Question

A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchange their places, find the number :

• A. $62$
• B. $34$
• C. $26$
• D. $43$

Answer 1

The correct option is B $34$

Let the units digit be x.
Hence tens digit will be $\frac{12}{x}$
Thus the number will be
$x+10\left(\frac{12}{x}\right)$
$=x+\frac{120}{x}$
Now applying the above condition, we get
$x+\frac{120}{x}+9=10\left(x\right)+\frac{12}{x}$
$x^2+120+9x=10x^2+12$
$9x^2-9x-108=0$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$x=4$ and $x=-3$
Considering the number to be positive, we get the number as
$4+\frac{120}{4}$
$=4+30$
$=34$