Question

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 $108$. What is the sum of the digits in the number?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (C)

$6$

Let the unit's digit be $x$

Then ten's digits $=x-2$
Therefore, original number $=10(x-2)+x=11x-20$
After reversing the digits the number is
i.e $10x+(x-2)=11x-2$
Given, $3(11x-20)+\frac{6}{7}(11x-2)=108$

$\Rightarrow 33x-60+\frac{66x-12}{7}=108$
$\Rightarrow 231x-420+66x-12=108\times 7=756$

$\Rightarrow 297x=1188$

$\Rightarrow x=4$
Thus sum of digits $=x+x-2=4+4-2=6$