Question 1184536: $19,670 is invested, part at 10% and the rest at 6%. If the interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 6% by $681.08, how much is invested at each rate? (Round to two decimal places if necessary.)
Answer by ikleyn(52864)   (Show Source):
You can put this solution on YOUR website! .
$19,670 is invested, part at 10% and the rest at 6%. If the interest earned from the amount invested at 10% exceeds
the interest earned from the amount invested at 6% by $681.08, how much is invested at each rate?
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0.1x - 0.06*(19670-x) = 681.08
x =  = 11,633.00.
ANSWER. $11,633 invested at 10%. The rest, (19670-11633) = 8037 dollars, invested at 6%.
CHECK. 0.1*11633 - 0.06*8037 = 681.08 dollars, the difference. ! Precisely 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.
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