Question

A woman has a total of $8000 to invest she invest part of the money in the account that pays 11% per year and the rest into account that pays 12% per year if the interest earned in the first year is $910 how much did she invest in each account?

Answer 1

Let's assume the woman invested x dollars in the account that pays 11% per year.

Since she invested a total of $8000, the amount invested in the account that pays 12% per year would be (8000 - x) dollars.

Now, let's calculate the interest earned from each investment:

Interest from the 11% account: 0.11x

Interest from the 12% account: 0.12(8000 - x)

According to the given information, the total interest earned in the first year is $910. Therefore, we can set up the following equation:

0.11x + 0.12(8000 - x) = 910

Let's solve this equation to find the value of x:

0.11x + 0.12 * 8000 - 0.12x = 910

0.11x - 0.12x = 910 - 0.12 * 8000

-0.01x = 910 - 960

-0.01x = -50

Dividing both sides by -0.01:

x = (-50) / (-0.01)

x = 5000

Therefore, the woman invested $5000 in the account that pays 11% per year.

The amount invested in the account that pays 12% per year would be 8000 - 5000 = $3000.

So, she invested $5000 in the 11% account and $3000 in the 12% account.