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Question 1125806: Last year, Elsa had $30,000 to invest. She invested some of it in an account that paid 6% simple interest per year, and she invested the rest in an account that paid 10% simple interest per year. After one year, she received a total of $2360 in interest. How much did she invest in each account?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52775)   (Show Source):
You can put this solution on YOUR website! .
Let x is invested at 10%. The annual interest of this investment is 0.1*x dollars.
The the investment at 6% is (30000-x), and its generates the annual interest of 0.06(30000-x) dollars.
Your equation is
interest + interest = total interest, or
0.1x + 0.06*(30000-x) = 2360 dollars.
0.1x + 0.06*30000 - 0.06x = 2360
0.1x - 0.06x = 2360 - 0.06*30000
x =  = 14000.
Answer. $14000 was invested at 10% and the rest, (30000-14000) = 16000 dollars were invested at 6%.
Check. 0.1*14000 + 0.06*16000 = 2360 dollars. ! Correct !
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It is a standard and typical problem on investments.
If you need more details, or if you want to see other similar problems solved by different methods, look into the lesson
- Using systems of equations to solve problems on investment
in this site.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
Here is a method for solving "mixture" problems like this which, if you understand it, will get you to the answer much faster and with far less work than the traditional algebraic method.
The key idea for this method is that the ratio in which the money was split between the two accounts is exactly determined by where the actual amount of interest lies between the amounts of interest that would have been earned at each of the two separate accounts.
The explanation in words sounds confusing; but the actual calculations are quick and simple. For this example:
30,000 at 6% = 1800; 30,000 at 10% = 3000
Actual interest = 2360
2360 is (560/1200) of the way from 1800 to 3000. (2360-1800 = 560; 3000-1800 = 1200).
That means 560/1200 of the money was invested at the higher rate.
Now convert that ratio into a ratio that is "convenient" for the total of $30,000 that was invested: 560/1200 = 56/120 = 14/30.
The amount invested at 10% was 14/30 of the $30,000, which is $14,000; the other $16,000 was invested at 6%.
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