Question 1190893: Gohan has $5000 to invest. He puts part in a savings
account that earns 6% interest and the rest in a high
interest savings account that pays 8%. If the total
interest for a year was $320, how much was put in each
account?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52871)   (Show Source):
You can put this solution on YOUR website! .
Gohan has $5000 to invest. He puts part in a savings
account that earns 6% interest and the rest in a high
interest savings account that pays 8%. If the total
interest for a year was $320, how much was put in each
account?
~~~~~~~~~~~~~~
x = invested at 8%; 5000-x invested at 6%.
Total annual interest equation
0.08x + 0.06*(5000-x) = 320 dollars.
Simplify and find x
0.08x + 300 - 0.06x = 320
0.08x - 0.06x = 320 - 300
0.02x = 20
x = 20/0.02 = 1000.
ANSWER. $1000 invested at 8%; the rest, $5000-$1000 = $4000 invested at 6%.
CHECK. 0.08*1000 + 0.06*4000 = 80 + 240 = 320, dollars, total annual interest. ! 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(13208)  (Show Source):
You can put this solution on YOUR website!
A quick and easy way to answer any 2-part "mixture" problem like this, if formal algebra is not required....
All $5000 invested at 6% would yield $300 interest; all invested at 8% would yield $400 interest.
The actual interest, $320, is 1/5 of the way from $300 to $400.
Therefore, 1/5 of the total was invested at the higher rate.
ANSWER: 1/5 of $5000, or $1000, was invested at 8%; the other $4000 at 6%
CHECK: .08(1000)+.06(4000) = 80+240=320
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