katie invested a total of $6000, part at 5% simple interest and part at 6% simple interest. at the end of 1…

katie invested a total of $6000, part at 5% simple interest and part at 6% simple interest. at the end of 1 year, the investments had earned $333 interest. how much was invested at each rate? katie invested $ at a rate of 5% and $ at a rate of 6%.

katie invested a total of $6000, part at 5% simple interest and part at 6% simple interest. at the end of 1 year, the investments had earned $333 interest. how much was invested at each rate? katie invested $ at a rate of 5% and $ at a rate of 6%.

Answer

Explanation:

Step1: Let the amount invested at 5% be $x$.

Then the amount invested at 6% is $(6000 - x)$ since the total investment is $6000$.

Step2: Calculate the interest from each investment.

The interest from the 5% - investment in 1 year is $I_1=0.05x$ (using the simple - interest formula $I = Prt$, where $t = 1$). The interest from the 6% - investment in 1 year is $I_2=0.06(6000 - x)$.

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

The total interest is $333$, so $0.05x+0.06(6000 - x)=333$.

Step4: Expand and simplify the equation.

$0.05x + 360-0.06x=333$. Combine like terms: $360-(0.06x - 0.05x)=333$, which gives $360 - 0.01x=333$.

Step5: Solve for $x$.

Subtract 360 from both sides: $-0.01x=333 - 360=-27$. Then divide both sides by $- 0.01$: $x=\frac{-27}{-0.01}=2700$.

Step6: Find the amount invested at 6%.

The amount invested at 6% is $6000 - x=6000 - 2700 = 3300$.

Answer:

Katie invested $2700$ at a rate of 5% and $3300$ at a rate of 6%.