Problem 41 - Intermediate Algebra

Question

Investing. A woman invested some money in a credit union paying \(5 \%\) annual simple interest and three times as much in a money market account paying \(4.25 \%\) annual simple interest. If she earned \(\$ 1,420\) interest in one year, how much did she invest in each account?

Step-by-Step Solution

Answer

The woman invested \( \$8000 \) in the credit union and \( \$24000 \) in the money market account.

Step 1: Define Variables

Let's denote the amount invested in the credit union as \( x \) and the amount invested in the money market account as \( 3x \) since she invested three times as much in the money market account.

Step 2: Write Equations for Interest

The interest from the credit union is calculated using the formula \( \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \). Thus, \( 0.05x \) for the credit union. For the money market account, the interest is \( 0.0425 \times 3x \).

Step 3: Set Up Interest Equation

According to the problem, the total interest earned from both investments is \( \$1420 \). So, the equation is: \( 0.05x + 0.0425 \times 3x = 1420 \).

Step 4: Solve for x

Simplify and solve the equation: 1. \( 0.05x + 0.1275x = 1420 \)2. \( 0.1775x = 1420 \)3. \( x = \frac{1420}{0.1775} \)4. \( x \approx 8000 \).

Step 5: Calculate Each Investment

Now we know \( x \), the credit union investment is \( \$8000 \), and the money market account investment is \( 3x = 24000 \).

Key Concepts

Investment Problems

Investment problems are an interesting topic in mathematics because they combine real-world scenarios with algebraic thinking. Often, they revolve around determining the allocations of funds in different investment avenues, each with distinct interest rates.

• An investment problem usually involves splitting a sum of money across several investments.
• Each investment earns interest at a specified rate over time.
• The goal is to determine how much was invested in each option based on the total interest earned.

In the given exercise, the task was to discover how much money was placed in two different accounts. One pays a higher interest rate but has a lower initial investment, while the other has a higher initial investment but offers a slightly lower interest rate.
Recognizing this setup helps us apply logical reasoning and algebraic techniques to find solutions.

Algebraic Equations

Algebraic equations are a powerful tool for solving mathematics problems involving investments. They help us establish relationships between known and unknown quantities.

• Identify variables: Here, the unknown amount invested in one account is represented by a variable, often denoted as \( x \).
• Create equations: By relating the variables to known outcomes (like total interest), we form equations.

In the exercise, setting the amount in the credit union as \( x \) and in the money market account as \( 3x \), allows the use of simple interest calculations to develop and simplify equations. The total interest earned equation, \( 0.05x + 0.0425 \times 3x = 1420 \), becomes a straightforward linear equation.
Using algebraic manipulation, we simplify this to \( 0.1775x = 1420 \) and solve by division, isolating \( x \). Thus, algebra becomes a bridge connecting the problem's setup to the solution.

Interest Calculations

Interest calculations are essential to understanding investment returns. With simple interest, the concepts are easier to grasp, as they rely on a direct formula.

• Simple interest is calculated as \( \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \).
• The principal is the initial amount of money invested, the rate is the interest earned per time period, and time is the duration for which the money is invested.

In this exercise, the formula is applied to each account separately. The credit union's interest uses \( 0.05 \times x \) because of its rate, while \( 0.0425 \times 3x \) applies to the money market account.
Combining these interests gives the total interest, in this case, \$1420. Simple interest forms the core mathematical underpinning that allows us to model and solve these financial problems effectively.