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.

Answer:

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 \[=10y+x\]

\[=10\times 5+3\]

\[=50+3\]

\[=53\]