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 $$\frac{6}{7}$$ times the number obtained by reversing the digits equals 108. The sum of the digits in the number is :

A. 6

B. 7

C. 8

D. 9

Answer: Option A

Solution(By Examveda Team)

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