The correct answer is option (A):
5
Let's break down this problem. We need to find two-digit numbers that, when their digits are reversed, result in a number that is 27 greater than the original number.
Let's represent the original two-digit number as 10a + b, where 'a' is the tens digit and 'b' is the units digit. When we reverse the digits, we get the number 10b + a.
The problem states that the reversed number is 27 greater than the original number. So, we can write the equation:
10b + a = 10a + b + 27
Now, let's simplify this equation:
9b - 9a = 27
Divide both sides by 9:
b - a = 3
This means the difference between the units digit (b) and the tens digit (a) must be 3.
Now, we need to find all possible pairs of digits (a, b) that satisfy this condition, keeping in mind that 'a' and 'b' are digits from 0 to 9, and 'a' cannot be 0 if the number is two-digit (otherwise it's single digit).
Let's list the possibilities:
* If a = 1, then b = 1 + 3 = 4. The number is 14, and its reverse is 41. (41 - 14 = 27). This matches the initial example.
* If a = 2, then b = 2 + 3 = 5. The number is 25, and its reverse is 52. (52 - 25 = 27).
* If a = 3, then b = 3 + 3 = 6. The number is 36, and its reverse is 63. (63 - 36 = 27).
* If a = 4, then b = 4 + 3 = 7. The number is 47, and its reverse is 74. (74 - 47 = 27).
* If a = 5, then b = 5 + 3 = 8. The number is 58, and its reverse is 85. (85 - 58 = 27).
* If a = 6, then b = 6 + 3 = 9. The number is 69, and its reverse is 96. (96 - 69 = 27).
If we try a = 7, 8, or 9, then b would be too large and exceed 9 (e.g. 7+3 = 10 and 10 is not a single digit). We cannot have a = 0 either, as the original problem specifies two-digit numbers.
We've found 5 other two-digit numbers, besides the original 14, that meet the condition. (25, 36, 47, 58, and 69).
Therefore, the correct answer is 5.