A typist starts to type the serial numbers of candidates in a list, upto 500. Minimum how many times he has to press the keys of numerals only ?

Options

A. 1389
B. less than 1000
C. 1392
D. can't say

Answer

To solve the problem of how many times the typist has to press the keys of numerals only while typing the serial numbers of candidates from 1 to 500, we can break it down into steps based on the number of digits in each range of numbers.

### Step-by-Step Solution:

1. **Identify the Ranges**:
- The numbers can be divided into three ranges:
- **1 to 9** (1-digit numbers)
- **10 to 99** (2-digit numbers)
- **100 to 500** (3-digit numbers)

2. **Count the Numbers in Each Range**:
- **1 to 9**: There are 9 numbers (1, 2, 3, 4, 5, 6, 7, 8, 9).
- **10 to 99**: To find the count, calculate \(99 - 10 + 1 = 90\) numbers (10, 11, ..., 99).
- **100 to 500**: To find the count, calculate \(500 - 100 + 1 = 401\) numbers (100, 101, ..., 500).

3. **Calculate the Total Digits for Each Range**:
- **1 to 9**: Each of these 9 numbers has 1 digit, so total digits = \(9 \times 1 = 9\).
- **10 to 99**: Each of these 90 numbers has 2 digits, so total digits = \(90 \times 2 = 180\).
- **100 to 500**: Each of these 401 numbers has 3 digits, so total digits = \(401 \times 3 = 1203\).

4. **Sum the Total Digits**:
- Now, add the total digits from all ranges:
\[ 9 + 180 + 1203 = 1392 \]

5. **Conclusion**:
- The minimum number of times the typist has to press the keys of numerals only is **1392**.

### Final Answer:
The typist has to press the keys a minimum of **1392** times.

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