Discussion
When a two-digit number 'X' is reversed, the number so formed is 63 more than 'X'. If the sum of digits of 'X' is 11, what is the value of 'X + 21' ?
Answer: D
X + 21 = 50.
Let X have tens digit \(a\) and units digit \(b\). Reversing adds 63:
\[ \begin{aligned} (10b + a) - (10a + b) &= 9(b - a) = 63 \\ b - a &= 7,\quad a + b = 11 \\ a &= 2,\ b = 9 \end{aligned} \]
So X = 29 and X + 21 = 50. The option 29 is X itself, not X + 21, and 113 adds 21 to the reversed number 92.
Check: 92 - 29 = 63 and 2 + 9 = 11. In general, reversing a two-digit number changes it by 9 times the difference of its digits, a useful shortcut for such questions.
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