Question: A two digit number is 4 times the sum of its digits. The sum of the number formed by reversing its digits and 9 is equal to 2 times the original number. Find the number.

Solution:

Let the digit and tens place of the original number be x and the digits at ones place be y. Then the required number is 10x +y.

According to the question,

Condition I,
A two digit number is 4 times the sum of its digits.
$or, 10x + y = 4(x+y)$
$or, 10x + y = 4x +4y$
$or, 10x - 4x = 4y - y$
$or, 6x = 3y$
$or, x = \frac{3}{6} y$
$or, x = \frac{1}{2} y$ - (i)

Condition II,
The sum of the number formed by reversing its digits and 9 is equal to 2 times the original number.

$or, (10y +x) +9 = 2(10x +y)$
$or, 10y +x +9 = 20x +2y$
$or, 10y -2y +9 = 20x -x$
$or, 8y  +9 = 19x$ - (ii)

Put value of x from equation (i) in equation (ii), we get,

$or, 8y + 9 = 19 \left (  \dfrac{1}{2} y \right )$
$or, 8y + 9 = \dfrac{19}{2}$
$or, 2(8y +9) = 19y$
$or, 16y +18 = 19y$
$or, 18 = 19y-16y$
$or, 3y = 18$
$or, y = \dfrac{18}{3}$
$\therefore y = 6$

Put value of y in equation (i), we get,

$or, x = \dfrac{1}{2} × 6$
$\therefore x = 3$

So, (x,y) = (3,6)
10x = 10×3 = 30
10x +y = 30 + 6 = 36

Therefore, the required two digit number is 36.