Tracy invested 6000 dollars for 1 year, part at 10% annual interest and the balance at 13% annual interest. Her total interest for the year is 712.50 dollars. How much money did she invest at each rate?

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Answer 1 · 2015-11-07T16:36:16

$2250 @10%
$3750 @13%

Let $x$ $x$ be the amount invested at 10%
$\Rightarrow 6000 - x$ $=> 6000 - x$ is the amount invested at 13%

$0.10 x + 0.13 (6000 - x) = 712.50$ $0.10x + 0.13(6000 -x) = 712.50$

$\Rightarrow 10 x + 13 (6000 - x) = 71250$ $=> 10x + 13(6000 -x) = 71250$

$\Rightarrow 10 x + 78000 - 13 x = 71250$ $=> 10x + 78000 - 13x = 71250$
$\Rightarrow - 3 x + 78000 = 71250$ $=> -3x + 78000 = 71250$

$\Rightarrow 3 x = 78000 - 71250$ $=> 3x = 78000 - 71250$

$\Rightarrow 3 x = 6750$ $=> 3x = 6750$
$\Rightarrow 2250$ $=> 2250$

$\Rightarrow 6000 - x = 3750$ $=> 6000 - x = 3750$