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Question 399228: Michael has $12,500 to invest. He invests part in an account which earns 4.2% annual interest and the rest in an account which earns 6.2% annual interest. He earns $669.50 in interest at the end of the year. How much was invested at each rate?
Found 2 solutions by stanbon, robertb:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Michael has $12,500 to invest. He invests part in an account which earns 4.2% annual interest and the rest in an account which earns 6.2% annual interest. He earns $669.50 in interest at the end of the year. How much was invested at each rate?
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Equations:
Quantity Eq.:: x + y = 12,500
Interest Eq.::0.042x + 0.062y = 669.50
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Multiply thru Quantity by 42.
Multiply thru Interest by 1000.
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42x + 42y = 42*12,500
42x + 62y = 669500
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Subtract 1st from 2nd and solve for "y":
20y = 144500
y = $7225 (amt. invested at 6.2%
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Solve for "x":
x+ y = 12,500
x + 7225 = 12,500
x = $5275 (amt. invested at 4.2%)
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Cheers,
Stan H.
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! 0.042x + 0.062(12,500 - x) = 669.50
==> 0.042x + 775 - 0.062x = 669.50
==> -0.02x = -105.5
==> x = $5,275, the amount invested in 4.2%
and 12,500 - x = $7,225, the amount invested in 6.2%.
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