Work Algebra Word Problem Game

Work Algebra Word Problem Quiz/Game
This game focuses on solving word problems involving Works. Work problems including a wide range of work applications: Inflow/Outflow Scenarios, includes both basic rational equations and those requiring quadratic factoring, problems require solving for the individual’s time and combined total. Scroll down the page for a more detailed explanation.

 

 

How to Play the Work Algebra Explorer Game

1. Look at the Problem: Read the problem carefully. Solve it and select one of the answers.
   
2. Check Your Work: If you selected the right answer, it will be highlighted in green. If you are wrong, it will be highlighted in red and the correct answer will be highlighted in green. A hint will be given to help you find the correct answer.
   
3. Get a New Problem: Click “Next Question” for a new problem.
   Your score is tracked, showing how many you’ve gotten right.
   
4. Finish Game When you have completed 10 questions, your final score will be displayed.
    
   

How to Solve Work Algebra Word Problems
Work word problems generally deal with how fast different people or machines can complete a task. The “secret” to solving them is to stop thinking about time and start thinking about rates.

The Core Formula The foundation of every work problem is the relationship between Work (W), Rate (r), and Time (t): W = r × t When we are dealing with a single completed job (like painting one room or filling one tub), we set W = 1. This means the rate is always the reciprocal of the time: \(r = \frac{1}{t}\) Example: If it takes you 5 hours to paint a room, your rate is \(\frac{1}{5}\) of a room per hour.

Setting Up the Equation When two people work together, their individual rates add up to a combined rate. If Person A takes a hours and Person B takes b hours, and together they take T hours, the equation looks like this: \(\frac{1}{a} + \frac{1}{b} = \frac{1}{T}\)

Step-by-Step Problem Solving Let’s solve a common scenario: Person A takes $6$ hours to mow a lawn. Person B takes $3$ hours. How long would they take if they do it together? Step 1: Identify the individual rates Rate of Person A: \(\frac{1}{6}\) Rate of Person B: \(\frac{1}{3}\)

Step 2: Set up the sum of rates \(\frac{1}{6} + \frac{1}{3} = \frac{1}{x}\) (where x is the total time together)

Step 3: Find a common denominator To add \(\frac{1}{6}\) and \(\frac{1}{3}\), use 6 as the denominator: \(\frac{1}{6} + \frac{2}{6} = \frac{3}{6}\)

Step 4: Simplify and solve for x \(\frac{3}{6} = \frac{1}{2}\) So, \(\frac{1}{2} = \frac{1}{x}\). By “flipping” both sides (taking the reciprocal), we find: x = 2 hours.}

Advanced Variations Sometimes the problem asks for something different:

Finding a “Missing Partner”: If you know the total time and one person’s time, you subtract: \(\frac{1}{\text{Total}} - \frac{1}{\text{Known}} = \frac{1}{\text{Unknown}}\).

Variable Relationships: If Person A is twice as fast as Person B, use x and 2x as their times. The rates would be \(\frac{1}{x}\) and \(\frac{1}{2x}\).

Inflow and Outflow: If a pipe fills a tub while a drain empties it, you subtract the rates: \(\text{Rate}{\text{in}} - \text{Rate}{\text{out}} = \text{Rate}_{\text{combined}}\).
 

This video gives a clear, step-by-step approach to explain how to solve work algebra word problems.

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