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Let the ones digit of the original number be x and the tens digit of the original number be y. So, the original number is 10y+x.

Now, it is given that the two digit number is 3 more than 4 times the sum of its digits, so if we represent this mathematically, we get

$ 10y+x=3+4\left( x+y \right) $

$ \Rightarrow 10y+x=3+4x+4y $

$ \Rightarrow 6y-3x=3.............(i) $

It is also given that when the digits are reversed the number we get is 10x+y which is equal to 18 added to the original number. So, if we represent this as an equation, we get

$ 10y+x+18=10x+y $

$ 9x-9y=18...........(ii) $

Now, we will multiply equation (i) by 3 and add it with equation (ii). On doing so, we get

$ 9x-9y-3\left( 6y-3x \right)=18-3\times 3 $

\[\Rightarrow -9y+18y=18+9\]

\[\Rightarrow y=\dfrac{27}{9}=3\]

If we substitute y in equation (i), we get

$ 6y-3x=3 $

$ \Rightarrow 6\times 3-3x=3 $

$ \Rightarrow 18-3=3x $

$ \Rightarrow x=\dfrac{15}{3}=5 $

Hence, the original number is 10y+x=30+5=35.

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