Question

wayne miller, an investment banker, invested 55% of the banks available cash in an account that earns 8.25% annual interest. the remainder of the cash was placed in an account that earns 10% annual simple interest. the interest earned in one year was $46,091.25. find the total amount invested.

Explanation:

Step1: Let the total amount invested be $x$.

The amount invested at 8.25% is $0.55x$, and the amount invested at 10% is $(1 - 0.55)x=0.45x$.

Step2: Calculate the interest from each account.

The interest from the 8.25% - account is $I_1 = 0.55x\times0.0825$, and the interest from the 10% - account is $I_2=0.45x\times0.1$.

Step3: Set up the equation for the total interest.

The total interest $I = I_1+I_2$. We know that $I = 46091.25$. So, $0.55x\times0.0825+0.45x\times0.1 = 46091.25$.
First, calculate the coefficients: $0.55\times0.0825 = 0.045375$ and $0.45\times0.1 = 0.045$.
The equation becomes $0.045375x+0.045x = 46091.25$.
Combine like - terms: $(0.045375 + 0.045)x=46091.25$, $0.090375x = 46091.25$.

Step4: Solve for $x$.

$x=\frac{46091.25}{0.090375}=510000$.

Answer:

$510000$