Jolene invests her savings in two bank accounts,

Question

Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 12 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 5412 dollars. How much did she invest at each rate?

Amount invested at 5 percent interest is $

Amount invested at 12 percent interest is $

Answer 1

Let x represent the amount invested at 12%. Since Jolene puts twice as much in the lower yielding account, it follows that, 2x is the amount invested at 5%. The total annual interest earned is the sum of the annual interest earned in both accounts.

Total Annual Interest = Annual Interest Earned at 12% + Annual Interest Earned at 5%

Since simple interest per year is given by amount invested times annual interest rate, we write and solve the following equation:

5412 = 0.12(x) + 0.05(2x)

5412 = 0.12x + 0.10x

5412 = 0.22x

x = 5412 ÷ 0.22

x = 24600 ⇒ 2x = 49200

So that:

The amount invested at 5% interest is $49,200.

The amount invested at 12% interest is $24,600.

Answer 2

Let x be the amount in the high yield acct and 2x the amount in the lower yielding account:

Her annual interest 5412 dollars, so

5412 = 2x(.05) + x(.12)

5412 = (.1+.12)x = .22x

x = 5412/.22 = 24600 in the higher yield account.

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