Question

A man invests $4100 in three accounts that pay 6 % ,8 % , and 9 % in annual interest, respectively. He has 3times as much invested at 9 % than he does at 6 % . If his total interest for the year is $334, how much isinvested at each rate?

Answer 1

Write out the given parameters

`\begin{gathered} \text{amount invested in 3 accounts is \$4100} \\ \text{rate from each bank is 6\%, 8\% and 9\%} \\ \text{Total interest for the year is \$334} \end{gathered}`

We are to find the amount invested at each rate.

Define the parameter of the amount

Let the amount invested at

`\begin{gathered} \text{ 6\% be y} \\ \text{Then} \\ \text{ 9\% will be 3y} \\ \text{and } \\ \text{ at 8\% we have \$4100-(3y}+y) \end{gathered}`

Since the total interset received is $ 334

The total interest received will be the addition of the interest received at each rate

Hence

`\begin{gathered} \text{ Recall that 6\%=0.06,8\%=0.08 and 9\%=0.09} \\ \text{Then} \\ 334=0.06y+0.09*3y+0.08(4100-4y) \\ 334=0.06y+0.27y+328-0.32y \\ \end{gathered}`

Then subtract 328 from both sides, we have

`\begin{gathered} 334-328=0.06y+0.27y-0.32y+328-328 \\ 6=0.01y \\ \text{divide bothe side by 0.01} \\ y=(6)/(0.01)=600 \end{gathered}`

Then the amount invested at each rate is

`\begin{gathered} \text{ at 6\% is y=600} \\ \text{ at 9\% is 3y=3(600)=1800} \\ \text{ and } \\ \text{ at 8\% is 4100-4y=4100-4(600)=4100-2400}=1700 \end{gathered}`

Therefore,

The amount invested at 6% is 600

The amount invested at 8% is 1700

The amount invested at 9% is 1800