Problem 75 - Algebra 1

Question

INVESTING MONEY You receive \(\$ 5000\). You decide to invest the money in a one-year bond paying \(2 \%\) interest and in a one-year certificate of deposit paying \(6 \%\) interest. \- Let \(m\) represent the amount of money invested in the one-year bond. Write a function that represents the total amount of money \(T\) that you have after one year. Simplify the function.

Step-by-Step Solution

Answer

The function that represents the total amount of money \(T\) that you have after one year is \(T = 0.04m + \$5300\)

Step 1: Formulate the problem variables

First, let's denote \( m \) as the amount of money invested in the one-year bond which is offering \(2 \% \) interest. The rest of the money, which is \( \$5000 - m \), would be invested in the one-year certificate of deposit which is offering \(6 \% \) interest.

Step 2: Write the function

The function can be obtained by summing the principal amount invested in each instrument with their respective interest earned. This can be represented as: \( T = m + 0.02m + (\$5000 - m) + 0.06 (\$5000 - m) \)

Step 3: Simplify the function

Simplifying the equation above gives the total money at the end of the year: \( T = 0.02m + m + \$5000 - m + 0.06\$5000 - 0.06m \). Simplifying further provides us the total money function \(T = 0.04m + \$5300\)

Key Concepts

Function Representation

When dealing with investment problems in algebra, creating a function representation is a crucial first step. By turning a word problem into a mathematical function, we can easily manipulate the numbers and better understand the relationships between different elements of the problem.
In this problem, we consider two investment products: a bond and a certificate of deposit. First, we define the variable to represent our funds in these instruments. Here, we use the variable \( m \) to signify the amount invested in the bond. The total amount invested is \( \\(5000 \), thus the remaining amount \( \\)5000 - m \) goes into the certificate of deposit.
This setup allows us to formulate a function for the total value of an investment. This function captures both the principal and the interest gained from both investments, helpfully summarizing everything into one equation.

Interest Calculation

Interest calculation is essential in determining the profitability of an investment. It essentially means figuring out how much extra money your investment will make over a period of time.
In our exercise, the bond gives a \( 2\% \) return, while the certificate of deposit returns \( 6\% \). To find out how much interest each investment earns, you multiply the principal with the interest rate.

• For the bond: The interest earned is \( 0.02m \).
• For the certificate of deposit: You calculate the interest as \( 0.06(\$5000 - m) \).

This straightforward multiplication helps us tally up the total interest from both sources, which we can then incorporate into our function for overall investments. Interest calculation represents the investment's growth, steering our understanding of profit generated over a given period.

Algebraic Simplification

After formulating a function with interest calculations, the next step is algebraic simplification. This involves combining like terms and reducing the equation to its simplest form.
Initially, the function was written as: \[ T = m + 0.02m + (\\(5000 - m) + 0.06 (\\)5000 - m) \]
By expanding and combining like terms, we perform a critical simplification process:

• First, distribute the interest rates: \( 0.02m \) and \( 0.06 \times \\(5000 \) .
• Combine all the \( m \) terms, including the \( m \) and \(-0.06m \). This leads to \( 0.04m \).
• Add the constants, especially the interests earned and principal from non-\( m \) amounts, leading to \( \\)5300 \).

Ultimately, this results in a neat and compact form of the function: \( T = 0.04m + \$5300 \). Simplification is the final step that makes complex expressions easier to comprehend and utilize for decision making.