Question

Peter has money in two savings accounts. One rate is 9% and the other is 11% . If he has $200 more in the 11% account and the total interest is $68 , how much is invested in each savings account?

A)$230 invested in the 9%,$430 invested in the 11% account.
B)$210 invested in the 9%,$420 invested in the 11% account.
C)$200 invested in the 9%,$430 invested in the 11% account.
D)$230 invested in the 9%,$400 invested in the 11% account.

Answer 1

After setting up a system of equations to represent the interest earned from two savings accounts and solving it, we find that $230 is invested in the 9% account and $430 in the 11% account, making option A correct.

Step-by-step explanation:

The problem states that Peter has money in two different savings accounts, one earning 9% interest and the other 11% interest, with the account at 11% having $200 more than the other. To find out how much is invested in each savings account, we set up a system of equations based on the total interest earned, which is $68.

Let the amount in the 9% account be x dollars. Therefore, the amount in the 11% account would be x + $200. The total interest from both accounts is given by:

0.09x + 0.11(x + 200) = 68

Solving this equation, we find:

1. 0.09x + 0.11x + 22 = 68
2. 0.20x = 46
3. x = 46 / 0.20
4. x = $230

Therefore, $230 is invested in the 9% account, and $230 + $200 = $430 is invested in the 11% account.

The correct answer is A) $230 invested in the 9% account, $430 invested in the 11% account.