Question

A number consists of two digits such that the digit in the unit’s place is more by 2 than the digit in the ten’s place. Three times the number added to $\frac{6}{7}$ times the number obtained by reversing the digits equals 108. Then the number is:

• A. 24
• B. 72
• C. 52
• D. 48

Answer 1

The correct option is A 24

Let the tens digit of the number be $x$.
Then the ones digit will be $x+2$

Therefore, the original number is
$10\left(x\right)+(x+2)=11x+2$

Number obtained by reversing the digits is
$10(x+2)+\left(x\right)=11x+20$

Given,
$3(11x+2)+\frac{6}{7}\times (11x+20)=108\frac{21(11x+2)+6(11x+20)}{7}=108231x+42+66x+120=756297x+162=756297x=594x=\frac{594}{297}=2$

Substituting this in this equation for the original number, we get,
$11x+2=11\left(2\right)+2=24$

Therefore, the original number is 24.