Question

Q4.The sum of the digits of a two-digit number is 9 .Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer 1

Solution :

Let the ten's place digit be r & let the one's place digit be m respectively;

$\boxed{\bf{Original\:number=10r+m}}}}}\\\boxed{\bf{Reversed\:number=10m+r}}}}}$

A/q

$\longrightarrow\sf{r+m=9}\\\\\longrightarrow\sf{r=9-m...................(1)}$

&

$\longrightarrow\sf{9(10r+m) = 2(10m+r)}\\\\\longrightarrow\sf{90r + 9m = 20m + 2r}\\\\\longrightarrow\sf{90r - 2r + 9m - 20m = 0}\\\\\longrightarrow\sf{88r - 11m = 0}\\\\\longrightarrow\sf{88(9-m) -11m=0\:\:[from(1)]}\\\\\longrightarrow\sf{792 - 88m - 11m = 0}\\\\\longrightarrow\sf{792 - 99m = 0}\\\\\longrightarrow\sf{792 = 99m}\\\\\longrightarrow\sf{m= \cancel{792/99}}\\\\\longrightarrow\bf{m=8}$

Putting the value of m in equation (1),we get;

$\longrightarrow\sf{r=9-8}\\\\\longrightarrow\bf{r=1}$

Thus;

$\boxed{\sf{The\:number=10r+m=[10(1)+8]=[10 + 8] = \boxed{\bf{18}}}}}$