Problem 78 - A Graphical Approach to College Algebra

Question

Investment problems such as those in Exercises \(75-80\) can be solved by using a method similar to the one explained in Example \(2,\) along with the simple- interest formula \(I=P R T\) where I is the interest earned, \(P\) is the initial amount of money deposited, \(R\) is the annual interest rate as a decimal, and \(T\) is the time the money is deposited in years. Solve each problem. Let \(T=1\) year for each exercise. Investing a Building Fund A church building fund has invested some money in two ways: part of the money at \(4 \%\) interest and four times as much at \(3.5 \% .\) Find the amount invested at each rate if the total annual income from interest is \(\$ 3600\).

Step-by-Step Solution

Answer

$20,000 at 4% and $80,000 at 3.5%.

Step 1: Define Variables

Let's assume the amount invested at 4% is \( x \). According to the problem, four times this amount is invested at 3.5%. So, the amount invested at 3.5% is \( 4x \).

Step 2: Write the Formula for Total Interest

The interest earned from the amount invested at 4% is \( 0.04x \), and the interest from the amount invested at 3.5% is \( 0.035 \times 4x = 0.14x \). According to the problem, the total interest is $3600. So, the equation for total interest is:\[ 0.04x + 0.14x = 3600 \]

Step 3: Solve the Equation for x

Combine like terms:\[ 0.18x = 3600 \]Solve for \( x \) by dividing both sides by 0.18:\[ x = \frac{3600}{0.18} \approx 20000 \]

Step 4: Calculate Amounts Invested

We have found that \( x = 20000 \). This is the amount invested at 4%. The amount invested at 3.5% is \( 4x = 4 \times 20000 = 80000 \).

Step 5: Verify the Solution

Calculate the interest for each amount:- Interest from \\(20,000 at 4%: \( 0.04 \times 20000 = 800 \)- Interest from \\)80,000 at 3.5%: \( 0.035 \times 80000 = 2800 \)The total interest is \( 800 + 2800 = 3600 \), which matches the given condition, confirming our solution is correct.

Key Concepts

Simple Interest

Simple interest is an easy way to calculate the interest earned or paid on an investment or loan. To calculate the simple interest, you use the formula: \( I = PRT \). Here:

• \( I \) is the interest earned.
• \( P \) is the principal amount (the initial amount of money).
• \( R \) is the interest rate expressed as a decimal.
• \( T \) is the time in years the money is invested or borrowed.

Understanding simple interest is important for solving basic investment problems, such as determining how much money you will earn from an account or owe on a loan over a given period.
Unlike compound interest, simple interest does not take into account the effect of additional earned interest beyond the initial amount. This makes calculations straightforward and is often used for shorter investment periods or simpler scenarios.
In the context of our exercise, simple interest helps us understand how much the church fund earns from its investments, allowing us to set up effective equations to find the amounts invested at different rates.

Interest Rates

Interest rates are percentages that indicate how much interest will be earned or paid over a certain period, based on the principal amount. They can greatly impact the total amount earned on an investment or paid over the life of a loan.
In the exercise, there are two different interest rates involved: 4% and 3.5%. Here, the church fund’s earnings are based on these rates, determining how much money is earned annually on each investment.
To use interest rates correctly in calculations, convert them from percentages to decimal form. This means dividing by 100. So:

• 4% becomes \(0.04\)
• 3.5% becomes \(0.035\)

Interest rates are particularly crucial when calculating simple interest as in this example. They directly affect the amount of money the church building fund earns.
For instance, a higher rate results in more interest. Conversely, a lower interest rate yields less. Proper conversion and usage of interest rates in formulas like \(I = PRT\) are essential for accurately solving investment problems.

Algebraic Equations

Algebraic equations help us solve problems involving unknown variables by establishing relationships between them. In investment problems, such as the one described, we use algebraic equations to determine unknown amounts based on given conditions.
The exercise involves setting up an equation from the simple interest formula for each investment. We define the variable \(x\) to represent the amount invested at 4%. Knowing that four times this amount is invested at 3.5%, we can express this as \(4x\). These expressions are crucial for setting up our equation.
We then calculate the total interest from both investments and set up the equation \[0.04x + 0.14x = 3600\]. This equation is derived from the combination of interest earned from each part of the investment.
By solving for the variable \(x\), we find its value and consequently determine the amounts invested at each interest rate. The ability to manipulate and solve such algebraic expressions is fundamental in solving real-world financial problems. It requires competency in combining like terms and applying inverse operations to isolate variables.