Time and Work — Study Notes

Overview

Time and Work is a high-yield topic in the UP Police Constable exam, typically contributing 2–4 questions in the Numerical & Mental Ability section. This chapter tests your ability to convert real-world productivity scenarios into mathematical equations and solve them efficiently.

The core principle is simple: **Work = Rate × Time**. Every person, machine or pipe has a rate of work (how much they complete per unit time), and questions revolve around finding time taken, work done or efficiency when multiple entities work together or in shifts. Mastery requires understanding three sub-areas: **basic work problems** (individuals working alone or together), **pipes and cisterns** (filling/emptying tanks), and **wages distribution** (payment proportional to work done). These problems appear frequently and are often solved faster using the "efficiency method" rather than algebraic equations.

Strong command over LCM, fractions and unitary method is essential. Practice converting "days to complete work" into "work done per day" and vice versa. With systematic practice, you can solve most Time and Work questions in under 90 seconds.

Key Concepts

• **Work is constant**: Total work is typically assumed as 1 unit or taken as LCM of all given days for easier calculation. This allows you to focus on rates rather than absolute quantities.

• **Rate (Efficiency) = 1/Time**: If A completes work in 10 days, A's one-day work = 1/10. If B does it in 15 days, B's rate = 1/15. Combined rate when working together = 1/10 + 1/15.

• **Time taken together = Total Work / Combined Rate**: When multiple workers collaborate, add their individual rates to get the combined efficiency, then divide total work by this sum.

• **Negative work for emptying**: In pipes and cisterns problems, outlet pipes (emptying) have negative rates. If inlet fills in 6 hours and outlet empties in 8 hours, net rate = 1/6 - 1/8.

• **Work done = Efficiency × Time**: If A works for 5 days at rate 1/10 per day, work done = 5 × (1/10) = 1/2. Remaining work = 1 - 1/2 = 1/2.

• **Wages ∝ Work done**: Payment is distributed in the ratio of work done. If A works 10 days at efficiency 2 units/day and B works 15 days at 3 units/day, ratio = (10×2):(15×3) = 20:45 = 4:9.

• **Man-days concept**: 10 men in 6 days = 60 man-days of work. Same work by 15 men = 60/15 = 4 days. Total work capacity (M₁D₁) = (M₂D₂) when work is equal.

• **Efficiency and time are inversely proportional**: If A is twice as efficient as B, A takes half the time B takes. More efficient workers complete work faster.

Formulas / Key Facts

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

• **Combined work (A and B together)**: Time = **1 / (1/a + 1/b)** = **ab/(a+b)** where a, b are individual times

• **Three workers together**: Time = **1 / (1/a + 1/b + 1/c)**

• **LCM method**: Assume total work = LCM of all given days. Calculate efficiency = (Total work)/(Days taken). Add efficiencies and find time.

• **Work left after n days**: Remaining work = **1 - n × (rate per day)**

• **Pipes formula (one inlet, one outlet)**: Net rate = **(1/fill time) - (1/empty time)**

• **Man-day formula**: **M₁D₁W₂ = M₂D₂W₁** (when work amounts differ) or **M₁D₁ = M₂D₂** (same work)

• **Wage distribution**: Wages ratio = **(E₁ × T₁) : (E₂ × T₂)** where E = efficiency, T = time worked

• **Alternate day work**: If A and B work on alternate days, calculate total work done in 2-day cycles, then handle remaining work separately.

Worked Examples

**Example 1**: A can complete a work in 12 days and B in 18 days. If they work together, in how many days will the work be completed?

*Solution*: A's 1-day work = 1/12 B's 1-day work = 1/18 Combined 1-day work = 1/12 + 1/18 = (3+2)/36 = 5/36 Time taken together = 1 ÷ (5/36) = 36/5 = **7.2 days or 7 days 4.8 hours**

*LCM Method (faster)*: LCM(12,18) = 36 (assume total work) A's efficiency = 36/12 = 3 units/day B's efficiency = 36/18 = 2 units/day Combined = 5 units/day Time = 36/5 = **7.2 days**

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**Example 2**: A pipe can fill a tank in 6 hours. Another pipe can empty it in 8 hours. If both are opened simultaneously, when will the tank be full?

*Solution*: Filling rate = 1/6 per hour Emptying rate = -1/8 per hour Net rate = 1/6 - 1/8 = (4-3)/24 = 1/24 per hour Time to fill = 1 ÷ (1/24) = **24 hours**

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**Example 3**: A can do work in 10 days, B in 15 days. They work together for 4 days, then A leaves. In how many more days will B finish the remaining work?

*Solution*: Combined 1-day work = 1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6 Work done in 4 days = 4 × (1/6) = 4/6 = 2/3 Remaining work = 1 - 2/3 = 1/3 B's 1-day work = 1/15 Time for B to finish = (1/3) ÷ (1/15) = (1/3) × 15 = **5 days**

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**Example 4**: 12 men can complete work in 8 days. How many men are needed to complete the same work in 6 days?

*Solution*: Total work = 12 × 8 = 96 man-days Men required = 96 / 6 = **16 men**

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**Example 5**: A and B together earn Rs. 1800 for a work. A worked 10 days and B worked 12 days. A alone can finish the work in 20 days and B in 30 days. What is A's share?

*Solution*: A's efficiency = 1/20, work done = 10 × (1/20) = 10/20 = 1/2 B's efficiency = 1/30, work done = 12 × (1/30) = 12/30 = 2/5 Ratio = (1/2) : (2/5) = 5:4 A's share = (5/9) × 1800 = **Rs. 1000**

Common Mistakes

• **Adding times instead of rates** → Wrong: A takes 10 days, B takes 15 days, together = 25 days. Correct: Add rates (1/10 + 1/15), then invert to get time (7.2 days).

• **Forgetting negative rate for outlet pipes** → In cistern problems, always subtract the emptying rate. An outlet reduces net filling speed.

• **Assuming equal efficiency when distributing wages** → Wages depend on (efficiency × time), not just time. Two workers for the same duration but different efficiencies get different pay.

• **Not converting fractional days to hours** → 7.2 days means 7 days + 0.2×24 = 7 days 4.8 hours, not 7 days 2 hours.

• **Ignoring the "remaining work" after partial completion** → When one person leaves midway, calculate leftover work accurately before applying the remaining worker's rate.

Quick Reference

• If A does work in *a* days and B in *b* days, together = **ab/(a+b)** days.
• Total work = LCM of individual days. Efficiency = LCM/Days. Sum efficiencies for combined rate.
• Pipes: Inlet rate (+), Outlet rate (−). Net rate = sum of all rates.
• Man-days: **M₁D₁ = M₂D₂**. More men → fewer days (inverse proportion).
• Wages split = ratio of **(efficiency × days worked)** for each person.
• Practice converting "work in *n* days" ↔ "1/*n* work per day" fluently for speed.