Question

$26,876 is invested, part at 9% and the rest at 5%. If the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 5% by $720.78, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

9514 1404 393

Answer:

• $14,747 at 9%
• $12,129 at 5%

Explanation:

Let x represent the amount invested at 9%. Then the difference in interest amounts is ...

(9%)x -(5%)(26876 -x) = 720.78 . . . . . assuming a 1-year investment

0.14x -1343.80 = 720.78 . . . . . . . . . simplify

0.14x = 2064.58 . . . . . . . . . . . . . . add 1343.80

x = 14,747 . . . . . . . . . . . . . . . . divide by 0.14

$14,747 is invested at 9%; $12,129 is invested at 5%.