- Trending Categories
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
A man invested an amount at $ 12 \% $ per annum simple interest and another amount at $ 10 \% $ per annum simple interest. He received an annual interest of Rs. $ 2600 . $ But, if he had interchanged the amounts invested, he would have received Rs. 140 less. What amounts did he invest at the different rates?
Given:
A man invested an amount at \( 12 \% \) per annum simple interest and another amount at \( 10 \% \) per annum simple interest.
He received an annual interest of Rs.\( 2600 . \)
If he had interchanged the amounts invested, he would have received Rs. 140 less.
To do:
We have to find the amounts he invested.
Solution:
Let's say he has invested amounts in two different schemes A and B.
Let the amount invested in scheme A be $Rs.\ x$ and the amount invested in scheme B be $Rs.\ y$.
SI on $Rs.\ x$ at 12% per annum for 1 year $=Rs.\ \frac{x\times12\times1}{100}=Rs.\ \frac{12x}{100}$
SI on $Rs.\ y$ at 10% per annum for 1 year $=Rs.\ \frac{y\times10\times1}{100}=Rs.\ \frac{10y}{100}$
SI on $Rs.\ x$ at 10% per annum for 1 year $=Rs.\ \frac{x\times10\times1}{100}=Rs.\ \frac{10x}{100}$
SI on $Rs.\ y$ at 12% per annum for 1 year $=Rs.\ \frac{y\times12\times1}{100}=Rs.\ \frac{12y}{100}$
According to the question,
$\frac{12x}{100}+\frac{10y}{100}=2600$
$\frac{12x+10y}{100}=2600$
$12x+10y=2600(100)$......(i)
$\frac{10x}{100}+\frac{12y}{100}=2600-140$
$\frac{10x+12y}{100}=2460$
$10x+12y=2460(100)$......(ii)
Multiplying equation (i) by 12 on both sides, we get,
$12(12x+10y)=12(2600)(100)$
$144x+120y=31200(100)$.....(iii)
Multiplying equation (ii) by 10 on both sides, we get,
$10(10x+12y)=10(2460)(100)$
$100x+120y=24600(100)$.....(iv)
Subtracting equation (iv) from equation (iii), we get,
$(144x+120y)-(100x+120y)=31200(100)-24600(100)$
$144x-100x+120y-120y=6600(100)$
$44x=6600(100)$
$x=\frac{6600(100)}{44}$
$x=150(100)$
$x=15000$
Substituting $x=15000$ in equation (i), we get,
$12(15000)+10y=2600(100)$
$180000+10y=260000$
$10y=260000-180000$
$10y=80000$
$y=\frac{80000}{10}$
$y=8000$
The money he invested at the different rates is Rs. 15000 and Rs. 8000 respectively.
167 Views
