Problem 92 - Algebra for College Students

Question

Solve each problem. Pickups and cars. The ratio of pickups to cars sold at a dealership is 2 to \(3 .\) If the dealership sold 142 more cars than pickups in \(2006,\) then how many of each did it sell?

Step-by-Step Solution

Answer

284 pickups and 426 cars.

Step 1: Define Variables

Let the number of pickups sold be denoted by \(P\) and the number of cars sold be denoted by \(C\).

Step 2: Set Up the Ratio

It is given that the ratio of pickups to cars is 2 to 3. Therefore, we have the equation: \[ \frac{P}{C} = \frac{2}{3} \] This can be rewritten as: \[ 3P = 2C \]

Step 3: Use Additional Information

The problem states that there were 142 more cars sold than pickups. This gives us another equation: \[ C = P + 142 \]

Step 4: Substitute Second Equation into First Equation

Substitute \(C = P + 142\) into the equation from Step 2: \[ 3P = 2(P + 142) \]

Step 5: Solve for Pickups \(P\)

Multiply out and solve for \(P\): \[ 3P = 2P + 284 \] Subtract 2P from both sides: \[ P = 284 \]

Step 6: Solve for Cars \(C\)

Use the value of \(P\) found in Step 5 to find \(C\): \[ C = P + 142 \] \[ C = 284 + 142 \] \[ C = 426 \]

Step 7: Verify the Solution

Verify that the ratio and difference are correct: \[ \frac{P}{C} = \frac{284}{426} = \frac{2}{3} \] and \(426 - 284 = 142\). Thus, the solution is correct.

Key Concepts

ratio and proportion

When dealing with problems that involve ratios and proportions, it's important to understand what these terms mean. A ratio is a comparison between two quantities. For example, if the ratio of pickups to cars sold at a dealership is 2 to 3, this tells you that for every 2 pickups sold, 3 cars were sold. In mathematical notation, you can write this as \[ \frac{P}{C} = \frac{2}{3} \]. To solve ratio problems, setting up an equation from the given ratio is crucial. In this case, we set up the equation \[ 3P = 2C \] to express the relationship between pickups and cars. This equation is a central piece in solving this problem because it ties the quantities of pickups and cars together in a straightforward way.

algebraic equations

Algebraic equations help us express relationships between variables. In the given problem, we have two algebraic equations: \[ C = P + 142 \] and \[ 3P = 2C \]. The first equation comes from the information that there were 142 more cars sold than pickups. The second equation comes from the given ratio of pickups to cars. By substituting the first equation into the second equation, you can solve for one variable, then use that solution to find the other variable. This process involves replacing one variable in terms of another, allowing you to simplify and solve the equation step by step. When you substitute \( C = P + 142 \) into \( 3P = 2C \), you get \[ 3P = 2(P + 142) \]. Solving this, we first distribute the 2 on the right side to get \[ 3P = 2P + 284 \]. Then, by isolating \(P\), we subtract 2P from both sides to get \[ P = 284 \].

problem-solving steps

Solving math problems boils down to following a series of logical steps. Let's break it down using our exercise:

• Define Variables: Start by defining what each variable represents. Here, let \( P \) be the number of pickups and \( C \) be the number of cars.
• Set Up the Ratio: Use the given ratio to set up an equation, \( \frac{P}{C} = \frac{2}{3} \) turns into \( 3P = 2C \)
• Incorporate Additional Information: Convert any extra details into equations. We used \( C = P + 142 \), which tells us there are 142 more cars than pickups.
• Substitution: Substitute the simpler equation into the ratio equation: \( 3P = 2(P + 142) \).
• Solve: Solve for one variable first. By simplifying, you get \( P = 284 \).
• Find the Other Variable: Once you have \(P\), use it to find \(C\): \( C = P + 142 \), resulting in \(C = 426 \).
• Verification: Always verify your answers. Check the ratio \( \frac{284}{426} \), which simplifies to \( \frac{2}{3} \), and the difference \( 426 - 284 = 142 \) to ensure accuracy.

Following these systematic steps is essential in breaking down and solving ratio problems effectively.