Time and Work — Study Notes

**Railway Group D (Level 1) Mathematics**

Overview

Time and Work is a high-yield topic in RRB Group D, typically contributing 2–4 questions per paper. The core principle is simple: work is the product of rate (efficiency) and time. Questions test your ability to relate individual work rates, combine efficiencies for joint work, and handle inverse proportions between workers and time.

Mastery requires understanding three problem types: (1) persons working together or in turns, (2) pipes filling or emptying tanks (cisterns), and (3) efficiency comparison between workers. The math is straightforward—mostly unitary method and fraction manipulation—but exam questions layer conditions (workers leaving mid-project, alternating work days, leaking tanks) to test clarity of thought. Strong fundamentals in LCM, fraction addition, and ratio help you solve these in under two minutes per question.

Expect word problems framed around construction, painting, typing, or pipe-and-cistern scenarios. The key mental shift: treat "work" as a measurable quantity (often set to 1 or 100 units) and "rate" as work per unit time. With this lens, every Time and Work problem becomes a rate equation you can solve systematically.

Key Concepts

• **Work as a quantity**: Total work can be assigned any convenient value—commonly 1 unit or 100%. If A finishes a job in 10 days, A's one-day work = 1/10 of the job.
• **Rate (efficiency) = Work / Time**: If efficiency is constant, work done is directly proportional to time spent. Doubling workers (assuming same efficiency) halves the time.
• **Man-Days or Work-Units**: Total work = Number of persons × Days × Hours per day. Adjust this product when comparing different groups: 6 men in 8 days = 48 man-days of work.
• **Inverse proportion for time**: If work is constant, Time ∝ 1/Number of workers. If 5 workers take 12 days, 10 workers take 6 days (assuming equal efficiency).
• **Combined work rate**: When A and B work together, their one-day combined work = (A's rate) + (B's rate). If A does 1/x per day and B does 1/y per day, together they do (1/x + 1/y) = (x+y)/(xy) per day.
• **Pipes and cisterns analogy**: Inlet pipe = positive work (filling); outlet/leak = negative work (emptying). Net rate = sum of all rates with appropriate signs.
• **Efficiency ratio**: If efficiencies are in the ratio m:n, then times taken are in the ratio n:m (inverse). If A is twice as efficient as B, A takes half the time B takes.
• **Partial work and remainder**: If part of the work is done, subtract that fraction from 1 to find remaining work. Then apply rates to the remainder.

Formulas / Key Facts

1. **One-day work formula**: If A completes work in *x* days, A's 1-day work = 1/*x*. 2. **Combined rate**: (A + B)'s 1-day work = 1/*a* + 1/*b* = (*a* + *b*) / (*a* × *b*), so time to finish together = (*a* × *b*) / (*a* + *b*) days. 3. **Work = Rate × Time**: Total work *W* = efficiency *e* × time *t*. Rearrange to find any unknown. 4. **Man-day equivalence**: *M₁* men in *D₁* days = *M₂* men in *D₂* days ⇒ *M₁* × *D₁* = *M₂* × *D₂* (for equal work and efficiency). 5. **Efficiency and time inverse**: If efficiency ratio = *m* : *n*, time ratio = *n* : *m*. If A:B efficiency = 2:3, time taken by A:B = 3:2. 6. **Pipes filling/emptying**: Inlet rate *I* (positive), outlet rate *O* (negative). Net rate = *I* − *O*. Time to fill = Capacity / Net rate. 7. **Alternate-day work**: If A and B work on alternate days, find work done in 2-day cycles, then handle the remainder separately. 8. **Fraction of work done in *t* days**: If rate = 1/*n* per day, work done in *t* days = *t*/*n*; remaining work = 1 − *t*/*n*.

Worked Examples

**Example 1: Basic joint work** *A can complete a task in 12 days, B in 15 days. Working together, how long to finish?*

• A's 1-day work = 1/12
• B's 1-day work = 1/15
• Combined 1-day work = 1/12 + 1/15 = (5 + 4)/60 = 9/60 = 3/20
• Time together = 1 ÷ (3/20) = 20/3 = 6⅔ days

**Answer**: 6 days 16 hours (since ⅔ day = 16 hours).

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**Example 2: Pipes and cisterns with leak** *Pipe A fills a tank in 6 hours, pipe B in 8 hours. A leak empties it in 12 hours. If all are open, time to fill the tank?*

• A's rate = 1/6 tank/hour (positive)
• B's rate = 1/8 tank/hour (positive)
• Leak's rate = 1/12 tank/hour (negative)
• Net rate = 1/6 + 1/8 − 1/12 = (4 + 3 − 2)/24 = 5/24 tank/hour
• Time to fill = 1 ÷ (5/24) = 24/5 = 4.8 hours = 4 hours 48 minutes

**Answer**: 4 hours 48 minutes.

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**Example 3: Work and wages by efficiency** *A is twice as efficient as B. Together they finish in 10 days. How long would B alone take?*

• Let B's efficiency = 1 unit/day, A's = 2 units/day.
• Combined efficiency = 3 units/day.
• Total work = 3 × 10 = 30 units.
• B alone does 1 unit/day, so time = 30 / 1 = 30 days.

**Answer**: B alone takes 30 days.

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**Example 4: Partial work then joint** *C completes a job in 20 days. After working 5 days, C is joined by D, and they finish in 5 more days together. How long would D alone take?*

• C's 1-day work = 1/20.
• Work done by C in 5 days = 5/20 = 1/4.
• Remaining work = 1 − 1/4 = 3/4.
• C and D together finish 3/4 in 5 days ⇒ 1-day combined work = (3/4)/5 = 3/20.
• D's 1-day work = 3/20 − 1/20 = 2/20 = 1/10.
• D alone: 1 ÷ (1/10) = 10 days.

**Answer**: D alone takes 10 days.

Common Mistakes

1. **Confusing rate and time**: Students add times instead of rates. Wrong: If A takes 4 days and B takes 6 days, together ≠ 10 days. Correct: Add rates (1/4 + 1/6) then invert. 2. **Forgetting to subtract for leaks/outlets**: In pipe problems, always assign negative rate to outlets. Omitting the minus sign gives an inflated net rate and wrong answer. 3. **Not accounting for remaining work**: After partial work, students apply the original time formula to the whole job. Correct: Calculate fraction completed, subtract from 1, then find time for the remainder. 4. **Ignoring efficiency differences**: Assuming equal efficiency when the problem states different skill levels. Correct: Use the given efficiency ratio to set individual rates before combining. 5. **Misapplying man-days in scaling**: When men or hours change, students forget to adjust both sides. Correct: Use *M₁D₁H₁* = *M₂D₂H₂* and solve for the unknown, keeping all units consistent.

Quick Reference

• **1-day work = 1 / (Total days for that person)**
• **Together time = (Product of individual times) / (Sum of individual times)** for two workers.
• **Pipes: Inlet (+), Outlet/Leak (−); Net rate = sum of signed rates.**
• **Efficiency ratio *m*:*n* ⇒ Time ratio *n*:*m* (inverse).**
• **Man-days: *M₁D₁* = *M₂D₂* for equal work and efficiency.**
• **Alternate days: Calculate 2-day cycle work, find number of full cycles, then add remainder work separately.**