Problem 37 - Intermediate Algebra

Question

Comparing Travel. A plane can fly 600 miles in the same time as it takes a car to go 240 miles. If the car travels 90 mph slower than the plane, find the speed of the plane.

Step-by-Step Solution

Answer

The speed of the plane is 150 mph.

Step 1: Define Variables

Let the speed of the plane be \( x \) mph. Then, the speed of the car is \( x - 90 \) mph because it's 90 mph slower than the plane.

Step 2: Write Time Equation

Time taken by both the plane and the car is equal. For the plane: \[ \text{Time} = \frac{600}{x} \]For the car: \[ \text{Time} = \frac{240}{x - 90} \]Since both times are equal, we have:\[ \frac{600}{x} = \frac{240}{x - 90} \]

Step 3: Solve the Equation

To solve \( \frac{600}{x} = \frac{240}{x - 90} \), cross multiply to get:\[ 600(x - 90) = 240x \]Expand and simplify:\[ 600x - 54000 = 240x \]Rearrange the terms:\[ 600x - 240x = 54000 \]\[ 360x = 54000 \]

Step 4: Solve for x

Solve for \( x \) by dividing both sides by 360:\[ x = \frac{54000}{360} \]Calculate the division:\( x = 150 \)

Step 5: Conclusion

The speed of the plane is 150 mph.

Key Concepts

Solving Equations

Solving equations is a fundamental part of algebra. This involves finding the value of the variable that satisfies the equation. In our problem, the equation stemmed from the concept that the time taken by the plane and the car is the same. We translated this scenario into a mathematical expression:

• The time for the plane is expressed as \( \frac{600}{x} \)
• The time for the car is \( \frac{240}{x - 90} \)

Setting these two expressions equal formed the equation \( \frac{600}{x} = \frac{240}{x - 90} \).
This is a rational equation due to the variables being in the denominator.
To solve it, we used cross-multiplication, transforming the equation into a linear form, which is 600 times the speed difference \((x - 90)\) equal to 240 times the plane's speed \(x\).
Expanding and simplifying leads us to solve a straightforward linear equation where we rearranged and isolated the variable on one side. This process allowed us to determine the plane's speed. This is a systematic approach to solving rational equations commonly encountered in algebra.

Rate and Distance Problems

Rate and distance problems are classic algebraic word problems that involve calculating the speed, distance, or time of travel. In this exercise, the main formula used is \( \text{Distance} = \text{Rate} \times \text{Time} \). Specifically, the problem asks for the speed of a plane given its distance and state that it takes the same time as a car covering a shorter distance.

• For the plane: It flies 600 miles, and the formula expressed time as \( \frac{600}{x} \) where \( x \) is the speed.
• For the car, moving 240 miles at a slower speed \((x - 90)\), its time is \( \frac{240}{x - 90} \).

Both times are set equal because the plane and car travel simultaneously within the same duration.
Solving rate and distance problems often involves summarizing what we know about distance, speed, and time, making these types of algebra problems practical and applicable, reflecting real-world scenarios.

Variables in Algebra

Understanding variables is crucial in solving algebra word problems, as they are symbols used to represent unknown numbers. In this exercise, the variable \( x \) was chosen to represent the unknown speed of the plane. This is a standard practice where variables act as placeholders to help set up equations.

• The speed of the car was described as \( x - 90 \) because the car travels 90 mph slower than the plane.

By defining variables in this manner, the relationship between the different moving objects can be accurately expressed mathematically. It simplifies handling the problem, as once the variable is isolated and determined, substituting back gives precise results. Variables are the backbone of translating verbal expressions into mathematical language, which is essential for both problem-solving and communicating complex algebraic concepts concisely.