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

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

On reversing digits the number is = 10 × y + x = 10y + x.

According to first condition, we have

$\Rightarrow \dfrac{10x + y}{10y + x} = 1\dfrac{3}{4} \\[1em] \Rightarrow \dfrac{10x + y}{10y + x} = \dfrac{7}{4} \\[1em] \Rightarrow 4(10x + y) = 7(10y + x) \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \\[1em] \Rightarrow 11(3x - 6y) = 0 \\[1em] \Rightarrow 3x - 6y = 0 ……..(i)$

According to second condition, we have

⇒ x + y = 12

⇒ x = 12 - y …….(ii)

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

⇒ 3(12 - y) - 6y = 0

⇒ 36 - 3y - 6y = 0

⇒ 36 - 9y = 0

⇒ 9y = 36

⇒ y = 4.

Substituting value of y in (ii) we get,

⇒ x = 12 - 4 = 8.

Number = 10 × x + y = 10 × 8 + 4 = 84.

Hence, number = 84.