Dana invested $10,000. Part of the money was invested at 10% and the remainder at 14%. The total yearly income from these investments was $1220. How much was invested at 14%?

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Answer 1 · 2016-11-13T18:21:09

$$ 770$ $$770$

This was a fun one, thanks!

You will want to set up two equations for the scenarios and then isolate one of the terms to find the other. Good times!

Let $x$ $x$ be the amount invested at 10%
Let $y$ $y$ be the amount invested at 14%

We know that $x$ $x$ and $y$ $y$ amount to $10,000 and that the interest generated equals $1,220, ergo:

$.1 x + .14 y = 1220$ $.1x + .14y = 1220$
$x + y = 10000$ $x + y = 10000$

Let's multiply the second line by .1 so that we can eliminate the $x$ $x$ and identify $y$ $y$ :

$.1 x + .14 y = 1220$ $.1x + .14y = 1220$
$.1 x + .1 y = 1000$ $.1x + .1y = 1000$

Subtract the second line from the first which leaves you with:

$.04 y = 220$ $.04y = 220$

Multipliy both sides by 100 (to eliminate the decimals and solve for $y$ $y$ ):

$4 y = 22000$ $4y = 22000$
$y = 5500$ $y = 5500$

Sub your $y$ $y$ value into the original statement to find $x$ $x$ :

$x + 5500 = 10000$ $x + 5500 = 10000$
$x = 4500$ $x = 4500$

Now double check your work by subbing these values in to the second statement:

$.1 x + .14 y = 1220$ $.1x + .14y = 1220$
$.1 (4500) + .14 (5500) = 1220$ $.1(4500) + .14(5500) = 1220$
$450 + 770 = 1220$ $450 + 770 = 1220$

Voila!