Problem 38 - Elementary Algebra

Question

A sum of \(\$ 2300\) is invested, part of it at \(10 \%\) interest and the remainder at \(12 \%\). If the interest earned by the \(12 \%\) investment is \(\$ 100\) more than the interest earned by the \(10 \%\) investment, find the amount invested at each rate.

Step-by-Step Solution

Answer

$800 at 10% interest and $1500 at 12% interest.

Step 1: Define Variables

Let's define the variables. Let \( x \) be the amount invested at \( 10\% \) interest and \( y \) be the amount invested at \( 12\% \) interest.

Step 2: Establish Equations

We have two pieces of information to create equations. First, the total sum of \( x+y=2300 \). Second, the interest at \( 12\% \) is \( 100 \) more than at \( 10\% \), so \( 0.12y = 0.10x + 100 \).

Step 3: Solve First Equation

From \( x+y=2300 \), we can express \( y \) in terms of \( x \): \( y = 2300 - x \).

Step 4: Substitute and Solve

Substitute \( y = 2300 - x \) into the second equation: \( 0.12(2300 - x) = 0.10x + 100 \). Expand and rearrange: \( 276 - 0.12x = 0.10x + 100 \).

Step 5: Simplify Equation

Combine like terms: \( 276 - 100 = 0.10x + 0.12x \), which simplifies to \( 176 = 0.22x \).

Step 6: Solve for x

Divide both sides by \( 0.22 \) to find \( x \): \( x = \frac{176}{0.22} \), leading to \( x = 800 \).

Step 7: Calculate y

Use \( y = 2300 - x \) to calculate \( y \): \( y = 2300 - 800 = 1500 \).

Step 8: Verify Solution

Check both conditions: The sum \( x + y = 2300 \) satisfies the initial condition. For interest, \( 0.12 \times 1500 = 180 \) and \( 0.10 \times 800 = 80 \), with \( 180 = 80 + 100 \), confirming the second condition.

Key Concepts

Simple Interest

Simple interest is a way to calculate the interest you earn or pay on a principal amount. Interest is calculated as a percentage of the initial sum and does not itself earn further interest. This makes it straightforward to calculate and understand.

Consider an example where you invest \( \\(1000 \) at an annual interest rate of \( 10\% \). The interest earned each year is \( 0.10 \times 1000 = \\)100 \). If you keep the money invested for three years, the total interest would be \( 3 \times 100 = \$300 \).

• The formula for simple interest: \( I = P \times r \times t \)
• \( I \) is the interest earned.
• \( P \) is the principal amount invested.
• \( r \) is the annual interest rate in decimal form.
• \( t \) is the time period in years.

This straightforward approach ensures that the interest rate is uniformly applied over time.

Variable Definition

In solving word problems, particularly those involving algebra, defining variables is crucial for translating a problem into equations. Variables represent unknown values that you aim to find.

For example, in the problem, two variables help clarify the financial situation:

• \( x \) represents the amount invested at \( 10\% \) interest.
• \( y \) represents the amount invested at \( 12\% \) interest.

By setting up these variables, we simplify the problem by replacing words with symbols, facilitating the formulation of equations. Choosing meaningful symbols or letters helps in remembering what each variable stands for, reducing confusion and errors in further steps.

Linear Equations

Linear equations are mathematical statements expressing a relationship between two variables with a constant rate of change. They appear frequently in algebra word problems, linking two variables through addition, subtraction, multiplication, or division.

In the original exercise, we derived two linear equations:

• The sum of the investments: \( x + y = 2300 \)
• The difference in interest: \( 0.12y = 0.10x + 100 \)

These equations are the key tools that we use to determine the values of \( x \) and \( y \). Solving linear equations requires steps such as substitution or elimination, with the goal of isolating one variable to find its value. Here, substitution entails expressing one variable in terms of another, inserting it into another equation, and then solving for the remaining unknown.

Verification of Solutions

Verification of solutions is an essential step to ensure the equations have been solved correctly. After calculating the values of the variables, it's important to check if these values satisfy both the original conditions set by the problem.

In our problem, we found \( x = 800 \) and \( y = 1500 \). Verification involves checking:

• The total investment: \( x + y = 800 + 1500 = 2300 \)
• The interest condition: \( 0.12 \times 1500 = 180 \) and \( 0.10 \times 800 = 80 \). \( 180 = 80 + 100 \), confirming the interest earned on the \( 12\% \) amount is indeed \( \$100 \) more.

Verification assures that the solution is not only mathematically correct but also contextually valid, fulfilling the problem's requirements. This step is crucial in gaining confidence in your problem-solving skills and ensuring no detail has been overlooked.