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19 February, 02:45

Jolene invests her savings in two bank accounts, one paying 6 percent and the other paying 10 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 7502 dollars. How much did she invest at each rate?

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  1. 19 February, 03:01
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    Answer: She invested $34100 at 10% and $68200 at 6%.

    Steps:

    Let x1 be the amount invested at 6%, x2 the amount at 10%

    After one year the simple interest is amount * (interest rate).

    The total interest after one year:

    7502 = x1 * 0.06 + x2 * 0.10

    and we know that x1 = 2*x2

    7502 = 2 * x2 * 0.06 + x2 * 0.10 = 0.22 * x2

    x2 = 34100

    x1 = 2*34100=68200
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