Question

When a number consisting of two digits is multiplied by $9,$ it becomes equal to twice the number obtained by reversing the digits of the original number. If the difference of two digits of the number is $7,$ find the number.

Answer 1

Let a two digit number be $10x+y$ in which x is tens place digit whereas y is units place digit.
According to question
$9(10x+y)=2(10y+x)$
$\Rightarrow 90x+9y=20y+2x$
$\Rightarrow 88x=11y$
$\Rightarrow 8x-y=0$ ......(i)
and $x-y=-7$ ....(ii) [$\therefore$ Reject +ve value]
Subtract eq(ii) from (i)

$7x=7$

we get $x=1$

From eq. (i)

$1-y=-7$

and $y=8$
Hence, required number is $10x+y=10\times 1+8=18$