Problem 29 - Intermediate Algebra

Question

Part of $$\$ 3000$$ is invested at \(4 \%\), another part at \(5 \%\), and the remainder at \(6 \%\). The total yearly income from the three investments is $$\$ 160$$. The sum of the amounts invested at \(4 \%\) and \(5 \%\) equals the amount invested at \(6 \%\). Determine how much is invested at each rate.

Step-by-Step Solution

Answer

$500 is invested at 4%, $1000 at 5%, and $1500 at 6%.

Step 1: Define Variables

Let \( x \) be the amount invested at \( 4\% \), \( y \) be the amount invested at \( 5\% \), and \( z \) be the amount invested at \( 6\% \). We know that the total amount invested is \$3000, so we can write an equation: \[ x + y + z = 3000 \]

Step 2: Set Up Income Equation

We also know the total yearly income from the investments is \$160. The income from each investment can be described by:1. Income from \( 4\% \) is \( 0.04x \).2. Income from \( 5\% \) is \( 0.05y \).3. Income from \( 6\% \) is \( 0.06z \).Adding these gives the total income: \[ 0.04x + 0.05y + 0.06z = 160 \]

Step 3: Introduce Investment Relationship

According to the problem, the sum of the amounts invested at \( 4\% \) and \( 5\% \) equals the amount invested at \( 6\% \):\[ x + y = z \]

Step 4: Solve the System of Equations

We now have a system of equations:1. \( x + y + z = 3000 \)2. \( 0.04x + 0.05y + 0.06z = 160 \)3. \( x + y = z \)Substituting equation 3 into equation 1 gives:\[ x + y + (x + y) = 3000 \Rightarrow 2x + 2y = 3000 \Rightarrow x + y = 1500 \] Now, use equation 3 in equation 2:\[ 0.04x + 0.05y + 0.06(x + y) = 160 \]Simplifying, we get:\[ 0.04x + 0.05y + 0.06x + 0.06y = 160 \]Combine like terms:\[ 0.1x + 0.11y = 160 \]Now solve for \( x \) and \( y \) using these two equations:1. \( x + y = 1500 \)2. \( 0.1x + 0.11y = 160 \)From \( x + y = 1500 \): \( y = 1500 - x \)Substitute \( y = 1500 - x \) into \( 0.1x + 0.11y = 160 \):\[ 0.1x + 0.11(1500 - x) = 160 \]Solve for \( x \):\[ 0.1x + 165 - 0.11x = 160 \]\[ -0.01x = -5 \]\[ x = 500 \]Substitute \( x = 500 \) back to find \( y \):\[ y = 1500 - 500 = 1000 \]Finally, find \( z \) using \( x + y = z \):\[ z = 500 + 1000 = 1500 \]

Step 5: Verify the Solution

Ensure the found values satisfy all original conditions:1. Total investment: \( 500 + 1000 + 1500 = 3000 \)2. Total income check: - At \( 4\% \), \\(500: \( 0.04 \times 500 = 20 \) - At \( 5\% \), \\)1000: \( 0.05 \times 1000 = 50 \) - At \( 6\% \), \$1500: \( 0.06 \times 1500 = 90 \)Total income: \( 20 + 50 + 90 = 160 \)All conditions are satisfied.

Key Concepts

Investment Problems

Investment problems typically involve determining where and how much to invest to achieve a specific financial goal. These problems require understanding how different investment vehicles work and how they contribute to overall returns. In this particular problem, we are dealing with three types of investments, each with its own return rate. Our goal is to distribute a total of \( \$ 3000 \) across these investments in such a way that maximizes income while satisfying the condition that the sum of certain investments equals another.

To tackle such problems, we usually set up equations based on:

• The total amount of money one has for investment.
• The incomes generated from each specific investment based on its interest rate.
• Any other conditions or relationships specified in the problem, such as a requirement that certain amounts equal each other.

These equations together form a system of equations, which we then solve simultaneously to find the values of the investment amounts that meet all criteria.

Interest Rates

Interest rates are crucial in investment problems because they determine how much income an investment will generate over a period of time. In this problem, we have different rates: \( 4\% \), \( 5\% \), and \( 6\% \), contributing to the total income of \( \$160 \).

Interest income for each type of investment can be calculated with a simple formula:
\[ \text{Interest Income} = \text{Principal Amount} \times \text{Interest Rate} \]

For example, the income generated by an investment of \( x \) at an interest rate of \( 4\% \) is expressed as \( 0.04x \).

Understanding how interest rates apply to investments is key because small differences in rates can significantly affect total income. Therefore, accurately determining which amounts are invested at which rates is essential in coming up with the correct solution in these types of problems.

Linear Equations

Linear equations are fundamental in solving investment problems as they help us set up and solve the relationships between different variables involved. In this problem, we use linear equations to denote total investments, income from investments, and any specified equality constraints.

We start by defining variables for the amounts invested at each interest rate:

• \( x \): amount invested at \( 4\% \)
• \( y \): amount invested at \( 5\% \)
• \( z \): amount invested at \( 6\% \)

With these variables, the total investment equation would be \( x + y + z = 3000 \), expressing the total amount initially invested. We also form a second equation to represent income: \( 0.04x + 0.05y + 0.06z = 160 \). Lastly, to reflect the problem constraint, we use: \( x + y = z \).

Solving these equations simultaneously requires algebraic manipulation to find the values of \( x \), \( y \), and \( z \). By substituting and rearranging these expressions, we simplify and determine the solution, ensuring they satisfy all given conditions.