Problem 116 - College Algebra with Modeling and Visualization

Question

From 2001 to 2005 the average number of hours that a user spent on the Internet each week increased by \(110 \% .\) This percent increase amounted to 11 hours. Find the average number of hours that a user spent on the Internet each week in 2001 and 2005.

Step-by-Step Solution

Answer

In 2001, users spent 10 hours, and in 2005, they spent 21 hours.

Step 1: Understanding Percent Increase

The problem states that the number of hours increased by \(110\%\), amounting to an 11-hour increase. This means the original number increased by its entire value plus an additional \(10\%\) of its value to total \(110\%\).

Step 2: Set Up the Equation for Percent Increase

Let \( x \) be the average number of hours in 2001. Since the increase was \(110\%\), this can be expressed as \(1.1x = 11\).

Step 3: Solve for the Initial Number of Hours in 2001

To find \(x\), solve the equation \(0.1x = 11\). This means \(x = 11 / 1.1\). Calculating this gives \(x = 10\). So, the average number of hours in 2001 was \(10\) hours.

Step 4: Calculate the Number of Hours in 2005

Using the original number \(x = 10\), the hours in 2005 was \(10 + 11 = 21\). Therefore, in 2005, a user spent an average of 21 hours per week on the Internet.

Key Concepts

Percent Increase

When we talk about a percent increase, we're looking at how much something has grown compared to its original value. In this case, the average number of hours spent on the internet each week increased by 110%. This means that the ending value was the original value plus 110% of that original value. To understand better, think of the original 100% as the total initial value, and the additional 10% beyond that represents the growth. This is why when it's said the increase amounts to an 11-hour increase, the 110% actually refers to the entire growth over the original number. This is critical when solving problems involving percent increase, as it helps differentiate between total growth versus just the added amount.

Equation Setup

Setting up the correct equation is essential in solving algebra problems efficiently. Here, we denote the average number of hours in 2001 as \( x \). According to the problem, the 110% increase amounted to an 11-hour gain. Hence, the equation can be set up as \( 1.1x = 11 \). This equation effectively captures the relationship established by the problem:

• \( 1.1x \) denotes the combined original value and the 110% increase.
• The right side of the equation "11" corresponds to the actual hours of increase.

These alignments ensure that the equation is properly configured to advance toward solving for the starting value \( x \).

Problem Solving Steps

Solving such algebra problems requires following a series of logical steps. Here's a simplified breakdown:

• **Understand the Percentages:** Identify the percent increase and its contribution to the overall equation.
• **Set Up the Equation:** Clearly translate the word problem into an equation using variables and known values.
• **Solve the Equation:** An equation like \( 1.1x = 11 \) means you solve for \( x \), which represents the hours in 2001.
• **Calculate Final Values:** Once you have \( x \), use it to find out how many hours in the later year, here 2005.

By methodically adhering to these steps, one can efficiently arrive at the solution in a structured and mistake-free manner.

Average Calculation

The notion of average plays an integral part in these types of problems. Here, we're dealing with averages to determine typical weekly internet usage in different years. After calculating that \( x = 10 \) hours in 2001, the 2005 usage was determined by adding the increase:

• Original average in 2001 was 10 hours.
• Additional 11-hour increase meant the average in 2005 was 21 hours.

This signifies that calculating averages often involves summing up initial and additional values, providing a clearer picture of the overall change over time.