Trevor invests $10000 for a year.he obtains 8% per year for a part of the money and 10%per an um for remainder.at end of the year he sill receive $950 in interest.how much does he invest at 10%
If $x at 10%, the rest $(1000-x) is at 8%. So, add up the interest
.10x + .08(10000-x) = 950
To find out how much Trevor invested at 10%, we need to set up a system of equations based on the given information.
Let's assume Trevor invested x dollars at 8% and (10000 - x) dollars at 10%.
First, let's calculate the interest earned from each investment:
Interest from the 8% investment = (x * 8%) = 0.08x dollars
Interest from the 10% investment = ((10000 - x) * 10%) = 0.1(10000 - x) dollars
According to the problem, the total interest earned is $950. So, we can write the equation:
0.08x + 0.1(10000 - x) = 950
Now, we can solve this equation to find the value of x:
0.08x + 1000 - 0.1x = 950
-0.02x = -50
x = -50 / -0.02
x = 2500
Therefore, Trevor invested $2500 at 10%.