An actor invests some money at 7%, and $44000 more than three times the amount at 10%. The total annual interest earned from the investment is $41770. How much did he invest at each amount
if $x at 7%, then
.07x + .10(3x+44000) = 41770
x = 101,000
Let's break down the problem into two parts to find the amount invested at each interest rate.
Part 1: Let x be the amount invested at 7% interest rate.
The interest earned from this investment can be calculated as 0.07x.
Part 2: The actor invests $44000 more than three times the amount invested at 7%.
So, the amount invested at 10% interest rate is (3x + $44000).
The interest earned from this investment can be calculated as 0.1(3x + $44000).
Now, we can set up an equation using the given information:
0.07x + 0.1(3x + $44000) = $41770
To solve this equation, we need to simplify and solve for x.
0.07x + 0.3x + $4400 = $41770
0.37x + $4400 = $41770
0.37x = $41770 - $4400
0.37x = $37370
To isolate x, we divide both sides of the equation by 0.37:
x = ($37370) / 0.37
x ≈ $101000
Therefore, the actor invested approximately $101000 at 7% interest rate.
To find the amount invested at 10% interest rate, we substitute the value of x into the equation:
Amount invested at 10% = 3x + $44000
Amount invested at 10% = 3($101000) + $44000
Amount invested at 10% ≈ $303000 + $44000
Amount invested at 10% ≈ $347000
Therefore, the actor invested approximately $101000 at 7% and $347000 at 10%.