Let's define the two-digit number as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.

We are given two conditions:

1. The sum of the digits is 5: $a + b = 5$

2. When the digits are reversed, the new number is 27 greater than the original number: $10b + a = 10a + b + 27$

Step 1: Simplify the second equation

Start by simplifying the second equation: $10b + a = 10a + b + 27$ Subtract $b$ and $a$ from both sides: $9b - 9a = 27$ Now, divide through by 9: $b - a = 3$

Step 2: Solve the system of equations

Now we have a system of two equations:

1. $a + b = 5$
2. $b - a = 3$

Add the two equations together to eliminate $a$: $(a + b) + (b - a) = 5 + 3$ $2b = 8$ $b = 4$

Step 3: Find $a$

Substitute $b = 4$ into the first equation: $a + 4 = 5$ $a = 1$

Final Answer:

The original number is $10a + b = 10(1) + 4 = 14$, and the reversed number is $10b + a = 10(4) + 1 = 41$.

Thus, the two numbers are $14$ and $41$.

Would you like further details on any step?

Here are five related questions to explore:

1. What happens if the sum of the digits is 7 instead of 5?
2. How does the equation change if the new number is 35 greater than the original?
3. What if the digits differ by 4 instead of 3?
4. Can you generalize the method for any two-digit number where the digits are reversed?
5. What if the number is three digits instead of two?

Tip: When working with digit-related problems, translating the digits into algebraic terms (like $10a + b$) can make the relationships between numbers easier to work with.