There is a three digit number such that the sum of its end digits (unit digit and hundredth place digit) is always a single digit number. Another three digit number is obtained by reversing the position of end digits of the original number. Then what can be the possible sum of the tens digits of both these numbers ?

Options

A. 5
B. 12
C. 15
D. 26

Answer

To solve the problem step by step, let's break it down:

### Step 1: Define the three-digit number
Let the three-digit number be represented as \( xyz \), where:
- \( x \) is the hundreds place digit,
- \( y \) is the tens place digit,
- \( z \) is the units place digit.

### Step 2: Understand the condition on the end digits
According to the problem, the sum of the end digits (hundreds place \( x \) and units place \( z \)) must be a single-digit number. This can be expressed mathematically as:
\[ x + z < 10 \]
This means \( x + z \) can take values from 0 to 9.

### Step 3: Form the reversed number
When we reverse the end digits, the new number becomes \( zyx \). In this number:
- The hundreds place is now \( z \),
- The tens place remains \( y \),
- The units place is now \( x \).

### Step 4: Identify the tens digits
In both numbers \( xyz \) and \( zyx \), the tens digit is \( y \) for both.

### Step 5: Calculate the sum of the tens digits
The sum of the tens digits of both numbers is:
\[ y + y = 2y \]

### Step 6: Determine the possible values for \( 2y \)
Since \( y \) is a digit, it can take values from 0 to 9. Therefore, \( 2y \) can take the following values:
- If \( y = 0 \), then \( 2y = 0 \)
- If \( y = 1 \), then \( 2y = 2 \)
- If \( y = 2 \), then \( 2y = 4 \)
- If \( y = 3 \), then \( 2y = 6 \)
- If \( y = 4 \), then \( 2y = 8 \)
- If \( y = 5 \), then \( 2y = 10 \)
- If \( y = 6 \), then \( 2y = 12 \)
- If \( y = 7 \), then \( 2y = 14 \)
- If \( y = 8 \), then \( 2y = 16 \)
- If \( y = 9 \), then \( 2y = 18 \)

### Step 7: Identify valid sums
However, since \( 2y \) must also be a single-digit number, the possible values for \( 2y \) are limited to:
- 0, 2, 4, 6, 8, 10 (not valid as a digit), 12, 14, 16, 18 (not valid as a digit).

From the above, the valid sums that are single-digit numbers are:
- 0, 2, 4, 6, 8, 10 (but 10 is not a single digit).

### Conclusion
The only valid sums that can be derived from \( 2y \) under the constraints given in the problem are:
- 0, 2, 4, 6, 8, 12, 14, 16, 18.

However, since \( y \) must be a digit (0-9), the only possible valid sum of the tens digits of both numbers is \( 12 \).

### Final Answer
Thus, the possible sum of the tens digits of both numbers is:
\[ \boxed{12} \]