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compulsory · CSS study notes

16. Work, Time and Pipes/Cisterns

CSS GSA — Paper 7, Part-II, Section VI, Sub-topic 16 (Work, Time and Pipes/Cisterns)

Past Paper Weightage

  • Very high in MPT/General Ability papers — work-rate and pipe-filling problems are among the most frequently repeated word-problem types; master the "rate" method below and most variations become fast, mechanical marks.

The Core Method — Work as a Rate

  • The foundation of every work/pipe problem: convert "time to complete a job" into a rate of work per unit time.
  • If a person/pipe completes a job in N days (or hours), their rate of work = 1/N of the job per day (or hour).
  • When multiple people/pipes work together, add their rates (not their times) to get the combined rate, then take the reciprocal to find the combined time.
  • Pipes filling vs draining: an inlet pipe has a positive rate (+1/N); an outlet/drain pipe has a negative rate (−1/N), since it undoes work already done.

Worked Example 1 (Two Workers)

  • A can complete a job in 12 days, B can complete it in 18 days. How long will they take working together?
    • Step 1: Find each person's rate: A's rate = 1/12 of the job per day; B's rate = 1/18 of the job per day.
    • Step 2: Add the rates to get the combined rate: 1/12 + 1/18. Find a common denominator (36): 3/36 + 2/36 = 5/36 of the job per day.
    • Step 3: Take the reciprocal of the combined rate to find the combined time: 36/5 = 7.2 days.

Worked Example 2 (Pipes — One Fills, One Drains)

  • Pipe A can fill a tank in 6 hours. Pipe B (a drain) can empty the same full tank in 9 hours. If both are opened together, how long to fill the tank?
    • Step 1: Rate of Pipe A (filling) = +1/6 of the tank per hour. Rate of Pipe B (draining) = −1/9 of the tank per hour.
    • Step 2: Combined rate = 1/6 − 1/9. Common denominator (18): 3/18 − 2/18 = 1/18 of the tank per hour.
    • Step 3: Time to fill = reciprocal of the rate = 18 hours.

Worked Example 3 (Partial Work Already Done)

  • A can finish a job in 10 days. He works alone for 4 days, then B joins him and together they finish the remaining work in 2 days. How long would B alone take to finish the entire job?
    • Step 1: A's rate = 1/10 per day. In 4 days alone, A completes 4 × (1/10) = 4/10 = 2/5 of the job.
    • Step 2: Remaining work = 1 − 2/5 = 3/5 of the job, completed by A and B together in 2 days, so their combined rate = (3/5) ÷ 2 = 3/10 of the job per day.
    • Step 3: A's rate is 1/10, so B's rate alone = combined rate − A's rate = 3/10 − 1/10 = 2/10 = 1/5 of the job per day.
    • Step 4: B's time alone = reciprocal of B's rate = 5/1 = 5 days.

Worked Example 4 (Three Pipes, One a Drain)

  • Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. A drain pipe C can empty the full tank in 15 minutes. If all three are opened together, how long to fill the tank (or does it fill at all)?
    • Step 1: Rates: A = 1/20, B = 1/30 (filling, positive); C = 1/15 (draining, negative).
    • Step 2: Combined rate = 1/20 + 1/30 − 1/15. Common denominator (60): 3/60 + 2/60 − 4/60 = 1/60 of the tank per minute.
    • Step 3: Since the combined rate is positive (1/60), the tank does fill, taking the reciprocal: 60 minutes.

Shortcut for Exam Speed

  • For two entities with times a and b working together:
Combined time=a×ba+b\text{Combined time} = \frac{a \times b}{a+b}

this LCM-style shortcut skips writing out fractions when you just need the final number quickly (e.g., Example 1: (12×18)/(12+18) = 216/30 = 7.2 days, matching Step 3 above)

  • For a drain pipe combined with fill pipes: treat the drain's time as a NEGATIVE quantity in the same shortcut formula, or add/subtract rates directly as shown above (the rate method is safer for 3+ pipes)

MCQ Traps

  • Never average the given TIMES directly (e.g., (12+18)/2 = 15 days is WRONG) — you must always work through RATES, then convert back to time at the end
  • A drain pipe's rate is subtracted, not added — forgetting this sign is the most common error in mixed fill/drain problems

Self-Test Questions

  1. A can complete a job in 15 days, B in 10 days. Find how long they take together, showing the rate-addition method.
  2. A tank can be filled by pipe X in 8 hours and emptied by pipe Y in 12 hours. If both are opened together, how long to fill the tank?
  3. A alone can finish a task in 20 days. After working alone for 5 days, B joins and they finish the remaining work together in 6 days. Find how long B alone would take, showing all steps.

