Question

Nelson has half of his investments in stock paying a 6% dividend and the other half in a stock paying 9% interest. if his total annual interest is $660 how much does he have invested?

Answer 1

Given:

percent of dividend = 6%

percent of interest = 9%

Total annual interest = $ 660

Let the amount invested in both stocks be x and y.

Annual interest on 6% dividend gives:

`\begin{gathered} =\text{ }(6)/(100)\text{ }*\text{ x} \\ =\text{ 0.06x} \end{gathered}`

Annual interest on 9% interest rate:

`\begin{gathered} =\text{ }\frac{9}{100\text{ }}*\text{ y} \\ =\text{ 0.09y} \end{gathered}`

The total annual interest is $ 660. We can write:

`0.06x\text{ + 0.09y =660}`

We are given that Nelson divided his investment in half. This implies:

`x\text{ = y}`

Substituting, we have:

`\begin{gathered} 0.06x\text{ + 0.09x = 660} \\ 0.15x\text{ = 660} \end{gathered}`

Divide both sides by 0.15:

`\begin{gathered} (0.15x)/(0.15)\text{ = }(660)/(0.15) \\ x\text{ = 4400} \end{gathered}`

Hence, the amount Nelson has invested:

`\begin{gathered} x\text{ = y = 4400} \\ \text{Amount invested = 4400 + 4400} \\ =\text{ 8800} \end{gathered}`

Answer:

Nelson has $4400 invested in each investment