Question

a professional athlete deposited an amount of money into a high - yield mutual fund that earns 14% annual simple interest. a second deposit, $2,300 more than the first, was placed in a certificate of deposit earning 8% annual simple interest. in one year, the total interest earned on both investments was $448. how much money was invested in the mutual fund?

Explanation:

Step1: Let the amount invested in mutual - fund be $x$.

The amount invested in the certificate of deposit is $x + 2300$.

Step2: Calculate the interest from each investment.

The interest from the mutual fund at 14% simple interest for 1 year is $I_1=0.14x$. The interest from the certificate of deposit at 8% simple interest for 1 year is $I_2 = 0.08(x + 2300)$.

Step3: Set up the equation based on the total interest.

The total interest $I_1+I_2 = 448$. So, $0.14x+0.08(x + 2300)=448$.

Step4: Expand and simplify the equation.

$0.14x+0.08x+0.08\times2300 = 448$. Which becomes $0.14x+0.08x+184 = 448$. Combining like - terms, we get $0.22x+184 = 448$.

Step5: Solve for $x$.

Subtract 184 from both sides: $0.22x=448 - 184=264$. Then divide both sides by 0.22: $x=\frac{264}{0.22}=1200$.

Answer:

1200