The sum of the digits of a two-digit number is 9 less than the number. Which of the following digits is at unit's place of the number?

### Concept & Logic Evaluating algebraic constraints to see if they yield a unique solution. When variables cancel out or leave multiple possibilities open, the data is insufficient. ### Step-by-Step Solution Let the ten's digit be $x$ and the unit's digit be $y$. The numerical value of the number is $10x + y$. The sum of its digits is $x + y$. According to the given condition: $\text{Sum of digits} = \text{Number} - 9$ $(x + y) = (10x + y) - 9$ Simplify the equation by subtracting $y$ from both sides: $x = 10x - 9$ $9x = 9$ $x = 1$ The algebra tells us definitively that the ten's digit ($x$) must be 1. The number is in the 10s (e.g., 10, 11, 12, ... 19). However, the unit's digit ($y$) completely canceled out of the equation. This means $y$ can be ANY digit from 0 to 9, and the original condition will still hold true. Since we cannot determine a single, unique value for the unit's place, the information provided is not enough. ### Exam Strategy & Shortcut Test numbers in the 10s to see if they fit. Try 12: Sum = 3. Number - 9 = $12 - 9 = 3$. (Matches) Try 15: Sum = 6. Number - 9 = $15 - 9 = 6$. (Matches) Since both 12 and 15 satisfy the condition, the unit digit could be 2, 5, or any other digit. Therefore, the answer is immediately "Data inadequate". ### Common Pitfall Solving for $x = 1$ and mistakenly thinking this 1 represents the unit's digit, thereby choosing option (a). It is crucial to remember which variable represents which place value. ### Final Answer Therefore, the correct answer is **Data inadequate**.