Question

a two digit number is such that the product of the digit is 12. When 36 is added to the digit they interchange their place. Formulate a quadratic equation​

Answer 1

QUESTION SHOULD BE:

• A two digit number is such that the product of the digits is 12. When 36 is added to the number, the digits interchange their places. Find the number.

ANSWER:

• Original number = 26

GIVEN:

• Product of digits of a two-digit number is 12.
• When 36 is added to the digit they interchange their place.

TO FIND:

• Original number

SOLUTION:

Let the digit at tens place be'x'.

Let the digit at once place be'y'.

Original number = 10x+y

Reversed number = 10y+x

CASE 1

=> xy = 12

=> x = 12/y. .....(i)

CASE 2

=> (10x+y) +36 = 10y+x

=> 10x +y -10y-x = (-36)

=> 9x-9y = (-36)

=> 9(x-y) = -36

=> x-y = -36/9

=> x-y = (-4)

=> x-y+4 = 0. ....(ii)

Putting x = 12/y from eq(i)

$\implies \: \dfrac{12}{y} - y + 4 = 0 \\ \implies \: \dfrac{12 - y {}^{2} + 4y}{y} = 0 \\ \implies \: - y {}^{2} + 4y + 12 = 0 \\ \implies \: - y {}^{2} + 6y - 2y + 12 = 0 \\ \implies \: ( - y {}^{2} + 6y) + ( - 2y + 12) = 0 \\ \implies \: - y(y - 6) - 2(y - 6) = 0 \\ \implies \: (y - 6)( - y - 2) = 0$

Either; (y-6) = 0

=> y-6 = 0

=> y = 6

Either; (-y-2) = 0

=> -y = 2

=> y = (-2). (Digit can't be negative)

Here ;

=> y = 6

Putting y = 6 in eq(i)

=> x = 12/6

=> x = 2

Original number = 10x+y

= 10*(2)+6

= 26