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Question 1207434: An investment adviser invested $14,000 in two accounts. One investment earned 6% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of interest earned for 1 year was $609. How much was invested in each account?
amount at 6% $
amount at 2.5%
Answer by ikleyn(52797)   (Show Source):
You can put this solution on YOUR website! .
An investment adviser invested $14,000 in two accounts.
One investment earned 6% annual simple interest, and
the other investment earned 2.5% annual simple interest.
The amount of interest earned for 1 year was $609.
How much was invested in each account?
amount at 6% $
amount at 2.5%
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Let x dollars are invested at 6%.
Then the amount invested at 2.5% is (14000-x) dollars.
First amount earns 0.06x dollars as a year interest.
Second amount earns 0.025(14000-x) dollars as a year interest.
Together, these amounts earn 0.06x + 0.025(14000-x) dollars as a year interest.
So, we write this equation
0.06x + 0.025(14000-x) = 609 dollars.
The setup is complete.
To solve this equation, simplify it step by step
0.06x + 0.025*14000 - 0.025x = 609
0.06x - 0.025 = 609 - 0.025*14000
0.035x = 259
x = 259/0.035 = 7400.
So, $7400 was invested at 6% and the rest, $14000 - $7400 = $6600 was invested at 0.025%. ANSWER
CHECK. 0.06*7400 + 0.025*6600 = 609 dollars as a total annual interest. ! precisely correct !
Solved.
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