Problem 77 - A Graphical Approach to Precalculus with Limits

Question

To obtain funds necessary for a planned expansion, a small company took out three loans totaling \(\$ 12,500 .\) The company was able to borrow some of the money at \(2 \% .\) It borrowed \(\$ 1000\) more than \(\frac{1}{2}\) the amount of the \(2 \%\) loan at \(3 \%\) and the rest at \(2.5 \% .\) The total annual interest was \(\$ 305 .\) How much did the company borrow at each rate?

Step-by-Step Solution

Answer

The company borrowed $5000 at 2%, $3500 at 3%, and $4000 at 2.5%.

Step 1: Define Variables

Let's denote the amount of money borrowed at \(2\%\) as \(x\). For the \(3\%\) loan, the company borrowed \(\frac{1}{2}x + 1000\). Let's denote the amount borrowed at \(2.5\%\) as \(y\). We can express the total amount of loans as an equation: \[ x + \left(\frac{1}{2}x + 1000\right) + y = 12500 \]

Step 2: Simplify the Total Loan Equation

Simplify the equation from Step 1: \[ x + \frac{1}{2}x + 1000 + y = 12500 \] This can be further simplified to: \[ \frac{3}{2}x + 1000 + y = 12500 \] Subtract 1000 from both sides to get: \[ \frac{3}{2}x + y = 11500 \]

Step 3: Formulate the Total Interest Equation

The total interest from all loans is \$305. This can be equated as: \[ 0.02x + 0.03\left(\frac{1}{2}x + 1000\right) + 0.025y = 305 \] Simplifying the middle term: \[ 0.015x + 30 \] So the equation becomes: \[ 0.02x + 0.015x + 30 + 0.025y = 305 \]

Step 4: Simplify and Rearrange Interest Equation

Combine like terms in the interest equation: \[ 0.035x + 0.025y + 30 = 305 \] Subtract 30 from both sides: \[ 0.035x + 0.025y = 275 \]

Step 5: Solve System of Equations

We now have a system of two equations: \(\frac{3}{2}x + y = 11500\) and \(0.035x + 0.025y = 275\). First, solve the first equation for \(y\): \[ y = 11500 - \frac{3}{2}x \] Substitute this expression for \(y\) into the second equation: \[ 0.035x + 0.025(11500 - \frac{3}{2}x) = 275 \]. Simplify and solve for \(x\).

Step 6: Calculate Values for x, y, and 3% Loan

Substituting into the second equation and simplifying gives: \[ 0.035x + 287.5 - 0.0375x = 275 \] \[ -0.0025x = -12.5 \] So \[ x = 5000 \] dollars. Use this value in \(\frac{1}{2}x + 1000\) to find the 3% loan: \[ \frac{1}{2}(5000) + 1000 = 3500 \] dollars. Substitute \(x = 5000\) into \(y = 11500 - \frac{3}{2}x\) to find \(y\): \[ y = 11500 - 7500 = 4000 \] dollars.

Key Concepts

Interest Rate

Interest rate is the amount charged by a lender to a borrower, expressed as a percentage of the loan amount. Understanding interest rates is crucial for any financial decision involving borrowing or lending. In the context of our problem, three different interest rates are described:

• 2% Interest Rate: This is the rate applied to the initial amount borrowed, labeled as \( x \). It indicates that for every \(1,000\) dollars borrowed, \( 20\) dollars will be charged annually as interest.
• 3% Interest Rate: This rate is applied to the second portion of the loan, calculated as \( \frac{1}{2}x + 1000 \). It shows that for every \(1,000\) dollars of this portion, \( 30 \) dollars will be accumulated as interest each year.
• 2.5% Interest Rate: Used for the remaining borrowed amount, \( y \). Here, \( 25 \) dollars of interest are charged annually for every \( 1,000 \) dollars loaned.

Each interest rate affects the total amount of interest paid over a year. Understanding these rates allows the borrower to estimate the cost of borrowing.

Loan Calculations

Loan calculations allow us to determine how much was borrowed and how interest will accumulate over time. They can assist in budgeting and financial planning. In the exercise, we deal with three parts of the loan:

• The company borrowed at three different interest rates: \(2\%, 3\%,\)and \(2.5\%\).
• To find what amounts were borrowed at each rate, a system of equations was used.

• The first equation, describing total amounts, is \( \frac{3}{2}x + y = 11500 \).
• The second equation, summarizing total interest, is \( 0.035x + 0.025y = 275 \).

By solving these equations, you find how much was borrowed at each rate.Understanding the process behind these calculations helps you grasp how different parts of a loan contribute to total borrowing and interest.

Algebraic Expressions

Algebraic expressions are mathematical phrases involving numbers, variables, and operation symbols. They are essential in solving real-life problems, like the one we encountered. Here, algebraic expressions helped represent different loan conditions.In the problem:

• \( x \) represents the amount borrowed at \(2\%\).
• The expression \( \frac{1}{2}x + 1000 \) defines the amount borrowed at \(3\%\).
• \( y \) stands for the remaining amount at \(2.5\%\).

Using algebraic expressions:

• We combined them into equations representing total loans and total interest.
• The equations were then solved simultaneously to find exact values of \( x \) and \( y \).

This demonstrates how algebra helps translate words into mathematical equations, enabling problem-solving in practical financial scenarios.