Question 1119152: Hi, I have a problem that I don't understand how to solve, where there are two different interest rates. Can you help me with this?
"Madge invested $10,000 for one year, part at 8% annual interest and the rest at 12% annual interest. Her total interest for the year was $944. How much did she invest at each rate?"
Thank you!
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Let x be the amount invested at 8% (in dollars).
Then the amount invested at 12% is (10000-x) dollars.
The amount x produces the interest equal 0.08*x.
The amount (10000-x) produces the interest equal 0.12*(10000-x).
Together, these investments produce the total interest of 0.08x + 0.12*(10000-x).
We are given that the total interest is 944 dollars.
It gives you an equation
0.08x + 0.12*(10000-x) = 944.
Simplify and solve it for x:
0.08x + 1200 - 0.12x = 944
-0.04x = 944 - 1200 = -256
x =  = 6400.
Answer. $6400 was invested at 8%. The rest (10000-6400) = 3600 dollars was invested at 12%.
Check. 0.08*6400 + 0.12*3600 = 944. ! 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(13200)  (Show Source):
You can put this solution on YOUR website!
Tutor @ikleyn showed a solution using the traditional algebraic method.
That is a perfectly good method; however, the answer to a problem like this can be obtained much faster and with less work by a different method.
By comparing the actual interest obtained to the amounts that could have been earned at the two separate interest rates, you can determine the ratio in which the money was split between the two investments.
The interest was $944; the interest if the whole amount had been invested at either of the two separate interest rates are $800 and $1200.
The actual interest of $944 is 144/400 (= 36%) of the distance from $800 to $1200
That means 36% of the $10,000, or $3600, was invested at the higher rate of 12%; the remaining $6400 was invested at 8%.
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