Time and Work — Study Notes

Overview

Time and Work is a high-scoring topic in SSC MTS Paper 1, appearing consistently with 2–4 direct questions every year. The core idea is simple: if you know how long a person (or machine, or pipe) takes to complete a task alone, you can calculate how long they'll take working together or under varying conditions. Mastering this topic builds a strong foundation for related areas like pipes and cisterns, which use identical logic but in a different scenario.

Students must develop comfort with two key skills: converting "days to complete work" into "work done per day" (the efficiency concept), and setting up equations when multiple workers collaborate. The arithmetic is straightforward—mostly fractions and LCM—but the variety of question styles (joint work, people leaving midway, alternate day work, pipes filling/emptying tanks) requires pattern recognition. Spend time on 15–20 varied problems to internalize the standard setups, and you'll handle any twist the exam throws at you.

Key Concepts

• **Work as a whole unit**: Treat the entire task as 1 unit of work. If A completes it in 10 days, A does 1/10 of the work per day. This fractional rate is called A's *efficiency* or *rate of work*.

• **Total work = LCM method**: To avoid fractions, assume total work = LCM of all individual times. If A finishes in 12 days and B in 15 days, take total work = LCM(12,15) = 60 units. Then A's efficiency = 60/12 = 5 units/day, B's = 4 units/day.

• **Combined efficiency**: When A and B work together, their one-day work = (1/A's time) + (1/B's time). Joint time = 1 / (combined one-day work). Using the LCM method, just add their efficiencies and divide total work by that sum.

• **Man-days concept**: Work ∝ (Number of workers) × (Number of days). If 5 men complete work in 8 days, the same work needs 40 man-days. So 10 men would finish in 4 days.

• **Pipes and cisterns analogy**: An inlet pipe filling a tank = a worker doing work. An outlet (drain) pipe emptying a tank = negative work. Net filling rate = (sum of inlet rates) − (sum of outlet rates).

• **Partial work and remaining work**: If someone works for a few days then leaves, calculate work done = (efficiency × days worked). Remaining work = Total work − Work done. The remaining people finish the leftover portion.

• **Alternate day work**: When A and B work on alternate days, calculate the work done in each 2-day cycle, then determine how many full cycles are needed and handle any leftover days separately.

Formulas / Key Facts

1. **One-day work**: If A finishes work in *n* days, A's 1-day work = 1/n.

2. **Combined work (two workers)**: Time together = 1 / (1/A + 1/B) = (A × B) / (A + B) when A, B are days to complete individually.

3. **LCM method for efficiency**: Total work = LCM(t₁, t₂, ...), then efficiency of each = Total work / Individual time. Combined time = Total work / Sum of efficiencies.

4. **Man-days or Work equivalence**: M₁ × D₁ = M₂ × D₂ (for same work). Extend to M₁ × D₁ × H₁ = M₂ × D₂ × H₂ if working hours per day differ.

5. **Pipes filling/emptying**: Net rate = Inlet rate − Outlet rate. If net rate is positive, tank fills; if negative, it empties. Time to fill/empty = Capacity / |Net rate|.

6. **Work done in *x* days**: Work = Efficiency × Number of days.

7. **Remaining work**: Remaining = Total − (Efficiency × Days already worked).

Worked Examples

**Example 1 (Basic Joint Work)**: A can do a work in 15 days, B can do it in 20 days. How long will they take working together?

*Solution*: A's 1-day work = 1/15, B's 1-day work = 1/20. Combined 1-day work = 1/15 + 1/20 = (4 + 3)/60 = 7/60. Time together = 1 ÷ (7/60) = 60/7 = 8⁴⁄₇ days.

*Alternative (LCM method)*: Total work = LCM(15, 20) = 60 units. A's efficiency = 60/15 = 4 units/day, B's = 60/20 = 3 units/day. Combined = 7 units/day. Time = 60/7 = 8⁴⁄₇ days.

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**Example 2 (Partial Work, One Leaves)**: A and B can complete work in 12 days together. A alone takes 20 days. A works for 4 days then leaves. How many more days does B need to finish?

*Solution*: Total work = LCM(12, 20) = 60 units (use joint time and A's time). A and B together: 1-day work = 1/12 ⇒ efficiency = 60/12 = 5 units/day. A alone: efficiency = 60/20 = 3 units/day. So B alone: 5 − 3 = 2 units/day. Work done by A in 4 days = 3 × 4 = 12 units. Remaining work = 60 − 12 = 48 units. B finishes in 48/2 = 24 days.

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**Example 3 (Pipes and Cistern)**: Pipe A fills a tank in 6 hours, pipe B empties it in 8 hours. If both are open, how long to fill the empty tank?

*Solution*: Let tank capacity = LCM(6, 8) = 24 units. A's rate = 24/6 = 4 units/hour (filling). B's rate = 24/8 = 3 units/hour (emptying). Net rate = 4 − 3 = 1 unit/hour. Time to fill = 24/1 = 24 hours.

Common Mistakes

1. **Adding times instead of rates**: Saying "A takes 10 days, B takes 15, together 25 days" is wrong. Always add the *work rates* (1/10 + 1/15), not the times themselves.

2. **Forgetting negative work for outlet pipes**: Treat an emptying pipe or a person destroying work as negative efficiency. Students often add all efficiencies as positive, leading to incorrect answers.

3. **Not accounting for partial days correctly**: If A and B together finish 7/10 of work in 7 days, don't assume they need 3 more days for the rest unless their combined rate stays constant. Check if anyone has left or conditions changed.

4. **Ignoring the "man-days" proportionality**: When men or hours change, remember work is proportional to (men × days × hours). Missing any one factor skews the calculation—especially when the problem states "working 8 hours/day" versus "10 hours/day."

5. **Misinterpreting alternate day work**: In alternate-day problems, students often multiply 1-day work by total days. Instead, calculate work per 2-day cycle, find number of full cycles, then manually add work on any remaining odd day.

Quick Reference

• **A alone in *a* days, B alone in *b* days ⇒ Together in (a×b)/(a+b) days.**
• **Total work = LCM of individual times** — cleanest method to avoid fraction errors.
• **Work done = Efficiency × Days worked.**
• **Pipes: Net rate = Inlets − Outlets.** Positive → fills, Negative → empties.
• **Man-days: M₁D₁ = M₂D₂** for same work.
• **Alternate days: Compute 2-day cycle work, then handle leftover day separately.**