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Question 1196056: Walt made an extra $10,000 last year from a part-time job. He invested part of the money at 4% and the rest at 3%. He made a total of $370 in interest. How much was invested at 3%?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website! .
Walt made an extra $10,000 last year from a part-time job.
He invested part of the money at 4% and the rest at 3%.
He made a total of $370 in interest. How much was invested at 3%?
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Let x be the amount (in dollars) invested at 3%.
Then the amount invested at 4% is (10,000-x) dollars (the rest).
Having it, write the money equation for the combined interest
0.03x + 0.04*(10000-x) = 370 dollars.
Simplify and find x
0.03x + 400 - 0.04x = 370
0.03x - 0.04x = 370 - 400
-0.01x = -30
x =  = 3000.
ANSWER. $3000 was invested at 3%.
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(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a quick and easy common-sense method for solving "mixture" problems like this where the numbers are "nice".
(1) All $10,000 invested at 3% would have earned $300 interest; all at 4% would have earned $400.
(2) The actual interest, $370, is 7/10 of the way from $300 to $400. (300 to 400 is a difference of 100; 300 to 370 is a difference of 70; 70/100 = 7/10.)
Therefore, 7/10 of the total was invested at the higher rate.
ANSWER: 7/10 of $10,000, or $7000, was invested at 4%; the other $3000 at 3%.
CHECK: .04(7000)+.03(3000)=280+90=370
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