Problem 104 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem involving rate of work. A couple is laying a tile floor. Working alone, one can do the job in 20 hours. If the two of them work together, they can complete the job in 12 hours. How long would it take the other one to lay the floor working alone?

Step-by-Step Solution

Answer

The other person would take 30 hours to lay the floor alone.

Step 1: Determine Rates of Work

Let's determine the work rates of the individuals involved in this exercise. If one person can do the job in 20 hours, their work rate is \( \frac{1}{20} \) of the job per hour. We let \( t \) be the time it takes for the other person to do the job alone, which means their work rate is \( \frac{1}{t} \) of the job per hour.

Step 2: Combine Work Rates

If both work together, they complete the job in 12 hours. Therefore, their combined work rate is \( \frac{1}{12} \) of the job per hour. Thus, we have the equation: \( \frac{1}{20} + \frac{1}{t} = \frac{1}{12} \).

Step 3: Solve the Equation for t

Simplify the equation to find \( t \). Start by eliminating the fractions: Multiply through by the least common multiple of 20, 12, and \( t \), which is 60 \( t \). This gives: \[ 60t \left( \frac{1}{20} + \frac{1}{t} \right) = 60t \cdot \frac{1}{12} \]Expanding the terms, we get:\[ 3t + 60 = 5t \]Rearrange this to solve for \( t \):\[ 60 = 5t - 3t \]\[ 60 = 2t \]Finally, divide both sides by 2:\[ t = 30 \]

Step 4: Interpret the Solution

The solution \( t = 30 \) means that the second person, working alone, would complete the job in 30 hours.

Key Concepts

Work Rate Calculation

In rate of work problems, determining individual work rates is essential. A work rate is simply the portion of a job done per unit of time. For instance, if someone can complete a task in a specific number of hours, their work rate is the reciprocal of that time. Thus, a person who finishes a job in 20 hours has a work rate of \( \frac{1}{20} \) of the job per hour.

• Work rate formula: \( \text{Work Rate} = \frac{1}{\text{Time}} \)
• Unit of work rate: fraction of a job per hour

It's important to first comprehend each individual's work capability before combining them in the following steps.

Combined Work Rate

Understanding how to calculate a combined work rate is the next step. When two individuals work together, their efforts are cumulative. If one person has a work rate of \( \frac{1}{20} \) and another's rate is \( \frac{1}{t} \), their combined rate is the sum of these rates.The equation that models this scenario is:

• Combined work rate: \( \frac{1}{20} + \frac{1}{t} = \frac{1}{12} \)

This equation specifies that together, they complete \( \frac{1}{12} \) of the job per hour. Comprehending combined work rates helps us solve for unknown variables, like how long it takes the other person to complete the job alone.

Equation Solving

Once you've set up the equation using work rates, solving for the unknown becomes your main goal. In our problem, the equation is \( \frac{1}{20} + \frac{1}{t} = \frac{1}{12} \). Solving this requires manipulating the equation to isolate \( t \).

• Expand terms: Multiply through by the least common multiple, such as 60 in this case.
• Equation simplifies to: \( 3t + 60 = 5t \)
• Isolate \( t \): Rearrange and solve \( 60 = 2t \)

Carefully handling these steps ensures you find the correct value for \( t \), which reflects real scenarios neatly.

Fraction Elimination

Fraction elimination is a crucial technique that simplifies the solving process of work rate equations. When you have equations comprising fractions, clearing them by multiplying by an appropriate value simplifies calculations and reduces error likelihood.

Multiplication by Least Common Multiple (LCM)

Multiplying every term by an LCM, such as 60 in this instance, clears all fractions. This step transforms the equation into integer terms.

• Eliminate fractions: Multiply \( 60t \left( \frac{1}{20} + \frac{1}{t} \right) = 60t \cdot \frac{1}{12} \)

By doing so, the equation simplifies to whole numbers, making it easier to solve accurately. Mastering this method helps in effectively managing similar mathematical problems without confusion.