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Jolene invests her savings in two bank accounts, one paying 6 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 7266 dollars. How much did she invest at each rate?

Amount invested at 6 percent interest is $
Amount invested at 9 percent interest is $

Respuesta :

Answer:

Amount invested at 6 percent interest is $69,200

Amount invested at 9 percent interest is $34,600

Step-by-step explanation:

Let the amount invested (or principal) at 6 percent interest be $2x

The principal(P) (amount invested) at 9 percent interest will be $x

Amount(A) = P(1 + rt)

Time(t) = one year = 1

Amount at 6% interest = 2x(1 + 0.06) = $2.12x

Interest accumulated = Amount - Principal = $2.12x - $2x = $0.12x

Amount at 9% interest = x(1 + 0.09) = $1.09x

Interest accumulated = Amount - Principal = $1.09x - $x = $0.09x

Total interest = $7,266

So, $0.12x + $0.09x = $7,266

$0.21x = $7,266

x = [tex]\frac{7,266}{0.21}[/tex] = $34,600

The principal (amount invested) at 9 percent interest rate is $34,600

The principal (amount invested) at 6 percent interest rate is 2 × $34,600 = $69,200

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