Question

$11,335

is invested, part at 9%
and the rest at 6%
. If the interest earned from the amount invested at 9%
exceeds the interest earned from the amount invested at 6%
by $865.35
, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

Let x be the amount invested at 9% and y be the amount invested at 6%

.According to the problem:

x + y = 11,335 ----(1) (the total amount invested is $11,335)0.09x - 0.06y = 865.35 ----(2) (the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 6% by $865.35)

We can use these two equations to solve for x and y.

Multiplying equation (1) by 0.06, we get

:0.06x + 0.06y = 680.10 ----(3)

Subtracting equation (3) from equation (2), we get:

0.03x = 185.25

x = 6,175

Substituting x = 6,175 into equation (1), we get:

y = 5,160

Therefore, $6,175 is invested at 9% and $5,160 is invested at 6%.

Explanation: