An investment banker received a bonus and invested that money in two investments. He put a certain sum of money in the first investment that returned 7% interest after one year. In the second investment he invested four times that of the first investment and got a 13% interest return after one year. If the total interest from his investments after one year was $7,965.00, find the amount invested at each interest rate.
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first investment --- x
2nd investment ----- 4x
.07x + .13(4x) = 7965
solve for x
To solve this problem, we can set up a system of equations based on the given information.
Let's say the amount invested in the first investment is x dollars. According to the problem, the second investment is four times the amount of the first investment, so the amount invested in the second investment is 4x dollars.
The interest on the first investment is 7% of x, which is 0.07x.
The interest on the second investment is 13% of 4x, which is 0.13 * 4x = 0.52x.
According to the problem, the total interest from both investments is $7,965.00. Therefore, we can write the equation:
0.07x + 0.52x = 7,965
Combining like terms, we get:
0.59x = 7,965
To solve for x, we divide both sides of the equation by 0.59:
x = 7,965 / 0.59
Using a calculator, we find:
x ≈ 13,500
So, the investment banker invested approximately $13,500 in the first investment.
To find the amount invested in the second investment, we can substitute the value of x in the equation:
4x = 4 * 13,500 = 54,000
Therefore, the investment banker invested approximately $54,000 in the second investment.
In summary, the amount invested at each interest rate is approximately $13,500 in the first investment and $54,000 in the second investment.