Problem 79 - 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. Lottery Winnings Nancy B. Kindy won \(\$ 200,000\) in a state lottery. She first paid income tax of \(30 \%\) on the winnings. Of the rest, she invested some at \(1.5 \%\) and some at \(4 \%,\) earning \(\$ 4350\) interest per year. How much did she invest at each rate?

Step-by-Step Solution

Answer

Nancy invested $50,000 at 1.5% and $90,000 at 4%.

Step 1: Calculate the remaining lottery winnings after tax

Nancy initially won \(200,000, but she had to pay 30% in taxes. First, we calculate the amount of tax she paid: \( 200,000 \times 0.3 = 60,000 \). This means she paid \)60,000 in taxes. Then we subtract this from the original amount to find how much she is left with: \( 200,000 - 60,000 = 140,000 \). So, Nancy has $140,000 to invest.

Step 2: Set up equations for the investment amounts

Let \( x \) be the amount invested at 1.5% and \( y \) be the amount invested at 4%. We know from Step 1 that the total amount invested is $140,000, so \( x + y = 140,000 \).

Step 3: Set up the interest equation

Nancy earns a total of $4350 in interest per year, given that the interest rates are 1.5% for \( x \) and 4% for \( y \). Using the simple interest formula \( I = PRT \), we know that \( 0.015x + 0.04y = 4350 \). We now have a system of linear equations to solve: \( \begin{cases} x + y = 140,000 \ 0.015x + 0.04y = 4350 \end{cases} \).

Step 4: Solve the system of equations

We can solve the system of equations using substitution or elimination. First, solve \( x + y = 140,000 \) for \( x \), giving \( x = 140,000 - y \). Substitute this into the second equation: \( 0.015(140,000 - y) + 0.04y = 4350 \). Simplifying, we get \( 2100 - 0.015y + 0.04y = 4350 \), or \( 0.025y = 2250 \). Solving for \( y \), we find \( y = 90,000 \). Then substitute back to find \( x: x = 140,000 - 90,000 = 50,000 \).

Step 5: Verify the solution

To verify, calculate the interest earned from each part. For \( x = 50,000 \) at 1.5%, the interest is \( 0.015 \times 50,000 = 750 \). For \( y = 90,000 \) at 4%, the interest is \( 0.04 \times 90,000 = 3600 \). Adding these, we get \( 750 + 3600 = 4350 \), which equals the given total interest, confirming the solution is correct.

Key Concepts

Simple Interest Formula

Understanding the simple interest formula is crucial for solving many investment problems. Simple interest is calculated on the principal amount, or the initial sum of money deposited or borrowed. The formula to calculate simple interest is: \[ I = PRT \] where

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

For example, if you invest \(1,000 at an annual interest rate of 3% for one year, the interest can be calculated by substituting these values into the formula: \[ I = 1000 \times 0.03 \times 1 = 30 \] So, the interest earned is \)30. This formula is handy for quick calculations and plays a key role in many financial decisions. Remember, since \(T\) is often one year, it sometimes affects calculations less directly. Nevertheless, it's essential in understanding how earnings are compounded over different periods.

Linear Equations

Linear equations are equations of the first degree, meaning they involve only the powers of one. In investment problems, they often emerge when setting up a scenario such as dividing funds between different accounts. For instance, if you have a total amount \(x\) and you know portions of it are invested at different rates, you might have something like \[ a + b = x \] where \(a\) and \(b\) represent different portions of funds.In problems such as Nancy's, we also set up interest equations—itself a form of linear equation—from the simple interest formula: \[ 0.015x + 0.04y = 4350 \] Here, each term represents the interest earned from different investments, and their sum equals total interest. Finding solutions to these equations involves methods such as substitution and elimination, facilitating the understanding of how funds are allocated or distributed across different investments.Linear equations provide a structured way to represent a financial situation with variables and constants, offering clarity and solvability to investment or financial planning problems.

Investment Problems

Investment problems typically deal with how to allocate funds across different investment options to achieve a desired financial outcome. They incorporate principles such as simple interest and concepts from algebra, such as linear equations. When solving these problems:

• Determine the total amount available to invest after any deductions, such as taxes.
• Identify different investment options and their interest rates.
• Set up equations based on the total sum and interest computations.
• Solve these equations to find out how much money should be invested in each option to meet specific goals.

In the problem from the exercise, Nancy must decide how to invest her post-tax lottery winnings: part at 1.5% and part at 4%, such that her total interest meets a target sum. Through a systematic approach using linear equations, she can accurately distribute her funds between the investments. Investment problems not only test our math skills but also our ability to think strategically about financial decisions in real-world contexts. Understanding them helps prepare us for similar scenarios involving financial planning and resource allocation.