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's place. Three times the number added to 6/7 times the number obtained by reversing the digits equals 108. The sum of the digits in the number is :

Answer: Option A

Let the unit's digit be x
Then, ten's digit = (x - 2)
$$\therefore 3\left[10\left(x-2\right)+x\right]+\frac{6}{7}$$     $$\left[10x+\left(x-2\right)\right]$$   $$=108$$
⇔ 231x - 420 + 66x - 12 = 756
⇔ 297x = 1188
⇔ x = 4
Hence, sum of the digits :
= x + (x - 2)
= 2x - 2
= 6