The problem written in the image states:

"A number consists of 2 digits. The digit in the 'Tens' place exceeds the 'Units' digit by 5. If 5 times the sum of the digits is subtracted from the number, the digits of the number are reversed. Find the number."

Let the number be $10x + y$, where:

• $x$ is the digit in the Tens place
• $y$ is the digit in the Units place

Step 1: Define the relationship between digits.

From the problem, the Tens digit exceeds the Units digit by 5: x = y + 5 \tag{1}

Step 2: Define the condition for the reversed number.

If 5 times the sum of the digits is subtracted from the original number, the digits of the number are reversed: (10x + y) - 5(x + y) = 10y + x \tag{2}

Step 3: Solve the system of equations.

Substitute $x = y + 5$ from equation (1) into equation (2): $(10(y + 5) + y) - 5((y + 5) + y) = 10y + (y + 5)$ Simplify the left-hand side: $(10y + 50 + y) - 5(2y + 5) = 10y + y + 5$ $(11y + 50) - 10y - 25 = 11y + 5$ $y + 25 = 11y + 5$ $25 - 5 = 11y - y$ $20 = 10y$ $y = 2$

Step 4: Find $x$.

From equation (1), $x = y + 5$, so: $x = 2 + 5 = 7$

Thus, the original number is $10x + y = 10(7) + 2 = 72$.

Final Answer:

The number is 72.

Would you like further explanation or have any other questions?

Related Questions:

1. How would the process change if the condition were different, such as if the difference between the digits was 4?
2. What if the subtraction condition involved multiplying the digits instead of summing them?
3. How would we solve this if the number had three digits instead of two?
4. Can you explain how reversing a three-digit number would affect the equation?
5. What mathematical strategies can be applied to solve similar digit-reversal problems?

Tip:

Always set up the problem with clear variables and ensure that all conditions are reflected in your equations. This will help avoid mistakes in complex word problems.