A number consists of two digits such that the digit in the ten's place is less by 2 than the digit in the unit place. Three times the number added to $$\frac{6}{7}$$ times the number obtained by reversing digits equals 108. The sum of digits in the number is -

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let the unit digit }} \cr & {\text{Ten digit }} = x - 2 \cr & \therefore {\text{ Number :}} \cr & = {\text{ }}10(x - 2){\text{ }} + {\text{ }}x \cr & = {\text{ }}10x{\text{ }} - {\text{ }}20{\text{ }} + {\text{ }}x \cr & = {\text{ }}11x{\text{ }} - {\text{ }}20 \cr} $$
New number obtained after reversing the digits
$$\eqalign{ & = 10x + x - 2{\text{ }} \cr & = {\text{ }}11x - 2 \cr & {\text{According to the question,}} \cr & {\text{3}}\left( {11x - 20} \right) + \frac{6}{7}\left( {11x - 2} \right) = 108 \cr & 7\left( {11x - 20} \right) + 2\left( {11x - 2} \right) = 36 \times 7 \cr & 77x - 140 + 22x - 4 = 252 \cr & 99x = 252 + 144 \cr & x = \frac{{396}}{{99}} = 4 \cr & {\text{Number :}} \cr & = {\text{ }}11x - 20{\text{ }} \cr & = {\text{ }}11 \times 4 - 20 \cr & = {\text{ }}24 \cr & {\text{Sum of digit }} = 2 + 4 = 6 \cr} $$