Question

Casey invests $9500 in two different accounts. The first account paid 10%, the second account paid 11% in interest. At the end of the first year, he had earned $986 in interest. How much was in each account?

A) $4500 at 10%, $5000 at 11%
B) $5000 at 10%, $4500 at 11%
C) $6000 at 10%, $3500 at 11%
D) $3500 at 10%, $6000 at 11%

Answer 1

Casey invested $5000 at 10% and $4500 at 11%.

Step-by-step explanation:

Let's assume that Casey invested x dollars in the first account and y dollars in the second account.

The first account pays 10% interest, so the interest earned from this account is 0.10x.

The second account pays 11% interest, so the interest earned from this account is 0.11y.

We can set up the following equation based on the given information: 0.10x + 0.11y = 986.

We also know that x + y = 9500. We can solve this system of equations to find the values of x and y.

Solving the system of equations gives us a solution of x = $5000 and y = $4500.

Therefore, the amount in each account is $5000 at 10% and $4500 at 11%.