Problem 33 - Prealgebra and Introductory Algebra

Question

A mechanic requires \(2 \mathrm{h}\) to repair a transmission, whereas an apprentice requires \(6 \mathrm{h}\) to make the same repairs. The mechanic worked alone for \(1 \mathrm{h}\) and then stopped. How long will it take the apprentice, working alone, to complete the repairs?

Step-by-Step Solution

Answer

The apprentice will take approximately 3 hours to complete the remaining repairs.

Step 1: Calculating the Rates

Firstly, we calculate the rate at which the mechanic and the apprentice perform the work. The mechanic can do the work in 2 hours, so his rate of work is \(1/2\) of the work per hour. Similarly, the apprentice can do the work in 6 hours, so his rate is \(1/6\) of the work per hour.

Step 2: Calculating amount of work done by the mechanic

We know, the mechanic worked alone for 1 hour. So, the amount of work done by the mechanic can be calculated by multiplying his rate of work with the time he/she worked, i.e., \(1/2 * 1 = 1/2\) of the work.

Step 3: Calculating the remaining work

Subtract the amount of work done by the mechanic from the total work to find the remaining work. Since the total work is 1 (as it is defined as such, the work that the mechanic or the apprentice can do is defined as 1 'unit' of work), subtracting the work done by the mechanic from it gives the remaining work that the apprentice has to do. So, work left = total work - work done by mechanic = \(1 - 1/2 = 1/2\) work.

Step 4: Calculating the time for the apprentice to complete the remaining work

Now calculate the time it will take the apprentice to complete the remaining half of the work. As the work done (W) equals rate (R) times time (T), we can rearrange this to calculate time as calculation of time taken is work done divided by rate of work, i.e., time taken by the apprentice = work left / apprentice's rate of work = \(1/2 / 1/6 = 3hrs \).

Key Concepts

Work-Rate Calculation

Understanding work-rate calculation is crucial when solving problems involving multiple workers or machines, each with different capabilities. In the given exercise, it's essential to calculate the rate of work for both the mechanic and the apprentice. The work-rate, often expressed as 'unit of work per unit of time', helps us understand how fast a person or machine can complete a given task.

For example, the mechanic's rate is calculated by understanding that the entire repair, represented as one unit of work, is completed in 2 hours. Hence, the mechanic's work-rate is \(1/2\) per hour. Similarly, for the apprentice, completing one unit of work in 6 hours gives a work-rate of \(1/6\) per hour. This fundamental calculation sets the stage for determining how much work is done over a given period, and subsequently, how much time is required to finish any remaining work.

Algebraic Word Problems

Algebraic word problems require translating a real-world situation into a mathematical model using variables and algebraic expressions. In the exercise, the work performed by the mechanic and the apprentice is modeled using fractions and the concept of work-rate. By establishing a relationship between work done, rate, and time - we can form an equation that allows us to solve for the unknown.

To make sense of the problem, break it down into smaller, understandable pieces. The mechanic works alone for 1 hour, which is a straightforward scenario to model. By multiplying the mechanic's rate of work (\(1/2\) per hour) by the time worked (1 hour), we determine the fraction of the total work completed. With the remaining work quantified, we proceed to determine how long it will take the apprentice to complete this using his individual work-rate.

Fraction Operations

Fraction operations are integral to solving work-rate problems, especially when individuals work at fractions of a unit per time. Adding, subtracting, and dividing fractions are common in these scenarios.

In our exercise, after calculating the amount of work done by the mechanic, we subtract this fraction from the 'whole' unit of work, highlighting subtraction of fractions. We then tackle a division of fractions when we need to find out how long it will take the apprentice to finish the remaining work. Remember, dividing fractions involves multiplying by the reciprocal, which is a key operation in the step-by-step solution. This process transforms \(1/2 \/ 1/6\) into \(1/2 * 6/1\), simplifying to 3, which represents the hours the apprentice needs to finish the job.