Question

Investment A clothing company borrows $$\$ 700,000$$. Some of the money is borrowed at \(8 \%\), some at \(9 \%\), and some at \(10 \%\) simple annual interest. How much is borrowed at each rate when the total annual interest is $$\$ 60,500$$ and the amount borrowed at \(8 \%\) is three times the amount borrowed at \(10 \%\) ?

Short answer

The amount borrowed at 8% interest is $150,000, at 9% interest is $500,000, and at 10% interest is $50,000.

Step-by-step solution

Formulation of equations from the given conditions

Let \( x \), \( y \), and \( z \) represent the amounts (in USD) borrowed at 8%, 9%, and 10% interest rates respectively. From the problem: the total borrowed is $700,000 so \( x + y + z = 700000 \), the total interest is $60,500 so \( 0.08x + 0.09y + 0.10z = 60500 \) and the amount borrowed at 8% is three times the amount borrowed at 10% so \( x = 3z \).

Substitute \( x = 3z \) into the other two equations

Replacing \( x \) with \( 3z \) in the total borrowed sum equation will provide \( 3z + y + z = 700000 \) which simplifies to \( 4z + y = 700000 \). Doing the same with the total interest equation will give \( 0.08 * 3z + 0.09y + 0.10z = 60500 \) which simplifies to \( 0.34z + 0.09y = 60500 \).

Solve for \( z \) and \( y \) by simultaneous equations

You now have two equations with two variables: \( 4z + y = 700000 \) and \( 0.34z + 0.09y = 60500 \). Multiply the second equation by \( 10 \) and subtract it from the first equation multiplied by \( 9 \). The result is \( 2.6z = 130000 \), which simplifies to \( z = 50000 \). Then substitute \( z = 50000 \) into the equation \( 4z + y = 700000 \) to find \( y = 500000 \).

Solve for \( x \) using the equation \( x = 3z \)

Substitute \( z = 50000 \) into the equation \( x = 3z \) to find \( x = 150000 \).

Key concepts

Simple Interest

Simple Interest is a method to calculate the interest charged or earned on a principal amount. It's simple because it is straightforward to calculate and does not compound.
To calculate simple interest, use the formula:

• Start with the Principal amount: the initial amount of money borrowed or invested.
• Multiply it by the Interest Rate (expressed as a decimal): the percentage of the principal that is paid in interest.
• Multiply by the Time period: the duration for which the money is borrowed or invested (usually in years).

This gives you the Simple Interest formula: \[ I = P \times R \times T \]where:

• \(I\) is the interest accrued.
• \(P\) is the principal amount.
• \(R\) is the annual interest rate in decimal form.
• \(T\) is the time period for which the interest is calculated.

In the given problem, multiple amounts are borrowed at different interest rates and it's important to set these out clearly when performing calculations. This allows for solving using Linear Equations effectively.

Linear Equations

Linear equations form the backbone of solving many algebraic problems. They are equations of the first degree, meaning they involve variables raised to the highest power of one.
Linear equations can be represented in the standard form: \[ax + by + cz = d\] where:

• \(a, b,\) and \(c\) are coefficients.
• \(x, y,\) and \(z\) represent variables.
• \(d\) is the constant term.

When dealing with problems like investments, these equations help represent total values and conditions set by the problem.
In the example provided, equations are formed using the conditions: the total amount borrowed and the total interest. These are converted into equations which can be later solved using simultaneous techniques.

Simultaneous Equations

Simultaneous equations involve solving two or more equations at the same time. They are extremely useful in problems that have multiple unknown variables. The goal is to find the values of the variables that satisfy all equations involved. There are several methods to solve simultaneous equations:

• Substitution Method: Solve one equation for one variable and substitute this value into the other equations.
• Elimination Method: Manipulate the equations such that adding or subtracting them removes one of the variables, making it easier to solve the remaining equation.
• Matrix Method: Use matrices and linear algebra to solve more complex sets of equations.

In the given problem, after forming the initial set of equations, the value for one variable is substituted into the other equations to simplify them. This involves reducing the number of variables gradually until each can be solved individually. Substitution and elimination are the primary techniques used here to resolve the unknown values of the amounts borrowed at different interest rates, like in the problem, where substitution allows us to express all amounts in terms of a single variable, leading ultimately to a full solution.