91.0k views
4 votes
$4,028

is invested, part at 14%
and the rest at 7%
. If the interest earned from the amount invested at 14%
exceeds the interest earned from the amount invested at 7%
by $313.39
, how much is invested at each rate? (Round to two decimal places if necessary.)

User Solotim
by
9.6k points

1 Answer

0 votes

Answer:

To solve this problem, let's assume that the amount invested at 14% is x dollars. Since the total amount invested is $4,028, the amount invested at 7% would be the remainder, which is (4,028 - x) dollars.

The interest earned from the amount invested at 14% can be calculated using the formula: (x * 14%) = 0.14x dollars.

Similarly, the interest earned from the amount invested at 7% can be calculated using the formula: ((4,028 - x) * 7%) = 0.07(4,028 - x) dollars.

According to the problem, the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 7% by $313.39. Therefore, we can write the equation:

0.14x - 0.07(4,028 - x) = 313.39.

Now, let's solve this equation:

0.14x - 0.07(4,028 - x) = 313.39

0.14x - 281.96 + 0.07x = 313.39

0.21x - 281.96 = 313.39

0.21x = 313.39 + 281.96

0.21x = 595.35

x = 595.35 / 0.21

x ≈ 2835

Therefore, approximately $2,835 is invested at 14% and the remainder, $4,028 - $2,835 = $1,193, is invested at 7%.

Explanation:

User Jjrdk
by
8.4k points