The sum of the digits in a two-digit number is 11. The number formed by interchanging the digits of the number will be 45 more than the original number. Find the original number. | Simultaneous Equations | SciPiPupil

Question: The sum of the digits in a two-digit number is 11. The number formed by interchanging the digits of the number will be 45 more than the original number. Find the original number.

Solution:

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

According to the question,

Condition I,

The sum of two digits in a two-digit number is 11.

$or, x + y = 11$

$or, x = 11 - y$ - (i)

Condition II,

The number formed by interchanging the digits of the number will be 45 more than the original number.

$or, 10y +x = 10x + y +45$

$or, 10y -y = 10x -x +45$

$or, 9y = 9x +45$

$or, 9(y) = 9(x +5)$

$or, y= x+5$ - (ii)

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

$or, y = 11- y +5$

$or, y + y = 16$

$or, 2y = 16$

$or, 2×y = 2×8$

$\therefore y = 8$

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

$or, x = 11 - 8$

$\therefore x = 3$

So, (x,y) = (3,8)

10x = 10×3 = 30

10x + y = 30+8 = 38

Therefore, the required two digit number is 38.