The digit at tens place of a two digit number is two times the digit at ones place. When 27 is subtracted from the number, its digit are reversed. Find the number. | Simultaneous Equation | SciPiPupil

Question: The digit at tens place of a two digit number is two times the digit at ones place. When 27 is subtracted from the number, its digit are reversed.

Solution:

Let the number at tens place be x and the number at ones place be y. The number is 10x+y.

According to the question,

Condition I,

The digit at tens place of a two digit number is two times the digit at ones place.

$or, x = 2y$ - (i)

Condition II,

When 27 is subtracted from the number, its digit are reversed.

$or, 10x + y -27 = 10y +x$

$or, 10x -x -27 = 10y -y$

$or, 9x -27 = 9y$ - (ii)

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

$or, 9(2y) -27 = 9y$

$or, 18y -9y = 27$

$or, 9y = 27$

$or, 9×y = 9×3$

$\therefore y = 3$

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

$or,x = 2y$

$or, x = 2×3$

$\therefore x = 6$

So, (x,y) = (6,3)

10x = 10×6 = 60

10x +y = 60+3 = 63

Therefore, the required two digit number is 63.