A man invests $\$ 3200$ in three accounts that pay $4 \%, 5 \%$, and $6 \%$ in annual interest, respectively. He has 3 times as much invested at $6 \%$ than he does at $4 \%$. If his total interest for the year is $\$ 170$, how much is invested at each rate?
He invested $\$ \quad$ at $4 \%, \$ \square$ at $5 \%$ and $\$ \square$ at $6 \%$.
Submit
Answer
So, the man invested \(\boxed{\$500}\) at 4%, \(\boxed{\$1200}\) at 5%, and \(\boxed{\$1500}\) at 6%.
Step 1 ：Let's denote the amount of money invested at 4%, 5%, and 6% as \(x\), \(y\), and \(z\) respectively.
Step 2 ：We know that the total amount of money invested is $3200, so we can write the equation \(x + y + z = 3200\).
Step 3 ：We also know that the amount of money invested at 6% is three times the amount invested at 4%, so we can write the equation \(z = 3x\).
Step 4 ：Finally, we know that the total interest for the year is $170, so we can write the equation \(0.04x + 0.05y + 0.06z = 170\).
Step 5 ：Now we have a system of three equations, and we can solve this system to find the values of \(x\), \(y\), and \(z\).
Step 6 ：The solution to the system is \(x = 500\), \(y = 1200\), and \(z = 1500\).
Step 7 ：So, the man invested \(\boxed{\$500}\) at 4%, \(\boxed{\$1200}\) at 5%, and \(\boxed{\$1500}\) at 6%.
