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

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) (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:

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.

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