Problem 21 - Algebra A Combined Function

Question

Smith Engineering found that an experienced surveyor surveys a roadbed in 4 hours. An apprentice surveyor needs 5 hours to survey the same stretch of road. If the two work together, find how long it takes them to complete the job.

Step-by-Step Solution

Answer

It takes them \( \frac{20}{9} \) hours, or about 2.22 hours, to complete the job together.

Step 1: Determine Surveyor Rates

To find the individual rates, first recognize that the experienced surveyor can complete 1 job in 4 hours, so their rate is \( \frac{1}{4} \) of the job per hour. The apprentice can complete 1 job in 5 hours, giving a rate of \( \frac{1}{5} \) of the job per hour.

Step 2: Add the Rates Together

Add the surveyors' rates together to find the rate at which they work together. The combined rate is \( \frac{1}{4} + \frac{1}{5} \). To add these fractions, first find a common denominator, which is 20. Thus, \( \frac{1}{4} = \frac{5}{20} \) and \( \frac{1}{5} = \frac{4}{20} \), so the combined rate is \( \frac{5}{20} + \frac{4}{20} = \frac{9}{20} \) of the job per hour.

Step 3: Solve for Time

The combined working rate is \( \frac{9}{20} \) of the job per hour. Let \( t \) be the time it takes to complete 1 job. Then \( t \times \frac{9}{20} = 1 \). Solving for \( t \), multiply both sides by 20 to get \( 20t = 9 \), and then divide by 9 to get \( t = \frac{20}{9} \) hours.

Key Concepts

Combined Work Problems

In combined work problems, two or more individuals or machines perform a task together. It’s crucial to determine how long it takes to complete a job when they join forces. By understanding each participant's contribution, you can figure out how efficiently they work together.
When tackling such problems, make sure you know the rate at which each worker can complete the job alone. Then think about how these rates interact when combined. This understanding is fundamental in solving real-world problems efficiently.
If you ever find yourself wondering how to split tasks among different workers to optimize time, combined work problems are your go-to tool!

Rate of Work

Rate of work is a central concept in work rate problems. It represents the amount of work completed in a specific unit of time, like an hour. To find a person's work rate, you can divide the entire job (usually considered as 1 complete task) by the amount of time it takes them to finish it.
For example, if an experienced surveyor can complete a survey in 4 hours, their rate of work will be \( \frac{1}{4} \) of the job per hour. Similarly, the apprentice's rate will be \( \frac{1}{5} \) of the job per hour.
Understanding rates helps in combining efforts, planning projects, and estimating completion times for various tasks effectively.

Addition of Fractions

In combined work problems, you often need to add fractions to determine the combined work rate. This process requires finding a common denominator, which is a shared multiple of the denominators involved.
For instance, considering rates \( \frac{1}{4} \) and \( \frac{1}{5} \), the smallest common denominator is 20. By rewriting \( \frac{1}{4} = \frac{5}{20} \) and \( \frac{1}{5} = \frac{4}{20} \), you can easily add them: \( \frac{5}{20} + \frac{4}{20} = \frac{9}{20} \).
Mastering the skill of adding fractions is helpful beyond math class, as it can apply to cooking recipes, money management, and even time scheduling!

Algebraic Equations

Algebraic equations are invaluable tools for solving work rate problems. They help you formalize and solve for unknown quantities, like the time to complete a job when working together.
In the example problem, once you have the combined rate \( \frac{9}{20} \) of the job per hour, you set up an equation: \( t \times \frac{9}{20} = 1 \). This equation represents the total job being completed.
Solving for \( t \), multiply each side by 20 to simplify, resulting in \( 20t = 9 \). Finally, divide by 9 to isolate \( t \), giving \( t = \frac{20}{9} \) hours.
Grasping algebraic equations allows you to tackle a wide range of mathematical and practical problems efficiently and confidently.