How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

- $5$
- $6$
- $8$
- $7$

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3
votes

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

- $5$
- $6$
- $8$
- $7$

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1
votes

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 $: \text{B}$

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