Question: A number consisting of two digits is three times the sum of its digits. If 45 is added to the number, the digits will be interchanged. Find the number.

Solution:

Let the digit at tens place of the two digit number be x and the digits at ones place be y. So, the required number is 10x+y.

According to the question,

Condition I,
The number is three times the sum of its digits.
$or, 10x +y = 3(x+y)$
$or, 10x +y =3x +3y$
$or, 10x -3x = 3y -y$
$or, 7x = 2y$
$or, x = \frac{2y}{7}$ - (i)

Condition II,
If 45 is added to the number, the digits are interchanged.
$or, 10x +y +45 = 10y +x$
$or, 10x -x +45 = 10y -y$
$or, 9x +45 = 9y$
$or, 9(x +5) = 9(y)$
$or, x +5 = y$ - (ii)

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

$or, \dfrac{2y}{7} + 5= y$

$or, \dfrac{2y}{7} = y -5$

$or, 2y = 7(y-5)$

$or, 2y = 7y -35$

$or, 35 = 7y-2y$

$or, 5y = 35$

$or, y = \dfrac{35}{5}$

$\therefore y = 7$

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

$or,  x = \dfrac{2×7}{7}$

$\therefore x = 2$

So, (x,y) = (2,7)
10x = 10×2 = 20
10x +y = 20+7 = 27

Therefore, the required two digit number is 27.