Question

The sum of a number of 2 digits and of the number formed by reversing the digits is 110,and the difference of the digits is 6. Find the number.

Answer 1

Solution :

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

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

A/q

$\longrightarrow\sf{r-m=6}\\\\\longrightarrow\sf{r=6+m....................(1)}$

&

$\longrightarrow\sf{(10r+m)+(10m+r)=110}\\\\\longrightarrow\sf{10r+m+10m+r=110}\\\\\longrightarrow\sf{11r+11m=110}\\\\\longrightarrow\sf{11(r+m)=110}\\\\\longrightarrow\sf{r+m=\cancel{110/11}}\\\\\longrightarrow\sf{r+m=10}\\\\\longrightarrow\sf{6+m+m=10\:\:[from(1)]}\\\\\longrightarrow\sf{6+2m=10}\\\\\longrightarrow\sf{2m=10-6}\\\\\longrightarrow\sf{2m=4}\\\\\longrightarrow\sf{m=\cancel{4/2}}\\\\\longrightarrow\bf{m=2}$

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

$\longrightarrow\sf{r=6+2}\\\\\longrightarrow\bf{r=8}$

Thus;

$\boxed{\sf{The\:number=10r+m=[10(8) + 2 ] = [80 + 2]=\bf{82}}}}$

Answer 2

Your answer is in the attachment

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