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?

Question

2483 views around the world

You can reuse this answer
Creative Commons License

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%

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

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

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

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

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

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