Question

Jolene invests her savings in two bank accounts, one paying 3 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 3120 dollars. How much did she invest at each rate?
Amount invested at 3 percent interest is $____
Amount invested at 9 percent interest is $___

Answer 1

Explanation:

X is the amount invested at 6%

Y is the amount invested at 9%

0.06X + 0.09Y = 4998

X = 2Y

0.06(2Y) + 0.09Y = 4998

.12Y + 0.09Y = 4998

0.21Y = 4998

21Y = 499800

Y = 499800/21 = 23800

So X = 2*23800 = 47600

$47,600 is invested at 6% and $23800 is invested at 9%

Answer 2

Let's denote the amount Jolene invested at 3 percent interest as 'x' dollars. Since she put twice as much in the lower-yielding account, the amount she invested at 9 percent interest would be '2x' dollars.

To calculate the interest earned from each account, we'll use the formula: Interest = Principal × Rate × Time.

For the 3 percent interest account:

Interest_3_percent = x × 0.03

For the 9 percent interest account:

Interest_9_percent = 2x × 0.09

We know that the total annual interest is $3120, so we can set up the equation:

Interest_3_percent + Interest_9_percent = 3120

Substituting the above equations, we have:

x × 0.03 + 2x × 0.09 = 3120

Simplifying the equation:

0.03x + 0.18x = 3120

0.21x = 3120

Dividing both sides of the equation by 0.21:

x = 3120 / 0.21

x = 14857.14

Therefore, Jolene invested approximately $14,857.14 at 3 percent interest and twice that amount, $29,714.29, at 9 percent interest.