Question 1199905: a person had $14,000 invested in two accounts, one paying 9% simple interest and one paying 10% simple interest. how much was invested in each account if the interest after 1 year is $1397?
Answer by ikleyn(52772) (Show Source):
You can put this solution on YOUR website! .
a person had $14,000 invested in two accounts, one paying 9% simple interest
and one paying 10% simple interest. how much was invested in each account
if the interest after 1 year is $1397?
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Let x be amount invested at 10% annually.
Then the amount invested at 9% is the rest (14000-x) dollars.
10% investment generates the annual interest of 0.1x dollars.
9% investment generates the annual interest of 0.09*(14000-x).
The totall annual interest equation is
0.1x + 0.09*(14000-x) = 1397 dollars.
Simplify this equation and find x
0.1x + 0.09*14000 - 0.09x = 1397,
0.1x - 0.09x = 1397 - 0.09*14000
0.01x = 137
x = 137/0.01 = 13700.
Thus $13700 invested at 10% and the rest, $14000 - $13700 = $300 invested at 0.09%. ANSWER
CHECK. 0.1*13700 + 0.09*300 = 1397 dollars, the total annual interest. ! correct !
Solved.
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- Using systems of equations to solve problems on investment
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