91Ó°ÊÓ

Harmonic Mean The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2 . The average is found as shown. Since Time \(=\) distance \(\div\) rate then Time \(1=\frac{100}{40}=2.5\) hours to make the trip Time \(2=\frac{100}{50}=2\) hours to return Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

Step by step solution

01

Define Problem (Part a)

Identify the problem requirements for part a. We need to find the average speed for a salesperson who drives 300 miles round trip at 30 mph to Chicago and returns at 45 mph. The total distance (round trip) is 300 miles.

02

Calculate Reciprocals (Part a)

First, calculate the reciprocals of the speeds for part a. The reciprocals are \( \frac{1}{30} \) for the trip to Chicago and \( \frac{1}{45} \) for the return.

03

Sum of Reciprocals (Part a)

Next, sum the reciprocals of the speeds: \( \frac{1}{30} + \frac{1}{45} = \frac{3}{90} + \frac{2}{90} = \frac{5}{90} \).

04

Apply Harmonic Mean Formula (Part a)

Now use the harmonic mean formula with \( n = 2 \): \( \text{HM} = \frac{2}{\frac{5}{90}} = \frac{2 \times 90}{5} = \frac{180}{5} = 36 \; \text{mph} \).

05

Define Problem (Part b)

For part b, find the average speed for a bus driver who drives 50 miles to West Chester at 40 mph and returns at 25 mph. The total trip distance is 100 miles.

06

Calculate Reciprocals (Part b)

Calculate the reciprocals of the speeds for part b: \( \frac{1}{40} \) for the trip to West Chester and \( \frac{1}{25} \) for the return.

07

Sum of Reciprocals (Part b)

Sum the reciprocals: \( \frac{1}{40} + \frac{1}{25} = \frac{5}{200} + \frac{8}{200} = \frac{13}{200} \).

08

Apply Harmonic Mean Formula (Part b)

Use the harmonic mean formula with \( n = 2 \): \( \text{HM} = \frac{2}{\frac{13}{200}} = \frac{2 \times 200}{13} = \frac{400}{13} \approx 30.77 \; \text{mph} \).

09

Define Problem (Part c)

For part c, find the average cost per pound when buying $500 worth of nails at $50 per pound and $500 worth at $10 per pound. Total cost is $1000, for 20 pounds total (10 pounds at each price).

10

Calculate Reciprocals (Part c)

Calculate the reciprocals of the costs per pound: \( \frac{1}{50} \) and \( \frac{1}{10} \).

11

Sum of Reciprocals (Part c)

Sum the reciprocals: \( \frac{1}{50} + \frac{1}{10} = \frac{1}{50} + \frac{5}{50} = \frac{6}{50} = \frac{3}{25} \).

12

Apply Harmonic Mean Formula (Part c)

Use the harmonic mean formula with \( n = 2 \): \( \text{HM} = \frac{2}{\frac{3}{25}} = \frac{2 \times 25}{3} = \frac{50}{3} \approx 16.67 \; \text{per pound} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Average Speed Calculation

When calculating the average speed for a journey involving different speeds over equal distances, the arithmetic mean can be misleading. Instead, the harmonic mean is used to find the correct average speed. This applies when the distance traveled is consistent in each part of the journey, but the speed varies.

To illustrate, imagine you drive a certain distance at different speeds. The traditional method of averaging (simply adding the speeds and dividing by two) does not account for the varying time spent at each speed. Instead, the harmonic mean provides a method that considers these time differences.

The formula for the harmonic mean, given two different speeds, is:

• HM = \( \frac{2}{\left( \frac{1}{speed1} + \frac{1}{speed2} \right)} \)

This approach results in a more accurate representation of your overall average speed.

Reciprocals

Reciprocals play a crucial role in calculating the harmonic mean. To find the harmonic mean, you need to work with the reciprocals of the speeds involved.

A reciprocal of a number is simply 1 divided by that number. For example, the reciprocal of 40 is \( \frac{1}{40} \). Understanding reciprocals is key to solving problems involving the harmonic mean.

In practical terms, when calculating the harmonic mean of multiple speeds, you first convert each speed to its reciprocal. This creates a new set of numbers that reflect the inverse relationship of their time intervals for a fixed distance. You then sum these reciprocal values to proceed with the harmonic mean calculation.

Mathematical Problems

Solving mathematical problems involving the harmonic mean can involve various scenarios. These problems are often seen in physics and economics, where rates and averages are frequently used.

For instance, consider a problem where you need to find the average speed over a journey with varying travel times. The harmonic mean provides a solution that gives a weighted average in terms of time spent rather than distance traveled.

• Example: Finding the average fuel efficiency when driving different sections at different speeds
• Example: Calculating the average cost per pound when paying different rates for equal amounts of an item

In all these cases, the harmonic mean considers the duration or quantity under each rate, providing a more accurate average.

Step by Step Solutions

Step-by-step solutions are beneficial for understanding and solving problems involving harmonic mean. Each step builds on the previous, ensuring clarity and understanding of the logic applied.

Here's how step-by-step solutions help:

• Break down complex problems into manageable parts
• Highlight the process of calculating reciprocals
• Demonstrate the application of the harmonic mean formula
• Provide error-checking at each step

By following a structured sequence, these solutions guide you through not only performing calculations but also truly understanding the underlying principles, making mathematical problems easier and more approachable.