Problem 48 - Elementary Algebra

Question

Joe assembles a computer by himself in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble 1 computer working alone?

Step-by-Step Solution

Answer

The assistant would take 1.5 hours to assemble one computer alone.

Step 1: Define Joe's Work Rate

Joe can assemble 1 computer in 1 hour, so his work rate is 1 computer per hour. This can be expressed as \( J = 1 \) computer/hour.

Step 2: Define Joint Work Rate

Together, Joe and his assistant can assemble 10 computers in 6 hours. So, their combined work rate is \( \frac{10}{6} \) computers per hour, which simplifies to \( \frac{5}{3} \) computers per hour.

Step 3: Define Assistant's Work Rate

Let \( A \) be the work rate of the assistant. The work rate of Joe and his assistant together is \( J + A = \frac{5}{3} \). Since we know Joe's work rate \( J = 1 \), the equation becomes \( 1 + A = \frac{5}{3} \).

Step 4: Solve for the Assistant's Work Rate

To find \( A \), we solve \( 1 + A = \frac{5}{3} \). Subtract 1 from both sides: \[ A = \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]. Thus, the assistant's work rate is \( \frac{2}{3} \) computers per hour.

Step 5: Calculate Time for Assistant to Assemble One Computer

The time \( t \) required for the assistant to assemble one computer working alone is the reciprocal of their work rate: \[ t = \frac{1}{\frac{2}{3}} = \frac{3}{2} \text{ hours} \]. Thus, it would take the assistant 1.5 hours to assemble 1 computer alone.

Key Concepts

Joint Work Rate

In work rate problems, it's vital to understand the concept of joint work rate when multiple people or agents work together. It is the combined rate at which they complete a task. For example, consider Joe and his assistant working together assembling computers. We calculate their joint work rate by considering how many computers they build together over a certain period.
In our exercise, they assembled 10 computers in 6 hours. The joint work rate is determined by dividing the total number of computers by the number of hours, which in this case is \( \frac{10}{6} = \frac{5}{3} \) computers per hour. This indicates that together, Joe and his assistant can assemble \( \frac{5}{3} \) computers every hour. Understanding this concept of joint work rate helps in determining the individual contributions to the task.

Assembly Work Rate

The assembly work rate is about understanding how fast someone can work on assembling a product or completing a task individually. Each person's work rate can be considered as their personal productivity level.
For Joe, his work rate is 1 computer per hour, as he can complete one computer by himself in an hour. His assistant, however, has a different work rate, which we need to find using the joint work rate. First, set up an algebraic equation as shown in the exercise. When Joe and his assistant work together, they assemble at a rate of \( \frac{5}{3} \) computers per hour. Given Joe's individual work rate, we subtract it from the joint rate to find the assistant's work rate.

• Joe's rate: 1 computer/hour.
• Assistant's rate: computed as \( \frac{5}{3} - 1 = \frac{2}{3} \) computers/hour.

Recognizing each person's work rate helps us determine individual capacity and efficiency.

Algebraic Equations

Algebraic equations play a key role in solving work rate problems by allowing us to model the shared and individual contributions mathematically. The goal is to let unknowns represent what we're trying to find—in our exercise, the assistant's work rate.
By setting up an equation, we combine known and unknown variables. Since the joint rate and Joe's rate are given, we can represent the assistant’s work rate with an equation. Once set up, solve step-by-step:

• The joint work rate equation: \( J + A = \frac{5}{3} \).
• Substitute Joe's work rate: \( 1 + A = \frac{5}{3} \).
• Solve for \( A \) (the assistant's work rate): \( A = \frac{5}{3} - 1 = \frac{2}{3} \).

These simple algebraic equations allow determination of key work rates, illustrating the power of algebra in practical scenarios.