Problem 21
Question
Norman and Suzanne Scarpulla own 35 shares of McDonald's stock and 69 shares of The Ohio Art Company stock (makers of Etch A Sketch and other toys). On a particular day in 2009 , their stock portfolio consisting of these two stocks was worth \(\$ 2360 .\) The McDonald's stock was \(\$ 60\) more per share than The Ohio Art Company stock. What was the price of each stock on that day?
Step-by-Step Solution
Verified Answer
The Ohio Art Company stock was $2.50, and McDonald's stock was $62.50 per share.
1Step 1: Identify Variables
Let the price of each share of The Ohio Art Company stock be \( x \) dollars. Since McDonald's stock is \( \$60 \) more per share, its price per share will be \( x + 60 \) dollars.
2Step 2: Set Up Equation
Write an equation based on the information provided: the total value of the stocks is \( \$2360 \). Thus, the equation is: \[ 35(x + 60) + 69x = 2360 \]
3Step 3: Expand and Simplify the Equation
Distribute and combine the terms:\[ 35x + 2100 + 69x = 2360 \]Combine like terms:\[ 104x + 2100 = 2360 \]
4Step 4: Solve for x
Isolate \( x \) by subtracting 2100 from both sides:\[ 104x = 260 \]Now, divide by 104:\[ x = \frac{260}{104} = 2.5 \]
5Step 5: Find McDonald's Stock Price
Now that we know \( x = 2.5 \), McDonald's stock price per share is \( x + 60 = 2.5 + 60 = 62.5 \).
Key Concepts
Solving EquationsVariables in AlgebraLinear Equations
Solving Equations
Solving equations is a fundamental skill in algebra that involves finding the value of unknown variables that make an equation true. When faced with a word problem like the one involving stock prices, the key to solving the equation is translating the problem into mathematical language.
Here's how we approach it:
Here's how we approach it:
- Begin by identifying what you know and what you want to find out. In this case, we know the total value of the stocks and the relationship between their prices.
- Formulate an equation that represents the situation. The equation might involve different operations like addition, subtraction, multiplication, or division.
- The next step is to simplify the equation if necessary, and then use algebraic operations to solve for the unknown variable. This often involves isolating the variable on one side of the equation by performing inverse operations.
- Verify your solution by plugging it back into the original equation to ensure it satisfies the given conditions.
Variables in Algebra
Variables are symbols used to represent unknown numbers or quantities in algebra. They are fundamental because they allow us to create equations and represent relationships between different quantities.
In our stock price problem:
It is essential to define your variables explicitly to avoid confusion, especially in more complex problems where multiple variables might be involved.
In our stock price problem:
- x is used to represent the price of each stock in The Ohio Art Company.
- This allows us to express other related quantities in terms of x, like the price of McDonald's stock, which is x + 60 per share.
It is essential to define your variables explicitly to avoid confusion, especially in more complex problems where multiple variables might be involved.
Linear Equations
Linear equations are a type of equation where the highest power of the variable is one. They appear frequently in algebra, especially in word problems like the one we're examining here.
Here's an overview of linear equations:
Here's an overview of linear equations:
- They typically have the form ax + b = c, where a, b, and c are constants.
- In our stock price problem, the equation is linear because it can be simplified to 104x + 2100 = 2360. This conforms to the typical linear structure.
- Solving a linear equation requires isolating the variable on one side. This can involve operations like adding, subtracting, multiplying, or dividing both sides by the same number.
Other exercises in this chapter
Problem 21
Solve each system of equations by the substitution method. $$ \left\\{\begin{aligned} x+2 y+5 &=-4+5 y-x \\ 2 x+x &=y+4 \end{aligned}\right. $$
View solution Problem 21
Solve each system of linear equations by graphing. See Examples 3 through \(6 .\) \(\left\\{\begin{array}{l}x+y=5 \\ x+y=6\end{array}\right.\)
View solution Problem 22
Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or dec
View solution Problem 22
Solve each system of equations by the substitution method. $$ \left\\{\begin{aligned} 5 x+4 y-2 &=-6+7 y-3 x \\ 3 x+4 x &=y+3 \end{aligned}\right. $$
View solution