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A certain number consists of two digits whose sum is $9$. It the order of digits is reversed, the new number is $9$ less than the original number. The original number is :

  1. $45$
  2. $36$
  3. $54$
  4. $63$
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Let digits of the number are $a$ and $b$ at tenth and ones place respectively.

Such that required number is $10a +b$

Given that $a+b = 9 \qquad \qquad \rightarrow(i)$

and if we replace the digits then the new number is 9 less than the original number,

$\begin{align} 10b + a +9 &= 10a+b \\ \implies 9a – 9b &= 9 \\ \implies ~~ a- b &=1 \qquad \qquad \rightarrow(ii)\end{align} $

$equation~(i)+(ii)~gives, \\ \implies 2a = 10 \implies a=5 \\ \> \\ equation~(i)-(ii)~gives, \\ \implies 2b = 8 \implies b=4 $

Hence, required number is $10\times 5 + 4 = 54$

Option (C) is correct.

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