Vic invested part of the 40,000 bonus that he received at 8%. The remainder was invested at 10%. If after 2 years, the earning from 8% investment is 1.2 times his earnings from 10% investment, how much was invested at each rate?
To determine how much Vic invested at each rate, let's use algebraic equations.
Let's assume the amount Vic invested at 8% is x dollars.Then, the amount invested at 10% would be the remaining amount, which is (40,000 - x) dollars.
The interest earned from the 8% investment over 2 years is calculated as:
Interest = Principle x Rate x Time
Therefore, the interest earned from the 8% investment would be 0.08x.
The interest earned from the 10% investment over 2 years is calculated as:
Interest = Principle x Rate x Time
In this case, the interest earned from the 10% investment would be 0.10(40,000 - x).
Based on the given information, we know that the earnings from the 8% investment are 1.2 times the earnings from the 10% investment.
So, we can set up the following equation:
0.08x = 1.2 * 0.10(40,000 - x)
Now, we can solve for x by simplifying and solving the equation:
0.08x = 0.12(40,000 - x) (multiplied 0.10 * 0.12)
0.08x = 4,800 - 0.12x (distributed the multiplication)
0.20x = 4,800 (combined like terms by adding 0.12x to both sides)
x = 4,800 / 0.20 (divided both sides by 0.20)
x = 24,000
Therefore, Vic invested $24,000 at 8% and the remaining amount ($40,000 - $24,000 = $16,000) was invested at 10%.