Time and Work — Study Notes

Overview

Time and Work is a fundamental topic in SSC GD Elementary Mathematics, accounting for 2–3 questions in every exam. The core principle is simple: if you know how long someone takes to complete a job, you can calculate their work rate per unit time. Most problems test your ability to combine work rates when multiple people or machines work together, or when they work in alternating patterns.

This topic appears in three main formats: basic joint work (two or more workers completing a task together), negative work (pipes filling and emptying tanks simultaneously), and alternating work patterns (workers taking turns on different days). Mastering the basic formula "Work = Rate × Time" and understanding the concept of man-days or unit work completed per day will solve 90% of questions. The remaining 10% require careful handling of fractions and the ability to track partial work completed over multiple cycles.

Students who practice 20–25 problems covering all three formats typically score full marks in this section. The mathematics is straightforward—no complex calculations—but attention to detail when setting up the equation separates successful candidates from those who make careless errors.

Key Concepts

• **Work as Unity**: The total work is always assumed to be 1 unit (or 100%) unless stated otherwise. This standardization makes calculations uniform across all problems.

• **Work Rate (Efficiency)**: If a person completes work in *n* days, their one-day work (rate) is 1/n. If A does a job in 10 days, A's rate = 1/10 per day.

• **Combined Work Rate**: When multiple people work together, their rates add. If A works at 1/10 per day and B at 1/15 per day, together they work at (1/10 + 1/15) per day.

• **Man-Days Concept**: Total work can be expressed as (number of people) × (number of days). If 5 men complete work in 12 days, the work = 60 man-days. This same work can be done by 10 men in 6 days.

• **Negative Work (Pipes)**: In pipe problems, filling pipes contribute positive work while emptying/leak pipes contribute negative work. Net rate = (sum of filling rates) – (sum of emptying rates).

• **Partial Work Tracking**: When workers alternate on different days, calculate work done in one complete cycle, then find how many full cycles occur, and handle the remaining work in the final partial cycle.

• **Efficiency Ratio**: If the ratio of time taken by two workers is a:b, their efficiency ratio is b:a (inverse relationship). A person who takes less time is more efficient.

• **Wages Distribution**: Wages are distributed in the ratio of work done, which equals the ratio of (efficiency × time worked) for each person.

Formulas / Key Facts

• **Basic Work Formula**: Work = Rate × Time, or equivalently, Rate = Work/Time and Time = Work/Rate.

• **One-Day Work**: If A completes work in *a* days, A's one-day work = 1/a.

• **Joint Work Time**: If A takes *a* days and B takes *b* days alone, together they take: Time = (a×b)/(a+b) days.

• **Three Workers Together**: If A, B, C take *a*, *b*, *c* days individually, together they take: Time = 1/(1/a + 1/b + 1/c) days.

• **Man-Days Formula**: M₁D₁ = M₂D₂ (when work efficiency is same). If work amounts differ: M₁D₁W₂ = M₂D₂W₁.

• **Pipe Formula (Two Pipes)**: If inlet fills in *a* hours and outlet empties in *b* hours, net time to fill = (a×b)/(b–a) hours (when b > a).

• **Work Left Formula**: If *n* days of work remain and a person works at rate 1/a per day, work left = n/a of total work.

• **Efficiency-Based Time**: If efficiency of A:B = m:n, then time taken by A:B = n:m.

Worked Examples

**Example 1: Basic Joint Work** A can complete a work in 15 days and B in 20 days. How long will they take working together?

*Solution*:

• A's one-day work = 1/15
• B's one-day work = 1/20
• Combined one-day work = 1/15 + 1/20 = (4+3)/60 = 7/60
• Time taken together = 1 ÷ (7/60) = 60/7 = 8(4/7) days

Alternatively using direct formula: Time = (15×20)/(15+20) = 300/35 = 60/7 days.

**Example 2: Pipes and Cisterns** A pipe fills a tank in 6 hours. Another pipe empties it in 10 hours. If both are opened together, when will the tank be full?

*Solution*:

• Filling rate = 1/6 per hour (positive)
• Emptying rate = 1/10 per hour (negative)
• Net rate = 1/6 – 1/10 = (5–3)/30 = 2/30 = 1/15 per hour
• Time to fill = 1 ÷ (1/15) = 15 hours

**Example 3: Alternating Work Pattern** A can do work in 12 days and B in 18 days. They work on alternate days starting with A. In how many days will the work be completed?

*Solution*:

• A's one-day work = 1/12, B's one-day work = 1/18
• Work in 2-day cycle (A works day 1, B works day 2) = 1/12 + 1/18 = (3+2)/36 = 5/36
• In 7 complete cycles (14 days): work done = 7 × 5/36 = 35/36
• Work remaining = 1 – 35/36 = 1/36
• On day 15, A works: A completes 1/12 in a day, so for 1/36 work, time = (1/36)/(1/12) = 1/3 day
• Total time = 14 + 1/3 = 14(1/3) days

**Example 4: Man-Days Problem** 12 men can complete a work in 15 days. After 5 days, 3 men leave. In how many more days will the work be completed?

*Solution*:

• Total work = 12 × 15 = 180 man-days
• Work done in 5 days = 12 × 5 = 60 man-days
• Work remaining = 180 – 60 = 120 man-days
• Men remaining = 12 – 3 = 9
• Days required = 120/9 = 13(1/3) days

Common Mistakes

**Mistake 1**: *Adding times instead of rates* → When A takes 10 days and B takes 15 days, students incorrectly add 10+15=25 days as joint time. **Fix**: Add rates (1/10 + 1/15), then take reciprocal for time.

**Mistake 2**: *Forgetting negative work in pipe problems* → Treating outlet pipes as additional filling pipes. **Fix**: Always subtract emptying rates from filling rates. Check whether the question asks "time to fill" or "time to empty" and adjust signs accordingly.

**Mistake 3**: *Incomplete cycle calculation in alternating work* → Stopping at full cycles and forgetting the partial day. **Fix**: After counting complete cycles, always check remaining work and calculate the fraction of the next day needed.

**Mistake 4**: *Inverting the efficiency ratio* → If A:B work in 4:5 time ratio, mistakenly using 4:5 as efficiency ratio. **Fix**: Efficiency ratio is inverse of time ratio. Less time = more efficiency, so efficiency ratio is 5:4.

**Mistake 5**: *Wrong man-days formula application* → Using M₁D₁ = M₂D₂ when work amounts are different. **Fix**: When work differs, use M₁D₁/W₁ = M₂D₂/W₂ or cross-multiply as M₁D₁W₂ = M₂D₂W₁.

Quick Reference

• If A takes *a* days alone, A's daily rate = 1/a; combined rates always add.
• Joint time for two workers = (product of individual times)/(sum of individual times).
• Total work = (number of workers) × (number of days) when efficiency is constant.
• In pipe problems: net rate = filling rate – emptying rate; watch for positive/negative.
• Alternating work: calculate full-cycle work first, then handle the remainder separately.
• Efficiency and time are inversely proportional: higher efficiency means less time needed.