Question

$18,389

is invested, part at 10%
and the rest at 8%
. If the interest earned from the amount invested at 10%
exceeds the interest earned from the amount invested at 8%
by $621.02
, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

Let x be the amount invested at 10% and y be the amount invested at 8%. We know that:

x + y = 18389 (equation 1)
0.10x - 0.08y = 621.02 (equation 2)

From equation 1, we can solve for y:

y = 18389 - x

Substituting this into equation 2, we get:

0.10x - 0.08(18389 - x) = 621.02

0.10x - 1471.12 + 0.08x = 621.02

0.18x = 2092.14

x = 11623

Substituting this back into equation 1, we get:

11623 + y = 18389

y = 6766

Therefore, $11623 is invested at 10% and $6766 is invested at 8%.