/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 A pilot flew a jet from Montreal... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 71

A pilot flew a jet from Montreal to Los Angeles, a distance of \(2500 \mathrm{mi}\). On the return trip, the average speed was \(20 \%\) faster than the outbound speed. The round-trip took 9 h 10 min. What was the speed from Montreal to Los Angeles?

Short Answer

Expert verified
The speed from Montreal to Los Angeles was approximately 518.5 mph.

Step by step solution

01

Define Variables

Let's start by defining the variables. Let \( x \) be the speed from Montreal to Los Angeles in miles per hour (mph). Since the return trip speed was 20% faster, the speed from Los Angeles to Montreal can be represented as \( 1.2x \).
02

Calculate Time for Each Leg of Trip

The time taken to travel from Montreal to Los Angeles is \( \frac{2500}{x} \) hours, and the time taken to travel back is \( \frac{2500}{1.2x} \) hours. This is because time is equal to distance divided by speed.
03

Convert Total Time and Set Up Equation

The total round-trip time is given as 9 hours and 10 minutes, which is 9.1667 hours when converted to decimal form. The equation representing the total trip time is: \[ \frac{2500}{x} + \frac{2500}{1.2x} = 9.1667 \]
04

Simplify and Solve the Equation

Start by finding a common denominator and solving the equation: \[ \frac{2500}{x} + \frac{2500}{1.2x} = \frac{2500 \times 1.2 + 2500}{1.2x} = 9.1667 \]Simplifying gives:\[ \frac{5000}{x} + \frac{5000}{1.2x} = 9.1667 \]Cross-multiplying gives:\[ 5000(1.2x) + 5000x = 9.1667(1.2x)x \]\[ 6000x + 5000x = 11,000x = 11x \times 9.1667 \]\[ 11,000 = 9.1667x \times 11 \]
05

Solve for x

Simplifying the previous equation, we have: \[ 1.1x \times 9.1667 = 11,000 \]\[ x \approx \frac{11,000}{10.0837} \]After calculating, we find \[ x \approx 518.5 \] mph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebra
Algebra is a branch of mathematics that deals with symbols and the rules governing their manipulation. It's essential for solving equations like the one in our exercise.
In our context, we define variables to symbolize unknown quantities. Here, the speed from Montreal to Los Angeles is denoted by the variable \( x \).
Using algebraic expressions, we can generalize the return speed as \( 1.2x \) since it is 20% faster than \( x \).
Algebra allows us to model real-world situations in the form of equations, which we can then solve to find solutions to complex problems.
Understanding how to set up the problem with variables is crucial, as it sets the stage for the rest of the solution.
Equation Solving
Equation solving is a critical aspect of algebra, where we work towards finding the value(s) of the variable(s) that satisfy the given equation.
In our problem, we have the equation:
  • \( \frac{2500}{x} + \frac{2500}{1.2x} = 9.1667 \)
We derive this equation based on the time taken for each leg of the trip and the total round-trip time.
This process involves rearranging the equation, finding a common denominator, and simplifying it.
Ultimately, we isolate the variable \( x \) by performing arithmetic operations like multiplication and division.
Solving equations requires logical steps to maintain mathematical balance, leading us to our desired result.
Rate of Travel
The rate of travel, or speed, is a measure of how fast something moves over a distance per unit of time.
In this exercise, we are dealing with two rates: the outbound speed \( x \) and the return speed \( 1.2x \).
Understanding the relationship between distance, speed, and time is crucial:
  • Time is calculated as \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).
This formula helps us understand how an increase in speed affects the time taken for a journey.
The exercise also highlights how changes in speed impact the overall travel time, allowing us to dissect complicated travel plans with relative ease.
Understanding these concepts helps in analyzing real-world travel scenarios and making logical travel plans efficiently.

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