Question

the sum of the two - digits numbers is 9. Also nine times this number is twice the number obtained by reversing the order of the digits . Find the number

Answer 1

S O L U T I O N :

Let the ten's digit number be r

Let the one's digit number be m

$\boxed{\bf{The\:original\:number\:=10r+m}}}}}\\\boxed{\bf{The\:reversed\:number\:=10m+r}}}}}$

A/q

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

&

$\longrightarrow\rm{9(10r+m)=2(10m+r)}\\\\\longrightarrow\rm{90r+9m=20m+2r}\\\\\longrightarrow\rm{90r-2r=20m-9m}\\\\\longrightarrow\rm{\cancel{88}r=\cancel{11}m}\\\\\longrightarrow\rm{m=8r...........(2)}$

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

$\longrightarrow\rm{r=9-8r}\\\\\longrightarrow\rm{r+8r=9}\\\\\longrightarrow\rm{9r=9}\\\\\longrightarrow\rm{r=\cancel{9/9}}\\\\\longrightarrow\bf{r=1}$

Putting the value of r in equation (2),we get;

$\longrightarrow\rm{m=8(1)}\\\\\longrightarrow\rm{m=8\times 1}\\\\\longrightarrow\bf{m=8}$

Thus;

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