in Quantitative Aptitude retagged by
287 views
3 votes
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?

  1. $5$
  2. $6$
  3. $8$
  4. $7$
in Quantitative Aptitude retagged by
13.4k points
287 views

1 Answer

1 vote
1 vote

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 $: \text{B}$

edited by
10.3k points
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true