• # question_answer The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.

 Let unit digit $=x$ Tens digit $=y$ So, original number $=\text{unit digit}+10\times \text{tens digit}$ $1=x+10y$ According to question, Sum of digits $=8$ so,        $x+y=8$                                                         ...(i) On reversing the digits, unit digit $=y$, Tens digit $=x$ So,   New number $=10x+y$ According to question, Difference $=18$ $\Rightarrow \,x+10y-(10x+y)=18$ $\Rightarrow \,x+10y-10x-y=18$ $\Rightarrow \,9y-9x=18$ $\Rightarrow \,y-x=2$                                                    ...(ii) By adding eq. (i) and (ii) $2y=10$ $y=\frac{10}{2}\Rightarrow y=5$ Put the value of y in eq. (i) $x+y=8$ $\Rightarrow x+5=8$ $\Rightarrow x=8-5$ $\Rightarrow x=3$ $\therefore$ Original number $=10y+x$ $=10\times 5+3$ $=50+3$ $=53$