SOLUTION: The annual interest on an $18,000 investment exceeds the interest earned on an
$8000 investment by $290.
The $18,000 is invested at a 0.5% higher rate of interest than the $80
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Question 1015187: The annual interest on an $18,000 investment exceeds the interest earned on an
$8000 investment by $290.
The $18,000 is invested at a 0.5% higher rate of interest than the $8000.
What is the interest rate of each investment?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The annual interest on an $18,000 investment exceeds the interest earned on an
$8000 investment by $290.
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18000(x/100) = 8000(y/100) + 290
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The $18,000 is invested at a 0.5% higher rate of interest than the $8000.
What is the interest rate of each investment?
x = y + 0.5
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Substitute for "x" and solve for "y":
180(y+0.5) = 80y + 290
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100y + 90 = 290
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100y = 200
y = 2% (interest on the $18000)
x = y+0.5 = 2.5% (interest on the $8000
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Cheers,
Stan H.
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