Problem 58 - Algebra and Trigonometry with Analytic Geometry

Question

A college student receives an interestfree loan of \(\$ 8250\) from a relative. The student will repay \(\$ 125\) per month until the loan is paid off. (a) Express the amount \(P\) (in dollars) remaining to be paid in terms of time \(t\) (in months). (b) After how many months will the student owe \(\$ 5000\) ? (c) Sketch, on a \(t P\)-plane, a graph that shows the relationship between \(P\) and \(t\) for the duration of the loan.

Step-by-Step Solution

Answer

(a) \( P = 8250 - 125t \); (b) 26 months; (c) A linear graph starting at (0, 8250) with a slope of -125.

Step 1: Understanding the Initial Situation

The problem starts with a student receiving an interest-free loan of \( \$8250 \). This amount represents the initial balance of the loan, and no interest will accumulate over time.

Step 2: Establishing the Decrease Per Month

Every month, the student repays \( \\(125 \) towards the loan. This means that the balance decreases linearly each month by \( \\)125 \).

Step 3: Expressing the Remaining Amount in Terms of Months

The amount remaining \( P \) after \( t \) months can be expressed as an equation that captures the initial amount minus \( \$125 \) for each month of repayment: \[ P = 8250 - 125t \].

Step 4: Calculating When \( P = \$5000 \)

To find the number of months until the student owes \( \\(5000 \), set \( P = 5000 \) and solve for \( t \):\[ 5000 = 8250 - 125t \]Solve the equation for \( t \):\[ 125t = 8250 - 5000 \]\[ 125t = 3250 \]\[ t = \frac{3250}{125} \]\[ t = 26 \]So it takes 26 months for the remaining amount to reach \( \\)5000 \).

Step 5: Sketching the Graph

To sketch a graph on the \( tP \)-plane, plot the points starting from \( (0, 8250) \), the initial point, decreasing linearly with each month by \( \$125 \) until it reaches 0. For simplicity, key points can be plotted, such as \( (26, 5000) \) found in the previous step, and eventually \( (66, 0) \) which represents the entire loan paid off, since:\[ 0 = 8250 - 125 \times 66 \].The line should have a slope of \(-125\), indicating the steady decrease in the loan amount each month.

Key Concepts

Interest-Free Loan

An interest-free loan means that you borrow money without having to pay any extra charges, known as interest, over time. In this case, the student received an interest-free loan of \( \$8250 \). This is advantageous for several reasons:

• Every payment made goes entirely to reduce the principal amount (the loan itself), not interest.
• The total repayment amount is equal to the borrowed amount.
• Simplifies budgeting as the total cost remains unchanged.

Understanding the terms of such a loan is crucial. You start with a known amount and need to calculate how this amount decreases over time solely by making regular payments.

Monthly Repayment

Monthly repayment refers to the amount paid back each month to diminish the loan balance. It's a fixed amount that, in this example, is \( \\(125 \) every month. This concept has several benefits and implications:

• Consistency: The same amount is paid every month, making it easier to manage monthly budgets.
• Predictability: You know exactly how long it will take to pay off the loan since no unexpected charges are added.
• Linearity: The decrease in loan balance is linear, meaning it decreases by an exact amount each month.

Remember, each payment reduces the principal amount by \( \\)125 \), leading us to the next topic, loan balance calculation.

Loan Balance Calculation

Loan balance calculation helps you determine how much you still owe at any point during repayment. In our example, to find the remaining amount after a certain number of months, we use the equation:\[ P = 8250 - 125t \]where:

• \( P \) is the loan balance remaining.
• \( 8250 \) is the initial loan amount.
• \( 125t \) represents total repayments made over \( t \) months.

This equation is a simple linear function where the remaining balance decreases by \( \\(125 \) each month. For instance, to check when the balance will reach \( \\)5000 \), set \( P = 5000 \) and solve the equation to find \( t = 26 \) months.

Graphing Linear Functions

Graphing linear functions is a powerful way to visualize how variables change over time. Here, we illustrate the relationship between the remaining loan balance \( P \) and time \( t \). You start by plotting the initial point \( (0, 8250) \), where \( t \) equals zero at the beginning of the loan. The function \( P = 8250 - 125t \) draws a straight line with a slope of \(-125\), showing a steady decrease.

• The line's downward slope reflects the \( \\(125 \) reduction in balance each month.
• Key points of interest include \((26, 5000)\), when the balance drops to \( \\)5000 \), and \((66, 0)\), when the loan is fully repaid.
• This visual representation confirms the linear nature of the repayment scheme.

By graphing this function, it becomes clear how the loan diminishes over time, offering a quick reference to gauge where you are in the process of repayment.