Work Rate Problems — Combined Work & Pipes

Work Rate Problems — Combined Work & Pipes

Work rate problems ask how quickly a job gets done — filling a pool, painting a wall, or completing a project. The key idea is simple and powerful: if we treat the complete job as one unit, we can add the rates of different workers. On this page we will learn the work rate formula, solve combined-work problems, and handle fill-and-drain pipes that work against each other.

Background and Basic Definitions

Work rate is the fraction of the job completed per unit of time:

\[ \text{rate} = \frac{\text{work}}{\text{time}} \]

Central convention: the complete job is denoted \(1\) (one full pool, one complete wall). Therefore, if a worker finishes the job in \(a\) hours, their rate is \( \frac{1}{a} \) of the job per hour.

Combined work: when two workers work together their rates add:

\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{T} \]

where \(T\) is the combined time. The shortcut formula is:

\[ T = \frac{a \cdot b}{a + b} \]

Part of the job completed in a given time: if they work together for \(t\) hours, the fraction completed is \( t \cdot \left( \frac{1}{a} + \frac{1}{b} \right) \).

Fill pipe vs. drain pipe: draining works in the opposite direction, so the rates are subtracted: \( \frac{1}{a} - \frac{1}{b} \). If the fill rate is greater the pool fills; if the drain rate is greater it empties; if equal, there is no change.

Solution Steps

  1. Step 1 — Label the complete job as \(1\), and write each worker's rate as \( \frac{1}{\text{their time}} \).
  2. Step 2 — Determine whether the workers act in the same direction (add rates) or in opposite directions — fill vs. drain (subtract rates).
  3. Step 3 — Add or subtract the rates to find the combined rate.
  4. Step 4 — The combined time is the reciprocal of the combined rate: \( T = \frac{1}{\text{combined rate}} \).
  5. Step 5 — To find the fraction completed, multiply the combined rate by the time worked.
  6. Step 6 — Sanity check: the combined time must always be less than the time of the fastest worker alone.

Worked Examples

Example 1: From Rate to Time

Problem: A worker paints \( \frac{1}{5} \) of a wall per hour. How long will it take to paint the entire wall?

Solution:

  1. The complete wall is \(1\), and the worker's rate is \( \frac{1}{5} \) per hour.
  2. Time = reciprocal of rate: \( T = \frac{1}{1/5} = 5 \).
  3. Check: in \(5\) hours the worker paints \( 5 \times \frac{1}{5} = 1 \), i.e. the full wall.

Answer: It will take \( 5 \) hours.

Example 2: Two Pipes Together

Problem: Pipe \(A\) fills a pool in \(3\) hours, and pipe \(B\) fills the same pool in \(6\) hours. How long do they take together?

Solution:

  1. Rates: \( \frac{1}{3} \) and \( \frac{1}{6} \) per hour.
  2. Combined rate: \( \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \) per hour.
  3. Combined time: \( T = \frac{1}{1/2} = 2 \) hours.
  4. Using the shortcut: \( T = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2 \).

Answer: Together they fill the pool in \( 2 \) hours.

Example 3: Fraction of the Job Completed

Problem: Worker \(A\) finishes a job in \(8\) hours and worker \(B\) finishes it in \(12\) hours. What fraction of the job will they complete together in \(3\) hours?

Solution:

  1. Combined rate: \( \frac{1}{8} + \frac{1}{12} = \frac{3}{24} + \frac{2}{24} = \frac{5}{24} \) per hour.
  2. Fraction completed in \(3\) hours: \( 3 \times \frac{5}{24} = \frac{15}{24} = \frac{5}{8} \).
  3. They completed \( \frac{5}{8} \) of the job; \( \frac{3}{8} \) remains.

Answer: They complete \( \frac{5}{8} \) of the job.

Example 4: Fill vs. Drain

Problem: A fill pipe fills a pool in \(4\) hours; a hole at the bottom drains a full pool in \(6\) hours. The pool is empty and both are open. How long does it take to fill?

Solution:

  1. Fill and drain act in opposite directions, so subtract: \( \frac{1}{4} - \frac{1}{6} \).
  2. Common denominator: \( \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \) per hour — positive net rate, so the pool does fill.
  3. Time: \( T = \frac{1}{1/12} = 12 \) hours.

Answer: The pool will be full in \( 12 \) hours.

Example 5: Number of Workers Required

Problem: \(6\) workers finish a job in \(15\) days. How many workers are needed to finish the same job in \(10\) days (at the same individual rate)?

Solution:

  1. Total work in worker-days: \( 6 \times 15 = 90 \) worker-days.
  2. To finish in \(10\) days requires \( \frac{90}{10} = 9 \) workers.
  3. This is an inverse proportion: fewer days \(\Rightarrow\) more workers, and the product (workers × days) remains constant.

Answer: \( 9 \) workers are required.

Common Mistakes

✗ Common mistake: Adding the times instead of the rates, e.g. claiming two pipes of \(3\) h and \(6\) h will fill in \(9\) h together.

✓ The correct way: Add rates, not times: \( \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \), giving \(2\) hours. The combined time is always less than the time of either worker alone.

✗ Common mistake: Adding rates for a fill-and-drain situation instead of subtracting.

✓ The correct way: Drain acts in the opposite direction, so subtract: \( \frac{1}{a} - \frac{1}{b} \). If the result is negative, the drain is stronger and the pool empties.

✗ Common mistake: Forgetting that the complete job equals \(1\) and trying to work with physical quantities (litres, square metres).

✓ The correct way: Always set the complete job to \(1\). Then every rate is a fraction of \(1\), the calculation is straightforward, and the physical size is irrelevant.

Practice Tips

  • Tip — Shortcut for two workers: \( T = \frac{a \cdot b}{a + b} \). Product over sum.
  • Tip — Sanity check: the combined time must be less than the time of the fastest worker alone.
  • Tip — For 'how much was done' problems use 'fraction = rate × time', then subtract from \(1\) to find what remains.
  • Tip — For inverse proportion (workers vs. days), the product workers × days stays constant — it equals the total amount of work.

Summary and Key Formulas

  • Work rate \( = \frac{\text{work}}{\text{time}} \); complete job \( = 1 \).
  • A worker who finishes in \(a\) hours \(\Rightarrow\) rate \( \frac{1}{a} \).
  • Combined work: \( \frac{1}{a} + \frac{1}{b} = \frac{1}{T} \), so \( T = \frac{ab}{a+b} \).
  • Fill vs. drain: subtract the rates, \( \frac{1}{a} - \frac{1}{b} \).
  • Fraction completed \( = t \cdot \big( \frac{1}{a} + \frac{1}{b} \big) \).