 \]

5. **Calculate the Number of 5-Digit Numbers that are Multiples of 7 but Not 5**:
\[ n = n_7 - n_{35} = 12857 - 2572 = 10285 \]

6. **Calculate the Probability \( p \)**:
\[ p = \frac{n}{\text{Total 5-digit numbers}} = \frac{10285}{90000} \]

7. **Calculate \( 9p \)**:
\[ 9p = 9 \times \frac{10285}{90000} = \frac{92565}{90000} = 1.0285 \]

### Final Answer:
Thus, \( 9p = 1.0285 \).