SOLUTION: Rodrigo invests $4000, part of it at 10% annual interest and the rest at 12% annual interest. If he receives $460 in interest at the end of one year, how much did he invest at each

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Rodrigo invests $4000, part of it at 10% annual interest and the rest at 12% annual interest. If he receives $460 in interest at the end of one year, how much did he invest at each      Log On


   

Question 1120806: Rodrigo invests $4000, part of it at 10% annual interest and the rest at 12% annual interest. If he receives $460 in interest at the end of one year, how much did he invest at each rate?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The ratio in which the money must be split between the two investments is exactly determined by where the actual interest lies between the amounts of interest that would have been received if the money had been invested all at the lower rate or all at the higher rate.

$4000 at 10% would yield $400 interest
the actual interest was $460
$4000 at 12% would yield $480 interest

$460 is three-fourths of the way from $400 to $480; that means 3/4 of the money needs to be invested at the higher rate.

So 3/4 of the $4000, or $3000, was invested at 12%; the other $1000 was invested at 10%.

Check:
(.12)($3000) + (.10)($1000) = $360+$100 = $460

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
Equation:


interest  + interest      = 460


0.10*x    + 0.12*(4000-x) = 460


Simplify and solve for x:


10x + 12*4000 - 12x = 460*100


-2x = 46000 - 12*4000


x      = %2846000-12%2A4000%29%2F%28-2%29 =  1000 dollars  (amount invested at 10%)


4000-x = 3000 dollars  (amount invested at 12%)


Check.  0.10*1000 + 0.12*3000 = 460  dollars.   ! Correct !


Answer.  $1000 invested at 10% and $3000 invested at 12%.

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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.