263 views

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. $63$

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.

626 points

1 vote
1
329 views
1 vote
2
420 views
1 vote