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(B) Let the number be $10x + y$,where $x$ is the ten's digit and $y$ is the unit's digit.According to the problem,$x = y - 2$.The number obtained by r

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$6$

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(B) Let the number be $10x + y$,where $x$ is the ten's digit and $y$ is the unit's digit.According to the problem,$x = y - 2$.The number obtained by reversing the digits is $10y + x$.Given the equation: $3(10x + y) + \frac{6}{7}(10y + x) = 108$.Substitute $x = y - 2$ into the equation:$3(10(y - 2) + y) + \frac{6}{7}(10y + (y - 2)) = 108$$3(11y - 20) + \frac{6}{7}(11y - 2) = 108$$33y - 60 + \frac{66y - 12}{7} = 108$Multiply the entire equation by $7$ to clear the denominator:$7(33y - 60) + 66y - 12 = 756$$231y - 420 + 66y - 12 = 756$$297y - 432 = 756$$297y = 1188$$y = \frac{1188}{297} = 4$.Since $x = y - 2$,then $x = 4 - 2 = 2$.The number is $24$.The sum of the digits is $x + y = 2 + 4 = 6$.

$A$ 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:

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