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Question 1162159: In investing $5,900 of a couple's money, a financial planner put some of it into a savings account paying 5% annual simple interest. The rest was invested in a riskier mini-mall development plan paying 10% annual simple interest. The combined interest earned for the first year was $460. How much money was invested at each rate?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website! .
x dollars invested at 10%
(5900-x) dollars invested at 5%
The total interest equation is
0.1x + 0.05*(5900-x) = 460 dollars.
From the equation,
x =  = 3300.
ANSWER. $3300 invested at 10%, and the rest, 5900-3300 = 2600 dollars invested at 5%.
CHECK. 0.1*3300 + 0.05*2600 = 460 dollars. ! Precisely correct !
Solved.
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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.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Here is a non-algebraic method for solving this kind of "mixture" problem.
The general method is often much faster and easier than the standard algebraic method -- although in this problem the difference in the amount of effort is not significant.
(1) All $5900 invested at 5% would have earned $295 interest; all at 10% would have earned $590 interest.
(2) Determine by whatever means you choose (maybe a number line?) that the actual interest of $460 is 165/295 = 33/59 of the way from $295 to $590.
(3) That means 33/59 of the total was invested at the higher rate.
ANSWER: 33/59 of $5900, or $3300, at 10%; the other $2600 at 5%.
CHECK: .10(3300)+.05(2600) = 330+130 = 460
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