287 views

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?

1. $5$
2. $6$
3. $8$
4. $7$

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 $: \text{B}$

10.3k points

1
430 views