SOLUTION: susan invest 2 times as much money at 9% as she does at 6%. If her total interest after 1 year is $1440, how much does she invest at each rate?

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Question 1116363: susan invest 2 times as much money at 9% as she does at 6%. If her total interest after 1 year is $1440, how much does she invest at each rate?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52770)   (Show Source): You can put this solution on YOUR website!
.
Let x be an amount invested at 6% (the lesser amount).

Then the amount invested at 9% is 2x, according to the condition.


Your equation to find x is


interest + interest   = total interest,    or


0.06x    + 0.09*(2x)  = 1440


0.06x + 0.18x = 1440  ====>  0.24x = 1440  ====>  x =  = 6000.


Answer.  $6000 was invested at 6%.  $12000 was invested at 9%.


Check.   0.06*6000 + 0.09*12000 = 1440 dollars.   ! Correct !

Solved.

--------------------
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(13198)   (Show Source): You can put this solution on YOUR website!


...or here is a way to solve the problem with logical analysis instead of formal algebra

The amount invested at 9% is 2 times the amount invested at 6%; and the 9% interest rate is 1.5 times the 6% interest rate. Together those two things mean the interest earned at 9% is 2*1.5 = 3 times the interest earned at 6%.

So 1/4 of the interest earned is from the 6% investment and 3/4 is from the 9% investment.

1/4 of the total interest of $1440 is $360, so the amount of interest from the 6% investment is $360.

And then the amount invested at 6% is $360/.06 = $6000.

Then the amount invested at 9% is 2*$6000 = $12000.
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