SOLUTION: $7,284
is invested, part at
13%
and the rest at
10%
. If the interest earned from the amount invested at
13%
exceeds the interest earned from the amount invested at
10%
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-> SOLUTION: $7,284
is invested, part at
13%
and the rest at
10%
. If the interest earned from the amount invested at
13%
exceeds the interest earned from the amount invested at
10%
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Question 1053505: $7,284
is invested, part at
13%
and the rest at
10%
. If the interest earned from the amount invested at
13%
exceeds the interest earned from the amount invested at
10%
by
$136.86
, how much is invested at each rate? (Round to two decimal places if necessary.) Found 2 solutions by jorel555, MathTherapy:Answer by jorel555(1290) (Show Source):
You can put this solution on YOUR website! Let n be the amount invested at 13%. Then:
13n-13686=10(728400-n)
13n-13686=7284000-10n
23n=7297686
n=317290.7
Then $3172.91 was invested at 13%, and $
4111.09 was invested at 10%. ☺☺☺☺
You can put this solution on YOUR website!
$7,284
is invested, part at
13%
and the rest at
10%
. If the interest earned from the amount invested at
13%
exceeds the interest earned from the amount invested at
10%
by
$136.86
, how much is invested at each rate? (Round to two decimal places if necessary.)
It can NEVER be what the other person stated. I never knew, even when I was a child that 412.4783 - 411.109 = $136.86
The correct answer is:
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