Let $`xy\text{’}$ be the two-digit number.
The two-digit number can be expressed as $10x+y,$ where $ y \neq 0 $
On interchanging the digits, the number will be $`yx\text{’},$ and can be expressed as $ 10y + x $
The two-digit number is more than thrice the number formed by interchanging the positions of its digits.
So, $ 10x + y > 3(10y + x) $
$ \Rightarrow 10x + y > 30y + 3x $
$ \Rightarrow \boxed{7x > 29y} \quad \longrightarrow (1) $
Minimum possible value of $ y = 1.$ So,
- $ x = 5 \Rightarrow 35 > 29 $
- $ x = 6 \Rightarrow 42 > 29 $
- $ x = 7 \Rightarrow 49 > 29 $
- $ x = 8 \Rightarrow 56 > 29 $
- $ x = 9 \Rightarrow 63 > 29 $
And, now we take $ y = 2, $ So,
- $ x = 9 \Rightarrow 63 > 58 $
For $ y = 3,$ no value of $`x\text{’}$ is possible.
$\therefore$ The two digits numbers are $: 51, 61, 71, 81, 91,92.$
Correct Answer $: \text{B}$