Question

A two digit number is such that the product of the digits is 12. When 36 is added to the number, the digits interchange their places. The number is _ _ _

Answer 1

Hint: The two digit number can be represented as $10y + x$ where ‘x’ is unit’s place digit and ‘y’ is ten’s place digit.
Solution of quadratic equation $a{x^2} + bx + c = 0$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ using quadratic formula or Sridhar Acharya’s Method.

Complete step-by-step answer:
Let unit place digit of two digit number be ‘x’ and
ten’s place digit of two digit number be ‘y’.
So, number is $10y + x$
If we reverse digits(unit place digit be ‘y’ and ten’s place digit be ‘x’), then number will be $10x + y$
Now, according to question
Product of digits is 12; $(xy = 12)$ and
$36 + (10y + x) = 10x + y$ (This is second condition given in question)
Now on simplifying it, we get: $9x - 9y = 36$
Taking 9 common from LHS; we get: $9(x - y) = 36$
On simplifying, we get: $x - y = 4$ ; $x = y + 4$
Put this value of ‘x’ in $(xy = 12)$; we get: $(y + 4)y = 12$
On simplifying, we get a quadratic equation: ${y^2} + 4y - 12 = 0$
Using middle term splitting, we get: $(y + 6)(y - 2) = 0$
So, we get two value of y: $y = - 6,y = 2$
But as we know digits cannot be negative so $y = - 6$ is rejected.
So, we get $y = 2$
On putting value of y in $(xy = 12)$ , we get value of x; $x = 6$
So, our number is $10y + x = 10(2) + 6 = 26$
So, our final answer is 26.

Note: There are many ways to solve a quadratic equation. Here we have solved using middle term splitting. You can use Sridhar Acharya’s Method or Method of completing the square to find roots of quadratic equations.