Problem 200
Question
Vern sold his 1964 Ford Mustang for \(\$ 55,000\) and wants to invest the money to earn him \(5.8 \%\) interest per year. He will put some of the money into Fund A that earns \(3 \%\) per year and the rest in Fund B that earns \(10 \%\) per year. How much should he invest into each fund if he wants to earn \(5.8 \%\) interest per year on the total amount?
Step-by-Step Solution
Verified Answer
Invest approximately \$33,714.29 in Fund A and \$21,285.71 in Fund B.
1Step 1: Define the Variables
Let x be the amount invested in Fund A and y be the amount invested in Fund B. Then, we know that x + y = 55,000.
2Step 2: Establish the Total Interest Equation
The total interest earned from both funds should equal 5.8% of the total amount. Therefore, 0.058 * 55,000 = 3/100 * x + 10/100 * y.
3Step 3: Simplify the Interest Equation
Simplify the total interest equation: 0.058 * 55,000 = 0.03x + 0.1y.
4Step 4: Solve the System of Equations
We have the following system of linear equations: 1. x + y = 55,000 2. 3x + 10y = 3,190,000. To eliminate y, multiply the first equation by 10: 10x + 10y = 550,000 Then subtract the simplified interest equation from it: (10x + 10y) - (3x + 10y) = 550,000 - 3,190 7x = 236,000.
5Step 5: Solve for x
Solving for x, we get: x = 236,000 / 7 x = 33,714.29 (approximately).
6Step 6: Find y
Substitute x back into the first equation to find y: x + y = 55,000 33,714.29 + y = 55,000 y = 55,000 - 33,714.29 y = 21,285.71 (approximately).
Key Concepts
Linear EquationsInterest CalculationSolving Systems of EquationsAlgebraic Expressions
Linear Equations
A linear equation is an algebraic statement where each term is either a constant or the product of a constant and a single variable. Linear equations form straight lines when plotted on a graph.
In this investment problem, we have two linear equations. The first is based on the total amount Vern invests:
Linear equations are the foundation for solving many algebra problems by representing real-world situations mathematically and helping us find unknowns.
In this investment problem, we have two linear equations. The first is based on the total amount Vern invests:
- x + y = 55,000
Linear equations are the foundation for solving many algebra problems by representing real-world situations mathematically and helping us find unknowns.
Interest Calculation
Interest calculation determines how much extra money an investment will earn over time. The formula for calculating simple interest is:
- I = P * R * T
- I = Interest earned
- P = Principal amount (initial investment)
- R = Interest rate (annual)
- T = Time (years)
- 0.058 * 55,000 = 3/100 * x + 10/100 * y
Solving Systems of Equations
To solve systems of equations, we find values for the variables that satisfy each equation simultaneously. Here, our system consists of two key equations:
- x + y = 55,000
- 0.03x + 0.1y = 0.058 * 55,000
- 10x + 10y = 550,000
- (10x + 10y) - (3x + 10y) = 550,000 - 3,190
7x = 236,000
Algebraic Expressions
An algebraic expression combines numbers, variables, and arithmetic operations to represent a mathematical phrase. For investment problems like this one, expressions help formulate and solve equations for the unknowns. An expression can be simplified to make solving easier. In our exercise, we establish and simplify the interest equation:
- 0.03x + 0.1y = 0.058 * 55,000
Other exercises in this chapter
Problem 198
Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs \(\$ 4.80\) per pound with Fescue
View solution Problem 199
Vartan was paid \(\$ 25,000\) for a cell phone app that he wrote and wants to invest it to save for his son's education. He wants to put some of the money into
View solution Problem 201
Stephanie inherited \(\$ 40,000\). She wants to put some of the money in a certificate of deposit that pays \(2.1 \%\) interest per year and the rest in a mutua
View solution Problem 202
Avery and Caden have saved \(\$ 27,000\) towards a down payment on a house. They want to keep some of the money in a bank account that pays \(2.4 \%\) annual in
View solution