Amanda invested a total of $3,600 into three separate accounts that pay 5%, 7%, and 9% annual interest. Amanda has four times as much invested in the account that pays 9% as she does in the account that pays 5%. If the total interest for the year is $282, how much did Amanda invest in each account?
let x be the amount at 5%
4x at 9%
3600-(x+4x) at 7%
.05x + .09*4x + .07(3600-5x) = 282
.05x + .36x + 252 - .35x = 282
.06x = 30
x = 500
so, she has
500 at 5%
2000 at 9%
1100 at 7%
To solve this problem, we need to set up a system of equations based on the information given.
Let's assume that Amanda invested x dollars in the account that pays 5%. Since the account that pays 9% has four times as much investment as the one that pays 5%, she invested 4x dollars in the account that pays 9%. The remaining amount of money, which is 3600 - x - 4x, is invested in the account that pays 7%.
Now, let's calculate the interest earned from each account. The interest earned from the account that pays 5% is (5/100) * x = 0.05x dollars. The interest earned from the account that pays 7% is (7/100) * (3600 - x - 4x) = 0.07(3600 - 5x) dollars. The interest earned from the account that pays 9% is (9/100) * 4x = 0.36x dollars.
According to the problem, the total interest earned for the year is $282. So, we can set up the following equation:
0.05x + 0.07(3600 - 5x) + 0.36x = 282
Now, let's solve this equation to find the value of x.
Multiplying out the equation, we get:
0.05x + 252 - 0.35x + 0.36x = 282
Combining like terms, we have:
0.06x + 252 = 282
Subtracting 252 from both sides, we obtain:
0.06x = 30
Dividing both sides by 0.06, we find:
x = 500
Therefore, Amanda invested $500 in the account that pays 5%. To find the investment in the account that pays 9%, we multiply this amount by 4, which gives us $2000. The remaining amount of $3600 - $500 - $2000 = $1100 is invested in the account that pays 7%.
So, Amanda invested $500 in the account that pays 5%, $2000 in the account that pays 9%, and $1100 in the account that pays 7%.