![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
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 two numbers is 16. The sum of their reciprocals is $\frac{1}{3}$. Find the numbers.
Given:
The sum of two numbers is 16. The sum of their reciprocals is $\frac{1}{3}$.
To do:
We have to find the numbers.
Solution:
Let one of the numbers be $x$.
This implies,
The other number $=16-x$.
According to the question,
$\frac{1}{x}+\frac{1}{16-x}=\frac{1}{3}$
$\frac{1(16-x)+1(x)}{x(16-x)}=\frac{1}{3}$
$\frac{16-x+x}{16x-x^2}=\frac{1}{3}$
$\frac{16}{16x-x^2}=\frac{1}{3}$
$3(16)=1(16x-x^2)$
$48=16x-x^2$
$x^2-16x+48=0$
Solving for $x$ by factorization method, we get,
$x^2-12x-4x+48=0$
$x(x-12)-4(x-12)=0$
$(x-12)(x-4)=0$
$x-12=0$ or $x-4=0$
$x=12$ or $x=4$
If $x=4$, then $16-x=16-4=12$.
If $x=12$, then $16-x=16-12=4$
The required numbers are $4$ and $12$.
Advertisements