Problem 56 - Intermediate Algebra

Question

Part of $$\$ 8400$$ is invested at 5\%, and the remainder is invested at \(8 \%\). The total yearly interest from the two investments is $$\$ 576$$. Determine how much is invested at each rate.

Step-by-Step Solution

Answer

\$3200 at 5% and \$5200 at 8%.

Step 1: Understanding the Problem

We have a total of \\( 8400 that is split into two investments. Part is invested at 5% interest and the remainder at 8%. The total yearly interest from both investments is \\) 576. We need to figure out how much was invested at each rate.

Step 2: Defining Variables

Let \( x \) be the amount invested at 5%, and \( 8400 - x \) be the amount invested at 8%. These expressions represent the two parts of the total investment.

Step 3: Setting Up the Interest Equations

The interest from the first investment is \( 0.05x \) and from the second is \( 0.08(8400 - x) \). The total interest from both is \$ 576, so we set up the equation: \[ 0.05x + 0.08(8400 - x) = 576 \].

Step 4: Solving the Equation

First, distribute the 0.08: \( 0.08 \times 8400 - 0.08x = 672 - 0.08x \). Substitute back into the equation to get \( 0.05x + 672 - 0.08x = 576 \). Simplify to \( -0.03x + 672 = 576 \).

Step 5: Isolate the Variable

Subtract 672 from both sides to get \( -0.03x = 576 - 672 \), which simplifies to \( -0.03x = -96 \).

Step 6: Solve for x

Divide both sides by -0.03 to isolate \( x \): \( x = \frac{-96}{-0.03} = 3200 \). This means \$3200 is invested at 5%.

Step 7: Determine the Other Amount

Using \( x = 3200 \), calculate the amount invested at 8%: \( 8400 - x = 8400 - 3200 = 5200 \). So, \$5200 is invested at 8%.

Key Concepts

Simple Interest

Simple interest is a fundamental concept in finance and mathematics used to calculate the interest earned or paid on an investment or loan. The interest is calculated on the original principal amount, meaning it does not compound over time.

To find simple interest, use the formula:

• Interest ( \( I \) ) = Principal ( \( P \) ) \( \times \) Rate ( \( r \) ) \( \times \) Time ( \( t \) )

Where

• Principal, is the initial sum of money
• Rate, is the percentage rate per period
• Time, is the duration for which the money is invested or borrowed

In the exercise, the simple interest for each investment is calculated separately using this formula. Remember, with simple interest, the total interest is directly proportional to time and rate applied.

Interest Rate Calculation

Calculating interest rates in word problems often involves understanding how different rates affect overall outcomes. In this particular problem, two different interest rates are applied to different portions of a single principal amount. The objective is to compute how much was invested at each rate given a total interest.

To solve this, you set up expressions where:

• \( 0.05x \) represents the interest gained from the portion invested at 5%.
• \( 0.08(8400 - x) \) represents the interest from the amount invested at 8%.

Given the total interest is \( 576 \), the equation set is:

• \( 0.05x + 0.08(8400 - x) = 576 \)

By solving this equation, you can find out exactly how much was invested at each rate. This problem showcases the calculation of interest based on different returns in one scenario.

Word Problems in Algebra

Word problems in algebra require translating a written description into a mathematical equation. These types of problems can sometimes be complex because it involves understanding the context and setting up correct mathematical expressions.

Here, interpreting the word problem as an algebraic expression involves:

• Identifying variables: In this case, let \( x \) be the amount invested at 5%, then \( 8400 - x \) is automatically the amount at 8%.
• Setting up equations: Writing all conditions described in the problem into algebraic form.
• Solve the equation: Simplifying and solving to find the values for the variables.

These steps help visualize and solve problems by breaking them down into smaller, manageable pieces. Algebraic word problems like these usually enhance logical thinking and problem-solving skills.