A man had $12,000 to invest. If his income from a savings account earing 7.5% is $280.00 greater than his income from another investment at 8%, how much did he invest at each rate?
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If he has $12,00 to invest, then the amount invested in the two accounts are X and 12000 - X.
.075 of one plus .08 of the other = $280
You should be able to work it from there. I hope this helps. Thanks for asking.
To solve this problem, we can set up a system of equations.
Let's assume that the man invested x dollars at 7.5% and (12000 - x) dollars at 8%. We can then set up the following equation to represent the interest earned from the two investments:
0.075x = 0.08(12000 - x) + 280
In this equation, 0.075 represents the interest rate of 7.5%, 0.08 represents the interest rate of 8%, and 280 represents the extra income earned from the first investment compared to the second investment.
To solve this equation, we can follow these steps:
1. Distribute 0.08 to the terms inside the parentheses:
0.075x = 960 - 0.08x + 280
2. Combine like terms:
0.155x = 1240
3. Divide both sides of the equation by 0.155:
x = 1240 / 0.155
x ≈ 8000
So, the man invested approximately $8000 at the 7.5% interest rate.
To find the amount invested at the 8% interest rate, we can subtract the amount invested at 7.5% from the total amount invested:
12000 - x = 12000 - 8000 = 4000
Therefore, the man invested $8000 at 7.5% and $4000 at 8%.