Investment problem

Question

Jolene invests her savings in two bank accounts, one paying 5 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 4161 dollars. How much did she invest at each rate?

Please have a full explanation and answer.

Answer 1

Let x be the amount invested at 9%.

0.05(2x) + 0.09x = 4161

From here you can solve for x and 2x to give the two amounts invested.

Answer 2

Investment problem: Jolene invests her savings in two bank accounts, one paying 5 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 4161 dollars. How much did she invest at each rate? Please have a full explanation and answer.

Detailed Solution:

First Account - Amount: $x & Interest: $0.05x Second Account - Amount: $y & Interest: $0.09y

Amount: x + y = Do not know

However, we know she invested in 1st account (5%) twice amount as in the 2nd account (9%). Therefore,

1st account (5%) twice amount as in the 2nd account (9%): x = 2y ==> x – 2y = 0 ==> Equation 1

Interest: 0.05x + 0.09y = 4161 ==> Equation 2

In TI-84 Plus, create 2 x 3 Matrix A of above Equations, do the math: Matrix ==> Math ==> rref([A])

rref([A]) gives you the solution: x = 43,800 and y = 21,900

First Account - Amount: x = $43,800 & Interest: $0.05(43,800) = $2190

Second Account - Amount: y = $21,900 & Interest: -$0.09(21,900) = $1971

Check:

Amount: x = 2y ==> $43,800 = 2($21,900) ==> $43,800 = $43,800

Interest: $0.05x + $0.09y = $4161 ==> $0.05(43,800) + $0.09(21,900) = $4161 ==> $4161 = $4161