Question

Tamara invests $8000 in two different accounts. The first account has a simple interest rate of 3% and the second account has a simple interest rate of 2%. How much did she invest in each account if the interest earned in them in the same at the end of one year?
PLEASE Show Work/How to solve it (: ​

Answer 1

Answer:

• $3200 at 3%
• $4800 at 2%

Step-by-step explanation:

Interest earned is proportional to the interest rate, so if the interest earned is the same, the amounts invested must be inversely proportional to the interest rates. That is, for the 3% and 2% accounts, the ratio of money invested is 2:3.

In other words, 2/5 of the money ($3200) was invested at 3%, and 3/5 of the money ($4800) was invested at 2%.

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If you need an equation, you can let x represent the amount invested at the highest rate. Then 8000-x is the amount invested at the lower rate. For the interest in the two accounts to be equal, we have ...

  3%·x = 2%·(8000-x) . . . . . the amounts of interest earned are the same

  3/2·x = 8000 -x . . . . . . . . divide by 2%

  5/2·x = 8000 . . . . . . . . . . . add x and simplify

  x = 8000·(2/5) = 3200  . . . multiply by the inverse of the x coefficient

  8000-x = 8000 -3200 = 4800 . . . . the amount invested at the lower rate

Tamara invested $3200 in the 3% account and $4800 in the 2% account.

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She earned $96 in each account for the year.