A two-digit number is such that the product of the digits is 24. When 45 is added to the number, the digits interchange their places. Find the number. | Quadratic Equation | SciPiPupil

Question: A two-digit number is such that the product of the digits is 24. When 45 is added to the number, the digits interchange their places. Find the number.

Solution:

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

According to the question,

Condition I,

The product of its digits is 24.

$or, xy = 24$ - (i)

Condition II,

When 45 is added to the number, the digits interchange their places.

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

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

$or, 9x +45 = 9y$

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

$or, x +5 = y$

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

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

$or, (y -5)y = 24$

$or, y² - 5y = 24$

$or, y² -5y -24 = 0$

$or, y² -(8-3)y -24 = 0$

$or, y² -8y +3y -24 = 0$

$or, y(y -8) +3(y -8) = 0$

$or, (y +3)(y -8) = 0$

Either,

$y +3 = 0$

$\therefore y = -3$

Or,

$y -8 = 0$

$\therefore y = 8$

Taking positive value of y only i.e. y = 8

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

$or, x = 8 -5$

$\therefore x = 3$

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

10x = 10×3 = 30

10x + y = 30+8 = 38

Therefore, the required two-digit number is 38.

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