Question

a dentist invested a portion of $11,000 in a 8% annual simple - interest account and the remainder in a 6.5% annual simple - interest government bond. the two investments earn $835 in interest annually. how much was invested in each account? at 8% $ at 6.5% $

Explanation:

Step1: Let the amount invested at 8% be $x$.

Then the amount invested at 6.5% is $(11000 - x)$.

Step2: Calculate the interest from each investment.

The interest from the 8% - investment is $I_1=0.08x$ (using the simple - interest formula $I = Prt$, where $t = 1$ year). The interest from the 6.5% - investment is $I_2=0.065(11000 - x)$.

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

The total interest is $835$, so $0.08x+0.065(11000 - x)=835$.

Step4: Expand and simplify the equation.

$0.08x + 715-0.065x=835$. Combine like terms: $(0.08x - 0.065x)+715 = 835$, which gives $0.015x+715 = 835$.

Step5: Solve for $x$.

Subtract 715 from both sides: $0.015x=835 - 715=120$. Then divide both sides by 0.015: $x=\frac{120}{0.015}=8000$.

Step6: Find the amount invested at 6.5%.

The amount invested at 6.5% is $11000 - x=11000 - 8000 = 3000$.

Answer:

at 8%: $8000
at 6.5%: $3000