The result of dividing a number of two digits by the number with the digits reversed is $\dfrac{5}{6}$. If the difference of digits is 1, 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,

$\Rightarrow \dfrac{10x + y}{10y + x} = \dfrac{5}{6} \\[1em] \Rightarrow 6(10x + y) = 5(10y + x) \\[1em] \Rightarrow 60x + 6y = 50y + 5x \\[1em] \Rightarrow 60x - 5x + 6y - 50y = 0 \\[1em] \Rightarrow 55x - 44y = 0 \\[1em] \Rightarrow 11(5x - 4y) = 0 \\[1em] \Rightarrow 5x - 4y = 0 …….(i)$

According to second condition we have,

⇒ y - x = 1

⇒ x = y - 1 …….(ii)

Substituting value of x in (i) we get,

⇒ 5(y - 1) - 4y = 0

⇒ 5y - 5 - 4y = 0

⇒ y = 5.

Substituting value of y in (ii) we get,

⇒ x = y - 1 = 5 - 1 = 4.

Number = 10 × x + y = 10 × 4 + 5 = 45.

Hence, number = 45.