- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Given :
The sum of the digits of a two-digit number is 9.
Nine times this number is twice the number obtained by reversing the order of the digits.
To do :
We have to find the given number.
Solution :
Let the two-digit number be $10x+y$.
$x + y = 9$
$x=9-y$.....(i)
The number formed on reversing the digits is $10y+x$.
Therefore,
$9(10x+y) = 2(10y+x)$
$90x+9y=20y+2x$
$90x-2x+9y-20y=0$
$88x-11y=0$
$11(8x-y) = 0$
$8x-y = 0$
$y=8x$
Substituting $y = 8x$ in equation (i), we get,
$x =9-8x $
$x+8x = 9$
$9x = 9$
$x=1$
This implies,
$y = 8x = 8(1)=8$
The original number is $10(1)+8 = 10+8 = 18$.
The original number is 18.