Question 91736: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the amount invested at 9%. The interest earned on this amount would then be (0.09)x.
The remainder (6000-x) is invested at 11%. The interest earned on this amount would be (0.11)(6000-x)
The sum (+) of these two amounts is $624.00
You can now write the equation needed to solve for x, the amount invested at 9%.
(0.09)x + (0.11)(6000-x) = $624.00 Simplify and solve for x.
0.09x+660-0.11x = $624.00 Combine like-terms.
(0.09-0.11)x+660 = $624.00 Subtract 660 from both sides.
-0.02x = -$36.00 Divide both sides by -0.02
x = $1800.00 and...
$6000-x = $6000-$1800 = $4200.00
Solution:
$1,800.00 was invested at 9%
$4,200 was invested at 11%
Check:
(0.09)($1,800.00)+(0.11)($4,200.00) = $162.00 + $462.00 = $624.00
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