⏱️ Time and Work

Day 2 | ⏱️ Part of 2-Hour Commercial Math Block | TCS Interview Prep

📌 Subtopics

• Individual work rate problems
• Combined work (two or more workers)
• Pipes and cisterns
• Alternate day work problems
• Work and wages
• Efficiency-based problems

📖 Comprehensive Notes

1. Individual Work Rate

Work done is the product of rate (efficiency) and time: $\text{Work} = \text{Rate} \times \text{Time}$

• If a person A can complete a piece of work in $n$ days, then: $\text{A's 1-day work} = \frac{1}{n}$
• LCM Method (Highly Recommended): Assume the Total Work is the LCM of the times taken by individuals. This avoids working with fractions.

Example: A can do a work in $10$ days and B in $15$ days. What is their daily efficiency if total work is assumed using the LCM method? $\text{Total Work} = \text{LCM}(10, 15) = 30\text{ units}$. $\text{A's efficiency} = 30 / 10 = 3\text{ units/day}$. $\text{B's efficiency} = 30 / 15 = 2\text{ units/day}$.

2. Combined Work

• If A takes $a$ days and B takes $b$ days to finish a work: $\text{Combined 1-day work} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$ $\text{Time taken together} = \frac{ab}{a+b}\text{ days}$
• Using the LCM Method: $\text{Time Taken} = \frac{\text{Total Work Units}}{\text{Sum of Efficiencies}}$

Example: Using the previous example, how long will A and B take together? $\text{Total Work} = 30\text{ units}$. $\text{Total Efficiency} = 3 + 2 = 5\text{ units/day}$. $\text{Time} = 30 / 5 = 6$ days.

3. Pipes and Cisterns

This is identical to Time and Work, but with negative work (leaks or outlet pipes).

• Inlet pipe (fills) $\rightarrow$ Positive work ($\text{Rate} = +1/t$).
• Outlet pipe (empties) / Leak $\rightarrow$ Negative work ($\text{Rate} = -1/t$).
• If an inlet pipe fills a tank in $x$ hours and an outlet empties it in $y$ hours (where $y > x$): $\text{Net rate of filling} = \frac{1}{x} - \frac{1}{y}$

Example: Pipe A can fill a tank in $6$ hours, and Pipe B can empty it in $8$ hours. If both are opened together, how long to fill the tank? Let total capacity $= \text{LCM}(6, 8) = 24\text{ units}$. Pipe A efficiency $= +4\text{ units/hour}$. Pipe B efficiency $= -3\text{ units/hour}$. Net efficiency $= +1\text{ unit/hour}$. Time $= 24 / 1 = 24$ hours.

4. Alternate Day Work

Workers work one after the other on alternate days.

• Calculate the work done in one full cycle (e.g., $2$ days if two workers).
• Find the number of complete cycles needed to get close to the total work, then calculate the remaining work and time for the next worker.

Example: A can do a work in $10$ days and B in $20$ days. If they work on alternate days starting with A, when is the work done? $\text{Total Work} = \text{LCM}(10, 20) = 20\text{ units}$. $\text{A's eff} = 2$, $\text{B's eff} = 1$. In $2$ days ($1$ cycle), work done $= 2 + 1 = 3\text{ units}$. In $6$ cycles ($12$ days), work done $= 3 \times 6 = 18\text{ units}$. Remaining work $= 20 - 18 = 2\text{ units}$. On day $13$, A works and does exactly $2\text{ units}$ (since his eff is $2$). Total time $= 13$ days.

5. Work and Wages

Wages are distributed in proportion to the actual amount of work done by each individual.

• If people work for the same duration, wages are divided in the ratio of their efficiencies.
• If they work for different durations, wages are divided in the ratio of the units of work completed ($\text{Efficiency} \times \text{Time}$).

Example: A can do work in $6$ days and B in $8$ days. They work together and earn $₹1400$. Find A’s share. Since they worked for same duration, wage ratio $=$ efficiency ratio $= \frac{1}{6} : \frac{1}{8} = 8 : 6 = 4 : 3$. A’s share $= \frac{4}{7} \times 1400 = ₹800$.

6. Efficiency-Based Problems

“A is $k$ times as efficient as B” $\implies \text{Rate of A} = k \times \text{Rate of B}$.

• Since efficiency is inversely proportional to time: $\frac{\text{Time taken by A}}{\text{Time taken by B}} = \frac{1}{k}$

Example: A is twice as efficient as B and takes $10$ days less than B to finish a work. Find B’s time. Let B take $x$ days, so A takes $x - 10$ days. Since A is twice as efficient, A takes half the time: $x - 10 = x/2 \implies x = 20$ days.

🌐 Learning Resources

✅ Practice Links

📝 Key Formulas

If A does work in 'a' days → A's 1 day work = 1/a
Combined rate: 1/a + 1/b = 1/T  →  T = ab/(a+b)

Pipes & Cisterns:
  Filling pipe rate = +1/t
  Draining pipe rate = -1/t
  Net rate = Sum of all rates

Tip: Use LCM method for faster solving — take LCM of days as total work units.

🔗 Related Topics

• Time Speed and Distance — Rate-based problems with similar structure
• Ratios and Proportion — Efficiency ratios and work sharing
• Equations — Setting up work rate equations
• Allegations and Mixtures — Similar ratio-based problem solving approach
• Percentages — Efficiency as a percentage