SOLUTION: Part of $6,000 is invested at 12%, another part at 14%, and the remainder at 15% yearly interest. The total yearly income from the three investments is $840. The sum of the amounts

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Part of $6,000 is invested at 12%, another part at 14%, and the remainder at 15% yearly interest. The total yearly income from the three investments is $840. The sum of the amounts      Log On

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Question 1118391: Part of $6,000 is invested at 12%, another part at 14%, and the remainder at 15% yearly interest. The total yearly income from the three investments is $840. The sum of the amounts invested at 12% and 14% equals the amount invested at 15%. How much is invested at each rate?
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Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


"The sum of the amounts invested at 12% and 14% equals the amount invested at 15%."

That means half of the $6000 was invested at 15%
15% of $3000 is $450, so the amount of interest from the other two investments is $840-$450 = $390.

If all of the remaining $3000 were invested at 12%, the interest would be $360; if all were invested at 14%, the interest would be $420.

The actual remaining amount of interest, $390, is exactly halfway between $360 and $420; that means the amounts invested at 12% and 14% are equal -- $1500 in each.

Answer: $3000 at 15%; $1500 at 14%; $1500 at 12%.

CHECK: .15(3000)+.14(1500)+.12(1500) = 450+210+180 = 840

Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
x    at   12%
y    at   14%
x+y  at   15%
------------------
------------------2x+2y=6000  or x+y=3000

Yearly Income, $840
12x%2B14y%2B15%28x%2By%29=84000, and x%2By=3000.
12x%2B14y%2B15%2A3000=84000
12x%2B14y=84000-45000
12x%2B14y=39000
12x%2B14%283000-x%29=39000------from hear, the algebra is a few short simple steps....