Past-Paper MCQs (FPSC)

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The FPSC MCQ papers you shared (2024 and 2025) had no direct work-time question.

Past-Paper Questions with Answers (FPSC)

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Every question below appeared in the FPSC papers you shared; the year and question number are in each title. Answers are worked out by me.

  • CSS 2020 Q6(a): Tariq does a job in 6 hours, Sajid in 4, Irfan in 8. Tariq and Sajid start; after 2 hours Sajid leaves and Irfan replaces him. How long to finish? (5)
    1. Rates per hour: Tariq 1/6, Sajid 1/4, Irfan 1/8
    2. First 2 hours (Tariq + Sajid): 2 × (1/6 + 1/4) = 2 × 5/12 = 5/6 of the job; 1/6 remains
    3. Tariq + Irfan: 1/6 + 1/8 = 7/24 per hour
    4. Time for 1/6: (1/6) ÷ (7/24) = 4/7 hour ≈ 34 minutes
    5. Total ≈ 2 hours 34 minutes
  • CSS 2019 Q6(b): One pipe fills a pool 1.25 times as fast as another. Together they fill it in 5 hours. How long does the slower pipe take alone? (5)
    1. Slower pipe rate = r; faster = 1.25r; together 2.25r = 1/5
    2. r = 1 ÷ (2.25 × 5) = 1/11.25
    3. Slower pipe alone takes 11.25 hours (11 h 15 min)
  • CSS 2024 Q8(b): Twelve men can do a job in 24 days. After 4 days, 4 men leave. How many more days will the remaining men take? (5)
    1. Total work = 12 × 24 = 288 man-days; done in 4 days = 48
    2. Remaining = 240 man-days; 8 men remain
    3. 240 ÷ 8 = 30 more days
  • CSS 2020 Q8(c): 50 men build 20 km of road in 40 days. How long will 70 men take? (5)
    1. Work = 50 × 40 = 2,000 man-days
    2. 2,000 ÷ 70 ≈ 28.6 days
  • CSS 2017 Q7(B): Ali takes 30 minutes to mark a paper and Aslam 25 minutes. Both start at 10:00 AM. When will they finish marking at the same time? (5)
    1. They finish together at the LCM of 30 and 25 = 150 minutes
    2. 150 minutes = 2 h 30 min after 10:00 AM → 12:30 PM
    3. By then Ali has marked 5 papers and Aslam 6
  • Speed, distance and time questions (CSS 2025 Q8(a), 2024 Q8(d), 2021 Q6(d))
    • CSS 2025 Q8(a): first five minutes: 68 + 127 + 208 + 312 + 535 = 1,250 m; next five minutes at 33 km/h: 33 × 5/60 = 2.75 km = 2,750 m. Total = 4,000 m (4 km) in 10 minutes, so average speed = 4 ÷ (1/6) = 24 km/h
    • CSS 2024 Q8(d): in 15 minutes A goes 15 km, B goes 25 km (opposite), C goes 20 km (at right angles). A–B = 40 km; A–C = √(15² + 20²) = 25 km; B–C = √(25² + 20²) ≈ 32.02 km
    • CSS 2021 Q6(d): walking 45 minutes at 3 km/h = 2.25 km; remaining 6 − 2.25 = 3.75 km in 0.5 hour → 7.5 km/h

CSS Exam Perspective

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  • Most repeated areas: combined work rates, workers leaving or joining, man-days, LCM of times, average speed
  • Tip: convert each worker into 'work per hour', add rates for people working together