Question

A two digit number is such that the product of its digits is 12 and when 36 is added to the number the digits interchange their places . Find the number. ​

Answer 1

$hello$

▪︎Solution :-

Let number = 10x + y

According to question,

xy = 12 --------(1)

Again, 10x + y + 36 = 10y + x

9x - 9y = 36

x - y = 4

x = 4 + y

Put this in equation (1)

y( y + 4) = 12

y² + 4y -12= 0

Hence, quadratic equation ,

is y² + 4y -12 = 0

Solve this y = 6, but y ≠ -2

Then , x = 2

So,number= 10×2 + 6 =26

Therefore, the answer is 26.

I hope my answer helps you....✌

Answer 2

$\huge{ \bold{ \mathscr{ \purple{Hey Mate....!!!}}}}$

♡ QUESTION ♡

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

♡ ANSWER ♡

• Required Quadratic Equation is ${x}^{2} + 4x - 12 = 0$

♡ EXPLANATION ♡

• Let the ten's digit of the number be x
• It is given that the product of digits is 12.

Unit's digit $= \frac{12}{x}$

Number $= 10x + \frac{12}{x}$

• If 36 is added to the number the digits interchange their places

then ,

$10x + \frac{12}{x} + 36 = 10 \times \frac{12}{x} + x$

$= > 10x + \frac{12}{x} + 36 = \frac{120}{x} + x$

$= > 9x - \frac{108}{x} + 36 = 0$

$= > {x}^{2} + 4x - 12 = 0$ ( divided throughout by 9 )

• Hence , the required Quadratic Equation is $\large \boxed{ {x}^{2} + 4x - 12 = 0}$