Question

A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is $6$ more than half of the original number. Find the difference of the digits of the given number.

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

$4$

Let the number be $10x+y$, where $x$ is the digit at ten's place and $y$ is the digit at ones place.
The number formed in revering the digits is $10y+x$.
Given, $10x+y=7(x+y)$
$\Rightarrow 10x+y=7x+7y$
$\Rightarrow 3x-6y=0$
$\Rightarrow x-2y=0$ ....(1)
And $10y+x=6+\frac{(10x+y)}{2}$
$\Rightarrow 20y+2x=12+10x+y$
$\Rightarrow 8x-19y=-12$ ....(2)
Solving equations (1) and (2), we get
$x=8;y=4$
So, the difference of the digits is $8-4=4$.