The result of dividing a two-digit number by the number with its digits reversed is $\Big(1\dfrac{3}{4}\Big)$. If the sum of the digits is 12, find the number.

Let the ten's and unit's digits of required number be x and y respectively.

Given,

Sum of the digits of the number is 12.

⇒ x + y = 12

⇒ x = 12 - y     …..(1)

Original number = 10x + y

Number obtained by reversing the digits = 10y + x

Given,

On dividing the number by the number with its digits reversed, the result is $\Big(1\dfrac{3}{4}\Big)$.

$\therefore \Big(\dfrac{10x + y}{10y + x}\Big) = \Big(1\dfrac{3}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{10x + y}{10y + x}\Big) = \dfrac{7}{4} \\[1em] \Rightarrow (10x + y) \times 4 = (10y + x) \times 7 \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \text{…..(2)}$

Substituting the value of x from equation (1) in (2), we get :

⇒ 33(12 - y) - 66y = 0

⇒ 396 - 33y - 66y = 0

⇒ 396 - 99y = 0

⇒ 99y = 396

⇒ y = $\dfrac{396}{99}$

⇒ y = 4.

Substituting value of y in equation (1), we get :

⇒ x = 12 - y

⇒ x = 12 - 4

⇒ x = 8.

Original number = (10x + y)

= 10 × 8 + 4

= 84.

Hence, the number is 84.