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 \[=10y+x\] \[=10\times 5+3\] \[=50+3\] \[=53\]
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