Question

Two times a two-digit number is 9 times the number obtained by reversing the digits and the sum of the digits is 9. The number is _____.

• A. 54
• B. 72
• C. 63
• D. 81

Hint:

When solving such problems, express the number in terms of its digits and use the given conditions to set up equations.

Answer 1

The Correct Option is D

Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit. The number obtained by reversing the digits is \( 10b + a \). We are given two conditions:
1. \( 2(10a + b) = 9(10b + a) \)
2. \( a + b = 9 \)
Solving the first equation: \[ 2(10a + b) = 9(10b + a) \quad \Rightarrow \quad 20a + 2b = 90b + 9a \] \[ 20a - 9a = 90b - 2b \quad \Rightarrow \quad 11a = 88b \quad \Rightarrow \quad a = 8b \] Substitute \( a = 8b \) into \( a + b = 9 \): \[ 8b + b = 9 \quad \Rightarrow \quad 9b = 9 \quad \Rightarrow \quad b = 1 \] So, \( a = 8 \). The number is \( 10a + b = 10(8) + 1 = 81 \). Thus, the number is 81.