SOLUTION: George invested $19,000 for one year, part at 11% and part at 12%. If the total amount of interest he earned was $2,200, how much money did he invest at each rate?

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Question 1115136: George invested $19,000 for one year, part at 11% and part at 12%. If the total amount of interest he earned was $2,200, how much money did he invest at each rate?
Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.
interest + interest       = total interest


0.11x    + 0.12*(19000-x) = 2200


0.11x + 0.12*19000 - 0.19x = 2200


-0.01x = 2200-0.12*19000 = -80  


x = %28-80%29%2F%28-0.01%29 = 8000.


Answer.  $8000 was invested at 11%;  the rest, 19000-8000 = 11000 dollars was invested at 12%.


Check.   0.11*8000 + 0.12*11000 = 22000.   ! Correct !

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It is a typical and standard problem on investment.

To see many other similar solved problems on investment,  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.