SOLUTION: Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 8 percent simple interest per year. She puts twice as much in the lower-yielding account

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 8 percent simple interest per year. She puts twice as much in the lower-yielding account       Log On


   



Question 1191434: Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 8 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 6498 dollars. How much did she invest at each rate?
Amount invested at 5 percent interest is $
Amount invested at 8 percent interest is

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


A standard straightforward algebraic solution....

let x = amount invested at 8%
then 2x = amount invested at 5%

The total interest was $6498:

.08%28x%29%2B.05%282x%29=6498
.08x%2B.10x=6498
.18x=6498
x=6498%2F.18=36100

ANSWER: x=$36,100 at 8%; 2x=$72,200 at 5%.

I personally would use a slightly different path which keeps the numbers I need to work with a bit smaller -- allowing a mental solution if your mental arithmetic is good.

Twice as much invested at 5% as at 8% means an average interest percentage rate of (2(5)+8)/3=18/3=6. $6498 interest at 6% means the total amount invested was $6498/.06 = $108,300.

The amount invested at 8% was one-third of that, which is $36,100; the amount invested at 5% was twice that, or $72,200.