Question

Jolene invests her savings in two bank accounts, one paying 4 percent and the other paying 9 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 5899 dollars. How much did she invest at each rate?

Amount invested at 4 percent interest is $ Amount invested at 9 percent interest is $​

Answer 1

Let's call the amount invested in the 4% account "x". According to the problem, Jolene invested twice as much in the 4% account, so the amount invested in the 9% account is "2x".

To calculate the interest earned on each account, we use the formula:

Interest = Principal × Rate × Time

For the 4% account, the interest earned is:

0.04 * x * 1 = 0.04x

For the 9% account, the interest earned is:

0.09 * 2x * 1 = 0.18x

The sum of the interest earned on both accounts is given as $5899:

0.04x + 0.18x = 5899

Combining like terms, we get:

0.22x = 5899

Solving for x, we get:

x = 26813.64

Jolene invested $26813.64 in the 4% account and twice as much in the 9% account, which is:

2x = 2 * 26813.64 = 53627.28

Therefore, Jolene invested $26813.64 at 4% and $53627.28 at 9%.