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Question 1196415: Two investments totaling $56,500 produce an annual income of $1835. One investment yields 4% per year, while the other yields 3% per year. How much is invested at each rate?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
Two investments totaling $56,500 produce an annual income of $1835.
One investment yields 4% per year, while the other yields 3% per year.
How much is invested at each rate?
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Basic equation expresses the combined annual interest
0.04x + 0.03*(56500-x) = 1835.
Simplify and find x, which is the amount invested at 4%
0.04x +0.03*56500 - 0.03x = 1835
0.01x = 1835 - 0.03*56500
0.01x = 140
x = 140/0.01 = 14000.
ANSWER. $14000 invested at 4% and $56500 - $14000 = $42500 invested at 3%.
CHECK. 0.04*14000 + 0.03*42500 = 1835. ! Correct !
Solved.
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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(13195)  (Show Source):
You can put this solution on YOUR website!
The other tutor showed a standard method for solving a problem like this. You should understand that method and know how to use it, as it is useful in many different kinds of problems.
This is essentially a 2-part mixture problem -- similar to mixing two antifreeze solutions containing different percentages of antifreeze.
Any such 2-part mixture problem can be solved using a non-algebraic approach, as shown below. If the numbers in the problem are "nice" (as they are in this problem), the solution involves only a bit of elementary arithmetic -- along with some powerful logical reasoning.
(1) All $56,500 invested at 3% would have yielded interest income of $1695.
(2) Investing it all at 4% would have yielded an additional $565 of income.
(3) The actual income, $1835, was $140 more than $1695.
(2) and (3) together mean that the fraction of the total invested at the higher rate was 140/565. Since the total amount invested was $56,500, the amount invested at the higher rate was 140/565 of $56,500, or $14,000.
ANSWERS: $14,000 was invested at 4%; the remaining $42,500 at 3%
CHECK: .04(14000)+.03(42500) = 560+1275 = 1835
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