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3 years ago

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Aleksandra Tomich invested $9,102 and part at 8% simple interest and part 4% simple interest for a period of 1 year. How much di

d she invest at each rate if each account earned the same interest?
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1 answer:

gayaneshka [121]3 years ago

6 0

Answer:

$3034 at 8%
$6068 at 4%

Step-by-step explanation:

The ratio of interest rates is 8% : 4% = 2 : 1. Since the amount of interest earned in each account is the same, the ratio of amounts invested will be the inverse of that, 1/2 : 1/1 = 1 : 2.

Then 1/(1+2) = 1/3 of the money is invested in the 8% account. That amount is ...

$9102/3 = $3034 . . . . invested at 8%

The remaining amount is invested at 4%:

$9102 - 3034 = $6068 . . . . invested at 4%

_____

The interest earned in 1 year in each account is

0.08·$3034 = 0.04·$6068 = $242.72

_____

If you really need an equation, you can let x represent the amount invested at the higher rate. (Using this variable assignment avoids negative numbers later.) Then 9102-x is the amount invested at the lower rate.

0.08x = 0.04(9102-x)

0.12x = 0.04·9102 . . . . . eliminate parentheses, add 0.04x

x = (0.04/0.12)·9102 = 9102/3 . . . . . divide by the coefficient of x. Same answer as above.

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